Algebra I

Lecture course in Bonn, SS 2025

1
 Contents

Chapter I. The Spectrum of a Ring 3

1. Rings, Ideals, Algebras 4
2. Common roots of multivariate polynomials: Hilbert’s Nullstellensatz v1 9
3. Reconstructing ideals from zero sets: Hilbert’s Nullstellensatz v2 13
4. Spectrum and Zariski topology 18
Chapter II. Modules 22
1. Modules 23
2. Hom, tensor, and change of scalars 27
3. Localizing rings and modules 33
Chapter III. Chain conditions 38
1. Noetherian spaces, rings, and modules 39
2. Artinian spaces, rings, and modules 44
3. Krull dimension and height 49
4. Nakayama’s lemma and the principal ideal theorem 51
Chapter IV. Integrality 55
1. Finite and integral ring maps 56
2. Lying over, going up, going down, and incomparability 60
3. Discrete Valuation Rings and Dedekind Domains 66
4. Noether normalization 69
Chapter V. Homological Algebra 71
1. Abelian Categories 72
2. Derived functors 76
3. Ext and Tor 81
Chapter VI. Regular local rings 84
1. Graded and filtered rings and modules 85
2. Inverse limits and the Artin–Rees lemma 89
3. Comparing a ring with its completion 95
4. Regular local rings 99
Chapter VII. Appendix 102
1. Categories 103
Bibliography 105

2
 CHAPTER I

The Spectrum of a Ring

3
4 I. THE SPECTRUM OF A RING

1. Rings, Ideals, Algebras

We start this lecture course by recalling basic definitions from commutative ring theory.
Definition 1.1: A ring is a set A together with two maps, called addition and multiplication
+: A×A →A
(a, b) 7→ a + b
·: A×A →A
(a, b) 7→ ab := a · b
satisfying the following properties:
(1) (A, +) is an Abelian group. We write 0 ∈ A for the neutral element of this group and call
it zero.
(2) Associativity of multiplication: For all a, b, c ∈ A, it holds that
a(bc) = (ab)c
(3) Commutativity of multiplication: For all a, b ∈ A, it holds that
ab = ba
(4) Existence of 1: There is an element 1 ∈ A with
1a = a
for all a ∈ A. We call this element one. 1
(5) Distributivity: For all a, b, c ∈ A, it holds that
a(b + c) = (ab) + (ac).
Convention 1.2: In particular:
All rings considered in this lecture are commutative rings with 1 unless explicitly stated
otherwise.

Definition 1.3: Let A be a ring.

(1) An element a ∈ A is called unit if there exists b ∈ A with ab = 1. We write A× for the
group (with respect to multiplication) of units in A.
(2) An element a ∈ A is called zero divisor if there exists 0 ̸= b ∈ A with ab = 0. 2
(3) A is called integral domain if 0 is its only zero-divisor.
(4) A is called field if A× = A − {0}.
Example 1.4: We assume some familiarity with the following basic examples of rings:
(1) The integers Z with the usual addition and multiplication are a ring. In fact (check!), Z is
an integral domain but not a field.
1One easily checks that 1 ∈ A is unique if it exists. Moreover, it holds that 0 ̸= 1 unless A = {0}, in which case
we call A the zero ring.
2It follows from the ring axioms that 0a = 0 for all a ∈ A. In particular, 0 is a zero divisor if and only if A is not
the zero ring.
 1. RINGS, IDEALS, ALGEBRAS 5

(2) For every integer n ≥ 1, the set Z/nZ of “residues modulo n” is a ring with addition and
multiplication modulo n. In fact (check!), the ring Z/nZ is an integral domain if and only
if n is a prime number.
(3) The sets Q, R, C and the finite fields Fq are fields.
(4) If A is a ring, then the set A[x] of polynomials in one variable over A is a ring. In fact
(check by comparing leading coefficients!), if A is an integral domain, then so is A[x].
(5) If A is a ring, then the set A[x, x−1 ] of Laurent polynomials in one variable over A is a
ring.
(6) More generally, for every set I, the set of A[{xi }i∈I ] of polynomials with variables indexed
by I is a ring.
(7) If A is a ring and B ⊆ A is a subset with 1 ∈ B, a − b ∈ B, and ab ∈ B for all a, b ∈ B,
then B is also a ring with addition and multiplication of A restricted to B. Such a B is
called subring of A.
(8) For a topological space X, the set C 0 (X) of continuous real-valued functions on X is a
ring with pointwise addition and multiplication.
Definition 1.5: Let A and B be rings and let f : A → B be a map. Then, f is called ring
homomorphism if the following conditions hold:
(1) f is a group homomorphism from (A, +) to (B, +).
(2) f (ab) = f (a)f (b) for all a, b ∈ A.
(3) f (1) = 1.
Example 1.6: You have already seen the following examples of ring homomorphisms:
(1) For every ring A, there is a unique ring homomorphism f : Z → A. Indeed, we have
f (1) = 1 by condition (3) and
n
X n
X n
X
f (n) = f ( 1) = f (1) = 1 =: n · 1.
i=1 i=1 i=1
(2) The reduction modulo n is the unique ring homomorphism f : Z → Z/nZ.
(3) Field extensions are ring homomorphisms.
(4) If g : X → Y is a continuous map of topological spaces, then
C 0 (Y ) → C 0 (X)
f 7→ f ◦ g
is a ring homomorphism.
Definition 1.7: Let A be a ring.
(1) An ideal I of A is a subgroup of (A, +) such that ab ∈ I for all a ∈ A and b ∈ I.
(2) Given a subset S ⊆ A, the ideal generated by S is
Xn
(S) := { λi ai | λi ∈ A, ai ∈ S}.3
i=1
If S = {a1 , . . . , an }, we write (S) = (a1 , . . . , an ).
3Exercise: Check that this is really an ideal.
6 I. THE SPECTRUM OF A RING

(3) A principal ideal of A is an ideal generated by one element.

(4) An ideal I of A is called prime ideal if I ̸= A and for all a, b ∈ A with ab ∈ I, we have
a ∈ I or b ∈ I.
(5) An ideal I of A is called maximal if I ̸= A and if J is an ideal with I ⊆ J ⊆ A, then
either J = I or J = A.
Example 1.8: We have some standard examples of ideals:
(1) Every ring A contains the zero ideal (0) = {0} ⊆ A and the unit ideal (1) = A.
(2) If I and J are ideals in a ring A, then I ∩ J is also an ideal.
(3) If I and J are ideals in a ring A, then the product
I · J = ({ij | i ∈ I, j ∈ J})
is also an ideal. Note that I · J ⊆ I ∩ J.
(4) All ideals of Z are of the form (n) for some n ≥ 0. The ideal (n) is maximal if and only
if n is a prime number. The ideal (0) is prime, but not maximal.
(5) If k is a field, then (x) ⊆ k[x] consists of polynomials with vanishing constant term. This
is an example of a maximal ideal.
Since an ideal I ⊆ A is a subgroup with respect to addition, we have an associated quotient
group A/I together with a group homomorphism π : A → A/I called quotient of A by I. We also
write a := π(a) for a ∈ A. One checks:
Lemma 1.9: Let A be a ring and I ⊆ A an ideal. Then, the quotient map π : A → A/I is
surjective and there is a unique multiplication on A/I such that π is a ring homomorphism.
For proofs of the following results on ring homomorphisms, please consult the lectures notes
for the course on introduction to Algebra:
Theorem 1.10: Let f : A → B be a ring homomorphism.
(1) The image f (A) of f is a subring of B.
(2) The kernel
ker(f ) := {a ∈ A | f (A) = 0}
of f is an ideal of A.
(3) If I ⊆ A is an ideal, then f factors through π : A → A/I if and only if I ⊆ ker(f ).
(4) (Homomorphism theorem) There is a unique ring homomorphism g : A/ ker(f ) → B
making the following diagram commute:
π/
A A/ ker(f )
g
$ 
f
B
Moreover, g is injective, hence induces an isomorphism of rings A/ ker(f ) ∼
= f (A).
We also assume that you are familiar with the following characterization of prime and maximal
ideals via the homomorphism theorem:
Lemma 1.11: Let A be a ring and I ⊆ A an ideal.
 1. RINGS, IDEALS, ALGEBRAS 7

(1) I is prime if and only if A/I is an integral domain.

(2) I is maximal if and only if A/I is a field.
This leads to the following observation, which is one of the key motivations to study the prime
spectrum of a ring that will introduce in the fourth lecture:
Corollary 1.12: Let f : A → B be a ring homomorphism. Let p ⊆ B be a prime ideal. Then,
f −1 (p) is prime.
P ROOF. The ideal f −1 (p) is the kernel of A → B → B/p, hence there is an inclusion
A/f −1 (p) ⊆ B/p. Since p is prime, B/p is an integral domain, hence so is A/f −1 (p) and thus
f −1 (p) is prime. □
Example 1.13: The above corollary fails if we replace prime ideals by maximal ideals. Consider
for example the inclusion Z ⊆ Q and the preimage of the zero ideal.
It will be useful to consider rings together with a scalar multiplication by another ring. These
are called algebras.
Definition 1.14: Let A be a ring.
(1) An A-algebra is a ring B together with a ring homomorphism A → B.
(2) A homomorphism of A-algebras is a ring homomorphism f : B → B ′ between A-
algebras such that the following diagram commutes
A /B

f

B′
Example 1.15: (1) Every ring is a Z-algebra in a unique way and every ring homomorphism
is a homomorphism of Z-algebras.
(2) If A is a ring, then any polynomial ring over A is an A-algebra via the ring homomorphism
A → A[{xi }i∈I ] sending elements of A to constant polynomials.
(3) If A is a ring, B is an a A-algebra via f : A → B, and M ⊆ B is any subset, there is a
unique A-algebra homomorphism
A[{xm }m∈M ] → B,
called evaluation map evM that sends xm to m ∈ B. The image of evM is denoted by
A[M ] (or A[b1 , . . . , bn ] if M = {b1 , . . . , bn }) and is called A-subalgebra of B generated
by M .
(4) In the above example, in the case where I consists of one element b ∈ B, the evaluation
at b is the map
A[x] → B
X X
λi xi 7→ f (λi )bi .
If f is clear from the context, we write g(b) := evb (g).
Definition 1.16: Let A be a ring and let B be an A-algebra.
8 I. THE SPECTRUM OF A RING

(1) A subset M ⊆ B is said to generate B as an A-algebra if B = A[M ].

(2) B is called finitely generated if it can be generated as an A-algebra by a finite subset of B.
From the homomorphism theorem, we deduce immediately:
Corollary 1.17: Let A be a ring and let B be an A-algebra. Then, B is finitely generated over
A if and only if there is an integer n ≥ 1, an ideal I ⊆ A[x1 , . . . , xn ], and an isomorphism of
A-algebras
A[x1 , . . . , xn ]/I → B.
Remark 1.18: More generally, every A-algebra is a quotient of a (not necessarily finitely gener-
ated) polynomial ring over A.
 2. COMMON ROOTS OF MULTIVARIATE POLYNOMIALS: HILBERT’S NULLSTELLENSATZ V1 9

2. Common roots of multivariate polynomials: Hilbert’s Nullstellensatz v1

Let us recall again some concepts from a basic course on Galois theory:
Definition 2.1: Let A be an integral domain.
(1) An element a ∈ A is called irreducible if a ̸= 0, a ̸∈ A× and for all b, c ∈ A with a = bc
either b ∈ A× or c ∈ A× .
(2) For two elements a, b ∈ A, we say that a divides b if b ∈ (a). We write a | b.
(3) An element a ∈ A is called prime if a ̸= 0, a ̸∈ A× and whenever a | (bc) for some
b, c ∈ A, then a | b or a | c.
Example 2.2: The following facts can be checked easily:
(1) Every prime number in Z is prime and irreducible.
(2) An element a of a ring A is prime if and only if (a) is a non-zero prime ideal.
Lemma 2.3: In an integral domain, every prime elements is irreducible.
Definition 2.4: Let A be an integral domain.
(1) A is called principal ideal domain (PID) if every ideal I of A is principal.
(2) A is called unique factorization domain (UFD) if every a ∈ A that is non-zero and not a
unit can be written as a product of irreducible elements, uniquely up to permutation and
units.
Lemma 2.5: Every PID is a UFD.
Example 2.6: The following rings are PIDs:
(1) The integers Z.
(2) The polynomial ring k[x] over a field k.
In fact, both of these rings are Euclidean domains, i.e., they admit a division with remainder and all
Euclidean rings are PIDs.
Theorem 2.7: Let A be a UFD. Then, A[X] is a UFD.
Lemma 2.8: Let A be a UFD and a ∈ A. Then, a is irreducible if and only if it is prime.
Lemma 2.9: Let A be a PID. For a ∈ A, the ideal (a) ⊆ A is maximal if and only if a is irreducible.
Corollary 2.10: Let A be a PID and let p ⊆ A be a non-zero prime ideal. Then, p is maximal.
P ROOF. We have p = (a) for some a ∈ A prime, by Example I.2.2. By Lemma I.2.8, a is
irreducible, so (a) is maximal by Lemma I.2.9. □
Example 2.11: Using Corollary I.2.10, we can describe all maximal ideals of some important rings:
(1) The maximal ideals of Z are exactly the (p) ⊆ Z with p a prime number.
(2) Let k be a field and 0 ̸= f ∈ k[x]. Then, f is a unit if and only if deg(f ) = 0. If
deg(f ) = 1, then f is irreducible.
(3) Recall that a field k is called algebraically closed if every non-constant polynomial f ∈
k[x] has a root in k. If deg(f ) > 1 and f has a root a, then f = (x − a)g with deg(g) >
10 I. THE SPECTRUM OF A RING

0, so f is not irreducible. Thus, k is algebraically closed if and only of the irreducible

polynomials in k[x] are exactly those of degree 1
(4) If k is not algebraically closed, there must be irreducible polynomials of higher degree.
For example, x2 + 1 ∈ R[x] is irreducible.
Generalizing the above examples, we prove:
Lemma 2.12: Let k be a field and P = (a1 , . . . , an ) ∈ k n . Then, the ideal
mP = (x1 − a1 , . . . , xn − an ) ⊆ k[x1 , . . . , xn ]
is maximal.
P ROOF. For f ∈ k[x1 , . . . , xn ], we have
f = f (a1 , . . . , an ) mod mP .
Thus, f ∈ mP if and only if f ∈ ker(φ), where φ is the evaluation map
φ = eva1 ,...,an : k[x1 , . . . , xn ] → k
f 7→ f (a1 , . . . , an ).
Since φ is clearly surjective (evaluate constant polynomials), we deduce from Theorem I.1.10 that
k[x1 , . . . , xn ]/mP ∼
= k, hence mP is maximal by Lemma I.1.11. □
We have seen in Example I.2.11 (3) that every maximal ideal is of the above form if k is
algebraically closed and n = 1 and that this is not true if k is not algebraically closed. Our next
goal is to show that this classification of maximal ideals also holds for higher n if k is algebraically
closed. For this, we need some preparations:
Definition 2.13: Let k be a field and A a k-algebra.
(1) An element a ∈ A is called algebraic over k if there is a non-zero polynomial f ∈ k[x]
with f (a) = 0. 4
(2) We say that A is algebraic over k if all its elements are algebraic over k.
(3) A finite subset {a1 , . . . , an } ⊆ A is called algebraically independent over k if eva1 ,...,an :
k[x1 , . . . , xn ] → A is injective. An arbitrary subset M ⊆ A is called algebraically inde-
pendent over k if all its finite subsets are algebraically independent over k.
(4) If A is a field, a transcendence basis for A over k is an algebraically independent subset
M ⊆ A such that A is algebraic over k(M ), where k(M ) ⊆ A is the subfield generated
by M over k.
Theorem 2.14 (Existence of transcendence basis): Let L/K be a field extension.
(1) A subset M ⊆ L is a transcendence basis for L/K if and only if it is a maximal alge-
braically independent subset.
(2) If M ′ ⊆ L is algebraically independent and M ′′ ⊆ L is such that L/K(M ′′ ) is algebraic,
then there exists a transcendence basis M with M ′ ⊆ M ⊆ M ′ ∪ M ′′ .
In particular, there exists a transcendence basis for L/K.
4In other words, an element a ∈ A is algebraic over k if and only if the evaluation map ev : k[x] → A is not
a
injective.
 2. COMMON ROOTS OF MULTIVARIATE POLYNOMIALS: HILBERT’S NULLSTELLENSATZ V1 11

Recall that the field of fractions Frac(A) of an integral domain A is the set of equivalence
classes of fractions ab with a, b ∈ A, 0 ̸= b, with the usual multiplication and addition.
Lemma 2.15 (Universal property of field of fractions): Let A be an integral domain and f : A → B
a ring homomorphism. Then, f factors through A → Frac(A) if and only if f (A − {0}) ⊆ B × .
The following lemma will turn out to be extremely useful:
Lemma 2.16: Let A be an algebra over a field k.
(1) If A is an integral domain and algebraic over k, then A is a field.
(2) If A is a field and contained in a finitely generated k-algebra, then A is algebraic over k.
P ROOF. (1): Let 0 ̸= a ∈ A. Replacing A by the image of φ = eva : k[x] → A, we may
assume that φ is surjective. Let I be the kernel of φ. Since A is an integral domain, I is a prime
ideal by Lemma I.1.11. Since a is algebraic over k, I is non-zero, hence maximal by Corollary
I.2.10, so A is a field by Lemma I.1.11.
(2): Seeking a contradiction, assume that there exists a1 ∈ A that is not algebraic over k. Let
B be a finitely generated k-algebra containing A. Let m ⊆ B be any maximal ideal and consider
L = B/m. Then, L is a field that is still finitely generated as a k-algebra and the morphism
A → B → B/m is injective, being a morphism of fields. We choose elements a2 , . . . , an ∈ L
and n such that L = k[a1 , . . . , an ]. After reordering the ai , we may assume that a1 , . . . , ar is
a transcendence basis of L over k. Let K := k(a1 , . . . , ar ), so that L/K is an algebraic field
extension, finitely generated by ar+1 , . . . , an . Thus, L/K is a finite field extension. We choose and
fix a K-basis e1 , . . . , em of L.
Using this basis, we get a homomorphism of non-commutative rings

φ : L → K m×m
b 7→ Ab ,

where Ab is the matrix representing multiplication by b on L ∼ = K m . Let g ∈ k[a1 , . . . , ar ] be a

common denominator of all the entries of all the φ(ai ). Then, φ(L) ⊆ k[a1 , . . . , ar , g −1 ]m×m .
Now, let p ∈ k[a1 ] be an irreducible element. Since k(a1 ) ⊆ K, the matrix φ(p−1 ) is a diagonal
matrix with all entries equal to p−1 . On the other hand, A is a field, so we have p−1 ∈ A ⊆ L,
so p−1 ∈ k[a1 , . . . , ar , g −1 ]. We conclude that there exists an integer s ≥ 0 and an element
f ∈ k[a1 , . . . , ar ] such that p−1 = g −s f , hence g s = pf .
Since the a1 , . . . , ar are algebraically independent, the rings k[a1 , . . . , ar ] and k[a1 ] are isomor-
phic to polynomial rings in r resp. 1 variables over k, hence both are UFDs by Theorem I.2.7. There
are infinitely many distinct irreducible elements in k[a1 ] 5 and they stay irreducible in k[a1 , . . . , ar ]
(Check this!). The previous paragraph shows that every irreducible element in k[a1 ] divides g s in
k[a1 , . . . , ar ]. Since the latter is a UFD, this implies that all irreducible elements in k[a1 ] divide g.
On the other hand, using a prime factorization of g, we see that there are only finitely many distinct
irreducible elements that divide g. Therefore, our assumption that a1 is not algebraic over k must
be false and we conclude. □

5If k is infinite, then every a − a with a ∈ k is irreducible. If k is finite, then there are finite field extensions of k
1
of arbitrarily large degree, hence infinitely many irreducible elements in k[x] ∼
= k[a1 ].
12 I. THE SPECTRUM OF A RING

Applying (2) of the above lemma to the case where A itself is finitely generated as a k-algebra
and using that finitely generated algebraic field extensions are finite, we obtain the following:
Corollary 2.17 (Zariski’s lemma): Let L/K be a field extension such that L is finitely generated
as a K-algebra. Then, L/K is a finite field extension.
Recall that the preimage of a maximal ideal under a ring homomorphism might not be maximal.
The situation is better for finitely generated algebras over fields:
Corollary 2.18: Let k be a field and let f : A → B be a morphism of k-algebras. Let m ⊆ B be a
maximal ideal. Assume that B is finitely generated over k. Then, f −1 (m) ⊆ A is a maximal ideal.
P ROOF. The ideal f −1 (m) is the kernel of the homomorphism A → B → B/m, hence
A/f −1 (m) is a k-subalgebra of the field (Lemma I.1.11) B/m. Since B is finitely generated over k,
so is B/m, hence algebraic over k by Lemma I.2.16 (2). As a k-subalgebra of B/m, the k-algebra
A/f −1 (m) is thus also algebraic over k. We already know that A/f −1 (m) is an integral domain,
hence a field by Lemma I.2.16. Thus, f −1 (m) is maximal by Lemma I.1.11. □
Corollary 2.19: Let k be an algebraically closed field and let m ⊆ k[x1 , . . . , xn ] be a maximal
ideal. Then, there exists a point P = (a1 , . . . , an ) ∈ k n with m = mP := (x1 − a1 , . . . , xn − an ).
P ROOF. By Corollary I.2.18, the intersection k[xi ] ∩ m, which is nothing but the preimage
of m under the inclusion k[xi ] → k[x1 , . . . , xn ], is maximal for each i. Since k[xi ] is a PID by
Example I.2.6, there exists an irreducible polynomial pi ∈ k[xi ] with (pi ) = m ∩ k[xi ]. Since k is
algebraically closed, pi = xi − ai for some ai ∈ k. We conclude that xi − ai ∈ m, hence mP ⊆ m
and since mP is already maximal, we have mP = m. □
Corollary 2.20 (Hilbert’s Nullstellensatz v1): Let k be an algebraically closed field and let I ⊊
k[x1 , . . . , xn ] be a proper ideal. Then, there exists a point P = (a1 , . . . , an ) ∈ k n such that
f (P ) = 0 for all f ∈ I.
P ROOF. Since I is a proper ideal, there exists a maximal ideal m containing I 6. By Corollary
I.2.19, there exists a point P ∈ k n with m = mP . Since I ⊆ mP and mP is the kernel of the
evaluation map at P , we get that f (P ) = 0 for all f ∈ I. □

6This is an easy consequence of Zorn’s lemma. See the lecture notes for “Einführung in die Algebra” for details.
 3. RECONSTRUCTING IDEALS FROM ZERO SETS: HILBERT’S NULLSTELLENSATZ V2 13

3. Reconstructing ideals from zero sets: Hilbert’s Nullstellensatz v2

We have seen in the previous lecture that every proper ideal in the polynomial ring has a com-
mon zero. Next, we investigate to what extend these common zero sets determine the ideal.
Definition 3.1: Let k[x1 , . . . , xn ] be a polynomial ring over an algebraically closed field k. Let
M ⊆ k[x1 , . . . , xn ] and X ⊆ k n be subsets.
(1) The affine algebraic set defined by M (or vanishing set of M ) is the set
V (M ) = {P ∈ k n | f (P ) = 0 ∀f ∈ M }.
An affine algebraic set is an affine algebraic set defined by some set of polynomials.
(2) The (vanishing) ideal of X is the set
I(X) := {f ∈ k[x1 , . . . , xn ] | f (P ) = 0 ∀P ∈ X}.
Observe the symmetry in the definitions of I(X) and V (M ). The goal of this lecture is to prove
that the maps
X →7 I(X)
V (I) ←[ I
are mutually inverse bijections between affine algebraic sets and so-called radical ideals in the poly-
nomial ring. First, we observe:
Lemma 3.2: The maps I(−) and V (−) are inclusion reversing in the following sense:
(1) If X ⊆ Y ⊆ k n , then I(Y ) ⊆ I(X).
(2) If M ⊆ M ′ ⊆ k[x1 , . . . , xn ], then V (M ′ ) ⊆ V (M ).
P ROOF. (1): Let f ∈ I(Y ). Then, f (P ) = 0 for all P ∈ Y , hence f (P ) = 0 for all P ∈ X,
and so f ∈ I(X).
(2): Let P ∈ V (M ′ ). Then, f (P ) = 0 for all f ∈ M ′ , hence f (P ) = 0 for all f ∈ M , and so
P ∈ V (M ). □
Using this, one direction of the bijection is easy to see.
Lemma 3.3: Let k be an algebraically closed field and let X ⊆ k n be an affine algebraic set. Then,
V (I(X)) = X.
P ROOF. Since X is an affine algebraic set, we have X = V (I) for some ideal I. By definition
of V (−) and I(−), we see that X ⊆ V (I(X)) and I ⊆ I(X). Applying V (−) to the latter inclusion
and using that it reverses inclusions, we obtain
V (I(X)) ⊆ V (I) = X ⊆ V (I(X)),
hence all inclusions are equalities. □
In other words, if X ⊆ k n is an affine algebraic set, then it is precisely the simultaneous
vanishing locus of all polynomials that vanish along X and not larger! The harder part is to describe
I(V (I)).
Remark 3.4: We make some preliminary observations:
14 I. THE SPECTRUM OF A RING

(1) If k is algebraically closed and (M ) ̸= k[x1 , . . . , xn ], then V (M ) ̸= ∅ by the first version

of Hilbert’s Nullstellensatz as in Corollary I.2.20
(2) The set I(X) is an ideal. Indeed, the sum of two polynomials vanishing at P also vanish
at P and if f vanishes at P , then so does gf for any polynomial g.
(3) In particular, M ⊆ (M ) ⊆ I(V (M )), so there is no hope to recover the set M from
V (M ).
(4) If f m vanishes at a point P , then so does f . In other words, f m ∈ I(X) implies f ∈ I(X).
The last observation above leads to the definition of radical.
Definition 3.5: Let A be a ring and I an ideal. The radical of I is the set
√
I := {f ∈ A | ∃m ∈ Z>0 such that f m ∈ I}.
√
The ideal I is called radical if I = I.
Remark 3.6: We observe:
√
(1) We always have the inclusion I ⊆ I. √ √
(2) If I and J are two ideals of A with I ⊆ J, then I ⊆ J.
(3) Every prime ideal is radical. Indeed, if m ≥ 2 and am−1 a = am ∈ I, then I being prime
means that apm−1 ∈ I or a ∈ I, so a ∈ I by induction on m.

(4) The radical (0) of the zero ideal (0) is the set of nilpotent elements of A. It is called
nilradical of A.
Lemma 3.7: Let A be a ring and I ⊆ A an ideal. Let M be any set of radical ideals containing I.
Then,
√ \
I⊆ J.
J∈M
√ √
P ROOF. By Remark I.3.6, the inclusion I ⊆ J implies I ⊆ J = J, so we are done. □

It will turn out that we get equality in the above lemma if we let J run over the set of all
prime ideals containing I. For finitely generated algebras over fields, we can even let J run over all
maximal ideals. To prove this, we use the following trick, which will imply both statements:
Proposition 3.8 (Rabinowitsch trick): Let A be a ring and I ⊆ A an ideal. Then,
√ \
I= p,
p∈M

where
M = {p ⊆ A | p = A ∩ m with m ⊆ A[x] is maximal and I ⊆ p.}
√
In particular, I is an ideal.
P ROOF. ⊆: By Corollary I.1.12, the ideal p = A ∩ m is prime and prime ideals are radical by
Remark I.3.6, soTthe inclusion holds by Lemma I.3.7.
⊇: Let a ∈ p∈M p and consider the ideal
J := (I ∪ {ax − 1}) ⊆ A[x].
 3. RECONSTRUCTING IDEALS FROM ZERO SETS: HILBERT’S NULLSTELLENSATZ V2 15

Claim: J = A[x].
Proof of claim: If J ̸= A[x], then there exists a maximal ideal m ⊆ A[x] with J ⊆ m. Since
I ⊆ J ∩ A ⊆ m ∩ A, we have m ∩ A ∈ M and thus a ∈ m. On the other hand, clearly ax − 1 ∈ m.
But then
1 = a · x − (ax − 1) ∈ m,
so m = A[x], a contradiction. Thus, we must have J = A[x]. ■

Since J = A[x], we can write

X
1= gi bi + g(ax − 1)

with g, gi ∈ A[x] and bi ∈ I. Consider the evaluation map of A-algebras

φ = evx−1 : A[x] → A[x, x−1 ]
f 7→ f (x−1 ).
Applying φ to our expression of 1 and multiplying by xm for some large enough m ≥ 1 to make
everything a polynomial, we get
X
xm = hi bi + h(a − x) ∈ A[x],

where hi = xm gi (x−1 ), h = xm−1 g(x−1 ), and note that the bi do not change since they do not
involve the variable x. Evaluating this expression at a, we obtain
X
am = hi (a)bi ,

so am ∈ I, as desired. □
Corollary 3.9: Let A be a ring and I ⊆ A an ideal. Then,
√ \
I= p.
I⊆p
p prime ideal

P ROOF. The inclusion ⊆ follows from Lemma I.3.7. In Proposition I.3.8, all members of M
are prime, so we get the other inclusion. □
Corollary 3.10: Let A be a finitely generated algebra over a field and I ⊆ A an ideal. Then,
√ \
I= m.
I⊆m
m maximal ideal

P ROOF. The inclusion ⊆ follows from Lemma I.3.7. In Proposition I.3.8, all members of M
are maximal by Corollary I.2.18, so we get the other inclusion. □
Theorem 3.11 (Hilbert’s Nullstellensatz v2): Let k be an algebraically closed field and I ⊆
k[x1 , . . . , xn ] an ideal. Then,
√
I(V (I)) = I.
16 I. THE SPECTRUM OF A RING
√
P ROOF. ⊇: Let f ∈ I and choose m ≥ 1 such that f m ∈ I. Let P = (a1 , . . . , an ) ∈ V (I) ⊆
k n . Then, f m (P ) = 0, so f (P ) = 0 and thus f ∈ I(V (I)).
⊆: Let f ∈ I(V (I)). By Corollary I.3.10, it suffices to show that f lies in every maximal ideal
m of k[x1 , . . . , xn ] with I ⊆ m. So, let m be such a maximal ideal. By Corollary I.2.19, there exists
a point P = (a1 , . . . , an ) with m = mP . Note that P = V (mP ) and mP is exactly the kernel of the
evaluation map at P , so I ⊆ mP implies P ∈ V (I). As f ∈ I(V (I)), we deduce that f (P ) = 0,
hence f ∈ mP , as desired. □
Summing up everything we obtained so far, we deduce:
Corollary 3.12: Let k be an algebraically closed field. Then, the maps
{X ⊆ k n | X affine algebraic set} ↔ {I ⊆ k[x1 , . . . , xn ] | I radical ideal}
X 7→ I(X)
V (I) ←[ I
are mutually inverse inclusion reversing bijections.
These bijections behave well with respect to intersections and unions:
Proposition 3.13: Let k[x1 , . . . , xn ] be a polynomial ring over an algebraically closed field k.
(1) For two ideals I, J ⊆ k[x1 , . . . , xn ], we have
V (I) ∪ V (J) = V (I ∩ J).
(2) For any non-empty set S of subsets of k[x1 , . . . , xn ], we have
\ [
V (M ) = V ( M)
M ∈S M ∈S

P ROOF. (1): The inclusion V (I)∪V (J) ⊆ V (I ∩J) is clear by Lemma I.3.2. For the converse,
let P ∈ V (I∩J) and assume without loss of generality that P ̸∈ V (I). Then, there exists f ∈ I with
f (P ) ̸= 0. Let g ∈ J. Since f g ∈ I ∩ J and P ∈ V (I ∩ J), we have 0 = (f g)(P ) = f (P )g(P ).
As f (P ) ̸= 0, we must have g(P ) = 0, hence P ∈ V (J), as desired.
(2): This is immediate from the definitions. □
If you have some background in topology, you will observe that the above result implies that
affine algebraic sets behave like closed subsets of a topology on k n :
Definition 3.14: Let X be a set. A set τ of subsets of X is called topology on X if it satisfies the
following three properties:
(1) ∅, X ∈ τ .
(2) Arbitrary unions of elements of τ are in τ .
(3) Finite intersections of elements of τ are in τ .
The elements of τ are then called open subsets of X (w.r.t. τ ) and complements of elements of τ
are called closed subsets of X.
With this definition, the following is immediate from Proposition I.3.13:
Corollary 3.15: Let τ be the set of complements of affine algebraic sets in k n , where k is an
algebraically closed field. Then τ is a topology on k n , called Zariski topology.
 3. RECONSTRUCTING IDEALS FROM ZERO SETS: HILBERT’S NULLSTELLENSATZ V2 17

Remark 3.16 (Geometric interpretation of maximal and prime ideals): We give three remarks that
are crucial to develop intuition for the topology on the spectrum of a ring which we will introduce
next lecture. In the following, k denotes an algebraically closed field:
(1) If P ∈ k n is a point, then I(P ) is, by definition, the kernel of the evaluation map at P ,
hence I(P ) = mP . By Corollary I.2.19, every maximal ideal arises in this way, so under
the correspondence of Corollary I.3.12, points in k n correspond to maximal ideals. In
particular, we can think of V (I) as the set of all maximal ideals containing the ideal I.
(2) We call a topological space X irreducible if X ̸= ∅ and whenever X = X ′ ∪ X ′′ for two
closed subsets X ′ and X ′′ , then X = X ′ or X = X ′′ . Equivalently, irreducible spaces are
exactly those that cannot be written as a finite union of proper closed subsets.
In the Zariski topology on k n , one can check quickly using Corollary I.3.12 that the
irreducible affine algebraic sets are exactly the affine algebraic sets corresponding to the
radical ideals that are prime.
(3) Let (X, τ ) be a topological space and let
X ′ = {Z ⊆ X | Z irreducible closed subset},
that is, the points of X ′ are the irreducible closed subsets of X. Then, there is a map
s : X → X′
x 7→ {x}.
We can define a topology τ ′ on X ′ by saying that U ′ ⊆ X ′ is open if and only if there
exists U ∈ τ such that
U ′ = {Z ∈ X ′ | Z ∩ U ̸= ∅}
One can check that this is indeed a topology on X ′ and that s is continuous with respect
to these topologies. 7
In particular, using this formalism we get a natural Zariski topology on the set of all
irreducible closed subsets of k n . Via the correspondence of Theorem I.3.11, this yields
a Zariski topology on the set of prime ideals of the polynomial ring over k. In the next
lecture, we will give an alternative description of this topology that works for arbitrary
rings.

7The map s is sometimes called “soberification”.

18 I. THE SPECTRUM OF A RING

4. Spectrum and Zariski topology

In this lecture, we mimic what we did in the previous two lectures for the polynomial ring for
arbitrary rings. Recall from Remark I.3.16 and Corollary I.3.15 that:
• Radical ideals in k[x1 , . . . , xn ] correspond to closed (in the Zariski topology) subsets of
kn .
• Prime ideals in k[x1 , . . . , xn ] correspond to irreducible closed subsets of k n .
• Maximal ideals in k[x1 , . . . , xn ] correspond to points in k n .
• The Zariski topology on k n extends in a natural way to the set of all irreducible closed
subsets of k n .
The naive approach to extend this to arbitrary rings would be look at the sets of maximal ideals
and define a topology on it. However, in contrast to Corollary I.2.18, preimages of maximal ideals
under arbitrary ring homomorphisms are not necessarily maximal, so associating to a ring the set
of maximal ideals is not a functorial operation. But preimages of prime ideals are always prime by
Corollary I.1.12 and it will turn out that the space of prime ideals is indeed a functorial invariant of
a ring.
Definition 4.1: Let A be a ring.
(1) The (prime) spectrum of A is the set
Spec A := {p ⊆ A | p prime ideal}
of prime ideals of A.
(2) The maximal spectrum of A is the set
Spec max A := {m ⊆ A | m maximal ideal }
of maximal ideals of A.
(3) For an ideal I ⊆ A, the vanishing set of I is the set
V (I) := {p ∈ Spec A | I ⊆ p} ⊆ Spec A.
(4) For a subset X ⊆ Spec A, the vanishing ideal of X is the ideal
\
I(X) := p ⊆ A.
p∈X

Remark 4.2: Note (and check!) that:

(1) Being an intersection of radical ideals, the ideal I(X) is always radical.
(2) If A = k[x1 , . . . , xn ] with k an algebraically closed field and we consider only maximal
ideals, then Definition I.4.1 (3) and (4) translate via the correspondence of Theorem I.3.11
to Definition I.3.1.
Remark 4.3 (Every ring is a ring of functions on its spectrum): Let A be a ring and a ∈ A. Then,
a defines a function
a
a : Spec A → A/p
p∈Spec A
p 7→ a ∈ A/p,
where the right-hand side is a disjoint union.
 4. SPECTRUM AND ZARISKI TOPOLOGY 19

Note that a(p) = 0 ∈ A/p if and only if a ∈ p. So, the vanishing set of an ideal I is indeed a
“vanishing set”
V (I) = {p ∈ Spec A | a(p) = 0 ∀a ∈ I}
and the vanishing ideal of a subset X of Spec A is indeed a “vanishing ideal”
I(X) = {a ∈ A | a(p) = 0 ∀p ∈ X}.
Thus, from now on, it will be very helpful to think of elements of an arbitrary(!) ring as functions.
Remark 4.4: If k is an algebraically closed field, A = k[x1 , . . . , xn ], and m is a maximal ideal,
then A/m ∼= k 8.
Thus, if we take f ∈ k[x1 , . . . , xn ], restrict the associated function
a
f : Spec A → A/p
p∈Spec A

to maximal ideals and identify all the copies of k that we obtain on the right-hand side, we recover
the interpretation of polynomials as functions on k n from Remark I.4.3.
Example 4.5: Some more examples on how to think of ring elements as functions:
(1) If k = k̄ and A = k[x1 , . . . , xn ], let p = (x1 ). Then, a(p) = a(0, x2 , . . . , xn ) ∈
k[x2 , . . . , xn ] ∼
= k[x1 , . . . , xn ]/p.
(2) Let A = Z and n an integer. We know that
Spec A ∼ = {p ∈ Z>0 | p prime } ∪ {(0)}.
For a prime p ∈ Z>0 , we have n((p)) = n ∈ Z/pZ and n((0)) = n ∈ Z. Thus, for n ̸= 0,
V (n) =∼ {p ∈ Z>0 | p prime and p | n}
corresponds to the set of prime divisors of n via the description of Spec A that we used
above.
Next, we want to define a topology on Spec A. For this, we need the analogues of Lemma I.3.2
and Proposition I.3.13. The following is an exercise on Sheet 3:
Proposition 4.6: Let A be a ring.
(1) The maps I(−) and V (−) are inclusion reversing.
(2) If X ⊆ Spec A is a vanishing set, then V (I(X)) = X.
(3) For two ideals I, J ⊆ A, we have
V (I) ∪ V (J) = V (I ∩ J).
(4) For any non-empty set S of subsets of A, we have
\ [
V (M ) = V ( M ).
M ∈S M ∈S

Corollary 4.7: Let A be a ring and let τ be the set of complements of V (I) in Spec A. Then, τ is
a topology on Spec A, called Zariski topology.
8We saw this when we described m as the kernel of an evaluation map, but you can also apply Zariski’s lemma
I.2.17.
20 I. THE SPECTRUM OF A RING

Definition 4.8: Let A be a ring and a ∈ A. The principal open subset associated to a is
D(a) := {p ∈ Spec A | a ̸∈ p} = Spec A − V ((a)).
Remark 4.9: Note that every open subset U of Spec A is a union of principal open subsets. Indeed,
\ [ [
U = Spec A − V (I) = Spec A − V ((a)) = (Spec A − V ((a)) = D(a),
a∈I a∈I a∈I
hence the name principal open. Instead of letting the union run over all elements of I, we can also
take any set of generators of I.
Remark 4.10: Note (and check!) that the defining property of prime ideals implies that for a, b ∈
A, we have D(ab) = D(a) ∩ D(b).
We also have the following analogue of Hilbert’s Nullstellensatz, proved on Sheet 3:
Theorem 4.11: Let A be a ring and I ⊆ A an ideal. Then,
√
I(V (I)) = I.
We immediately get the following correspondence between closed subsets and radical ideals:
Corollary 4.12: Let A be a ring. Then, the maps
{X ⊆ Spec A | X vanishing set} ↔ {I ⊆ A | I radical ideal}
X 7→ I(X)
V (I) ←[ I
are mutually inverse inclusion reversing bijections.
Again, we can ask what happens to prime and maximal ideals under this correspondence:
Proposition 4.13: Let A be a ring and I ⊆ A a radical ideal.
(1) If I is prime, then V (I) = {I}.
(2) I is prime if and only if V (I) is irreducible.
(3) I is maximal if and only if V (I) is a point.
P ROOF. (1): Using Proposition I.4.6, we compute
\ \ [
{I} = V (J) = V (J) = V ( J) = V (I).
I∈V (J) J⊆I J⊆I

(2): If I is prime, then V (I) = {I} by (1). Since closures of irreducible spaces are irreducible
by Sheet 1 and singleton sets are clearly irreducible, we deduce that V (I) is irreducible.
Conversely, assume that V (I) is irreducible. Let a, b ∈ A with ab ∈ I and recall that D(ab) =
D(a)∩D(b). Since every prime ideal that contains I also contains ab, we have that V (I)∩D(ab) =
∅. Thus,
∅ = V (I) ∩ D(ab) = V (I) ∩ D(a) ∩ D(b) = (V (I) ∩ D(a)) ∩ (V (I) ∩ D(b)).
In other words, V (I)∩D(a) and V (I)∩D(b) are two open subsets of V (I) with empty intersection.
By Sheet 1, the irreduciblity of V (I) implies that one of them is empty, say V (I) ∩ D(a). This
 4. SPECTRUM AND ZARISKI TOPOLOGY 21

√ that a is contained in every prime ideal containing I. By Corollary I.3.9, we deduce that
implies
a ∈ I = I. Thus, I is prime.
(3): If I is maximal, then V (I) consists only of I.
√ p with I ⊆ p. In particular, p
Conversely, if V (I) is a point, then there is a unique prime ideal
is maximal. By Corollary I.3.9 and since I is radical, we have I = I = p. □
As mentioned in the beginning of this lecture, one of the main reasons for considering prime
ideals instead of maximal ideals is the following functoriality property of the spectrum (see Sheet
4):
Proposition 4.14: Let f : A → B be a ring homomorphism. Then, the map
f ∗ : Spec B → Spec A
p 7→ f −1 (p)
is continuous. More precisely, for every ideal I of A, it holds that (f ∗ )−1 (V (I)) = V (f (I)).
As an example, we can describe the case of quotient maps:
Proposition 4.15: Let A be a ring, let I ⊆ A be an ideal, and let π : A → A/I be the quotient
map. Then, the map
π ∗ : Spec A/I → Spec A
factors through a homeomorphism Spec A/I → V (I) with inverse
ψ : V (I) → Spec A/I
p 7→ p/I.
P ROOF. First, we show that π ∗ factors through V (I): Since π −1 (0) = I, we have I ⊆
π −1 (q) = π ∗ (q) for all q ∈ Spec A/I, so π ∗ indeed factors through V (I).
Next, one easily deduces from I ⊆ p and using the surjectivity of π that p/I is a prime ideal. 9
Similarly, it is easy to see that ψ is an inverse of π as a map of sets.
It remains to show that ψ is continuous. For this, it suffices to show that π ∗ is an open map and
for this it suffices to show that principal open sets map to principal open sets, since every open set
is a union of these. Now, for a ∈ A with image a ∈ A/I, we compute that
π ∗ (D(a)) = {π −1 (q) ⊆ A | q ∈ Spec A/I, a ̸∈ q}
= {p ∈ Spec A | I ⊆ p, a ̸∈ p}
= V (I) ∩ D(a),
as desired. □

9Caution: For general ring maps, images of ideals are not even necessarily ideals. Consider for example the inclusion
Z → Q and any non-zero ideal in Z. Also, even for surjective ring maps, images of prime ideals are not necessarily prime.
Consider for example the image of the prime ideal (x2 − y) under k[x, y] → k[x, y]/(y). So, here both surjectivity of π
and I ⊆ p are important.
CHAPTER II

Modules

22
 1. MODULES 23

1. Modules
After rings, the other main objects of this lecture are modules. Modules are to rings what vector
spaces are to fields:
Definition 1.1: Let A be a ring. An A-module (or module over A) is an Abelian group (M, +)
together with a scalar multiplication by A, which is a map
·:A×M → M
(a, m) 7→ a · m =: am
satisyfing the following two properties:
(1) Distributivity w.r.t. module addition: For all a ∈ A and m, n ∈ M : a(m + n) = am + an.
(2) Distributivity w.r.t. ring addition: For all a, b ∈ A and m ∈ M : (a + b)m = am + bm.
(3) Compatibility of multiplications: For all a, b ∈ A and m ∈ M : a(bm) = (ab)m.
(4) Compatibility with 1: For all m ∈ M : 1m = m.

Example 1.2:
(1) Modules over fields are nothing but vector spaces.
(2) Every ring A is a module over itself by taking the multiplication on A as scalar multipli-
cation.
(3) If f : A → B is a ring homomorphism, then B is a module over A via
a · b := f (a)b
with a ∈ A, b ∈ B.
(4) Every Abelian group M is a module over Z by setting
Xa
a·m= m
i=1
for a ≥ 0 and
−a
X
a·m=− m
i=1
for a < 0.
Definition 1.3: Let A be a ring and let M, N be A-modules. A map f : M → N is called A-
module homomorphism (or homomorphism of A-modules, or A-linear) if it satisfies the following
properties:
(1) For all m, m′ ∈ M : f (m + m′ ) = f (m) + f (m′ ).
(2) For all a ∈ A, m ∈ M : f (am) = af (m).
An A-module homomorphism f is called isomorphism if f is bijective 1. If there is an isomorphism
between two A-modules M and N , we write M ∼ = N.

Example 1.4:
1One easily checks that then f −1 is also A-linear
24 II. MODULES

(1) Morphisms of modules over a field k are nothing but k-linear maps.
(2) A map f : A → B of Abelian groups is a group homomorphism if and only if it is
a homomorphism of Z-modules. Indeed, in this case, Property (2) in Definition II.1.3
follows from Property (1) with the definition of scalar multiplication on Abelian groups
given in Example II.1.2 (4).
Many of the constructions we know from linear algebra work in the context of modules:
Definition 1.5: Let A be a ring and let M and M ′ be A-modules.
(1) An A-submodule (or submodule) of M is a subgroup N ⊆ M of (M, +) such that for all
a ∈ A and n ∈ N , it holds that an ∈ N . 2
(2) If N1 , N2 are two submodules of M , the (internal) sum of N1 and N2 is defined as
N1 + N2 := {n1 + n2 | n1 ∈ N1 , n2 ∈ N2 } ⊆ M.
(3) If N ⊆ M is a subset, the A-submodule generated by N is defined as
X
⟨N ⟩A := { ai ni | ai ∈ A, ni ∈ N }.
f inite

If N = {n1 , . . . , nk }, we also write ⟨n1 , . . . , nk ⟩A := ⟨N ⟩A . If A is clear from the

context, we drop it from the notation.
(4) We say that M is finitely generated if there is a finite subset N ⊆ M with ⟨N ⟩ = M .
Occasionally, we call such a module finite.
(5) If N ⊆ M is a submodule, the quotient of M by N is the quotient group M/N together
with the scalar multiplication
A × M/N → M/N
(a, m) 7→ am := (am).
(6) If f : M → M ′ is an A-linear map, the kernel of f is
ker(f ) := {m ∈ M | f (m) = 0} ⊆ M,
the image of f is
im(f ) := {f (m) ∈ N | m ∈ M } ⊆ M ′ ,
and the cokernel of f is
coker(f ) := M ′ /im(f )
(7) A sequence of A-linear maps
f g
. . . → M ′ → M → M ′′ → . . .
is called exact at M if ker(g) = im(f ). It is called exact if it is exact everywhere. A short
exact sequence is an exact sequence of the form
f g
0 → M ′ → M → M ′′ → 0
2With this module structure, the inclusion N ⊆ M is A-linear.
 1. MODULES 25

(8) If {Mi }i∈I is a set of A-modules, the product of the Mi is defined as the Cartesian product
Y
Mi
i∈I

with componentwise addition and scalar multiplication.

(9) If {Mi }i∈I is a set of A-modules, the direct sum of the Mi is defined as
M Y
Mi := {(mi )i∈I | mi = 0 for all but finitely many i ∈ I} ⊆ Mi
i∈I i∈I
3
with componentwise addition and scalar multiplication.
Remark 1.6: One easily checks that all the sets defined above are in fact modules, all subsets are
submodules, and all obvious maps are A-linear. In particular, if f : M → N is an A-linear map,
the isomorphism of Abelian groups M/ ker(f ) → im(f ) given by the homomorphism theorem in
group theory is an A-linear isomorphism.
Example 1.7: If A is a ring, the A-submodules of A are exactly the ideals.
Caution 1.8: If B is an A-algebra, then we have seen that B is also an A-module. We thus have
two competing notions of B being “finitely generated”, as an A-module and as an A-algebra. These
two notions do not agree. For example, if A is a field, then B is finitely generated as an A-module
if and only if it is finite-dimensional, while A[x] is finitely generated as A-algebra, but not finite-
dimensional.
Example 1.9: If A is a ring, M is an A-module, and I ⊆ A is an ideal, then
IM := ⟨{am | a ∈ I, m ∈ M }⟩ ⊆ M
is an A-submodule, hence M/IM is an A-module. In fact, the scalar multiplication A × M/IM →
M/IM induces a well-defined scalar multiplication A/I × M/IM → M/IM : Let b ∈ A/I and
let a, a′ ∈ A be two lifts of b to A, then for every m ∈ M/IM , we have
am = am = bm = a′ m = a′ m.
In particular, if m is a maximal ideal of A, then M/mM is a vector space over the field A/m. It may
be thus be helpful to think of M as a collection of vector spaces over the various fields A/m (and,
as we will see using localization later, over the fields Frac(A/p)), but be warned that the module
M carries more data than such a collection.
Remark 1.10 (Some universal properties): Let us record some universal properties of the objects
we defined above4:
(1) Given an A-linear map f : M → M ′ , let g : ker(f ) → M be the inclusion and g ′ : M ′ →
coker(f ) the quotient by the image. Then, the following hold:
(a) An A-linear map h : N → M factors (uniquely) through g if and only if f ◦ h = 0.
(b) An A-linear map h : M ′ → N factors (uniquely) through g ′ if and only if h ◦ f = 0.

3In particular, note that direct sum and product are equal if I is finite, but not in general.
4It is a good exercise to check at least some of them
26 II. MODULES
L
(2) Given a set {Mi }i∈I ofQ A-modules, let ιi : Mi → i∈I Mi be the inclusion of the i-
th factor and let πi : i∈I Mi → Mi be the projection onto the i-th factor. Then, the
following hold:
(a) Given an A-module N and L A-linear maps gi : Mi → N for all i ∈ I, there exists a
unique A-linear map g : i∈I Mi → N such that g ◦ ιi = gi .
(b) Given an A-module N and A-linear
Q maps gi : N → Mi for all i ∈ I, there exists a
unique A-linear map g : N → i∈I Mi such P that πi ◦ g = gi .
In (a), the map g is given by sending (mi )i∈I to gi (mi ), which is well-defined since
almost all mi are zero.
 2. HOM, TENSOR, AND CHANGE OF SCALARS 27

2. Hom, tensor, and change of scalars

As for vector spaces over fields, there are tensor products and duals of modules.
Definition 2.1: Let A be a ring and let M, M ′ , N be A-modules.
(1) The module of A-linear maps from M to N is the module
HomA (M, N ) := {f : M → N | f A − linear}
with pointwise addition and multiplication, that is,
f + g : a 7→ f (a) + g(a)
and
af : b 7→ af (b).
(2) The dual of M is M ∗ := HomA (M, A).
(3) A map f : M × M ′ → N is called A-bilinear if for all m ∈ M and m′ ∈ M ′ , the two
maps f (m, −) : M ′ → N and f (−, m′ ) : M → N are A-linear.
(4) A tensor product of M and M ′ is a pair (T, f ), where T is an A-module and f : M ×M ′ →
T is an A-bilinear map satisfying the following universal property:
Given any A-module N and any A-bilinear map g : M × M ′ → N , there exists a
unique A-linear map h : T → N such that h ◦ f = g.
Proposition 2.2: Let A be a ring and M, M ′ be A-modules. Then, the tensor product of M and
M ′ exists and is unique up to unique isomorphism.
P ROOF. By the usual uniqueness arguments for universal objects, it suffices to prove that M ⊗A
M ′ exists. 5 L
Consider the A-module M ×M ′ A consisting of finite L tuples of elements a(m,m′ ) ∈ A indexed
by elements of M and M ′ . Consider the submodule of M ×M ′ A generated by the following
elements, where a(m,m′ ) ∈ A, m, m1 , m2 ∈ M, m′ , m′1 , m′2 ∈ M ′ are arbitrary:
a(m,m′ ) − 1(am,m′ )
a(m,m′ ) − 1(m,am′ )
1(m1 ,m′ ) + 1(m2 ,m′ ) − 1(m1 +m2 ,m′ )
1(m,m′1 ) + 1(m,m′2 ) − 1(m,m′1 +m′2 ) .

Let M ×M ′ A → M ⊗A M ′ be the a(m,m′ ) m⊗m′ for the

L P
quotient by this submodule. We write
′ . Note that using the first two relations,
L
image of the tuple ((a(m,m′ ) )) ∈ M ×M ′ A in PM ⊗A M
′
every element in M ⊗A M can be written as m ⊗ m ′

By taking the quotient by the relations above, we made the map

f : M × M ′ → M ⊗A M ′
(m, m′ ) 7→ m ⊗ m′

5We will never use the explicit construction of M ⊗ M ′ described below again, so feel free to forget it after reading
A
this proof.
28 II. MODULES

A-bilinear. Moreover, one checks that given any A-bilinear map g : M × M ′ → N to some
A-module N , the map
M ⊗ M′ → N
X A X
m ⊗ m′ 7→ g(m, m′ )
is well-defined and the unique A-linear map with h ◦ f = g. □
Notation 2.3: From now on, we write M ⊗A M ′ for the tensor product of M and M ′ constructed
above and we let m ⊗ m′ be the image of (m, m′ ) in M ⊗A M ′ . Elements of the form m ⊗ m′
are called elementary tensors. Note that not every element of the tensor product is necessarily an
elementary tensor!
Lemma 2.4 (Identities for tensor products): Let A be a ring and let M, M1 , M2 , M3 , N be A-
modules.
(1) Tensoring is commutative: There is an A-linear isomorphism
M ⊗A N → N ⊗A M
X X
(mi ⊗ ni ) 7→ (ni ⊗ mi )
(2) Tensoring with the ring itself does not change the module: There is an A-linear isomor-
phism
A ⊗A M → M
a ⊗ m 7→ am
(3) Tensoring is distributive: There is an A-linear isomorphism
(M1 ⊕ M2 ) ⊗A N → (M1 ⊗A N ) ⊕ (M2 ⊗A N )
X X
(mi , m′i ) ⊗ n 7→ ((mi ⊗ n), (m′i ⊗ n))

(4) Tensoring is associative: There is an A-linear isomorphism 6

(M1 ⊗A M2 ) ⊗A M3 → M1 ⊗A (M2 ⊗A M3 )
X X
(mi ⊗ m′i ) ⊗ m′′i 7→ mi ⊗ (m′i ⊗ m′′i )
(5) Tensoring is functorial: If f : M1 → M2 is an A-linear map, then
f ⊗ id : M1 ⊗A N → M2 ⊗A N
X X
mi ⊗ ni 7→ f (mi ) ⊗ ni
is A-linear.
Using tensor products, we can “extend scalars”:
Definition 2.5: Let f : A → B be a ring homomorphism.
(1) For a B-module N , the restriction of scalars of N along f is N considered as an A-module
with scalar multiplication a · n := f (a)n. We write A N for this A-module.
6In particular, we can define M ⊗ M ⊗ M := (M ⊗ M ) ⊗ M and the way we set the brackets does
1 A 2 A 3 1 A 2 A 3
not matter.
 2. HOM, TENSOR, AND CHANGE OF SCALARS 29

(2) For an A-module M , the extension of scalars

P of M alongP f is M ⊗A B considered as a
B-module with scalar multiplication b · ( m ⊗ b′ ) := m ⊗ bb′ . We write MB for this
B-module.
For a ring A, let ModA be the collection of all A-modules. Then, for an A-module N , the
associations
ModA → ModA
M 7 → HomA (M, N )
M 7→ HomA (N, M )
M 7→ M ⊗A N
are natural in M , in the sense that if f : M → M ′ is A-linear, there are induced maps
◦f
HomA (M ′ , N ) → HomA (M, N )
f◦
HomA (N, M ) → HomA (N, M ′ )
f ⊗id
M ⊗A N → M ′ ⊗A N
the induced map of a composition is the composition of the induced maps and the induced map is
the identity if f is. We say that HomA (−, N ), HomA (N, −), (−) ⊗A N are functors from ModA to
ModA . Note that the maps induced by HomA (−, N ) go in the opposite direction than the original
f . Such functors are called contravariant. If the direction is not reversed, the functor is called
covariant. Given a functor ModA → ModA , a natural question to ask is what it does to exact
sequences.
Proposition 2.6 (Exactness properties of Hom and tensor): Let A be a ring, let N be an A-module.
(1) If
f g
M ′ → M → M ′′ → 0
is an exact sequence of A-modules, then the sequence
◦g ◦f
0 → HomA (M ′′ , N ) → HomA (M, N ) → HomA (M ′ , N )
is exact. One says that HomA (−, N ) is a left-exact contravariant functor.
(2) If
f g
0 → M ′ → M → M ′′
is an exact sequence of A-modules, then the sequence
f◦ g◦
0 → HomA (N, M ′ ) → HomA (N, M ) → HomA (N, M ′′ )
is exact. We say that HomA (N, −) is a left-exact (covariant) functor.
(3) If
f g
M ′ → M → M ′′ → 0
is an exact sequence of A-modules, then the sequence
f ⊗id g⊗id
M ′ ⊗A N → M ⊗A N → M ′′ ⊗A N → 0
is exact. We say that (−) ⊗A N is a right-exact (covariant) functor.
30 II. MODULES

P ROOF. We prove only (1). (2) is similar and (3) is on Sheet 5.

First, we need to check that ◦g is injective. So, let h : M ′′ → N be an A-linear map with
h ◦ g = 0, i.e., h is the 0-map on the image of g. Since g is surjective, h = 0.
Next, we need to check that ker(◦f ) = im(◦g).
⊇: This is clear, since g ◦ f = 0.
⊆: Let h : M → N with h ◦ f = 0. Since g : M → M ′′ is surjective with kernel equal to
the image of f , the homomorphism theorem (Remark II.1.6) implies that g satisfies the universal
property of the cokernel of f (see Remark II.1.10). Thus, h ◦ f = 0 implies that there exists a
unique A-linear map h′ : M ′′ → N such that h′ ◦ g = h. This is exactly saying that h has a (unique)
preimage under ◦g, that is, h ∈ im(◦g). □
Example 2.7: A functor ModA → ModA is called exact if it sends short exact sequences to short
exact sequences. None of the above three functors are exact in general. For this, consider the short
exact sequence
·2
0 → Z → Z → Z/2Z → 0
and observe that in ModZ :
(1) Applying HomZ (−, Z), we get a commutative diagram
◦(·2)
0 / HomZ (Z/2Z, Z) / HomZ (Z, Z) / HomZ (Z, Z)

∼
= id7→1 id7→1
  
0 /0 /Z ·2 /Z

where the vertical arrows are isomorphisms. In particular, ◦(·2) is not surjective.
(2) Applying HomZ (Z/2Z, −), we get a commutative diagram
0 / HomZ (Z/2Z, Z) / HomZ (Z/2Z, Z) / HomZ (Z/2Z, Z/2Z)

∼
= ∼
= id7→1
  
0 /0 /0 / Z/2Z

where the vertical arrows are isomorphisms. In particular, postcomposing with Z → Z/2Z
is not surjective.
(3) Applying (−) ⊗Z Z/2Z, we get a commutative diagram
(·2)⊗id
Z ⊗Z Z/2Z / Z ⊗Z Z/2Z / Z/2Z ⊗Z Z/2Z /0

a⊗b7→ab a⊗b7→ab a⊗b7→ab

 (·2)=0  
Z/2Z / Z/2Z / Z/2Z /0

where the vertical arrows are isomorphisms. In particular, (·2) ⊗ id is not injective.
This non-exactness observation leads to the definition of five natural classes of modules that we
will encounter again later this term:
Definition 2.8: Let A be a ring and M an A-module.
(1) M is called flat if (−) ⊗A M is exact.
 2. HOM, TENSOR, AND CHANGE OF SCALARS 31

(2) M is called projective if HomA (M, −) is exact.

(3) M is called injective if HomA (−, M ) is exact.
M is called free if there exists a set I and A-linear M ∼ L
(4) an isomorphism = i∈I A.
(5) M ∼ L
is called free of finite rank if M = i∈I A with I a finite set.
The reason why we mention free modules in this context is that they give us easy examples of
projective and flat modules.
Proposition 2.9: Let A be a ring and M a free A-module. Then, the following hold:
(1) M is projective.
(2) M is flat.
P ROOF. (1): If M ∼
L
= i∈I A, then, by the universal property of direct sums, there are natural
isomorphisms
HomA (M, N ) ∼ = i∈I HomA (A, N ) ∼
Q Y
= N
i∈I
g 7→ (g ◦ ιi )i∈I ,
(gi )i∈I 7→ (gi (1))i∈I .
Now, the claimed right-exactness of HomA (M, −) follows from
Q the simple
Q observation that for a
given surjective map g : N → N ′′ , the induced map i∈I g : i∈I N → i∈I N ′′ is surjective.
Q
(2): This is analogous to (1), using Lemma II.2.4 (3) (which also holds for infinite direct sums).
□
Remark 2.10:
(1) By Example II.2.7 (1), free modules, and even free modules of finite rank, are not neces-
sarily injective.
(2) We will see later that in fact all projective modules are flat.
Remark 2.11:
(1) Let M be an A-module. By the universal property of direct sums, a choice of generators
I ⊆ M uniquely determines a surjective A-linear map
M
A → M
i∈I
X
(ai )i∈I 7→ ai · i.
i∈I
Thus, every module is a quotient of a free module.
(2) By (1), an A-module M is finitely generated if and only if it is a quotient of a free module
of finite rank.
(3) If F : ModA → ModB is a right-exact functor that commutes (naturally) with taking
direct sums and such that F (A) is a finitely generated B-module, then F sends finitely
generated modules to finitely generated modules. Indeed, take any finitely generated mod-
ule M and an exact sequence
M
K→ A→M →0
i∈I
32 II. MODULES

with I finite, then

M
F (K) → F (A) → F (M ) → 0
i∈I
L
is exact and i∈I F (A) is a finite direct sum of finitely generated B-modules, hence
finitely generated.
As an immediate consequence of the third remark above, we obtain:
Corollary 2.12: Let A be a ring and M, N finitely generated A-modules. Then, M ⊗A N is a
finitely generated A-module.
 3. LOCALIZING RINGS AND MODULES 33

3. Localizing rings and modules

Let A be a ring and let X ⊆ Spec A be a subset. Thinking of elements of A as functions on
Spec A, is there a natural way of constructing functions on X from A? In Proposition I.4.15, we
have seen that the ring A/I is in a natural way a ring of functions on the closed subset V (I). To get
functions on other subsets, such as principal opens, we use the process of localization.
Definition 3.1: Let A be a ring and M an A-module.
(1) A multiplicative subset of A is a subset U ⊆ A such that 1 ∈ U and for any two u, u′ ∈ U
also uu′ ∈ U .
(2) Given a multiplicative subset U of A, we define an equivalence relation ∼ on the set of
pairs (u, m) with u ∈ U and m ∈ M via
(u, m) ∼ (u′ , m′ ) : ⇐⇒ ∃v ∈ U : vu′ m = vum′ .
7
We write m
u for the equivalence class of (u, m) under ∼ and call m the numerator and u
the denominator of (u, m).
(3) Let U ⊆ A be a multiplicative subset. The localization of M with respect to U is the
A-module
U −1 M := (U × M )/ ∼
with addition
U −1 M × U −1 M → U −1 M
m m′ mu′ + m′ u
( , ′ ) 7→
u u uu′
and scalar multiplication
A × U −1 M → U −1 M
m am
(a, ) 7→ .
u u
8

(4) Let U ⊆ A be a multiplicative subset. The canonical map to the localization with respect
to U is the map
εM : M → U −1 M
m
m 7→
1
9

Remark 3.2: Let A be a ring and U ⊆ A a multiplicative subset. One checks:

(1) U −1 A is a ring if we define multiplication by multiplying numerators and denominators.
(2) If M is an A-module, then U −1 M is an U −1 A-module if we define scalar multiplication
by multiplying numerators and denominators.

7It is straightforward to check that this indeed defines an equivalence relation on U × M .

8Again, it is straightforward to check that these maps are well-defined and make U −1 M into an A-module.
9And again, it is easy to check that this map is a well-defined A-linear map.
34 II. MODULES

(3) If f : M → N is an A-linear map, then

U −1 f : U −1 M → U −1 N
m f (m)
7→
u u
is A-linear (and even U −1 A-linear). From this definition, it is clear that U −1 (f ◦ g) =
U −1 f ◦ U −1 g and that U −1 id = id, so U −1 (−) is a functor from ModA to ModA (or even
ModU −1 A ).
Lemma 3.3 (Localization is exact): Let A be a ring and U ⊆ A a multiplicative subset. Let
f g
M ′ → M → M ′′
be an exact sequence of A-modules. Then,
U −1 f U −1 g
U −1 M ′ → U −1 M → U −1 M ′′
is exact.
P ROOF. Since U −1 (g) ◦ U −1 (f ) = U −1 (f ◦ g) = U −1 0 = 0, we have im(U −1 (f )) ⊆
ker(U −1 (g)).
Conversely, let m −1 (g)). Then, g(m) = 0 ∈ U −1 M ′′ . This means that there exists
u ∈ ker(U u 1
v ∈ U such that
vg(m) = vu0 = 0 ∈ M ′′ .
Since g is A-linear, we have vg(m) = g(vm), hence vm ∈ ker(g). As the original sequence is
exact, we deduce that there exists n ∈ M ′ with f (n) = vm. Now,
n f (n) mv m
U −1 f ( )= = = ,
uv uv uv u
hence m
u ∈ im(U −1 f ), as desired. □

In particular, if M ⊆ N is a submodule, we can and will consider U −1 M as a submodule of

U −1 N . Using this, one checks:
Lemma 3.4: Let A be a ring, M an A-module, and N, P ⊆ M submodules. Then:
(1) U −1 (N + P ) = U −1 N + U −1 P
(2) U −1 (N ∩ P ) = U −1 N ∩ U −1 P
(3) U −1 (M/N ) ∼= U −1 M/U −1 N as U −1 A-modules.
Lemma 3.5 (Universal property of localization): Let f : A → B be a ring homomorphism and
U ⊆ A a multiplicative subset with f (U ) ⊆ B × . Then, there exists a unique ring homomorphism
g : U −1 A → B with f = g ◦ εA .
P ROOF. Let g : U −1 A → B be a ring homomorphism with f = g ◦ εA , then for all a ∈ A and
u ∈ U it holds that
a a u
g( ) = g( ) · g( )−1 = f (a)f (u)−1 .
u 1 1
 3. LOCALIZING RINGS AND MODULES 35

Hence, g is unique if it exists. In fact, we can define g using the above formula if we can show that
′
it is well-defined. So, let ua′ ∈ U −1 A such that there exists v ∈ U with vu′ a = vua′ . Then,
f (a)f (u)−1 − f (a′ )f (u′ )−1 = f (au′ − a′ u)f (uu′ )−1
= f (v(au′ − a′ u))f (vuu′ )−1 = f (0) = 0.
□
Example 3.6: Let A be a ring and M an A-module.
(1) If A is an integral domain and U = A − {0}, then U −1 A = Quot(A).
(2) If 0 ∈ U , then U −1 M = 0. Indeed, for all u ∈ U and m ∈ M , we have 0um = u0m,
hence
m 0m 0
= = .
u 0u 0
In particular, since U is a multiplicative subset, we get U −1 M = 0 as soon as U contains
some nilpotent element.
(3) The set U = {a ∈ A | a is not a zero divisor} ⊆ A is a multiplicative subset. The
localization U −1 A is called total ring of fractions and one can check that U is the largest
multiplicative subset such that the canonical map εA : A → U −1 A is injective.
(4) If a ∈ A, the set U = {an | n ∈ Z≥0 } is a multiplicative subset. We write Ma := U −1 M
for the localization with respect to this subset and call it localization of M at a.
(5) If p ∈ Spec A, the set U = A − p is a multiplicative subset. We write Mp := U −1 M for
the localization with respect to this subset and call it localization of M at p.
(6) If A = Z and p is a prime number, then
a
Zp = { n | a ∈ Z, n ∈ N≥0 } ⊆ Q
p
while
a
Z(p) = { | a, b ∈ Z, p ∤ b} ⊆ Q.
b
Since we have a canonical map A → U −1 A, we can ask what it does on spectra, similar to what
we did in Proposition I.4.15. We give more refined information in this case:
Proposition 3.7: Let A be a ring and U ⊆ A a multiplicative subset.
(1) Every ideal J ⊆ U −1 A satisfies J = U −1 (ε−1 −1
A (J)) = (εA (εA (J))).
Every ideal I ⊆ A satisfies ε−1 −1 I) =
S
(2) A (U u∈U (I : (u)).
(3) For an ideal I ⊆ A, we have U I = U −1 A if and only if U ∩ I ̸= ∅.
−1

(4) Let
X = {p ∈ Spec A | U ∩ p = ∅}.
Then, the map
ε∗A : Spec U −1 A → Spec A
factors through a homeomorphism Spec U −1 A → X with inverse
ψ : X → Spec U −1 A
p 7→ U −1 p
36 II. MODULES

P ROOF. (1): Given an ideal J ⊆ U −1 A, set I = ε−1 A (J). We claim that J = (εA (I)). The
inclusion “⊇” is clear. For “⊆”, let ua ∈ J. Then, a1 = u1 · ua ∈ J, hence a ∈ I, and so ua ∈ (εA (I)),
as claimed.
Next, we claim U −1 I = (εA (I)). For “⊆”, it suffices a 1 a
Pn atoi observe that u = u · 1 ∈ (εA (I)) for
a ∈ I, U ∈ U . For “⊇”, given an arbitrary element i=1 ui ∈ (εA (I)), one can pass to a common
denominator.
(2): “⊆”: If a ∈ ε−1A (U
−1 I), there exists b ∈ I and u ∈ U with a = b ∈ U −1 A. Thus, there
1 u
exists v ∈ U withSvua = vb ∈ I, hence a ∈ (I : (uv)).
“⊇”: If a ∈ u∈U (I : (u)), then there exists u ∈ U with ua ∈ I. Thus, a1 = ua −1 I,
u ∈ U
hence a ∈ ε−1A (U
−1 I).

(3): We have U −1 I = U −1 A if and only if ε−1A (U

−1 I) = A. By (2), this happens if and only if

there exists u ∈ U with 1 ∈ (I : (u)), hence if and only if there exists u ∈ U ∩ I.

(4): First, we show that ψ is well-defined:
Let p ∈ X. By Lemma II.3.4, we have U −1 A/U −1 p ∼ = U −1 (A/p) as U −1 A-modules, hence as
rings, since U A → U A/U p is surjective. By (3), the ring U −1 A/U −1 p ∼
−1 −1 −1
= U −1 (A/p) is not
the zero ring. Since p is prime, A/p is an integral domain. By Example II.3.6 (3), the localization
U −1 (A/p) is contained in the field of fractions of A/p, hence an integral domain, and so U −1 p is a
prime ideal.
Next, we show that ε∗A and ψ are inverse to each other. For p ∈ X, we have by (2):
[
ε∗A (ψ(p)) = ε−1
A (U
−1
p) = (p : (u))
u∈U

Since p is prime and U ∩ p = ∅, every element of A that multiplies u ∈ U into p must already be in
p, hence ε∗A (ψ(p)) = p.
For q ∈ Spec U −1 A, we have by (1):

ψ(ε−1
A (q)) = U
−1 −1
(εA (q)) = q.

It remains to show that ψ is continuous, or, equivalently, that ε∗A sends closed sets to closed sets.
So, let J ⊆ U −1 A be an ideal. By (1), we know that J = U −1 I for some ideal I ⊆ A. Then,

ε∗A (V (U −1 I)) = {ε−1

A (q) | q ∈ Spec U
−1
A, U −1 I ⊆ q}
= {p | p ∈ Spec A, U ∩ p = ∅, U −1 I ⊆ U −1 p}

We claim that this is nothing but V (I)∩X, hence closed. If I ⊆ p, then U −1 I ⊆ U −1 p by exactness
of localization. Conversely, if U −1 I ⊆ U −1 p, then

I ⊆ ε−1
A (U
−1
I) ⊆ ε−1
A (U
−1
p) = p.

Thus, ε∗A (V (U −1 I)) = V (I) ∩ X. □

In general, the set X is neither open nor closed. For localization at single elements, it is:
Corollary 3.8: Let A be a ring and a ∈ A. Then, the canonical map ε : A → Aa induces a
homeomorphism
Spec Aa → D(a).
 3. LOCALIZING RINGS AND MODULES 37

P ROOF. Let U = {an }n∈Z≥0 . If p ∈ Spec A is a prime ideal, then U ∩ p ̸= ∅ if and only if
an ∈ p for some n ≥ 0. Since p is prime, this is equivalent to a ∈ p and thus to p ∈ V ((a)). Hence,
the X of Proposition II.3.7 is nothing but D(a). □
Recall the following definition from the exercise sheets:
Definition 3.9: A ring A is called local if it has a unique maximal ideal.
Corollary 3.10: Let A be a ring and p ∈ Spec A. Then, Ap is a local ring and the image of
Spec Ap → Spec A
consists of all prime ideals q with p ∈ V (q). The maximal ideal of Ap is pAp := (εA (p))
P ROOF. Again, we determine the X from Proposition II.3.7. We have U = A − p. Hence, for
a prime ideal q ∈ Spec A, we have U ∩ q = ∅ if and only if q ⊆ p, or, equivalently, p ∈ V (q).
To show that Ap is a local ring, we have to show that its spectrum contains a unique closed
point. Equivalently, since the two spaces are homeomorphic, we have to show that X contains a
unique closed point. The closed subsets of X are those of the form
X ∩ V (I) = {q ∈ Spec A | I ⊆ q ⊆ p}
for some ideal I. If X ∩ V (I) ̸= ∅, then X ∩ V (I) contains p and if I = p, then X ∩ V (I) = {p}.
Thus, p is the unique closed point of X and hence its preimage under ε∗A is the unique maximal
ideal of Ap . □
Example 3.11: The name “localization” is motivated by Corollary II.3.8 and Corollary II.3.10:
By localizing at an element a, we forget all points of our space Spec(A) that lie outside D(a).
By localizing at a prime p, we replace the space Spec(A) by the subspace Spec(Ap ) that consists
exactly of all points that lie in every open neighborhood of p. From this, it is clear that localizing at
a prime ideal is “more local” than localizing at an element.
If A = k[x] with k an algebraically closed field, a = x and p = (x), then Spec(Aa ) =
Spec(A) − {(x)} while Spec(Ap ) = {(0), (x)}. In pictures (the little circle in the second picture
signifies that we remove the point (x)):
 CHAPTER III

Chain conditions

38
 1. NOETHERIAN SPACES, RINGS, AND MODULES 39

1. Noetherian spaces, rings, and modules

The topological spaces that we obtain as spectra of rings can be quite wild: For example, we
saw points that are not closed. The goal of this chapter is to understand how putting conditions on
chains of closed subsets (resp. on chains of ideals on the ring-theoretic side) of the spectrum yields
more controllable classes of rings.
Definition 1.1: Let X be a topological space. X is called Noetherian if for every descending chain
of closed subsets
Z1 ⊇ Z2 ⊇ . . . ⊇ Zn ⊇ . . .
there exists n0 > 0 such that Zn = Zn′ for all n, n′ ≥ n0 .
Obviously, every closed subspace of a Noetherian topological space is again Noetherian. One
of the key features of Noetherian spaces is that they can be decomposed into irreducible subspaces:
Proposition 1.2: Let X be a Noetherian topological space. Then, the following hold:
(1) There exist finitely many closed irreducible subsets Z1 , . . . , Zn ⊆ X with
n
[
X= Zi
i=1
and Zi ̸⊆ Zj for i ̸= j.
(2) For any collection of Zi as in (1), every irreducible subset Z ⊆ X is contained in some
Zi .
(3) For any collection of Zi as in (1), the Zi are exactly the maximal irreducible subsets of X.
In particular, the Zi are uniquely determined up to reordering.
The Zi are called irreducible components of X.
P ROOF. Observe: Since X is Noetherian, every non-empty set of closed subsets of X has a
minimal element (otherwise we can produce an infinite descending chain).
(1): Let M be the set of closed subsets of X for which a decomposition as in (1) does not exist.
We want to show that M = ∅. If M ̸= ∅, we find a minimal element Y ∈ M . Then, Y is not
irreducible, so we find Y1 , Y2 ⊊ Y closed with Y = Y1 ∪ Y2 . Since Y was minimal in M , both Y1
and Y2 have a decomposition as in (1), hence we can write Y as a finite union of irreducible closed
subsets. Deleting the subsets that are contained on others, we get a decomposition as in (1) for Y ,
contradicting our assumption. Hence, M = ∅.
(2): Fix a collection of Zi as in (1) and let Z ⊆ X be irreducible. Then,
[n
Z= (Zi ∩ Z).
i=1
Since Zi ∩ Z ⊆ Z is closed and Z is irreducible, there exists an i with Zi ∩ Z = Z, hence Z ⊆ Zi ,
as claimed.
(3): Fix a collection of Zi as in (1).
If Z ⊆ X is a maximal irreducible subset, then Z ⊆ Zi for some i by (2), hence Z = Zi by
maximality. Thus, every maximal irreducible subset of X is one of the Zi .
Moreover, the Zi are maximal: If Zi ⊆ Z with Z irreducible, then by (2) there exists a Zj with
Z ⊆ Zj , hence Zi ⊆ Zj , so i = j and so Z = Zi . □
40 III. CHAIN CONDITIONS

Using the bijection of Corollary I.4.12, we can directly translate this to a condition on chains
of radical ideals in rings. We obtain a slightly stronger notion by putting a condition on chains
of arbitrary ideals. Since ideals of a ring are exactly the submodules, this allows us to extend the
definition of Noetherian also to modules:
Definition 1.3: Let A be a ring and M an A-module.
(1) M is called Noetherian (as an A-module) if for every ascending chain of submodules
M 1 ⊆ M 2 ⊆ . . . ⊆ Mn ⊆ . . .
there exists n0 > 0 such that Mn = Mn′ for all n, n′ ≥ n0 . If such an n0 exists, we say
that the chain stabilizes at Mn0 .
(2) A is called Noetherian (as a ring) if it is so as a module over itself.
With this terminology, we immediately deduce the following algebraic properties of Noetherian
rings:
Corollary 1.4: Let A be a Noetherian ring. Then, the following hold:
(1) Spec A is Noetherian.
(2) A contains only finitely many
√ minimal prime ideals.
(3) If I ⊆ A is an ideal, then I is the intersection of finitely many prime ideals.
P ROOF. (1): The condition on chains of ideals in A implies the corresponding condition for
radical ideals, hence this follows directly from Corollary I.4.12.
(2): By Proposition I.4.13, Corollary I.4.12 translates prime ideals to irreducible closed subsets,
so this follows from (1) and Proposition III.1.2.
(3): By Corollary I.3.9, we have
√ \
I= p.
p∈V (I)

As a subspace of the Noetherian space Spec A, the topological space V (I) is Noetherian, hence
contains finitely many maximal irreducible subsets. By Proposition I.4.15, these correspond to
prime ideals that are minimal among those containing I, so we are done. □
Example 1.5: Let us check whether the rings we know are Noetherian:
(1) Every field is Noetherian, since its only ideals are the trivial ones.
(2) Every PID is Noetherian. 1 In particular, Z and k[x] are Noetherian.
(3) The ring k[{xi }i∈N ] is not Noetherian. Indeed, the chain
(x1 ) ⊆ (x1 , x2 ) ⊆ . . .
does not stabilize.
Remark 1.6: The converse of Corollary III.1.4 (1) is false. For example:
Let k be a field and A = k[{xi }i∈N ]/(x21 , x22 , . . .). Then, the ascending chain of ideals
(x1 ) ⊊ (x1 , x2 ) ⊊ . . .

1See Theorem 3.13 in the notes on “Einführung in die Algebra”

 1. NOETHERIAN SPACES, RINGS, AND MODULES 41
√
does not stabilize and so A is not √ Noetherian. However, note that√ the nilradical of A is 0 =
(x1 , x2 , . . .), Spec A ∼
= Spec A/ 0 by Proposition I.4.15, and A/ 0 ∼= k, hence Spec A is just a
single point and so clearly Noetherian as a topological space.
We have the following alternative characterization for Noetherian modules:
Theorem 1.7: Let A be a ring and M an A-module. The following are equivalent:
(1) M is Noetherian.
(2) Every submodule of M is finitely generated.
P ROOF. (1) ⇒ (2) : We show that if M contains a submodule N ⊆ M such that N is not
finitely generated, then M is not Noetherian.
Choose n1 ∈ N . Since N is not finitely generated, we have ⟨n1 ⟩A ̸= N , hence there exists
n2 ∈ A − ⟨n1 ⟩A . Now, we have
⟨n1 ⟩A ⊊ ⟨n1 , n2 ⟩A ⊊ N.
Again, the inclusion on the right is strict because N is not finitely generated. Repeating this process,
we find a non-stabilizing chain
⟨n1 ⟩A ⊊ . . . ⊊ ⟨n1 , . . . , nk ⟩A ⊊ . . . ,
so M is not Noetherian.
(2) ⇒ (1) : Let
N1 ⊆ N2 ⊆ . . . ⊆ Nn ⊆
S
be a chain of submodules of M . Then, N := i>0 Ni is also a submodule of M : Indeed, since
every finite set of elements is contained in a common Ni , so N is closed under addition and scalar
multiplication. By assumption, N is finitely generated, say by n1 , . . . , nk . By the same observation
as above, the finite set {n1 , . . . , nk } is contained in some common Ni , hence Ni = N and so the
chain stabilizes. Thus, M is Noetherian. □
One of the reasons why we consider the Noetherian property is that behaves well with respect
to many of the operations we saw so far. As a first example, we study sub- and quotient modules:
Proposition 1.8: Let A be a ring, M an A-module, and N ⊆ M a submodule. Then, the following
are equivalent:
(1) M is Noetherian.
(2) N and M/N are Noetherian.
P ROOF. (1) ⇒ (2): Assume that M is Noetherian. Every ascending chain in N is an ascending
chain in M , so N is Noetherian. If
U1 ⊆ . . . ⊆ Un ⊆ . . .
is an ascending chain of submodules in M/N and π : M → M/N is the quotient map, then
π −1 (U1 ) ⊆ . . . ⊆ π −1 (Un ) ⊆ . . .
is an ascending chain in M , hence our assumption that M is Noetherian guarantees that it stabilizes.
Since π is surjective, we have π(π −1 (Un )) = Un , so the original chain stabilizes as well.
(2) ⇒ (1): Let
M1 ⊆ M2 ⊆ . . . ⊆ M n ⊆ . . .
42 III. CHAIN CONDITIONS

be an ascending chain of submodules of M . The induced sequence

π(M1 ) ⊆ π(M2 ) ⊆ . . . ⊆ π(Mn ) ⊆ . . .
of submodules of M/N stabilizes, since M/N is Noetherian. Similarly, the sequence
M1 ∩ N ⊆ M2 ∩ N ⊆ . . . ⊆ Mn ∩ N ⊆ . . .
stabilizes. Choose n0 > 0 such that Mn ∩ N = Mn′ ∩ N and π(Mn ) = π(Mn′ ) for all n, n′ ≥ n0 .
We claim that then also Mn = Mn′ . By symmetry, it suffices to prove the inclusion ⊆:
If m ∈ Mn , then π(m) ∈ π(Mn ) = π(Mn′ ), so there exists m′ ∈ Mn′ such that π(m) =
π(m′ ). Thus, π(m−m′ ) = 0. The kernel of π|Mn is Mn ∩N , hence m−m′ ∈ Mn ∩N = Mn′ ∩N .
Thus,
m = (m − m′ ) + m′ ∈ Mn′ ,
as desired. □
Using this, we can show that finitely generated modules and algebras over Noetherian rings are
again Noetherian. First, we treat the case of modules:
Theorem 1.9: Let A be a Noetherian ring and M an A-module. Then, the following are equivalent:
(1) M is Noetherian as an A-module.
(2) M is finitely generated.
In particular, submodules of finitely generated modules over Noetherian rings are finitely generated.
P ROOF. (1) ⇒ (2) : This is a special case of Theorem III.1.7.
(2) ⇒ (1) : First, assume that M is free of finite rank, i.e., there is an isomorphism M ∼
=
⊕n
A . Then, the claim follows by induction on n from Proposition III.1.8 by using the short exact
sequences
0 → A⊕(n−1) → A⊕n → A → 0.
For a general M , the finite generation implies that there is a short exact sequence of the form
0 → K → A⊕n → M → 0,
so M is Noetherian by Proposition III.1.8 and the free case treated above. □
Theorem 1.10: Let A be a Noetherian ring and B a finitely generated A-algebra. Then, B is
Noetherian.
P ROOF. Since every finitely generated A-algebra is a quotient of a polynomial ring A[x1 , . . . , xn ]
in finitely many variables, we may assume by Proposition III.1.8 that B = A[x1 , . . . , xn ]. 2 By in-
duction on n, we may assume that B = A[x].
We want to show that if A[x] is not Noetherian, then neither is A. So, assume that A[x] is not
Noetherian so that, by Theorem III.1.7, there exists an ideal I ⊆ A[x] that is not finitely generated.
Choose an element f1 ∈ I of the smallest degree d1 . Recursively, choose fi ∈ I − (f1 , . . . , fi−1 )
of the smallest degree di . Note that I − (f1 , . . . , fi−1 ) is non-empty in every step, because I is not

2Here, we are using that if A → A/I is a quotient map and A/I is Noetherian as an A-module, then also as an
A/I-module. This is because every A-submodule M of A/I is also an A/I-submodule, since the scalar multiplication
A × M → M factors through A/I × M → M .
 1. NOETHERIAN SPACES, RINGS, AND MODULES 43

finitely generated. Let ai be the leading coefficient of fi . This way, we have constructed a chain of
ideals
(1) (a1 ) ⊆ (a1 , a2 ) ⊆ . . .
in A. We claim that this sequence does not stabilize, which will finish the proof.
So, assume that there exists a k ≥ 1 such that (a1 , . . . , ak ) = (a1 , . . . , ak+1 ). Then, there exist
b1 , . . . , bk ∈ A with ak+1 = ki=1 bi ai . Consider the polynomial
P

k
X
(2) g = fk+1 − bi xdk+1 −di fi .
i=1
Then, since fk+1 ̸∈ (f1 , . . . , fk ), we have g ̸∈ (f1 , . . . , fk ). Note that in Equation (2), all polynomi-
als on the right-hand side have degree dk+1 , and coefficient of xdk+1 is exactly ak+1 − ki=1 bi ai =
P
0. Thus, deg(g) < dk+1 , contradicting the minimality in the choice of fi+1 . Hence, in Sequence
(1) all inclusions are strict, hence A is not Noetherian. □
Using our results from Chapter 1, Theorem III.1.10 has the following geometric significance:
Corollary 1.11 (Hilbert’s basis theorem): Let k be an algebraically closed field and X ⊆ k n an
affine algebraic set. Then, there exist finitely many polynomials f1 , . . . , fm ∈ k[x1 , . . . , xn ] such
that X = V ((f1 , . . . , fm , )).
Remark 1.12: The slogan of this section could be:
“Every sub- or quotient module of anything finitely generated over a Noetherian ring is again Noe-
therian.”
In particular:
(1) Every finitely generated Abelian group is Noetherian as a Z-module.
(2) For every field k, every finitely generated module over every finitely generated k-algebra
A = k[x1 , . . . , xn ]/(I) is Noetherian over A.
However, be warned that we did not prove that every subalgebra of a finitely generated algebra
over a Noetherian ring is again Noetherian and this is in fact false:
For example, take A = k, B = k[x, y] and C = k[x, xy, xy 2 , . . .]. By Theorem III.1.10, the
A-algebra B is Noetherian. However, in C, the chain of ideals
(x) ⊆ (x, xy) ⊆ . . . ⊆ (x, xy, . . . , xy n ) ⊆ . . .
does not stabilize. Indeed, you can check directly that xy n ̸∈ (x, xy, . . . , xy n−1 ) for every n ≥ 1. In
particular, C is not Noetherian. As a consequence, by Theorem III.1.10, C is not finitely generated
as a k-algebra.
44 III. CHAIN CONDITIONS

2. Artinian spaces, rings, and modules

We try to mimic the previous section by looking at chains that go in the opposite direction:
Definition 2.1: Let X be a topological space. X is called Artinian if for every ascending chain of
closed subsets
Z1 ⊆ Z2 ⊆ . . . ⊆ Zn ⊆ . . .
there exists N > 0 such that Zn = Zn′ for all n, n′ ≥ N .
Again, we obtain a slightly stronger condition by putting chain conditions on chains of arbitrary
ideals and generalize to modules.
Definition 2.2: Let A be a ring and M an A-module.
(1) M is called Artinian if for every descending chain of submodules
M1 ⊇ M2 ⊇ . . . ⊇ M n ⊇ . . .
there exists N > 0 such that Mn = Mn′ for all n, n′ ≥ n0 . If such an n0 exists, we say
that the chain stabilizes.
(2) A is called Artinian if it is so as a module over itself.
And as before, the following is an immediate consequence of Corollary I.4.12.
Corollary 2.3: Let A be an Artinian ring. Then, Spec A is Artinian.
Example 2.4: Let us check whether the rings we know are Artinian:
(1) Every field is Artinian, since its only ideals are the trivial ones.
(2) If k is a field and n > 0 is an integer, then A = k[x]/(xn ) is Artinian, since A has only
finitely many ideals, namely the (xi ) with 0 ≤ i ≤ n.
(3) The ring Z is not Artinian, as the chain
(2) ⊇ (4) ⊇ . . . ⊇ (2n ) ⊇ . . .
shows. Similarly, k[x] can be seen not to be Artinian by considering the chain of ideals
(xn ).
In particular, not every Noetherian ring is Artinian.
As for the Noetherian property, there are non-Artinian rings with Artinian spectrum, in fact the
same example as in Remark III.1.6 works.
Remark 2.5: Before we continue, let us note that being Artinian must be a very restrictive property
for rings: Indeed, for two ideals I and J in a ring A, we can use the products I n J m to construct
descending chains of ideals in A. If A is Artinian, all such chains stablize, so there must be some
interaction between any two ideals of A.
As an example of the above observation, we have the following finiteness result, which states
that the spectrum of an Artinian ring is just a finite union of closed points.
Theorem 2.6: Let A be an Artinian ring. Then, the following hold:
(1) A has only finitely many maximal ideals m1 , . . . , mn .
(2) Every prime ideal of A is maximal.
 2. ARTINIAN SPACES, RINGS, AND MODULES 45

P ROOF. (1): Seeking a contradiction, assume that there exists a sequence {mi }i∈N of pairwise
distinct maximal ideals. Set Ij := ji=1 mi . Clearly, Ij ⊆ Ij ′ for j > j ′ , so we get a descending
T
chain of ideals. Since A is Artinian, we deduce that there exists n0 > 0 such that
n\
0 +1 n0
\
mi = mi ,
i=1 i=1
Tn0
hence i=1 mi ⊆ mn0 +1 , and so there exists i ≤ n0 with mn0 +1 = mi 3, contradicting our assump-
tion.
(2): Let m1 , . . . , mn be the finitely many maximal ideals of A. 4 Set I = ni=1 mi and consider
Q
the chain of ideals I i with i ∈ N. Since A is Artinian, this sequence stabilizes. Let J = I i for
i ≫ 0.
Assume we can show that J = 0. Then, if p ∈ Spec A is a prime, we have J ⊆ p, hence
p contains a (non-empty) product of maximal ideals. Since p is prime, this means it contains a
maximal ideal, so it is itself maximal and we are done.
Thus, it remains to show that J = 0. Assume J ̸= 0 and consider the set
M = {J ′ ⊆ A | J ′ ideal with J ′ J ̸= {0}}.
Since J ̸= 0, M is non-empty. By the chain condition, M must contain a minimal element J.
b Pick
an element x ∈ J with xJ ̸= {0}. By minimality, we have J = (x) and x. Since J = I = I i =
b b 2 2i

J for i ≫ 0, we have
(x)J · J = (x)J 2 = (x)J ̸= 0,
hence (x)J = (x), again by minimality. Thus, there exists y ∈ J with xy = x, or, equivalently,
x(y − 1) = 0. Since J is contained in every maximal ideal, so is y. But then (y − 1) is contained in
no maximal ideal (otherwise 1 would be in a maximal ideal, so that A = 0). Thus, (y − 1) is a unit
and then x(y − 1) = 0 implies x = 0 and in turn J = 0, contradicting our assumption. □
There is no characterization for Artinian modules analogous to Theorem III.1.7. However, the
analogue of Proposition III.1.8 holds with essentially the same proof (just invert the inclusions):
Proposition 2.7: Let A be a ring, M an A-module, and N ⊆ M a submodule. Then, the following
are equivalent:
(1) M is Artinian.
(2) N and M/N are Artinian.
Next, we want to compare the Noetherian and Artinian properties. To do this conceptually, we
first introduce the notion of simple modules, length, and composition series:
Definition 2.8: Let A be a ring and M an A-module.
(1) M is called simple if M ̸= 0 and its only A-submodules are 0 and M .
(2) For a finite chain of A-submodules
M = M0 ⊋ . . . ⊋ Mn
3Recall what this means geometrically: We have a closed point V (m Sn0
n0 +1 ) contained in a finite union i=1 V (mi ) =
Tn0
V ( i=1 mi ) of closed points, so it has to be equal to one of these points.
4We assume wlog that A ̸= 0.
46 III. CHAIN CONDITIONS

(strict inclusions!) the integer n is called length of the chain.

(3) A composition series for M is a chain of A-submodules of the form
M = M 0 ⊋ . . . ⊋ Mn = 0
such that each Mi /Mi+1 is a simple A-module.
(4) The length lengthA (M ) of M over A is the length of any composition series of M (and
we set lengthA (M ) := ∞ if M does not admit a composition series).
We need to check that the length is well-defined:
Proposition 2.9: Let A be a ring and M an A-module with a composition series of finite length.
Then, every composition series of M has the same length and every chain of submodules can be
extended to a composition series.
P ROOF. Write ℓA (M ) for the minimal length of a composition series of M (with ℓA (M ) := ∞
if there is composition series). We first prove two preliminary claims:
Claim 1: If N ⊊ M is a proper submodule, then ℓA (N ) < ℓA (M ).
Proof of Claim 1: Let
M = M 0 ⊋ . . . ⊋ Mn = 0
be a composition series for M and set Ni = N ∩ Mi . Then, Ni /Ni+1 ⊆ Mi /Mi+1 . The latter
module is simple, hence either Ni = Ni+1 or Ni /Ni+1 = Mi /Mi+1 . Thus, removing repetitions,
the chain
N = N0 ⊋ . . . ⊋ Nn′ = 0
is a composition series of length n′ ≤ n. If Ni /Ni+1 = Mi /Mi+1 for all i, then N = M by
induction on n, contradicting our assumption. Hence, n′ < n and thus ℓA (N ) < ℓA (M ). ■
Claim 2: Any chain in M has length at most ℓA (M ).
Proof of Claim 2: Let
M0 ⊋ . . . ⊋ M k
be a chain of length k. By Claim 1, we have
ℓA (Mk ) < . . . < ℓA (M0 ) ≤ ℓA (M ).
But ℓA (Mk ) ≥ 0, hence k ≤ ℓA (M ). ■

Now, if
M = M 0 ⊋ . . . ⊋ Mn = 0
is a composition series for M , then n ≤ ℓA (M ) by Claim 2, hence n = ℓA (M ) by definition of
ℓA (M ). Thus, any two composition series for M have the same length. The claim about extensions
follows from a straightforward downward induction on the length of the chain. □
Proposition 2.10: Let A be a ring and M an A-module. Then, the following are equivalent:
(1) M is both Noetherian and Artinian.
(2) lengthA (M ) < ∞.
 2. ARTINIAN SPACES, RINGS, AND MODULES 47

P ROOF. (2) ⇒ (1) : By Proposition III.2.9, the length of every finite chain is bounded by
lengthA (M ), hence all ascending and descending chains stabilize.
(1) ⇒ (2) : If M = 0, we are done. If not, then, since M is Noetherian, the set of proper
submodules of M has a maximal element M1 ⊊ M . Repeating this for M1 (which is Noetherian
by Proposition III.1.8), we find a chain
M ⊋ M1 ⊋ . . . ⊋ Mn ⊋ . . . .
Since M is Artinian, this process must terminate, that is, we must have Mn = 0 for some n ≫ 0.
Since we chose the submodules maximal in every step, the quotients Mi /Mi+1 are simple, so this
is a composition series of finite length, as desired. □
For a vector space M over a field A, the situation is simpler. You can check that dimA M =
lengthA M 5 by observing that M is simple if and only if dimA M = 1. Moreover, both chain
conditions coincide in this case:
Lemma 2.11: Let M be a vector space over a field A. Then, the following are equivalent:
(1) M is Artinian.
(2) M is Noetherian.
(3) lengthA M < ∞.
P ROOF. By Proposition III.2.10, (3) implies (1) and (2).
For the converse statements, assume that lengthA M = ∞, so that we find an infinite sequence
m1 , . . . , mn , . . . of linearly independent elements of M . Then, the chains
⟨m1 ⟩ ⊊ . . . ⊊ ⟨m1 , . . . , mn ⟩ ⊊ . . .
and
⟨{mi }i>0 ⟩ ⊋ . . . ⊋ ⟨⟨{mi }i>n ⟩⟩ ⊋ . . .
are both infinite, so M violates both chain conditions. □
Theorem 2.12: Let A be a ring. The following are equivalent:
(1) A is Artinian.
(2) A is Noetherian and every prime ideal of A is maximal.
P ROOF. We first prove the theorem under the following extra assumption:
(∗) : There exist finitely many maximal ideals mi such that ni=1 mi = 0.
Q

Assuming (∗), let Ij = ji=1 mi . This yields the chain of ideals

Q

(0) = In ⊆ . . . ⊆ I0 = A.
By Propositions III.1.8 and III.2.7, the ring A is Noetherian (resp. Artinian) if and only if all the
Ij /Ij+1 are Noetherian (resp. Artinian) as A-modules. Now, every A-submodule of Ij /Ij+1 is a
A/mj+1 -module, since Ij · mj+1 ⊆ Ij+1 , so every chain of A-submodules is, in particular, a chain
of A/mj+1 -submodules. Since mj+1 is maximal, A/mj+1 is a field, so ascending and descending
chain conditions coincide by Lemma III.2.11. Thus, (1) ⇔ (2) holds if (∗) holds. It remains to
show that both (1) and (2) imply condition (∗).
5In the sense that both numbers agree if they are finite and that one of them is infinite if and only if the other is.
48 III. CHAIN CONDITIONS

If (1) holds, we have seen in the proof of Theorem III.2.6 that the zero ideal can be written as a
product of finitely many maximal ideals. This is condition (∗).
If (2) holds, let p1 , . . . , pn be the (finitely many, by Corollary III.1.4) minimal primes of A. By
Corollary I.3.9, we have
p n
\ Yn
(0) = pi ⊇ pi =: I.
i=1 i=1
Thus, for every a ∈ I, there exists m > 0 with am = 0. Since A is Noetherian, I is finitely
generated, hence I N = 0 for some N > 0. By assumption, the pi are maximal, so we wrote the
ideal (0) as a product of finitely many maximal ideals as in condition (∗). □
 3. KRULL DIMENSION AND HEIGHT 49

3. Krull dimension and height

Our next goal is to define a notion of dimension for topological spaces and rings.
Definition 3.1: Let X be a topological space.
(1) For a finite chain of subsets of X
X0 ⊊ . . . ⊊ Xn
(strict inclusions!), the integer n is called length of the chain. The length of the empty
chain is defined to be −1.
(2) The Krull dimension dim(X) of X is the supremum of the lengths of all finite and strictly
increasing chains of irreducible closed subsets of X.
(3) If X = Spec A for a ring A, we set dim(A) := dim(Spec A).
(4) If M is a module over a ring A, we set dim(M ) := dim(A/Ann(M )).
Note that in the definition of dim(X), we consider only irreducible subsets. Otherwise, for
example every topological space with infinitely many closed points would have infinite dimension.

Example 3.2:
(1) If X is a singleton, then dim(X) = 0.
(2) If A is a field, then dim(A) = 0.
(3) For Z, the only strictly increasing chains of prime ideals are those of the form (0) ⊊ (p),
so dim(Z) = 1.
(4) Similarly, if A is any principal ideal domain and not a field, then dim(A) = 1.
(5) In particular, if k is a field, then dim(k[x]) = 1. If k is algebraically closed, this fits with
our intuition that Specmax (k[x]) ∼ = k is a 1-dimensional object.
(6) If k is a field, then dim(k[{xi }i∈N ]) = ∞.
Remark 3.3: If X is Noetherian and Z1 , . . . , Zn ⊆ X are its irreducible components, then
dim(X) = max{dim(Z1 ), . . . , dim(Zn ), −1}.
Indeed, by Proposition III.1.2, every chain of irreducible subsets of X is contained in a single
irreducible component (apply (2) to the largest subset in the chain).
Caution 3.4: Note that this does not imply that Noetherian rings have finite dimension. For an
example, see the bonus exercise on Sheet 8.
Restricting to chains starting at certain primes, we get the notion of height:
Definition 3.5: Let A be a ring.
(1) Let p ∈ Spec A be a prime ideal. The height ht(p) of p is the supremum over all lengths
of finite strictly increasing chains of prime ideals of the form
pn ⊊ . . . ⊊ p0 = p.
(2) Let I ⊆ A be an ideal. Then, the height ht(I) of I is defined as
ht(I) = min{ht(p) | I ⊆ p, p ∈ Spec A}.
50 III. CHAIN CONDITIONS

Remark 3.6: By Corollary II.3.10, we have ht(p) = dim(Ap ).

Proposition 3.7: Let A be a ring and p ∈ Spec A. Then,
dim(Ap ) + dim(A/p) ≤ dim(A).
P ROOF. By Proposition I.4.15, a chain of prime ideals of length n in A/p yields a chain of
prime ideals
p ⊆ p0 ⊊ . . . ⊊ pn .
By Proposition II.3.10, a chain of prime ideals of length m in Ap yields a chain of prime ideals
qm ⊊ . . . ⊊ q0 ⊆ p.
Concatenating the two chains, we get a chain of prime ideals in A. Thus, if dim(Ap ) or dim(A/p)
is infinite, so is dim(A). If both are finite, we can take n = dim(A/p) and m = dim(Ap ). By
maximality, we must have q0 = p = p0 , so the length of the concatenated chain is m + n. By
definition of Krull dimension, this is bounded above by dim(A), as desired. □
In rings A where equality holds in Proposition III.3.7 (we will encounter some of these later,
for example finitely generated integral domains over fields and local Cohen–Macaulay rings), we
can interpret ht(p) as the codimension of V (p) in Spec A.
Example 3.8: In general, the inequality in the above proposition is strict, because the left-hand side
of the inequality measures only chains of prime ideals that contain p:
Let k be a field and A = k[x, y]/(x, y)·(x−1). Then, A/p ∼ = k and Ap ∼ = k[x, y](x,y) /(x, y) ∼
=
k, where the first isomorphism holds because (x − 1) is a unit in the localization. Hence, in this
case dim(Ap ) + dim(A/p) = 0. However, A contains the chain
(x − 1) ⊊ (x − 1, y),
so dim(A) > 0.
We can rephrase Theorem III.2.12 as follows:
Theorem 3.9: Let A be a ring. Then, the following are equivalent:
(1) A is Artinian.
(2) A is Noetherian and dim(A) = 0.
 4. NAKAYAMA’S LEMMA AND THE PRINCIPAL IDEAL THEOREM 51

4. Nakayama’s lemma and the principal ideal theorem

Now that we have a notion of dimension, we would like to be able to compute it. One of the
key tools for this is the principal ideal theorem that we will prove in this lecture. On the way, we
will prove Nakayama’s lemma, which will turn out to be an extremely useful tool throughout all of
commutative algebra.
We begin with some results on adjugate matrices:
Lemma 4.1: Let A be a ring and B = (bi,j )i,j ∈ An×n a square matrix with entries in A. Let ci,j
be the determinant of the matrix obtained from B by deleting the i-th row and j-th column and let
C = ((−1)i+j ci,j )i,j ∈ An×n , called adjugate matrix of B. Then
CB = det(B)In .
P ROOF. You know this from linear algebra if A is a field. Now, assume that A′ = Z[{xi,j }ni,j=1 ],
B ′ = (xi,j )i,j and K = Q({xi,j }ni,j=1 ). For this matrix B ′ , the adjugate C ′ lies in (A′ )n×n and the
equality
(3) C ′ B ′ = det(B ′ )In
holds in K n×n by the field case. Since both sides lie in (A′ )n×n ⊆ K n×n , equality holds over
A′ . Now, there is a unique ring homomorphism φ : A′ → A that sends xi,j to bi,j . Applying
φ componentwise to (3), we obtain the result, since the determinants that appear are polynomial
expressions in the entries of the matrix B ′ , so forming them commutes with φ. □
This has the following consequence:
Lemma 4.2: Let A be a ring, M = ⟨m1 , . . . , mn ⟩A a finitely generated A-module, and bi,j ∈ A
with i, j = 1, . . . , n with
X n
bij mj = 0
j=1
for i = 1, . . . , n. Write B = (bij )i,j ∈ An×n . Then, det(B) ∈ Ann(M ).
P ROOF. Write m = (m1 , . . . , mn )t and let C be the adjugate of B. Then,
det(B)mi = (det(B)In m)i = (CBm)i = (0, . . . , 0)ti = 0.
□
Recall from the exercises that the Jacobson radical J(A) of a ring A is the intersection of all
its maximal ideals.
Theorem 4.3 (Nakayama’s lemma, version 1): Let A be a ring and let M be a finitely generated
A-module. Let I ⊆ A be an ideal.
(1) If IM = M , then there exists a ∈ 1 + I with aM = 0.
(2) If IM = M and I ⊆ J(A), then M = 0.

Pn P ROOF. (1): Choose generators m1 , . . . , mn ∈ M . Since IM = M , we can write mi =

n×n and let a = det(B). If
b m
j=1 ij j with bij ∈ I. Consider the matrix B = (δij − b )
ij i,j ∈ A
we reduce the entries of B modulo I, we obtain the identity matrix, hence a ∈ 1 + I. By Lemma
III.4.2, we have aM = 0, as desired.
52 III. CHAIN CONDITIONS

(2): Since I lies in all maximal ideals, a lies in none of them, for otherwise they would contain
1. Thus, a is a unit and hence aM = 0 implies M = 0. □
Nakayama’s lemma has surprisingly many consequences. We give only one immediate such
consequence here. Recall that we can think of a module M over a ring A as a collection of vector
spaces M/mM over the quotients A/m together with extra data. A natural question in this context
is whether a basis of M/mM can be extended to a generating set of M in a neighborhood of
m ∈ Spec A. This is the geometric interpretation of the special case where A is local with maximal
ideal m in the following corollary:
Corollary 4.4 (Nakayama’s lemma, version 2): Let A be a ring and let M be a finitely generated
A-module. Let I ⊆ A be an ideal and m1 , . . . , mn ∈ M .
(1) If M/IM = ⟨m1 , . . . , mn ⟩A , then there exists a ∈ 1+I such that Ma = ⟨ m11 , . . . , m1n ⟩Aa .
(2) If M/IM = ⟨m1 , . . . , mn ⟩A and I ⊆ J(A), then M = ⟨m1 , . . . , mn ⟩A .
P ROOF. (1): Let N = ⟨m1 , . . . , mn ⟩A ⊆ M . Then, M/N is a finitely generated A-module.
Observe that since M/IM = ⟨m1 , . . . , mn ⟩A , we can write every m ∈ M as m = m′ + ai mi
P
with m′ ∈ IM and so the class of m in M/N is in IM/N . So, we can apply Theorem III.4.3 to
M/N to obtain an a ∈ 1 + I with a(M/N ) = 0, or, equivalently, aM ⊆ N . Since localization
is a functor, this implies a(Ma ) ⊆ Na . But since a is a unit in Aa , we have a(Ma ) = Ma , hence
Ma ⊆ Na , as claimed.
(2): As in Theorem III.4.3, I ⊆ J(A) implies that a is a unit, hence Ma = M and Aa = A, so
(2) follows from (1). □
Now we can come back to Krull dimensions by proving the principal ideal theorem:
Theorem 4.5 (Principal ideal theorem): Let A be a Noetherian ring and a ∈ A. Let p ∈ Spec A
be a prime ideal which is minimal with the property that a ∈ p. Then, ht(p) ≤ 1. In particular, if a
is not a unit, then ht((a)) ≤ 1.
P ROOF. Using Remark III.3.6 and Corollary II.3.10, we can localize at p to assume that A is
local with maximal ideal p and we have to show that dim(A) ≤ 1. 6 By Proposition I.4.15, the
quotient ring A/(a) has a unique prime ideal, hence A/(a) is Artinian by Theorem III.2.12. We
claim that it suffices to show that the localization Aa is also Artinian. Indeed, if this holds and
q ∈ Spec A is strictly contained in p, then a ̸∈ q, hence q ∈ D(a). Since Aa is Artinian, we have
dim(D(a)) = 0 by Theorem III.3.9. Thus, q is a minimal prime and hence the chain q ⊊ p cannot
be extended and so dim(A) ≤ 1.
It remains to show that Aa is Artinian. For this, consider the natural maps ε : A → Aa and
π : A → A/(a). Let I, J ⊆ Aa be ideals with J ⊆ I and π(ε−1 (I)) = π(ε−1 (J)). We want to
show that I = J, for then the length of every chain of ideals in Aa is at most the length of a chain
of ideals in A/(a), which is finite by Proposition III.2.10 since A/(a) is an Artinian ring. To show
that I = J, it suffices to show that ε−1 (I) ⊆ ε−1 (J) by Proposition II.3.7 (1).
So, let x ∈ ε−1 (I). Since π(ε−1 (I)) = π(ε−1 (J)), there exists y ∈ ε−1 (J) and z ∈ A with
x = y + az. Thus, az ∈ ε−1 (I). Since a is invertible in Aa , this implies that z maps into I under
ε, hence z ∈ ε−1 (I). We have therefore shown that ε−1 (I) = ε−1 (J) + (a)ε−1 (I).
6Note that localizations of Noetherian rings are Noetherian. This follows immediately from the description of ideals
in the localized ring given in Proposition II.3.7.
 4. NAKAYAMA’S LEMMA AND THE PRINCIPAL IDEAL THEOREM 53

Thus, setting M = ε−1 (I)/ε−1 (J), we have (a)M = 0. The ring A is local with maximal
ideal p and a ∈ p. Thus, by Nakayama’s lemma III.4.3, we have M = 0 and so ε−1 (I) = ε−1 (J),
as desired. □
Remark 4.6: Note that if A is a Noetherian integral domain, then the unique prime ideal of height
0 in A is the zero ideal. Thus, in such a ring, if 0 ̸= a ̸∈ A× , then ht((a)) = 1.
Remark 4.7: Let us explain what the principal ideal theorem means for the Zariski topology on k n
using the correspondence of Corollary I.4.12:
Every irreducible component of the vanishing set V (a) ⊆ k n of a single non-zero non-constant
polynomial a ∈ k[x1 , . . . , xn ] has codimension exactly 1.
Theorem 4.8 (Krull’s height theorem): Let A be a Noetherian ring and a1 , . . . , an ∈ A. Let
p ∈ Spec A be a prime ideal which is minimal with the property that a1 , . . . , an ∈ p. Then,
ht(p) ≤ n.
P ROOF. We prove this by induction on n. The case n = 0 is clear. Localizing at p, we may
assume that A is local with maximal ideal p.
Let q ⊊ p be a prime ideal strictly contained in p and such that there is no prime ideal strictly
between q and p. We want to show that ht(q) ≤ n − 1. By minimality of p among prime ideals
containing the ai , we may assume that a1 ̸∈ q. Then, p is minimal
p among the ideals containing
q + (a1 ). Since p is also the unique maximal ideal, we have p = q + (a1 ). Thus, for every i ≥ 2,
we find ki > 0, bi ∈ q, and ci ∈ A such that

aki i = bi + ci a1 .

Hence, (a1 , ak22 , . . . , aknn ) = (a1 , b2 , . . . , bn ). Thus, p is minimal over (a1 , b2 , . . . , bn ), so by Propo-
sition I.4.15, the ideal p/(b2 , . . . , bn ) ⊆ A/(b2 , . . . , bn ) is minimal over the principal ideal (a1 ). By
Theorem III.4.5, we conclude that ht(p/(b2 , . . . , bn )) ≤ 1. Since q/(b2 , . . . , bn ) is strictly con-
tained in p/(b2 , . . . , bn ), we conclude that ht(q/(b2 , . . . , bn )) = 0. Thus, q is minimal among all
primes containing (b2 , . . . , bn ), hence ht(q) ≤ n − 1 by the induction hypothesis. This is what we
had to show. □

This allows us to finally compute the dimension of some more rings:

Corollary 4.9: Let k be an algebraically closed field and A = k[x1 , . . . , xn ]. Then, dim A = n.
P ROOF. The chain
0 ⊊ (x1 ) ⊊ . . . ⊊ (x1 , . . . , xn )
shows that dim A ≥ n. By Corollary I.2.19, every maximal ideal m in A is generated by n elements,
hence has height at most n by Theorem III.4.8. Thus, dim A ≤ n. □
Definition 4.10: Let A be a Noetherian local ring with maximal ideal m. The embedding dimension
edim(A) of A is dimA/m m/m2 .
Corollary 4.11: Let A be a Noetherian local ring. Then,

dim(A) ≤ edim(A) < ∞.

54 III. CHAIN CONDITIONS

P ROOF. Let m be the maximal ideal of A. Since A is Noetherian, m is finitely generated, hence
so is m/m2 and thus edim(A) = dimA/m m/m2 is finite. By Nakayama’s lemma III.4.4, the ideal
m can be generated by edim(A) elements. So, by Theorem III.4.8, we have dim(A) = ht(m) ≤
edim(A). □
In the context of the principal ideal theorem, it is useful to control when an element a ∈ A is
contained in a minimal
√ prime. The following general fact can be used, where a ring A is called
reduced if 0 = 0, i.e., if it has no non-zero nilpotent elements:
Lemma 4.12: Let A be a ring.
(1) Every minimal prime ideal of A consists of zero divisors.
(2) If A is reduced, every zero divisor of A is contained in a minimal prime ideal.
P ROOF. (1): Let p ∈ Spec A with ht(p) = 0. Then, pAp is the only prime ideal in Ap , hence
equal to its nilradical. So, if a ∈ p, then a1 ∈ Ap is nilpotent, so there exists b ∈ A − p and n > 0
such that an b = 0, hence a is a zero divisor.
T (2): Let a ∈ A be a zero divisor and b ̸= 0 with ab = 0. Since A is reduced, b ̸∈ 0 =
p∈Spec A,ht(p)=0 p, so there exists a minimal prime p not containing b. Since p is prime and 0 ∈ p,
we deduce that a ∈ p. □
Example 4.13: Lemma III.4.12 (2) can fail without the assumption that A is reduced. For example,
if k is a field and A = k[x, y]/(x2 , xy), then the nilradical of A is (x), which is itself a prime ideal,
hence the unique minimal prime ideal of A. But x + y ∈ A is a zero divisor that is not contained in
(x).
Remark 4.14: Thus, if we assume in Theorem III.4.5 that a is not a unit and not a zero divisor,
then ht(a) = 1. The corresponding assumption in Theorem III.4.8 is a bit more subtle and will lead
to the notion of regular sequence that we will study later.
We finish this lecture with another useful lemma:
Lemma 4.15 (Prime avoidance): Let A Sbe a ring, let I ⊆ A be an ideal and p1 , . . . , pn ⊆ A ideals
such that pi is prime for i ≥ 2. If I ⊆ ni=1 pi , then there exists an i such that I ⊆ pi .
P ROOF. We prove this by induction on n, the case n = 1 being clear. S
Seeking a contradiction, assume that for every i = 1, . . . , n, there exists ai ∈ I − j̸=i pj ⊆ pi .
Then, the element a := ( n−1
Q
i=1 ai ) + an lies in I but in none of the pi : Indeed, if a ∈ pi for some
Qn−1
i < n, then an ∈ pi , contradicting our assumption. If a ∈ pn , then i=1 ai ∈ pn and as pn is prime
S ai ∈ pn , again contradicting our assumption.
this implies that for some i < n we have
Thus, there exists an i with I ⊆ j̸=i pj and we are done by induction. □
Remark 4.16 (Geometric interpretation of prime avoidance): The contrapositive of our statement
of prime avoidance has the following geometric interpretation, assuming that p1 is prime as well:
Let p1 , . . . , pn ∈ Spec A be points
S and V (I) ⊆ Spec A a closed subset with pi ̸∈ V (I) for all
i. Then, by prime avoidance, I ̸⊆ ni=1 pi , so there exists an a ∈ I that does not lie in any pi , or
equivalently pi ∈ D(a) for all i. In other words, all the p1 , . . . , pn lie in a common principal open
in the complement of V (I).
CHAPTER IV

Integrality

55
56 IV. INTEGRALITY

1. Finite and integral ring maps

The goal of this chapter is to study ring maps that satisfy integrality properties, generalizing the
notion of being “algebraic” from field theory.
Definition 1.1: Let A be a ring and B an a A-algebra, say via f : A → B.
(1) Recall that a polynomial g ∈ A[x] is called monic if its leading coefficient is 1.
(2) Given b ∈ B, an integral equation for b over A is a monic polynomial g ∈ A[x] with
g(b) = 0.
(3) An element b ∈ B is called integral over A if there exists an integral equation for b over
A.
(4) The A-algebra B is called integral (over A) (or f is called an integral ring map) if every
element of B is integral over A.
(5) The A-algebra B is called integral ring extension if B is integral over A and f is injective.
(6) The A-algebra B is called finite if it is finitely generated as an A-module.

Example 1.2:
(1) If A and B are fields, being √ A is the same as being algebraic over A.
√ integral over
(2) For A = Z and B =√Z[ 2] := {a + b 2 | a, b ∈ Z} ⊆ R, the ring B is integral over A.
Indeed, given (a + b 2), the polynomial
g = x2 − 2ax + a2 − 2b2 ∈ Z[x]
√
is an integral equation for (a + b 2) over A.
√ −1
(3) The element 2 ∈ R is not integral over Z (even though it is algebraic!). To see this,
assume that there exist ai ∈ Z with
√ −n √ −n+1
2 + a1 2 + . . . + an = 0.
√ n √
Multiplying by 2 and using that 1 and 2 are linearly independent over Q, we can
compare the rational parts on both sides of the equation to obtain the equality
1 + 2a2 + 4a4 + . . . = 0.
Since the left-hand
√
side is an odd integer, this is impossible.
1+ 5
(4) Let s = 2 ∈ R be the golden ratio. Then s2 − s − 1 = 0, so s is integral over Z. Now
√ √
let A = Z[ √5] := {a + b 5√| a, b ∈ Z} ⊆ R. Then, the following hold:
• s ∈ Q( 5) = Quot(Z[ 5]).
• s is integral over A with integral equation x2 − x − 1 of degree 2.
• s is not in A, hence it admits no integral equation of degree 1 over A.
• s is a root of the non-monic polynomial 2x − 2x ∈ A[x] of degree 1.
Recall that a tower of finite field extensions is again finite. With the same proof (products of
generators are generators), one shows:
Lemma 1.3: Let A → B and B → C be ring homomorphisms such that C is finite over B and B
is finite over A. Then, C is finite over A.
 1. FINITE AND INTEGRAL RING MAPS 57

There is a very close connection between integral and finite ring homomorphisms. This is
similar to the fact that in field theory the finite field extensions are exactly the finitely generated
algebraic extensions. One direction is easy.
Lemma 1.4: Let B be an A-algebra. If B is finite over A, then it is integral.

Pn P ROOF. Choose b1 , . . . , bn such that B = ⟨b1 , . . . , bn ⟩A . Given b ∈ B, we write bbi =

j=1 ci,j bj . By Lemma III.4.2, we conclude that det((bIn − ci,j )i,j ) kills B as B-module. Define
the monic polynomial g := det((xIn − ci,j )i,j ) ∈ A[x]. Then, 0 = g(b) · 1 = g(b), so g is an
integral equation for b over A. □

For the converse, we need the following:

Lemma 1.5: Let B be an A-algebra. Let b1 , . . . , bn ∈ B. Then, the bi are all integral over A if
and only if there exists an A-subalgebra B ′ ⊆ B such that bi ∈ B ′ and B ′ is finite over A.
P ROOF. By the previous lemma, we only have to prove that integrality of the bi implies the
existence of B ′ .
By induction, and since finiteness behaves well in towers of algebras by Lemma IV.1.3, we may
assume that n = 1 and we write b1 = b. Since b is integral over A, there is a monic polynomial
g ∈ A[x] with g(b) = 0, so we can write
m−1
X
bm = ai bi
i=0

for some ai ∈ A. Thus, the A-subalgebra A[b] ⊆ B generated by b equals the finite submodule
⟨1, . . . , bm−1 ⟩A ⊆ B and we are done. □

Combining the above two lemmas, we obtain:

Corollary 1.6: Let B be an A-algebra. Then, the following are equivalent:
(1) B is finite over A.
(2) B is integral over A and finitely generated as an A-algebra.
(3) B is generated as an A-algebra by finitely many integral elements.
Since integrality can be tested using finitely generated algebras, we deduce that integrality be-
haves well in towers of algebras:
Corollary 1.7: Let A → B and B → C be ring homomorphisms such that C is integral over B
and B is integral over A. Then, C is integral over A.
P ROOF. For c ∈ C, choose a monic g ∈ B[x] with g(c) = 0. Let B ′ ⊆ B be the A-subalgebra
generated by the coefficients of g. By Lemma IV.1.5, B ′ is finitely generated as an A-module. Also
by Lemma IV.1.5, there exists a B ′ -subalgebra C ′ ⊆ C that is finitely generated as a B ′ -module
and such that c ∈ C ′ . Applying Lemma IV.1.3 to A → B ′ → C ′ , we obtain that C ′ is finite as an
A-module. Since c ∈ C ′ , c is integral over A by Lemma IV.1.4 and we are done. □

The following is an easy consequence of Corollary IV.1.6.

58 IV. INTEGRALITY

Corollary 1.8: Let B be an A-algebra. Then, the set

B ′ := {b ∈ B | b integral over A} ⊆ B
of integral elements is an A-subalgebra of B.
This leads to the following definitions:
Definition 1.9: Let B be an A-algebra.
(1) The integral closure of A in B is the A-subalgebra B ′ ⊆ B of elements that are integral
over A.
(2) If A ⊆ B, we say that A is integrally closed in B if A = B ′ .
(3) If A is an integral domain, the normalization of A is the integral closure of A in Quot(A).
(4) An integral domain is called normal if it equals its normalization.
We already know one class of rings that is certainly normal:
Proposition 1.10: Let A be a UFD. Then, A is normal.
P ROOF. Let ab ∈ Quot(A) be integral over A. Using a prime factorization in A, we may assume
that a and b are coprime. Then, the integrality implies that there is an equation of the form
an an−1
+ a 1 + . . . + an = 0
bn bn−1
with ai ∈ A. Multiplying by bn , we see that
an = −(a1 an−1 + . . . + an bn−1 )b,
so b | an . Thus, every prime factor of b divides a. Since a and b are coprime, we conclude that b is
a unit in A, hence ab ∈ A ⊆ Quot(A), as desired. □
Example 1.11:
(1) By Proposition IV.1.10, the rings Z and √ k[x1 , . . . , xn ] where k is a field are normal.
(2) By Example IV.1.2, the ring A = Z[ 5] is not integrally closed in its field of fractions
√ √
1+ 5
√ √
Q( 5), since s = 2 ∈ Q( 5) − Z[ 5] is integral over A. Hence, A is not normal.
√ √
We claim that the normalization of A is A′ := Z[s]. To see this, let a + b√ 5 ∈ Q( 5)
be integral over A. By Corollary IV.1.7 and√ since A is integral over Z, a √+ b 5 is integral
over Z. Applying the involution in Gal(Q( 5)/Q), we see that a − √ b 5 is integral
√ over
Z as well, via the same integral
√ equation.
√ Thus, the sum 2a = a + b 5 + a − b 5 and the
product a2 − 5b2 = (a + b 5)(a − b 5) are integral over Z. But these numbers are both
rational and since Z is normal, they are in fact integers.
′
Since 2a ∈ Z, we either have a ∈ Z or a = a2 with a′ ∈ Z an odd integer. In the
√
first case, a2 − 5b2 ∈ Z implies that b ∈ Z and so a + b 5 ∈ A′ . In the second case,
′
a2 − 5b2 ∈ Z implies that b = b2 with b′ ∈ Z an odd integer. Thus,
√ √
√ a′ + b′ 5 a′ − b′ ′ 1+ 5
a+b 5= = +b( ) ∈ A′ .
2 2 2
(3) Let k be an algebraically closed field and A = k[x, y]/(y 2 −x3 ). The polynomial y 2 −x3 ∈
k[x][y] is easily seen to be irreducible, hence A is an integral domain. Consider the element
 1. FINITE AND INTEGRAL RING MAPS 59

y
z := x ∈ K := Quot(A). This element is integral over A, since
y2 y 2 − x3
z2 − x = − x = = 0 ∈ A.
x2 x2
Thus, A is not normal.
We claim that the integral extension A ⊆ A[z] ⊆ K is the normalization of A. Note
that x, y ∈ k[z] ⊆ K, since x = z 2 and y = xz. Thus, A ⊆ k[z] and thus A[z] ⊆ k[z].
As clearly k[z] ⊆ A[z], we deduce that k[z] = A[z]. Since z ̸∈ k and k is algebraically
closed, the element z ∈ K is not algebraic over k, so the subalgebra k[z] ⊆ K generated
by z is isomorphic to a polynomial ring in one variable over k. Thus, k[z] is normal by
(1). Since k[z] is also integral over A and has the same field of fractions as A, it must be
the normalization of A.
Finally, we note that taking integral closures commutes with localization:
Lemma 1.12: Let B be an A-algebra and let U ⊆ A be a multiplicative subset. Let A′ ⊆ B be
the integral closure of A in B and let (U −1 A)′ ⊆ U −1 B be the integral closure of U −1 A in U −1 B.
Then,
U −1 A′ = (U −1 A)′ .
In particular, integral domains that are localizations of normal integral domains are normal.
P ROOF. Since localization is exact, we have U −1 A′ ⊆ U −1 B, so we can check the equality as
subrings of U −1 B.
“⊆”: Take b ∈ A′ and u ∈ U . Since b is integral over A, there exist ai ∈ A such that
d
Pd−1 i
b + i=0 ai b = 0. Then
d
b d X ai b d−i
( ) + ( ) =0
u ui u
i=1
is an integral equation for ub over U −1 A, so ub ∈ (U −1 A)′ , as desired.
“⊇”: Take ub ∈ (U −1 A)′ . Then, there exist ai ∈ A, ui ∈ U with
d
b X ai b
( )d + ( )d−i = 0
u ui u
i=1
in U −1 B. This means that there exists a v ∈ U such that
d
X
(vu1 · · · ud b)d + v i ui1 · · · ui−1
i · · · uid · (vu1 · · · ud b)d−i = 0
i=1
vu1 ···ud b
in B. Thus, vu1 · · · ud b ∈ A′ , hence b
u = vu1 ···ud u ∈ U −1 A′ , as claimed. □
60 IV. INTEGRALITY

2. Lying over, going up, going down, and incomparability

If X is a topological space and x, x′ ∈ X are two points such that x′ ∈ {x}, we say that x′ is
a specialization of x, or equivalently that x is a generalization (or generization) of x′ , and we write
x ⇝ x′ . Thus, for example if A is a ring and p, p′ ∈ Spec A are prime ideals, then p ⇝ p′ if and
only if p′ ∈ V (p) = {p} if and only if p ⊆ p′ . In this case, we say that p′ is a specialization of p
and p is a generalization and write p ⇝ p′ .
Example 2.1:
(1) Let A be a local ring with maximal ideal m. Then, m is a specialization of every p ∈
Spec A and every p ∈ Spec A is a generalization of m.
(2) Let A be an integral domain. Then, the zero ideal of A is a prime ideal p. Every prime
ideal of A is a specialization of p and p is a generalization of every prime ideal of A.
The question how these specializations behave under morphisms of spectra leads to the follow-
ing notions:
Definition 2.2: Let f : X → Y be a continuous map of topological spaces.
(1) We say that specializations lift along f if for all y, y ′ ∈ Y such that y ⇝ y ′ and all x ∈ X
with f (x) = y, there exists an x′ ∈ X with f (x′ ) = y ′ and x ⇝ x′ .
(2) We say that generalizations lift along f if for all y, y ′ ∈ Y such that y ⇝ y ′ and all x′ ∈ X
with f (x′ ) = y ′ , there exists an x ∈ X with f (x) = y and x ⇝ x′ .
Remark 2.3: Consider the following picture of the continuous map f : X → Y :

Given y ⇝ y ′ in Y , we can either first lift y to x ∈ X and ask for an x′ ∈ X with f (x′ ) = y ′
and x ⇝ x′ (lifting the specialization), or we can first lift y ′ to x′ ∈ X and ask for an x ∈ X with
f (x) = y and x ⇝ x′ (lifting the generalization).
We recall the following notions from the exercises:
Definition 2.4: Let f : A → B be a ring homomorphism, let f ∗ : Spec B → Spec A be the
induced map on spectra and let p ∈ Spec A.
(1) The residue field of A at p is the field k(p) = Ap /pAp ∼
= Quot(A/p).
 2. LYING OVER, GOING UP, GOING DOWN, AND INCOMPARABILITY 61

(2) The fiber of f ∗ over p is (f ∗ )−1 (p) ⊆ Spec B with its subset topology.
The following was proved on Sheet 7:
Lemma 2.5: Let f : A → B be a ring homomorphism and let p ∈ Spec A. Let f ∗ : Spec B →
Spec A be the induced map on spectra. Then, (f ∗ )−1 (p) is homeomorphic to Spec (B ⊗A k(p))
and B ⊗A k(p) ∼= Bp /pBp is called the fiber ring over p.
Definition 2.6: Let f : A → B be a ring homomorphism and let f ∗ : Spec B → Spec A be the
induced map on spectra. We say that:
(1) f satisfies lying over if f ∗ is surjective.
(2) f satisfies going up if specializations lift along f ∗ .
(3) f satisfies going down if generalizations lift along f ∗ .
(4) f satisfies incomparability if all points in the fibers of f ∗ are closed (in the fiber).
Remark 2.7: Let us translate these four properties back to rings and ideals:
(1) f satisfies lying over if and only if for every prime ideal p ⊆ A, there exists a prime ideal
q ⊆ B with f −1 (q) = p. We say that q lies over p. If f is an inclusion of rings, then
f −1 (q) = q ∩ A, which explains the name “lying over”. From now on, we will write
q ∩ A := f −1 (q) when f is injective.
(2) f satisfies going up if and only if for every inclusion of prime ideals p ⊆ p′ and every
prime ideal q ⊆ B lying over p, there exists a prime ideal q′ that contains q and lies over
p′ . In particular, ascending chains of prime ideals in A can be lifted to ascending chains of
prime ideals in B, hence the name “going up”.
(3) f satisfies going down if and only if for every inclusion of prime ideals p ⊆ p′ and every
prime ideal q′ ⊆ B lying over p′ , there exists a prime ideal q contained in q′ that lies over
p. In particular, descending chains of prime ideals in A can be lifted to descending chains
of prime ideals in B, hence the name “going down”.
(4) f satisfies incomparability if and only if for all prime ideals p ⊆ A and any two prime
ideals q, q′ of B lying over p, having an inclusion q ⊆ q′ implies that q = q′ .In particular,
any two prime ideals in B lying over the same prime ideal in A cannot be compared (via
the inclusion relation), hence the name.
Lemma 2.8 (Integral ring extensions satisfy lying over): Let f : A ,→ B be an integral ring
extension. Then, f ∗ : Spec B → Spec A is surjective.
P ROOF. Pick p ∈ Spec A. Consider the commutative diagram
BO / Bp
O
f fp

A / Ap

Since integral closure and localization commute by Lemma IV.1.12 and since localization is exact
(so f stays injective), fp is an integral ring extension. Since p lies in the image of f ∗ if and only
if it lies in the image of (fp )∗ , we may therefore replace A and B by Ap and Bp , respectively, and
assume that A is a local ring with maximal ideal p and B = Bp . By Lemma IV.2.5, it suffices to
show that the fiber ring is non-zero, or equivalently that pB ̸= B.
62 IV. INTEGRALITY
P
Seeking a contradiction, assume that pB = B. Then, we can write 1 = ai bi with ai ∈ p
and bi ∈ B. Let B ′ ⊆ B be an A-subalgebra P that is finite as an A-module and that contains all the
bi (this exists by Lemma IV.1.5). Since 1 = ai bi , we deduce that also pB ′ = B ′ . But now B ′
is a finite A-module, so Nakayama’s lemma III.4.3 implies that B ′ = 0, hence 1 = 0 also in B, so
B = 0. This is absurd, since f is injective. □
Lemma 2.9 (Integral ring maps satisfy incomparability): Let B be an A-algebra.
(1) If B is integral over A, then A → B satisfies incomparability.
(2) If B is finite over A, then the fibers of Spec B → Spec A are finite (and discrete).
P ROOF. (1) Let p ∈ Spec A. By Lemma IV.2.5, we have to show that every prime ideal q in
the fiber ring C = B ⊗A k(p) is maximal. Since the fiber ring does not change if we localize at p 1,
we may assume that A is local with maximal ideal p.
Now, consider the integral domain C/q = B/π −1 (q), where π : B → B/pB = C is the
quotient map. Since π is surjective and B is integral over A, every element in C is integral over
A (use an integral equation for a lift to B). The morphism A → C factors through k(p) = A/p,
so every element in C/q is algebraic over k(p). By Lemma I.2.16, this implies that C/q is itself a
field, hence q is maximal, as claimed.
(2) was proved on Sheet 7. □
Lemma 2.10 (Integral ring maps satisfy going up): Let B be an integral A-algebra. Then, A → B
satisfies going up.
P ROOF. Let p ⊆ p′ be prime ideals in A and let q ∈ Spec B with f −1 (q) = p. Then, we have
a commutative diagram
πB
BO / B/q
O
f g
πA
A / A/p

where g is induced by f and injective. Since A → B → B/q is integral, so is g, hence by Lemma

−1 ′′
IV.2.8, there exists a prime ideal q′′ ∈ Spec B/q with g −1 (q′′ ) = p′ /p. Setting q′ = πB (q ), we
′
have q ⊆ q and we deduce that
−1 ′′ −1 −1 ′′ −1 ′
f −1 (q′ ) = f −1 (πB (q )) = πA (g (q )) = πA (p /p) = p′ .
□
As a consequence of the above three results, we can compare dimensions of integral ring exten-
sions. Note that the following equality holds for all ring maps that satisfy lying over, incomparabil-
ity, and going up or going down:
Corollary 2.11: Let f : A ,→ B be an integral ring extension. Then,
dim(A) = dim(B).
1For example, use

B ⊗A k(p) ∼
= B ⊗A (Ap /pAp ) ∼
= B ⊗A Ap ⊗Ap (Ap /pAp ) ∼
= (B ⊗A Ap ) ⊗Ap (Ap /pAp ) ∼
= Bp ⊗Ap k(p)
and observe that these are isomorphisms of rings.
 2. LYING OVER, GOING UP, GOING DOWN, AND INCOMPARABILITY 63

P ROOF. If
p0 ⊊ . . . ⊊ pn
is a chain of prime ideals in A, we can pick a prime lying over p0 by lying over IV.2.8 and extend
the above chain to a chain of prime ideals in B
q0 ⊆ . . . ⊆ qn
with qi ∩ A = pi by going up IV.2.10. In particular, the inclusions stay strict. Thus, dim(A) ≤
dim(B).
Conversely, given a chain of prime ideals in B
q0 ⊊ . . . ⊊ qn ,
incomparability IV.2.9 guarantees that if we intersect with A we get a strict chain of prime ideals in
A, hence dim(B) ≤ dim(A). □
Note that the above argument does not give control over the behaviour of heights of primes
under integral ring extensions. For this, we need the going down property, which, unfortunately,
does not hold for general integral ring extensions (see Sheet 10).
To prove the going down theorem under extra assumptions, we need some preparations.
Lemma 2.12: Let L/K be a finite normal (not necessarily separable) field extension. Let G =
AutK (L) be the group of K-linear field automorphisms of L. Then, G is finite, L/LG is separable
and LG /K is purely inseparable. In particular, if p := char(K) ≥ 0, then for all a ∈ LG , there
n
exists n ≥ 0 such that ap ∈ K.
P ROOF. From an introductory course on algebra we know that |G| is at most the degree of
separability of L/K, hence finite. Also, we know that L/LG is a Galois extension. To show that
LG /K is purely inseparable, let a ∈ LG and let f be its minimal polynomial over K. Since L/K
is normal, f splits over L into linear factors. As in the proof of the main theorem of Galois theory,
the group G acts transitively on the roots of f . On the other hand, G fixes a, so a is the only root of
n
f , hence f = xp − b for some b ∈ K and we are done. □
Using this, we can show that the Galois group acts transitively on the fibers of an integral
extension of normal domains whose fraction field extension is normal:
Lemma 2.13: Let L/K be a finite normal field extension. Let G = AutK (L). Let A ⊆ K be a
normal subring with fraction field K and let B ⊆ L be the integral closure of A in L. Then, for two
prime ideals q, q′ ∈ Spec B with q ∩ A = q′ ∩ A, there exists g ∈ G with g(q) = q′ .
P ROOF. Let p = char(K) ≥ 0. Note that for every g ∈ G, we have g(b) ∈ B for all b ∈ B.
Indeed, applying g to an integral equation for b over A, we see that g(b) ∈ L is also integral over A,
hence g(b) ∈ B, since B is the integral
Q closure of A in L.
Now, let b ∈ q′ . The product g∈G g(b) is G-invariant, so by Lemma IV.2.12, there exists
n ≥ 0 such that
n
Y
c := g(b)p ∈ K ∩ B = A.
g∈G
Since c is a multiple of b, we also know that c ∈ q′ , hence
c ∈ A ∩ q′ = A ∩ q ⊆ q.
64 IV. INTEGRALITY

Now, since q is prime, it follows that there exists a g ∈ G with g(b) ∈ q. Since b is arbitrary, we
conclude that
[
q′ ⊆ g(q).
g∈G

Note that g(q) is again a prime ideal, since, as we explained in the first paragraph, g restricts to an
automorphism of the ring B. By prime avoidance III.4.15, we deduce that there exists a g ∈ G with
q′ ⊆ g(q). Since g fixes A ⊆ K elementwise, we have
g(q) ∩ A = q ∩ A = q′ ∩ A,
so by incomparability IV.2.9, the inclusion q′ ⊆ g(q) must be an equality. □
Theorem 2.14 (Going down holds for finite extensions of normal domains): Let f : A → B be a
ring homomorphism. Assume that:
(1) f is injective.
(2) A and B are integral domains.
(3) A is normal.
(4) B is a finite A-algebra.
Then, f satisfies going down.
P ROOF. Let p ⊆ p′ be prime ideals in A and let q′ ∈ Spec B with q′ ∩ A = p′ . We are looking
for a prime ideal q ⊆ q′ with q ∩ A = p.
Let K = Quot(A) and L = Quot(B). Since B is finite over A, the field extension L/K is
algebraic, being generated by integral, hence algebraic, elements. As B is finitely generated as an
A-algebra, the field extension L/K is algebraic and finitely generated, hence finite. Thus, there
exists a normal closure, i.e., a field extension N/L such that N/K is normal and finite. Let C ⊆ N
be the integral closure of A in N , so that in particular B ⊆ C.
By lying over IV.2.8 and going up IV.2.10, we find r, r′ ∈ Spec C with r ⊆ r′ , r ∩ A = p and
r ∩ A = p′ . The problem is that we cannot guarantee that r′ ∩ B = q′ . The situation is summarized
′

in the following picture:

Spec C  / Spec B 
(−)∩B (−)∩A
/ Spec A

rO ′  / r′ ∩ B =?? ′  / p′
O
q O

?  ? ?
r / r∩B  /p

To fix this, we want to correct our choice of r and r′ by an automorphism in G := AutK (N )

using Lemma IV.2.13. By lying over IV.2.8, we find an s′ ∈ Spec C with s′ ∩ B = q′ (but not
necessarily containing r). However, since s′ ∩ A = p′ = r′ ∩ A Lemma IV.2.13 implies that we find
a g ∈ G with g(r′ ) = s′ . Set s := g(r) and q := s ∩ B. Then, since G fixes A pointwise, we have
q ∩ A = s ∩ A = g(r) ∩ A = r ∩ A = p
 2. LYING OVER, GOING UP, GOING DOWN, AND INCOMPARABILITY 65

and
q = s ∩ B = g(r) ∩ B ⊆ g(r′ ) ∩ B = s′ ∩ B = q′ .
This finishes the proof. □
Then, a proof analogous to the one for Corollary IV.2.11 yields the following corollary, which
applies equally well to ring maps that satisfy lying over, incomparability, and going down.
Corollary 2.15: In the setting of Theorem IV.2.14, let q ∈ Spec B and p = q ∩ A. Then,
ht(p) = ht(q).
66 IV. INTEGRALITY

3. Discrete Valuation Rings and Dedekind Domains

Recall that the Noetherian rings of Krull dimension 0 are exactly the Artinian rings. What about
dimension 1? It turns out that, with some extra properties (integral, Noetherian, local, normal), such
rings can be characterized by the existence of discrete valuations.
Definition 3.1: Let K be a field. A discrete valuation on K is a surjective map ν : K → Z ∪ {∞}
with the following properties:
(1) ν(ab) = ν(a) + ν(b) for all a, b ∈ K.
(2) ν(a + b) ≥ min(ν(a), ν(b)) for all a, b ∈ K.
(3) ν(a) = ∞ ⇐⇒ a = 0.
The set of all a ∈ K with ν(a) ≥ 0 is a subring of K called valuation ring of ν. A ring A is called
discrete valuation ring (DVR) if it is the valuation ring of a discrete valuation on some field.
Example 3.2: The standard examples are:
(1) Take K = Q and fix a prime number p ∈ Z>0 . Any non-zero rational number a can be
written as a = pn b for some n ∈ Z and some b ∈ Q whose numerator and denominator are
coprime to p. The p-adic valuation of a is defined as νp (a) := n. The associated valuation
ring consists of rational numbers that can be written as a fraction whose denominator is
not divisible by p, hence it equals the localization Z(p) at the prime ideal (p).
(2) Let K = k(x) with k some field. Fix a non-zero irreducible polynomial f ∈ k[x]. Then,
we can define νf (a) analogous to (1) and obtain the localization k[x](f ) as the associated
valuation ring.
Remark 3.3: Let us note some properties of discrete valuation rings A ⊆ K associated to a valua-
tion ν:
(1) We have
A× = {a ∈ K | ν(a) = 0} − {0},
since ν(a) = 0 implies ν(a−1 ) = 0 and if a ∈ A× , then 0 = ν(a) + ν(a−1 ), so ν(a) =
ν(a−1 ), because both terms on the right hand side are non-negative.
(2) The set
m = {a ∈ K | ν(a) > 0}
is an ideal. By Exercise 4 on Sheet 6, this implies that A is local and m is its maximal
ideal.
(3) If I ⊊ A is a proper ideal, then there exists an element a ∈ I with minimal valuation
ν(a) = n > 0. For b ∈ I, we thus have ν( ab ) = ν(b) − ν(a) ≥ 0, hence ba−1 ∈ A and so
b = ba−1 a ∈ (a), hence I = (a). Thus, A is a principal ideal domain.
(4) A generator of the maximal ideal of m is also called uniformizer of the DVR A. Since ν is
surjective, the uniformizers in A are exactly those a ∈ A with ν(a) = 1. Now, if (b) ⊆ A
is any ideal, say with ν(b) = n, then ν(ba−n ) = 0, hence (b) = (an ). Thus, the ideals of
A are exactly the (an ) with n ∈ Z≥0 .
Proposition 3.4: Let A be a Noetherian local integral domain with dim(A) = 1 and maximal ideal
m. Then, the following are equivalent:
(1) A is a discrete valuation ring.
 3. DISCRETE VALUATION RINGS AND DEDEKIND DOMAINS 67

(2) A is normal.
(3) m is a principal ideal.
(4) dimA/m m/m2 = 1.
P ROOF. (1) ⇒ (2): By the above remark, we know that DVRs are PIDs, hence UFDs, and
UFDs are normal by Proposition IV.1.10.
(2) ⇒ (3): Take 0 ̸= a ∈ m. By the principalp ideal theorem III.4.5, we have ht(a) = 1, hence
m is the unique prime ideal containing (a) and so (a) = m.
Take an ideal p that is maximal among all colon ideals (a) : (b) with b ∈ A − (a). Note that p
exists because A is Noetherian. We claim that p is a prime ideal: Since b ̸∈ (a), we have p ̸= A.
Moreover, if c, d ∈ A − p, then bc ̸∈ (a) and (a) : (b) ⊆ (a) : (bc), hence (a) : (b) = (a) : (bc) by
maximality of p. Thus, d ̸∈ (a) : (bc) and so cd ̸∈ (a) : (b) = p.
But now (a) ⊆ p and since m was the only prime ideal containing (a), we have p = m. Write
c := ab ∈ Quot(A). Consider the A-submodule

I := c−1 · m ⊆ Quot(A).
As m = (a) : (b), we have that in fact I ⊆ A, so I is an ideal of A.
We claim that I = A. Seeking a contradiction, assume that I ⊆ m. Write m = (a1 , . . . , an )
and
Xn
c−1 · ai = ci,j aj .
j=1

As in Lemma IV.1.4, we can use Lemma III.4.2 to deduce that

g(x) = det((xIn − ci,j )i,j ) ∈ A[x]
is an integral equation for c−1 over A. Since A is normal, we thus have c−1 ∈ A, so b ∈ (a),
contradicting our assumption.
Now, c−1 · m = A implies that c ∈ m and multiplying the equality by c shows that m = cA =
(c), as desired.
(3) ⇔ (4) : One direction is clear and the other follows easily from Nakayama III.4.4.
(3) ⇒ (1) : Set K = Quot(A) and let m = (x). First, note that mi ̸= mi+1 for all i ≥ 0,
for otherwise m · mi = mi , hence mi = 0 by Nakayama’s lemma, hence m consists of nilpotent
elements and so m = 0, as A isTan integral T domain, a contradiction to dim(A) = 1.
Then, we claim that m · i≥0 m = i≥0 mi , from which it follows that i≥0 mi = 0 by
i
T
Nakayama’s lemma.T
For this, let a ∈ i≥0 mi . Then, certainly a = bx for some b ∈ A. If b ∈ i≥0 mi , we are done.
T

If not, then there exists a smallest s ≥ 0 such that b ̸∈ ms+1 . Thus, b = uxs for some unit u ∈ A× .
But then a = uxs+1 , hence (a) = (xs+1 ) = ms+1 ̸⊆ ms+2 , a contradiction.
Hence, for every a ∈ A which is a non-zero non-unit, there exists a unique i ≥ 1 such that
a ∈ (xi ), a ̸∈ (xi+1 ).
We set ν(a) := i and extend this to K by ν( ab ) = ν(a) − ν(b) and ν(0) := ∞. It is straight-
forward to check that this defines a discrete valuation in the sense of Definition IV.3.1 and that A is
the associated discrete valuation ring. □

As a consequence, we get a structure result in the non-local setting.

68 IV. INTEGRALITY

Theorem 3.5: Let A be a Noetherian integral domain with dim(A) = 1. Then, the following are
equivalent:
(1) A is normal.
(2) For all p ∈ Spec A with p ̸= (0), Ap is a discrete valuation ring.
P ROOF. (1) ⇒ (2) : Follows from Lemma IV.1.12 and Proposition IV.3.4.
(2) ⇒ (1) : Let f : A → A′ be the morphism from A to the integral closure A′ of A in Frac(A).
By Proposition IV.3.4 and Lemma IV.1.12, the morphisms fp are surjective for all p ∈ Spec A.
Since surjectivity is a local property, we conclude that f is surjective, hence an isomorphism. 2 □
Definition 3.6: Let A be a ring and k, K fields.
(1) A is called Dedekind domain if A is Noetherian, integral, normal, and with dim(A) = 1.
(2) K is called number field if K is a finite field extension of Q.
(3) A is called ring of integers in K if A is the integral closure of Z in the number field K.
(4) A is called affine algebraic curve over k if A is the integral closure of k[x] in a finite field
extension of k(x).
Both rings of integers and affine algebraic curves turn out to be Dedekind domains. More
generally, the following is true:
Theorem 3.7: Let A be a Dedekind domain and L a finite field extension of Quot(A). Then, the
integral closure B of A in L is a Dedekind domain.
Note that B is clearly an integral normal domain and dim(B) = dim(A) = 1 by Corollary
IV.2.11. The hard part is showing that B is Noetherian. This is a consequence of the following
theorem of Krull–Akizuki:
Theorem 3.8 (Krull–Akizuki): Let A be a Noetherian integral domain with dim(A) = 1. Let L be
a finite field extension of Quot(A). Let B be a ring with A ⊆ B ⊆ L. Then, B is Noetherian.
For time reasons, we skip the proof of Theorem IV.3.8 even though we could give it with the
methods we have developed so far. See [Sta25, Tag 00P7] for a proof.
Remark 3.9: Ideals in Dedekind domains are well-behaved. For example, one can show that every
ideal in a Dedekind domain has a unique factorization as a product of prime ideals. For time reasons,
we will omit the discussion of this subject and refer to [AM69, Chapter 9] and [Kem11, Section
14.2, 14.3] instead.

2More generally, this argument shows that being normal is a local property of integral domains.
 4. NOETHER NORMALIZATION 69

4. Noether normalization
Now that we know that the Krull dimension does not change in integral extensions (Corollary
IV.2.11), we can try to use this to compute the Krull dimension of some more rings. The goal of this
lecture is to carry this out in the case of finitely generated algebras over a field. As a consequence,
we will see that there is a connection between Krull dimension and transcendence degrees.
Recall that elements a1 , . . . , an of a k-algebra A are called algebraically independent over k if
the k-subalgebra k[a1 , . . . , an ] ⊆ A generated by the ai is isomorphic over k to a polynomial ring
in n variables.
Theorem 4.1 (Noether normalization): Let A be a non-zero finitely generated algebra over a field
k. Then, there exist algebraically independent elements a1 , . . . , an ∈ A such that A is finite over
k[a1 , . . . , an ].
P ROOF. Write A = k[x1 , . . . , xm ]/I for some ideal I. We prove the claim by induction on m.
If m = 0, there is nothing to show. If I = (0), we can choose ai = xi , so we may assume
I ̸= (0). Pick 0 ̸= f ∈ I and write
X
f= αi1 ,...,im xi11 · · · ximm
(i1 ,...,im )∈S

for some finite set S of m-tuples of natural numbers. For any d > 0 consider the function
s : S → N0
m
X
(i1 , . . . , im ) 7→ ij dj−1 .
j=1

Note that s is injective if we choose d large enough and we will assume this from now on.3 For
i−1
i > 1, set yi = xi − xd1 . Then,
m−1
f = f (x1 , y2 + xd1 , . . . , ym + x1d )
s(i ,...,i )
X
= αi1 ,...,im (x1 1 m + gi1 ,...,im (x1 , y2 , . . . , ym ))
(i1 ,...,im )∈S

where degx1 (gi1 ,...,im ) < s(i1 , . . . , im ). Since s is injective, there is a unique (i1 , . . . , im ) ∈ S such
that N := s(i1 , . . . , im ) is maximal. Since A ̸= 0, our f is not constant, so we have N > 0. We
can write
f = αi1 ,...,im xN1 + h(x1 , y2 , . . . , ym )
for some polynomial h with degx1 (h) < N . We conclude that
−1
(4) xN
1 + αi1 ,...,im h(x1 , y2 , . . . , ym ) ∈ I.

Now, let B = k[y2 + I, . . . , ym + I] ⊆ A, so that A = B[x1 + I]. Equation (4) shows that
x1 + I ∈ A is integral over B, hence A is finite over B. Moreover, B is generated over k by the
(m − 1) elements y2 + I, . . . , ym + I, hence, by induction hypothesis, B is finite over a polynomial
ring. We conclude by Lemma IV.1.3. □
3More precisely, it suffices to choose d so that d > deg (f ) for all i.
xi
70 IV. INTEGRALITY

Remark 4.2: We can give a geometric interpretation of Noether normalization in the case where k
is algebraically closed:
If A = k[x1 , . . . , xm ]/I, then the affine algebraic set V (I) ⊆ k m can be realized as a finite (by
IV.2.9) cover (by IV.2.8) of some k n .
Recall that we have already computed the dimension of the polynomial ring over algebraically
closed fields in Corollary III.4.9. Motivated by Theorem IV.4.1, let us extend this to arbitrary fields.
Lemma 4.3: Let k be a field and A = k[x1 , . . . , xn ]. Let m ⊆ A be a maximal ideal. Then, m can
be generated by n elements. In particular, dim(A) = n.
P ROOF. By Theorem III.4.8, it suffices to prove that m can be generated by n elements. By
Exercise 2 on Sheet 2, we know that L = k(m) = A/m is a finite extension of k.
Let αi be the image of xi in L and let Li = k(α1 , . . . , αi ) ⊆ L. Since L/k is finite, we have
Li = k[α1 , . . . , αi ] ⊆ L. In other words, Li is the image of the k-homomorphism
k[x1 , . . . , xi ] → k[x1 , . . . , xn ] → L.
Let fi ∈ k[x1 , . . . , xi ] ∩ m be a monic polynomial of minimal degree whose image in Li−1 [xi ] is
the minimal polynomial of αi over Li−1 . The above morphism thus factors through a k-algebra
morphism
ψi : Bi := k[x1 , . . . , xi ]/(f1 , . . . , fi ) → L.
We claim that ψi is an isomorphism onto Li ⊆ L. We argue by induction on i. The case i = 0
is clear. For the induction step, note that the morphism ψi factors through
ψi+1
Li ∼
= k[x1 , . . . , xi ]/(f1 , . . . , fi ) → Bi+1 → L.
Now, we chose fi+1 in such a way that it is irreducible over Li , hence Bi+1 is a field. Thus, ψi+1 is
injective and we are done. □
Corollary 4.4: Let A be a finitely generated integral domain over a field k. Then,
dim(A) = tr.degk (Quot(A)).
P ROOF. By Noether normalization IV.4.1, we can write A as a finite extension of a polynomial
ring B. By Corollary IV.2.11, we have dim(A) = dim(B). As in the proof of Theorem IV.2.14, the
extension Quot(A)/Quot(B) is finite, and thus tr.degk (Quot(A)) = tr.degk (Quot(B)).
Finally, we have dim(B) = tr.degk (Quot(B)) by Lemma IV.4.3. □
 CHAPTER V

Homological Algebra

71
72 V. HOMOLOGICAL ALGEBRA

1. Abelian Categories
In this section, we introduce the notion of an Abelian category, which turns out to be the natural
setting to do (classical) homological algebra in. The key example is the category ModA for some
ring A. We start with the following general notions:
Definition 1.1: Let C be a category and A a ring.
(1) C is called ModA -enriched if the class HomC (X, Y ) of morphisms with source X and
target Y has the structure of an A-module and the composition
HomC (X, Y ) × HomC (Y, Z) → HomC (X, Z)
is bilinear. A ModZ -enriched category is called preadditive.
(2) An object X in C is called initial if for every object Y of C, there exists a unique morphism
X →Y.
(3) An object X in C is called terminal if for every object Y of C, there exists a unique
morphism Y → X.
(4) An object X in C is called zero object if it is both initial and terminal. The zero object is
usually denoted by 0.
(5) Given a family of objects {Xi }i∈I of C index by a set I, a product of the Xi is an object X
together with morphisms πi : X → Xi such that for every object Y of C and every family
of morphisms
Q fi : Y → Xi , there exists a unique f : Y → X with πi ◦ f = fi . We write
X = i∈I Xi . If I = {1, . . . , n}, we also write X = X1 × . . . × Xn .
(6) Given a family of objects {Xi }i∈I of C index by a set I, the coproduct of the Xi is an
object X together with morphisms ιi : Xi → X such that for every object Y of C and
morphisms fi : Xi → Y , there exists a unique f : X → Y with f ◦ιi = fi .
every family of `
We write X = i∈I Xi . If I = {1, . . . , n}, we also write X = X1 ⨿ . . . ⨿ Xn .
(7) A morphism f : X → Y is called monomorphism (or “left-cancellative”) if for all objects
Z and all morphisms g1 , g2 : Z → X, the implication
f ◦ g1 = f ◦ g2 ⇒ g1 = g2
holds.
(8) A morphism f : X → Y is called epimorphism (or “right-cancellative”) if for all objects
Z and all morphisms g1 , g2 : Y → Z, the implication
g1 ◦ f = g2 ◦ f ⇒ g1 = g2
holds.
Example 1.2: Let C = Ring be the category of rings. As an exercise, you can check:
(1) C is not preadditive.
(2) Z is an initial object in C.
(3) The zero ring is a terminal object in C.
(4) C has no zero object.
(5) The Cartesian product of rings is a product in C.
(6) The tensor product of rings (over Z) is a coproduct in C.
(7) The monomorphisms in C are exactly the injective ring maps.
 1. ABELIAN CATEGORIES 73

(8) Every surjective ring map is an epimorphism in C. The inclusion Z → Q is an epimor-

phism, but not surjective.
The category ModA of modules over a ring A is preadditive. We can talk about kernels and
cokernels in a preadditive category.
Definition 1.3: Let C be a preadditive category, X, Y objects, and f : X → Y a morphism in C.
(1) f is called zero morphism if it is 0 in Hom(X, Y ). We write f = 0.
(2) A kernel of f is a pair (Z, g), where Z is an object of Z and g is a morphism g : Z → X
with f ◦ g = 0 and such that for every object Z ′ and every morphism g ′ : Z ′ → X with
f ◦ g ′ = 0, there exists a unique morphism h : Z ′ → Z with g ′ = g ◦ h.
We denote a kernel of f by ker(f ) → X.
(3) A cokernel of f is a pair (Z, g), where Z is an object of Z and g is a morphism g : Y → Z
with g ◦ f = 0 and such that for every object Z ′ and every morphism g ′ : Y → Z ′ with
g ′ ◦ f = 0, there exists a unique morphism h : Z → Z ′ with g ′ = h ◦ g.
We denote a cokernel of f by Y → coker(f ).
(4) If ker(f ) exists, then a coimage of f is defined as a cokernel of ker(f ) → X.
We denote a coimage of f by X → coim(f ).
(5) If coker(f ) exists, then an image of f is defined as a kernel of Y → coker(f ).
We denote an image of f by im(f ) → Y .
Initial and terminal objects are unique up to unique isomorphism if they exist, and so are kernel,
cokernel, image and coimage.
Lemma 1.4: In a preadditive category:
(1) Kernels and images are monomorphisms.
(2) Cokernels and coimages are epimorphisms.
P ROOF. We prove only that kernels are monomorphisms, the other statements can be proved
similarly. So, let f : X → Y be a morphism and g : ker(f ) → X its kernel. Let h1 , h2 : Z →
ker(f ) be two morphisms with g ◦ h1 = g ◦ h2 . In particular, 0 = f ◦ g ◦ h1 = f ◦ g ◦ h2 . By the
uniqueness in the definition of kernel, we get that h1 = h2 . □
Lemma 1.5: Let C be a preadditive category and f : X → Y a morphism in C such that kernel,
cokernel, coimage, and image of f exist. Then, there is a unique morphism coim(f ) → im(f ) such
that f equals
X → coim(f ) → im(f ) → Y.
P ROOF. Since ker(f ) → X → Y is 0, there is a unique morphism g : coim(f ) → Y such that
f = X → coim(f ) → Y.
Then, the morphism
coim(f ) → Y → coker(f )
must be 0, because it is so when precomposed with X → coim(f ), and coimages are epimorphisms
by Lemma V.1.4. Hence, g factors uniquely through im(f ) → Y , as claimed. □
Using these concepts, we can define Abelian categories.
74 V. HOMOLOGICAL ALGEBRA

Definition 1.6: An Abelian category is a category C with the following properties:

(0) C has a zero object.
(1) C is preadditive.
(2) All finite products and coproducts exist.
(3) All kernels and cokernels exist.
(4) For all morphisms f , the morphism coim(f ) → im(f ) in Lemma V.1.5 is an isomorphism.
Remark 1.7: Let C be an Abelian category and f : X → Y a morphism in C. You can check the
following:
(1) f is a monomorphism if and only if ker(f ) = 0. In this case, we write X ⊆ Y and
say that X is a subobject of Y . For two subobjects X1 , X2 ⊆ Y , we write X1 ⊆ X2 if
the monomorphism X1 → Y factors through a morphism (necessarily a monomorphism)
X1 → X2 . We write X1 = X2 if X1 ⊆ X2 and X2 ⊆ X1 . For a subobject X ⊆ Y , we
write X → Y /X for the cokernel of the monomorphism X → Y .
(2) f is an epimorphism if and only if coker (f ) = 0.
Theorem 1.8: Let A be a ring. The category ModA of modules over A is an Abelian category.
P ROOF. The 0-module is a 0-object. In lecture II.2, we have seen that ModA is preadditive. In
Remark II.1.10, we have seen that kernels, cokernels, products, and coproducts exist in ModA . If
f : M → N is an A-linear map, then coim(f ) = M/ker(f ), which is isomorphic to im(f ) by
the homomorphism theorem, and this isomorphism is exactly the map we constructed in Lemma
V.1.5. □
Remark 1.9: Many of the notions we studied for modules generalize to an arbitrary Abelian cate-
gory C.
(1) Since we know what it means for subobjects to be equal, we can talk about exact se-
quences.
(2) The notion of exact functor extends to functors between Abelian categories.
(3) For an object X in C, we can consider the functor HomC (X, −) : C → ModZ , which turns
out to be a left-exact functor by the same proof as in Proposition II.2.6.
(4) An object X of C is called projective if HomC (X, −) is exact.
(5) The opposite category of C is again an Abelian category. In particular, the contravariant
Hom functor HomC (−, X) is a contravariant left-exact functor and we call X injective if
this functor is exact. Note that X is projective (resp. injectve) if and only if X is injective
(resp. projective) considered as an object of the opposite category.
(6) Finite products and coproducts are canonically isomorphic and usually denoted by ⊕.
Example 1.10:
(1) By Proposition II.2.9, a free module M over a ring A is a projective object in ModA . By
Remark II.2.11, every object of ModA is thus a quotient of a projective module.
(2) An Abelian group A is called divisible if for every a ∈ A and every positive integer n,
there exists a b ∈ A with nb = a. As you will see on Sheet 12, the injective objects in
ModZ are exactly the divisible Abelian groups and this can be used to show that every
object of ModA for an arbitrary ring A is a subobject of an injective object.
 1. ABELIAN CATEGORIES 75

A key notion that we will come back to in the next section is the following:
Definition 1.11: Let C be an Abelian category.
(1) For an object X of C an injective resolution of X is an exact sequence I• of the form
I0 → I1 → . . .
where each Ii is injective together with a map
X → I0
making the resulting sequence
(5) 0 → X → I0 → . . .
exact. We also say that (5) is an injective resolution of X. For the sake of brevity, we
sometimes write
X → I•
for an injective resolution as above. 1
(2) For an object X of C a projective resolution is an exact sequence of the form
. . . → P1 → P0
where each Pi is projective together with a map
P0 → X
making
. . . → P0 → X → 0
exact.
(3) C is said to have enough injectives (resp. projectives) if every object of C has an injective
(resp. projective) resolution.
Example 1.12: Let A be a ring.
(1) The category ModA has enough projectives. Indeed, if M is an A-module, then M is the
quotient of a free module, say via f : P0 → M . Repeating this for ker(f ), we find a
projective resolution of M .
(2) By Sheet 12, the category ModZ also has enough injectives. From now on, we will use the
more general fact that ModA has enough injectives.

1If the I are not necessarily injective, we call such an exact sequence a “resolution” of A.
i
76 V. HOMOLOGICAL ALGEBRA

2. Derived functors
2
Let F := F 0 : C → D be a left-exact functor between Abelian categories, for example
HomC (X, −). In particular, F sends short exact sequences to left-exact sequences. In this lecture,
we will learn how to extend F 0 to a collection of functors that send short exact sequences to (long)
exact sequences. Formally:
Definition 2.1: Let C and D be Abelian categories.
(1) A functor F : C → D is called additive if the induced map
F : HomC (X, Y ) → HomD (F (X), F (Y ))
is a homomorphism of Abelian groups for all objects X and Y of C.
(2) A (cohomological) δ-functor from C to D is the following data:
(a) A collection of additive functors F i : C → D for all i ≥ 0.
(b) For all short exact sequences
0 → X → Y → Z → 0,
a collection of δ i : F i (Z) → F i+1 (X) called boundary maps.
These data are assume to satisfy the following properties:
(a) For every short exact sequence
0 → X → Y → Z → 0,
the sequence
0 / F 0 (X) / F 0 (Y ) / F 0 (Z)

δ0

/ F 1 (X) / F 1 (Y ) / F 1 (Z)

δ1
/ ...
is exact.
(b) For every commutative diagram with exact rows
0 /X /Y /Z /0

  
0 / X′ / Y′ / Z′ /0
the diagrams
δi / F i+1 (X)
F i (Z)

 
δi / F i+1 (X ′ )
F i (Z ′ )
commute for all i.
2For detailed proofs, please see [Wei94, Chapter 1 and 2]
 2. DERIVED FUNCTORS 77

(3) Let F = {(F i , δFi )}i≥0 and G = {(Gi , δGi )}

i≥0 be two δ-functors from C to D. A
morphism of δ-functors from F to G is a collection of natural transformations ti : F i →
Gi such that for all short exact sequences
0 → X → Y → Z → 0,
the diagrams
i
δF
F i (Z) / F i+1 (X)

 i
δG 
Gi (Z) / Gi+1 (X)

commute for all i.

(4) A δ-functor F = {(F i , δFi )}i≥0 from C to D is called universal if for every δ-functor
G = {(Gi , δG i )} 0 0
i≥0 from C to D and every natural transformation t : F → G , there
i
exists a unique morphism of δ-functors {t }i≥0 : F → G with t = t .0

As usual, universal δ-functors are unique up to unique isomorphism of δ-functors if they exist.
There is a useful criterion to check whether a given δ-functor is universal, due to Grothendieck (we
omit the proof, see [Gro57, II.2.2.1]):
Theorem 2.2 (Grothendieck): Let F = {(F i , δ i )}i≥0 : C → D be a δ-functor between Abelian
categories. If for every object X of C and every i > 0, there exists a monomorphism u : X → I
with F i (u) = 0, then F is universal.
Derived functors, which we will define next, are examples of universal δ-functors.
Definition 2.3: Let F : C → D be a left-exact additive functor between Abelian categories. As-
sume that C has enough injectives. For every object X of C, choose an injective resolution
dX dX
X → I0X →
0
I1X →
1
I2X → . . .
and for every morphism f : X → Y of objects with injective resolutions X → I•X , Y → I•Y choose
a morphism of exact sequences f• : I•X → I•Y lifting f . 3
For i ≥ 0, the i-th right-derived functor of F is the functor Ri F where
Ri F (X) := H i (F (I•X )) := ker(F (dX X
i ))/im(F (di−1 ))

is the i-th cohomology of the complex F (I•X ) and where

Ri F (f ) : Ri F (X) → Ri F (Y )
is induced by F (fi ) 4.
Remark 2.4: A priori, the above definition depends on the choice of injective resolution. That this
is not so is a consequence of the fundamental lemma of homological algebra that you prove on Sheet
12. Indeed, one can check that two morphisms f• , f•′ that are chain homotopy equivalent induce the
same map between the cohomologies of F (I•X ) and F (I•Y ).
3This is possible by Sheet 12
4Spell out how exactly F (f ) induces this arrow.
i
78 V. HOMOLOGICAL ALGEBRA

Example 2.5: Given an explicit injective resolution, computing Ri F is straightforward:

For example, consider the short exact sequence of Abelian groups
0 → Z → I0 := Q → I1 := Q/Z → 0.
Since I0 and I1 are divisible, they are injective in ModZ . Now, say we want to compute Ri Hom(Z/pZ, Z).
To do this, we apply Hom(Z/pZ, −) to the resolution I0 → I1 → 0 and obtain
Hom(Z/pZ, Q) → Hom(Z/pZ, Q/Z) → 0.
Now, Hom(Z/pZ, Q) = 0, since Q is torsion-free, and Hom(Z/pZ, Q/Z) ∼ = Z/pZ, since Q/Z has
exactly p elements of order p. Taking cohomology of the above complex, we obtain

0 if i = 0

R Hom(Z/pZ, Z) ∼
i
= Z/pZ if i = 1

0 if i ≥ 2.


One corollary to Remark V.2.4 is the following:

Lemma 2.6: In the setting of Definition V.2.3, let I be an injective object of C. Then, Ri F (I) = 0
for all i > 0.
P ROOF. Use the injective resolution I to compute Ri F (I). □
Next, we want to turn the Ri F into a δ-functor. For this, we need the following key fact.
Lemma 2.7: Let C, D be Abelian categories and F : C → D a left-exact additive functor. Let
0→X→Y →Z→0
be a short exact sequence. If X is injective, then
0 → F (X) → F (Y ) → F (Z) → 0
is exact.
P ROOF. We only give a sketch: We use the following two general facts about Abelian categories
that you can prove as an exercise:
(1) Additive functors send finite direct sums to finite direct sums.
(2) (Splitting lemma) Given a short exact sequence
0 → X → Y → Z → 0,
the following are equivalent:
(a) X → Y has a left-inverse.
(b) Y → Z has a right-inverse.
(c) There is an isomorphism Y → X ⊕ Z, compatibly with the maps to and from X and
Z.
Now, for the lemma, since X is injective, the map X → Y has a left-inverse, hence Y ∼
= X ⊕ Z.
∼
Applying F , we obtain F (Y ) = F (X) ⊕ F (Z), so F (Y ) → F (Z) is an epimorphism. □
Using the Horseshoe lemma (Sheet 12), we can then turn the collection of Ri F into a δ-functor
as follows:
 2. DERIVED FUNCTORS 79

Given a short exact sequence

0 → X → Y → Z → 0,
choose compatible injective resolutions
0 0 0

  dX
 dX
0 /X / IX 0
/ IX 1
/ ...
0 1

f0 f1
  dY
 dY
0 /Y / IY 0
/ IY 1
/ ...
0 1

g0 g1
  dZ
 dZ
0 /Z / IZ 0
/ IZ 1
/ ...
0 1

  
0 0 0
Now, forget the first column and apply F . By Lemma V.2.7, the columns are still exact and the rows
are chain complexes (i.e. the composition of two consecutive arrows is 0). In particular, as
X
F (Ii−1 ) → F (IiX ) → F (Ii+1
X
)
is 0, the universal property of cokernels induces an arrow
coker(F (dX X
i−1 )) → F (Ii+1 ).
Since the composition
coker(F (dX X X
i−1 )) → F (Ii+1 ) → F (Ii+2 )
is 0, the universal property of kernels induces an arrow
DiX : coker(F (dX X
i−1 )) → ker(F (di+1 )).

You can check5 that Ri F (X) ∼

= ker(DiX ) ∼ X ). Applying the Snake Lemma (Sheet 6 6)
= coker(Di−1
twice7, we obtain a diagram with exact rows
coker(F (dX / coker(F (dY )) / coker(F (dZ )) /0
(6) i−1 )) i−1 i−1

DiX DiY DiZ

  
0 / ker(F (dX )) / ker(F (dY )) / ker(F (dZ ))
i+1 i+1 i+1

and one can check that this diagram commutes. Applying the Snake lemma again, we obtain a map
δ i
Ri F (Z) ∼
= ker(DiZ ) → coker(DiX ) ∼
= Ri+1 F (X).
5It may be helpful to carry this out in the category of modules over a ring.
6You can prove this in an arbitrary Abelian category
7To a diagram with two exact columns, once with the horizontal maps F (d
i−1 ), and once with the horizontal maps
F (di ).
80 V. HOMOLOGICAL ALGEBRA

Theorem 2.8: Let F : C → D be a left-exact additive functor. Assume that C has enough injectives
and consider the collection {(Ri F, δ i )}i≥0 constructed above. Then, this is a universal δ-functor
with R0 F ∼
= F.
P ROOF. If
d
0 → X → I0 →0 . . .
is an injective resolution of an object X, then the left-exactness of F implies that
0 → F (X) → F (I0 ) → F (I1 )
is still exact, hence
R0 F (X) := ker(d0 ) ∼= F (X).
i
From the construction of δ , it is clear that the complex
δi
. . . → Ri F (Y ) → Ri F (Z) → Ri+1 F (X) → Ri+1 F (Y ) → . . .
is exact. Property (2) (b) of Definition V.2.1 follows from the naturality of the boundary map that
appears in the Snake Lemma (the last part of the corresponding exercise on Sheet 6).
The universality follows from Theorem V.2.2 and Lemma V.2.6. □
Remark 2.9: In a completely analogous way, one can define the left-derived functors Li F : C → D
of a right-exact additive functor F : C → D by using projective resolutions.
By passing to opposite categories, we can also define right-derived (resp. left-derived) functors
of contravariant left-exact (resp. right-exact) additive functors.
Remark 2.10: For a left-exact (an analogous definition works for right-exact functors) additive
functor F : C → D, an object X of C is called F -acyclic if Ri F (X) = 0 for all i > 0. If
0→X→Y →Z→0
is a short exact sequence and Y is F -acyclic, then the associated long exact sequence shows that
Ri F (Z) ∼= Ri+1 F (X) holds for all i > 0.
This can be used to show that derived functors can be computed using acyclic resolutions.
 3. EXT AND TOR 81

3. Ext and Tor

Applying the formalism of the previous section to the functors we already know and using
Example V.1.12, we obtain the following definitions:
Definition 3.1: Let A be a ring and M an A-module.
(1) The (covariant) i-th Ext-functor ExtiA (M, −) is the i-th right-derived functor of HomA (M, −) :
ModA → ModA .
(2) The (contravariant) i-th Ext-functor ExtiA (−, M ) is the i-th right-derived functor of HomA (−, M ) :
ModopA → ModA .
(3) The i-th Tor-functor TorA i (M, −) is the i-th left-derived functor of M ⊗A (−) : ModA →
ModA .
Remark 3.2: We now have two a priori different definitions of the modules ExtiA (M, N ). But in
fact, these two definitions coincide. Similarly, one can show that TorA ∼ A
i (M, N ) = Tori (N, M ). We
omit the proof of these two facts for time reasons and instead refer to [Wei94, Section 2.7].
Remark 3.3: Note that we can similarly define the Ext functors in arbitrary Abelian categories
(with enough injectives resp. projectives).
Using the Tor functors, we can characterize flat modules. We start with a simple observation:
Lemma 3.4: Let A be a ring, M an A-module, and I ⊆ A an ideal. Then,
TorA ∼
1 (M, A/I) = ker(M ⊗A I → M ).

P ROOF. Tensoring the short exact sequence

0 → I → A → A/I → 0
with M , we obtain a long exact sequence
. . . → TorA A
1 (M, A) → Tor1 (M, A/I) → M ⊗A I → M → M/IM → 0.

Since A is free, it is projective by Proposition II.2.9, hence tensor-acyclic by Lemma V.2.6 and thus
TorA
1 (M, A) = 0 and we are done. □
Proposition 3.5: Let A be a ring and M an A-module. The following are equivalent:
(1) M is flat.
(2) TorA
i (M, N ) = 0 for all i > 0 and all A-modules N .
(3) TorA
1 (M, A/I) = 0 for all ideals I ⊆ A.

P ROOF. (1) ⇒ (2) : Choose a projective resolution

P• → N → 0.
Since M ⊗A (−) is exact, the resulting sequence
M ⊗A P• → M ⊗A N → 0
is again exact. Equivalently, its cohomology groups are 0, hence TorA
i (M, N ) = 0 for all i > 0.
(2) ⇒ (3) is clear.
82 V. HOMOLOGICAL ALGEBRA

(3) ⇒ (1): Let

0 → N1 → N2 → N3 → 0
be a short exact sequence of A-modules. It suffices to show that TorA 1 (M, N3 ) = 0. One can
show (and if there is time we will see later) that Tori (M, N3 ) = 0 if and only if TorA
A ′
i (M, N3 ) for
′
every finitely generated submodule N3 ⊆ N3 , so we assume for simplicity here that N3 is finitely
generated.
Now, if n ∈ N3 is part of a minimal generating set, we have a short exact sequence
0 → A/Ann(n) → N3 → N3 /⟨n⟩A → 0,
and tensoring with M and we get an associated long exact sequence
. . . → TorA A A
1 (M, A/Ann(n)) → Tor1 (M, N3 ) → Tor1 (M, N3 /⟨n⟩A ) → . . .
By assumption, TorA A A
1 (M, A/Ann(n)) = 0 and so Tor1 (M, N3 ) ⊆ Tor1 (M, N3 /⟨n⟩A ). Since
N3 /⟨n⟩A can be generated by fewer elements than N3 , we are done by induction. □
Similar to how the Tor-functors detect flatness, the Ext-functors can be used to detect the exis-
tence of regular sequences.
Definition 3.6: Let A be a ring and M an A-module. A sequence of elements a1 , . . . , an ∈ A is
called M -regular sequence if the following two conditions hold:
(1) For all i = 1, . . . , n, the element ai is not a zero divisor in M/(a1 , . . . , ai−1 )M . 8
(2) (a1 , . . . , an )M ̸= M .
By Krull’s height theorem, modding out by A-regular sequences in a Noetherian ring A changes
the dimension in a controllable way. The maximal length of an M -regular sequence can be com-
puted using Ext-functors.
Proposition 3.7: Let A be a Noetherian ring, I ⊆ A an ideal, and M a finitely generated A-
module. For every n > 0, the following are equivalent:
(1) There exists an M -regular sequence a1 , . . . , an ∈ I.
(2) ExtiA (N, M ) = 0 for all i < n and all finitely generated A-modules N with Supp(N ) ⊆
V (I).
(3) ExtiA (A/I, M ) = 0 for all i < n.
(4) There exists a finitely generated A-module N with Supp(N ) = V (I) and such that
ExtiA (N, M ) = 0 for all i < n.
P ROOF. (1) ⇒ (2) : We prove this by induction on n. The sequence a2 , . . . , an is of length
(n − 1) and M/a1 M -regular, so by induction hypothesis (or obviously in the case n = 1) we have
ExtiA (N, M/a1 M ) = 0 for i < n − 1. Consider the exact sequence
1 a ·
0→M → M → M/a1 M → 0
which is exact on the left because a1 is not a zero divisor. Applying HomA (N, −) and looking at
the associated long exact sequence of Ext, we see that multiplication by a1 induces an injective
map
ExtiA (N, a1 ·) : ExtiA (N, M ) ,→ ExtiA (N, M ).
8Here, a ring element is called zero divisor in a module if it annihilates a non-zero element of the module.
 3. EXT AND TOR 83
p
However, as V (Ann(N )) = Supp(N ) ⊆ V (I), we have a1 ∈ I ⊆ Ann(N ), so there exists
m > 0 such that am
1 · : N → N is the 0 map. But then
Exti (N, a1 ·)m = Exti (N, am i m
1 ·) = Ext (a1 ·, M ) = 0
9
and this is injective if and only if Exti (N, M ) = 0, as desired.
(2) ⇒ (3) ⇒ (4) are clear.
(4) ⇒ (1): First, we have to construct
S an element a1 ∈ I which is not a zero Sdivisor in M . Seek-
ing a contradiction, assume that I ⊆ 0̸=m∈M Ann(m). By Sheet 9, we have 0̸=m∈M Ann(m) =
S
p∈AssA (M ) p. Thus, by prime avoidance III.4.15, there is an associated prime p with I ⊆ p, i.e.,
p ∈ V (I). Since Supp(N ) = V (I), we have Np ̸= 0, so 0 ̸= N ⊗A k(p) by Nakayama’s lemma.
The latter is a finitely generated vector space over the field k(p), we find a surjective Ap -linear map
Np → k(p)⊕j → k(p). On the other hand, p is an associated prime of M , so there is an injec-
tion A/p → M , hence, after localizing, an injective Ap -linear map k(p) → Mp . In particular, the
composition
Np → k(p) → Mp
is non-zero, hence
HomA (N, M )p = HomAp (Np , Mp ) ̸= 0
and so HomA (N, M ) ̸= 0, contradicting our assumption.
This settles the case n = 1 and now we argue by induction. The exact sequence
1 a ·
0→M → M → M/a1 M → 0
yields an exact sequence
Exti (N, M ) → Exti (N, M/a1 M ) → Exti+1 (N, M ).
For i < n − 1, the objects on the left and right vanish, hence Exti (N, M/a1 M ) = 0 for i < n − 1.
By the induction hypothesis, we find an M/a1 M -regular sequence a2 , . . . , an ∈ I. The sequence
a1 , . . . , an ∈ I is then an M -regular sequence of length n. □
This leads to the following definition:
Definition 3.8: Let A be a Noetherian ring, I ⊆ A an ideal, and M a finitely generated A-module.
The I-depth of M is defined as
depthI (M ) := min{i | Exti (A/I, M ) ̸= 0}.
If A is a local ring and I is its maximal ideal, we simply write depth(M ) instead.
Remark 3.9: By Krull’s height theorem III.4.8 and the connection between depth and regular se-
quences in Proposition V.3.7, a Noetherian local ring A satisfies
depth(A) ≤ dim(A).
If equality holds, we call A a Cohen–Macaulay ring.

9Here, we are implicitly using that the for any three modules X, Y, Z ∈ Mod , the morphism
A

HomA (X, Y ) → HomA (ExtiA (Z, X), ExtiA (Z, Y ))

is A-linear (and the analogous statement for the contravariant versions) and then apply Remark V.3.2. You can check the
former by following through the construction of derived functors.
 CHAPTER VI

Regular local rings

84
 1. GRADED AND FILTERED RINGS AND MODULES 85

1. Graded and filtered rings and modules

Many of the rings we encountered have a natural notion of grading (e.g. the degree for poly-
nomial rings). In this section, we formalize this observation to the concept of graded rings and
modules. Then, we go on to associated graded rings to filtrations. This will be used later to study
regular local rings.
Definition 1.1: Let A be a ring and Γ an Abelian group.
L
(1) A Γ-grading on A is a decomposition A = d∈Γ Ad of the group (A, +) such that for all
d, d′ ∈ Γ, the multiplication in the ring satisfies
Ad · Ad′ ⊆ Ad+d′ .
(2) A Γ-graded ring is a ring with a Γ-grading.
(3) A graded ring is a ring with a Z-grading such that Ad = 0 for d < 0.
(4) A morphism of Γ-graded rings is a ring homomorphism f : A → B such that f (Ad ) ⊆ Bd
for all d ∈ Γ.
L
Definition 1.2: Let A = d≥0 Ad be a graded ring.
(1) Non-zero elements of Ad are called homogeneous of degree d.
(2) An ideal I ⊆ A is called homogeneous if
M
I= (I ∩ Ad ).
d∈Z

Example 1.3: Let A be a ring and I ⊆ A an ideal.

(1) For every n ≥ 1, the polynomial ring B := A[x1 , . . . , xn ] is a graded ring with Bd
consisting
L of (0 and) homogeneous polynomials of degree Ld.
(2) If A = d≥0 Ad is a graded ring, then the subset A+ := d>0 Ad ⊆ A is a homogeneous
ideal, called irrelevant ideal.
(3) If A = k[x], then A+ = (x).
The ideal (1 + x) ⊆ k[x] is not homogeneous, since (1 + x) ∩ k[x]0 = (0) and 1 + x ̸∈
(4) L
d>0 k[x]d .
(5) The Rees algebra (or Blow-up algebra) of I in A is the graded ring
M
BlI (A) = I d.
d≥0
L
Remark 1.4: Let A = d≥0 Ad be a graded ring.
P
(1) By definition, every element a ∈ A has a unique decomposition a = d∈Z ad with ad
zero or homogeneous of degree d. If ad is non-zero, it is called homogeneous component
of degree d of a.
(2) The subgroup A0 ⊆ A is in fact a subring and 1 ∈ A0 . Moreover, every Ad is a module
over A0 through the ring multiplication on A.
(3) The i-th Veronese subring of A is
M
A(i) := Aid ⊆ A.
d∈Z
86 VI. REGULAR LOCAL RINGS

(4) Exercise: An ideal I ⊆ A is homogeneous if and only if it is generated by homogeneous √

elements. Moreover, if I and J are homogeneous ideals, then so are I + J, I · J, and I.
(5) Exercise: To test whether a homogeneous ideal p is prime, it suffices to check that for two
homogeneousL elements a, b with ab ∈ p, we have a ∈ p or b ∈ p. If p ⊆ A is a prime
ideal, then d≥0 (p ∩ Ad ) is a homogeneous prime ideal.
(6) A is Noetherian if and only if A0 is Noetherian and A is finitely generated over A0 .
Indeed, if A is finitely generated over A0 , then it is Noetherian by Theorem III.1.10.
Conversely, if A is Noetherian, then so is its quotient A0 ∼ = A/A+ . Moreover, the ideal
+
A is generated by finitely many elements a1 , . . . , an and one can show that these generate
A as an A0 -algebra.
Next, we extend the notion of grading to modules.
L
Definition 1.5: Let A = d∈Z Ad be a Z-graded ring.
L
(1) A graded module over A is an A-module M with a decomposition M = d∈Z Md of its
underlying Abelian group and such that the scalar multiplication on M satisfies
Ad · Md′ ⊆ Md+d′
for all d, d′ ∈ Z. 1
(2) A morphism of graded modules over A is an A-linear map f : M → N with f (Md ) ⊆ Nd
for all d ∈ Z.
L
Remark 1.6: Let A = d≥0 Ad be a graded ring and M a graded module over A. If U ⊆ A is a
multiplicative subset consisting of homogeneous elements, then we can write
M
U −1 M = (U −1 M )d ,
d∈Z
where
m
(U −1 M )d := { | ∃i such that m ∈ Md+i , a ∈ U ∩ Ai } ⊆ U −1 M.
a
For A = M , this turns U −1 A into a Z-graded ring and, for general M , U −1 M into a graded module
over U −1 A. 2
(1) If p ⊆ A is a homogeneous prime ideal, we set U = {a ∈ A − p | a homogeneous} and
call
M(p) := (U −1 M )0
the homogeneous localization of M at p.
(2) If a ∈ A is a homogeneous element, set U = {ai | i ≥ 0} and call
M(a) := (U −1 M )0
3
the homogeneous localization of M at a.
Given one graded module, we can produce many new ones by shifting the grading:
1Note that we do not ask that M = 0 for d < 0!
d
2Note: Even if A = 0 for d < 0, it does not necessarily hold that (U −1 A) = 0 for d < 0. For example, if
d d
A = k[x] with the usual grading and U = {xn }n∈N , then 1
x
∈ (U −1 A)−1 .
3Please be aware of the overloaded notation M !
(a)
 1. GRADED AND FILTERED RINGS AND MODULES 87
L L
Definition 1.7: Let A = d∈Z Ad be a Z-graded ring and M = d∈Z Md a graded A-module.
For i ∈ Z, the i-th twist of M is the graded A-module
M
M (i) := M (i)d
d∈Z

with M (i)d := Md+i .

Next, we want to obtain graded objects from filtrations:
Definition 1.8: Let A be a ring and I an ideal of A. Let M be an A-module.
(1) An I-filtration M• of M is a chain of submodules
M = M0 ⊇ M1 ⊇ . . .
with
I · Md ⊆ Md+1
for all d. If I = A, we simply call this a filtration.
(2) An I-filtration M• of M is called stable if IMd = Md+1 for d ≫ 0.
Example 1.9: Let A be a ring, I ⊆ A an ideal, M an A-module. The filtrations
A ⊇ I ⊇ I2 ⊇ . . .
M ⊇ IM ⊇ I 2 M ⊇ . . .
are called I-adic filtrations (of A and M , respectively). They are the key examples of stable I-
filtrations.
Using filtrations, we can construct new graded rings and modules:
Definition 1.10: Let A be a ring, I ⊆ A an ideal, M an A-module.
(1) The associated graded ring of A with respect to I is
M
grI (A) := I d /I d+1 ,
d≥0
′
where the multiplication is given on homogeneous elements a, b with a ∈ I d , b ∈ I d by
′ ′ ′ ′
I d /I d+1 × I d /I d +1 → I d+d /I d+d +1
(a, b) 7→ ab.
(2) If M• is an I-filtration of M , the associated graded module of M with respect to M• is
M
grM• (M ) := Md /Md+1 .
d∈Z

Note that this is a module over grI (M ) if we define scalar multiplication on homogeneous
elements a, m with a ∈ I d , m ∈ Md′ by
I d /I d+1 × Md′ /Md′ +1 → Md+d′ /Md+d′ +1
(a, m) 7→ am,
which is well-defined since the filtration M• is an I-filtration.
88 VI. REGULAR LOCAL RINGS

We finish this lecture by studying how passing to associated graded objects preserves finite
generation. First, observe that finite generation is not preserved by passing to associated graded
objects for general I-filtrations.
Example 1.11: Consider A = Z, I = (0), p > 0 a prime, and M = Z with the (non-stable)
I-filtration
M = Z ⊇ (p) ⊇ (p2 ) ⊇ . . .
Then, as an Abelian group, we have
M
grM• (M ) = Z/pZ.
d≥0
This is a non-finitely generated Abelian group, even though M was finitely generated.
This issue goes away if we focus on I-stable filtrations:
Lemma 1.12: Let A be a Noetherian ring, I ⊆ A and ideal and M a finitely generated A-module
with a stable I-filtration M• . Then, the following hold:
(1) grI (A) is Noetherian.
(2) grM• (M ) is a finitely generated grI (A)-module.
P ROOF. (1): Since A is Noetherian, we find finitely many a1 , . . . , an with I = (a1 , . . . , an ).
Let ai be the image of ai in I/I 2 . Then, the ai generate grI (A) as an A/I-algebra. Since A is
Noetherian, so is A/I, hence so is grI (A) by Theorem III.1.10.
(2): Since M• is I-stable, we find D > 0 such that MD+d = I d MD for all d ≥ 0. Thus,
grM• (M ) is generated by D
L
d=0 Md /Md+1 as a grI (A). Since A is Noetherian and M is finitely
generated, Md is finitely generated as an A-module by Theorem III.1.9, hence so is its quotient
Md /Md+1 . Since M• is an I-filtration, Md /Md+1 is annihilated by I, hence Md /Md+1 is a finitely
generated A/I = grI (A)0 -module. L
Now, take finitely many homogeneous elements m1 , . . . , mn ∈ d≤D Md /Md+1 ⊆ grM• (M )
L
that generate d≤D Md /Md+1 as grI (A)0 -module. Then, these will generate grM• (M ) as a
grI (A)-module. □
 2. INVERSE LIMITS AND THE ARTIN–REES LEMMA 89

2. Inverse limits and the Artin–Rees lemma

The goal of this section is to mimic how one passes from the polynomial ring to the power series
ring and extend this to arbitrary rings, ideals, and modules. Then, we compare the resulting notion
with the material of the previous lecture.
Definition 2.1: Let C be a category.
(1) A partially ordered set (poset) (I, ≤) is called directed if for all i, j ∈ I, there exists k ∈ I
with i, j ≤ k.
(2) An inverse system in C (indexed by a directed partially ordered set (I, ≤)) is a collection
of objects {Ai }i∈I together with morphisms θij : Aj → Ai for all i ≤ j such that
(a) θii = idAi for all i ∈ I.
(b) θik = θij ◦ θjk for all i ≤ j ≤ k in I.
(3) An inverse limit of the inverse system ({Ai }i∈I , {θij }i≤j ) is an object limi∈I Ai in C to-
←−
gether with morphisms πn : limi∈I Ai → An such that θij ◦ πj = πi for all i ≤ j and such
←−
that for any other object B with morphisms ψn : B → An satisfying θij ◦ ψj = ψi , there
exists a unique morphism f : B → limi∈I Ai with ψi = πi ◦ f for all i ∈ I.
←−
Lemma 2.2: Inverse limits exist in the categories of sets, rings, and modules.
P ROOF. For an inverse system ({Ai }i∈I , {θij }i≤j ), we define
Y
lim Ai := {(ai )i∈I ∈ Ai | θij (aj ) = ai for all i ≤ j}.
←−
i∈I i∈I
Q Q
In the category of modules, we can also describe this as ker(Ψ), where Ψ : i∈I Ai → i≤j Aij
Q
with Aij := Ai and Ψ is induced by the maps θij ◦ πj : i∈I Ai → Aij The universal property
follows easily from the universal property of the product (and kernel). □
Example 2.3: Let (I, ≤) be N with the usual ordering.
(1) If A is a ring, M an A-module, and J ⊆ A an ideal, then the rings Ai := A/J i form
an inverse system of rings and the modules Mi := M/J i M form an inverse system of
A-modules with the quotient maps as transition maps. The associated inverse limits are
called J-adic completion of A resp. M and are denoted by A b resp. M
c. Note that the
completion of course depends on the ideal J, even though we omit it from the notation.
(2) If A = k[x1 , . . . , xn ] and J = (x1 , . . . , xn ), then the map

k[[x1 , . . . , xn ]] → A
b

induced by the quotient maps

k[[x1 , . . . , xn ]] → k[[x1 , . . . , xn ]]/(x1 , . . . , xn )i ∼

= k[x1 , . . . , xn ]/J i
is an isomorphism, as can be checked using the explicit description of the inverse limit
above.
(3) If A = Z and J = (p), then the ring

Zp := Z
b
90 VI. REGULAR LOCAL RINGS

is called ring of p-adic integers. An element (ai )i≥0 ∈ Zp is usually denoted by ∞ i

P
i=0 ai p
and the multiplication works the same way as in the power series ring, but with carry (for
example, p · p = p2 ).
Remark 2.4: Given a directed partially ordered set (I, ≤), a subset J ⊆ I is called cofinal if for
all i ∈ I, there exists j ∈ J with i ≤ j. In particular, J is also a directed partially ordered set.
If ({Ai }i∈I , {θij }i≤j ) is an inverse system indexed by I and J ⊆ I is cofinal, then one can
show that the induced map limj∈J Aj → limi∈I Ai is an isomorphism.
←− ←−
In particular, if I is infinite, we can remove finitely many elements from I and obtain isomorphic
inverse limits.
Assume we are in the category of modules over a ring. Given a directed partially ordered index
set (I, ≤) and three inverse systems {Ai }i∈I , {Bi }i∈I , {Ci }i∈I , where we omit the transition maps
from the notation, assume that we have maps fi : Ai → Bi and gi : Bi → Ci making the following
diagrams commutative with exact rows:
0 / Aj / Bj / Cj /0

  
0 / Ai / Bi / Ci /0

With notation as in the proof of Lemma VI.2.2, the fi and gi define maps of products such that the
diagram
/ / / /0
Q Q Q
0 i∈I Ai i∈I Bi i∈I Ci

/
Q  /
Q  /
Q  /0
0 i≤j Aij i≤j Bij i≤j Cij
is commutative with exact rows. By the Snake lemma, we get an induced exact sequence
0 → lim Ai → lim Bi → lim Ci
←− ←− ←−
i∈I i∈I i∈I
We have the following very general criterion for right-exactness of inverse limits.
Lemma 2.5 (Mittag–Leffler): In the above setting, let θij be the transition maps for the system
{Ai }i∈I . Assume that I is countable. If for all i ∈ I, there exists i ≤ j such that θik (Ak ) = θij (Aj )
for all j ≤ k 4, then the sequence
0 → lim Ai → lim Bi → lim Ci → 0
←− ←− ←−
i∈I i∈I i∈I
is exact.
P ROOF. We only need to prove exactness on the right. Let (ci )i∈I ∈ limi∈I Ci . Let Ei =
←−
gi−1 (ci ), which
is not empty, since gi is surjective. The transition maps φij : Bj → Bi make
{Ei }i∈I into aninverse system of sets and it suffices to show that limi∈I Ei is non-empty, for then
←−
any element in this system will be a preimage of (ci )i∈I .
4One says that the inverse system ({A } , {θ }
i i∈I ij i≤j ) satisfies the Mittag–Leffler condition.
 2. INVERSE LIMITS AND THE ARTIN–REES LEMMA 91

We consider Ai as a subset of Bi via fi . Then, for every i ∈ I, there exists i ≤ j with

φij (Aj ) = φik (Ak ) for all j ≤ k. We claim that then also φij (Ej ) = φik (Ek ). The inclusion ⊇ is
clear. For the inclusion ⊆, let ej ∈ Ej . Let e′k ∈ Ek be any element and set e′j := φjk (e′k ). Then,
gj (ej − e′j ) = cj − cj = 0, so there exists aj ∈ Aj with ej − e′j = aj . By the Mittag–Leffler
condition, there exists ak ∈ Ak with φik (ak ) = φij (aj ). Then,
φik (e′k + ak ) = φij (φjk (e′k + ak )) = φij (e′j + aj ) = φij (ej )
hence φij (ej ) ⊆ φik (Ek ), as desired.
Now we have shown that the inverse system {Ei }i∈I satisfies the Mittag–Leffler condition
and we want to show that lim Ei is non-empty. Replacing Ei by the φij (Ej ) that appear in the
←−
Mittag–Leffler condition, we may assume that the transition maps of {Ei }i∈I are surjective. If I
is finite, then lim Ei is easily seen to be non-empty. If I is infinite, then, because it is countable,
←−
it has a cofinal countable totally ordered subset, so we may assume I = N. In this case, we can
find an en element of lim Ei by picking any e1 ∈ E1 and recursively choosing ej ∈ Ej with
←−
φ(j−1)j (ej ) = ej−1 . □

If
0 → M1 → M 2 → M 3 → 0
is a short exact sequence of modules over a ring A and J ⊆ A is an ideal, then we get induced exact
sequences
M1 /J i M1 → M2 /J i M2 → M3 /J i M3 → 0.
The image of the first map is M1 /(J i M2 ∩ M1 ) and for i ≤ j, the resulting diagrams with exact
rows
0 / M1 /(J j M2 ∩ M1 ) / M2 /J j M2 / M3 /J j M3 /0

  
0 / M1 /(J i M2 ∩ M1 ) / M2 /J i M2 / M3 /J i M3 /0

commute. Note that the system {M1 /(J i M2 ∩M1 )}i∈N satisfies the Mittag–Leffler condition, since
the transition maps are surjective. Thus, applying Lemma VI.2.5, we obtain an exact sequence
0 → lim(M1 /(J i M2 ∩ M1 )) → M
c2 → M
c3 → 0.
←−
i∈N

Next, we want to understand the natural map M c1 → lim (M1 /(J i M2 ∩ M1 )). We have the
←−i∈N
following general criterion for this to be an isomorphism:
Lemma 2.6: Let A be a ring, I ⊆ A an ideal, and M an A-module. Let M• be an I-filtration.
Then:
(1) The inclusions I i M ⊆ Mi define surjections M/I i M → M/Mi which in turn define an
A-linear map
φ : lim M/I i M → lim M/Mi .
←− ←−
i∈N i∈N
(2) If M• is stable, then φ is an isomorphism.
92 VI. REGULAR LOCAL RINGS

P ROOF. (1) is immediate.

(2): For an arbitrary inverse system ({Mi }i≥0 , {θij }i≤j ) indexed by N and a given d ≥ 0,
the d-th shifting map limi≥d Mi → limi≥0 Mi induced by θi(i+d) (and 0 in small degree) is an
←− ←−
isomorphism, as you can check easily from the explicit description in Lemma VI.2.2.
If M• is I-stable, then there exists d ≥ 0 with Md+i = I i Md for all i ≥ 0. Then, φ is induced
by M/I i M → M/Mi , which fits into the sequence
M/I d+i M → M/Md+i = M/I i Md → M/I i M → M/Mi .
Passing to inverse limits, we obtain morphisms
lim M/I i M → lim M/Mi → lim M/I i M → lim M/Mi ,
←− ←− ←− ←−
i≥d i≥d i≥0 i≥0

where the composition of any two arrows is a shifting isomorphism and the last arrow is φ. Thus, φ
is an isomorphism, as desired. □
Thus, we want to understand the stability of (J i M2 ∩ M1 ) the filtration. This follows from the
following version of the Artin–Rees lemma:
Proposition 2.7 (Artin–Rees lemma): Let M be a finitely generated module over a Noetherian ring
A. Let I ⊆ A be an ideal and M• a stable I-filtration on M . If M ′ ⊆ M is an A-submodule, then
the I-filtration
Md′ := Md ∩ M ′
on M ′ is stable.
P ROOF. First, note that the Md′ indeed define an I-filtration, since
IMd′ = I(Md ∩ M ′ ) ⊆ IMd ∩ M ′ = Md+1 ∩ M ′ = Md+1 ′
.
Recall the definition of the blow-up algebra BlI (A) = d≥0 I d . By the above, the graded BlI (A)-
L
module M M
M ′∗ := Md′ ⊆ Md =: M ∗
d∈Z d∈Z
is a BlI (A)-submodule.
Since M is finitely generated over A and A is Noetherian, so is every Md . For n ≥ 0, define
Mn
Qn := Md .
d=0
This is a finitely generated A-submodule of M ∗. It generates the BlI (A)-submodule
n
M
Q∗n := ( Md ) ⊕ IMn ⊕ I 2 Mn ⊕ . . . .
d=0
Since Qn is finitely generated as A-module, Q∗n is finitely generated as BlI (A)-module. The Q∗n
form an ascending chain of BlI (A)-submodules of M ∗ whose union is M ∗ .
Since BlI (A) is Noetherian (finitely generated over the Noetherian ring A), we have the follow-
ing equivalences:
III.1.9
M ∗ is finitely generated as BlI (A)-module. ⇐⇒ The chain of Q∗n stabilizes. ⇐⇒ There
exists d ≥ 0 such that I i Md = Md+i for all i ≥ 0. ⇐⇒ M• is stable.
 2. INVERSE LIMITS AND THE ARTIN–REES LEMMA 93

Thus, since M• is stable by assumption, M ∗ is finitely generated over the Noetherian ring
BlI (A) by Lemma VI.1.12, hence M ′ is finitely generated over BlI (A) by Theorem III.1.9, hence
M•′ is stable by the above chain of equivalences applied to M ′ . □

We immediately obtain the following exactness result for completions:

Corollary 2.8: Let A be a Noetherian ring, I ⊆ A an ideal, and
0 → M1 → M 2 → M 3 → 0
a short exact sequence of finitely generated A-modules. Then,
0→M
c1 → M
c2 → M
c3 → 0

is exact.
As another application of the Artin–Rees lemma, we obtain the following result that could be
phrased independently of the terminology of this section:
Corollary 2.9 (Krull’s intersection theorem): Let A be a Noetherian local ring, M a finitely gen-
erated module over A, and I ⊊ A a proper ideal. Then,
\
I i M = 0.
i≥0

In particular, if m is the maximal ideal of A, then

\
md = 0.
d≥0

P ROOF. Consider the submodule N = d≥0 I d M ⊆ M and the I-adic filtration on M . Since
T

N ∩ I i M = N , the induced filtration of N is constant. By Artin–Rees, it is also stable, hence

I i N = N for some i > 0. Since I is a proper ideal, it is contained in the Jacobson radical of A,
hence N = 0 by Nakayama’s lemma III.4.3. □

As a final corollary, we deduce the flatness of completions:

Proposition 2.10: Let A be a Noetherian ring, M an A-module, and I ⊆ A an ideal with respect
to which we take completions. Then, the following hold:
(1) There is a natural morphism of A-modules
b M ⊗A A
b→M c which is an isomorphism if M
is finitely generated.
(2) The natural map A → A b is flat.

P ROOF. (1) : The map φM : M ⊗A A

b→M
c is induced by the A-bilinear map

(m, (ai )i≥0 ) 7→ (ai m)i≥0

and we see that this is in fact A-linear

b and natural in the A-module M .
To prove that φM is an isomorphism if M is finitely generated, choose an exact sequence
0 → K → A⊕n → M → 0.
94 VI. REGULAR LOCAL RINGS

Tensoring with A
b and taking completions, we get a commutative diagram

K ⊗A A
b / A⊕n ⊗A A
b / M ⊗A A
b /0

φK φA⊕n φM
  
0 /K
b /A
d ⊕n /M
c / 0,

where the first row is exact by right-exactness of ⊗ and the second row is exact by Corollary VI.2.8.
The map φA is clearly an isomorphism, and from this it follows that φA⊕n is an isomorphism for
arbitrary n by induction. Hence, φM is surjective by the Snake lemma. Since M is an arbitrary
finitely generated module and K is finitely generated (because A is Noetherian), this implies that
φK is also surjective. Thus, φM is an isomorphism by the Snake lemma.
(2): Let J ⊆ A be an ideal. Then, tensoring the exact sequence
0 → J → A → A/J → 0
with A
b yields (by (1) and Corollary VI.2.8) the exact sequence

0 → Jb → A b → A/J
d → 0.

Thus, TorA
1 (A, A/J) = 0 and so A is flat over A by Proposition V.3.5.
b b □
 3. COMPARING A RING WITH ITS COMPLETION 95

3. Comparing a ring with its completion

In the previous lecture, we have learned that if A is a Noetherian ring and I ⊆ A is an ideal,
then the ring map f : A → A b is flat. We call an A-module M I-adically complete if the natural
A-linear map M → M is an isomorphism. In this lecture, we will study some properties that we
c
can translate back and forth between A and its completion A.b

Lemma 3.1: Let A be a Noetherian ring and I ⊆ A an ideal. Then, the following hold:
(1) Ib = I · Ab∼= I ⊗A A. b
(2) d n b n
(I) = (I) for all n ≥ 0.
(3) \
(A/I n) ∼ b Ibn ∼
= A/ = A/I n .
(4) A is I-adically
b and I-adically complete.
(5) I /I n+1 ∼
n
= Ibn /Ibn+1 .
(6) grI (A) ∼= grIb(A).
b
(7) Ib is contained in the Jacobson radical of A.
b

P ROOF. (1): By Proposition VI.2.10, the natural map

b → Ib ⊆ A,
I ⊗A A b

whose image is I · A,b is an isomorphism.

(2): Since the powes of I n are cofinal in the powers of I, Remark VI.2.4 guarantes that the
I-adic and I n -adic completion coincide. Thus, we can apply (1) to I n and get

(I
d n) = I n · A b n = (Ibn ).
b = (I · A)

(3): We have a short exact sequence

0 → I n → A → A/I n → 0.

Taking completions (which results in an exact sequence by Proposition VI.2.10), we obtain the short
exact sequence

0 → Ibn → A \
b → (A/I n ) → 0,

which yields the first isomorphism. Now,

\
(A/I n ) = lim(A/I n )/(I i · (A/I n )) ∼
= A/I n ,
←−
i≥0

since this is the constant inverse system with value A/I n for i ≥ n.
(4): The I-adic
b completion of A b can be described as the map on inverse limits induced by the
natural isomorphisms A/I → A/I n , hence is an isomorphism. Since I n · A
n b b b = Ibn , the I-adic and
I-adic
b completions of A b coincide.
(5): Take quotients in (3).
(6): Follows from (5).
96 VI. REGULAR LOCAL RINGS

b consider the sequence (Pi−1 aj )i≥0 ∈ lim A/

(7): For a ∈ I, b Ibi and lift it to an element b ∈ A
b
j=0 ←−
via the map φ : A b Ibi which is an isomorphism by (4). Then, (1 − a) · b = 1, since
b → lim A/
←−
Xi−1 i−1
X i−1
X
φ((1 − a) · b) = (1 − a) · φ(b) = (1 − a)( aj )i≥0 = (1, . . . , aj − aj , . . .) = 1.
j=0 j=0 j=1

Thus (1 − a) is a unit. Now, if there is a maximal ideal m and an a ∈ Ib with a ̸∈ m, then

m + (a) = A, so there exists b ∈ m, c ∈ A with b + ac = 1. But bc ∈ I, b hence b = (1 − ac) is a
unit, a contradiction. Hence, I is contained in every maximal ideal of A.
b b □

Proposition 3.2: Let A be a Noetherian local ring with maximal ideal m and A → A
b the natural
map to the m-adic completion. Then, the following hold:
(1) A b∼
b = mA
b is a local ring with maximal ideal m = m ⊗A A.
b
(2) A is m-adically
b b complete.
(3) Ab is Noetherian.

P ROOF. (1): By Lemma VI.3.1 (7), we know that m b is contained in the Jacobson radical. On
the other hand, by Lemma VI.3.1 (3), the ring A/m b ∼
b = A/m is a field, hence mb is maximal.
(2): This is part of Lemma VI.3.1.
(3): Choose generators a1 , . . . , an of m and consider the map
A[[x1 , . . . , xn ]] → A
b
xi 7→ ai .
Using the explicit description of A,
b one checks that this is a well-defined surjective ring homo-
morphism. Since A is Noetherian, so is A[[x1 , . . . , xn ]] (Sheet 7), hence so is its quotient A
b by
Proposition III.1.8. □

In fact, for a Noetherian local ring A, the map A → A

b is not only flat, but faithfully flat.

Definition 3.3: Let A be a ring. An A-module M is called faithfully flat if a complex (i.e. with
g ◦ f = 0) of A-modules
f g
0 → M1 → M2 → M 3 → 0
is exact if and only if
0 → M1 ⊗A M → M2 ⊗A M → M3 ⊗A M → 0
is exact.
We have the following characterization of faithfully flat modules:
Lemma 3.4: Let A be a ring and M a flat A-module. The following are equivalent:
(1) M is faithfully flat.
(2) For every non-zero A-module N , the tensor product N ⊗A M is non-zero.
(3) For all p ∈ Spec A, the tensor product k(p) ⊗A M is non-zero.
(4) For all maximal ideals m ∈ Spec A, the tensor product k(m) ⊗A M is non-zero.
 3. COMPARING A RING WITH ITS COMPLETION 97

P ROOF. (1) ⇒ (2) : Follows from tensoring idN : N → N with M .

(2) ⇒ (3) ⇒ (4) : These are clear.
(4) ⇒ (1) : Let
f g
0 → M1 → M 2 → M 3 → 0
be a complex such that
0 → M1 ⊗A M → M2 ⊗A M → M3 ⊗A M → 0
is exact. Let H = ker(g)/im(f ). By flatness of M , H ⊗A M ∼
= ker(g ⊗A M )/im(f ⊗A M ) = 0.
Seeking a contradiction, assume H ̸= 0 and pick 0 ̸= h ∈ H and set I = Ann(h). Then,
A/I ⊆ H, hence M/IM ∼ = A/I ⊗A M ⊆ H ⊗A M = 0 by flatness of M . Choose a maximal
ideal m with I ⊆ m. Then, 0 = M/IM → M/mM is surjective, hence k(m) ⊗A M = 0, a
contradiction. □
Applying this to the case where M is an A-algebra, the conditions (3) and (4) become a condi-
tion on fiber rings, which translate to the following:
Corollary 3.5: Let f : A → B be a flat ring homomorphism. Then, the following are equivalent:
(1) f is faithfully flat.
(2) f ∗ : Spec B → Spec A is surjective.
(3) Every closed point of Spec A is in the image of f .
Corollary 3.6: Let f : A → B be a flat ring homomorphism. Then, f satisfies going down.
P ROOF. Let p ⊆ p′ ⊆ A be a chain of prime ideals and let q′ ⊆ B be a prime ideal with
f −1 (q′ ). Consider the flat ring map g : Ap′ → Bq′ . Since the unique maximal ideal p′ Ap′ is in the
image of g ∗ , g is faithfully flat by Corollary VI.3.5. Thus, there is a prime ideal of Bq′ mapping to
pAp′ . Its preimage in B is a preimage of p under f ∗ . □

Corollary 3.7: Let A be a Noetherian local ring and f : A → A

b its completion. Then, the following
hold:
(1) f is faithfully flat.
(2) dim(A) = dim(A). b

P ROOF. (1): Since A is local and f is flat, it suffices to check that the unique maximal ideal of
A is in the image of f ∗ : Spec A
b → Spec A. The commutative diagram
f
A /A
b

 
∼
A/m
= / A/
b m b

shows that f −1 (m)

b = m.
(2): Since A and A b are local, we have to show that the heights of their maximal ideals are the
∗ −1
same. Note that (f ) (m) = m. b From the fact that f satisfies going down, we obtain that
ht(m) ≤ ht(m).
b
98 VI. REGULAR LOCAL RINGS

On the other hand, by induction on d := dim(A), one can prove that there exist elements a1 , . . . , ad
such that V (a1 , . . . , ad ) = V (m). 5 We conclude that
b = (f ∗ )−1 (V (m)) = (f ∗ )−1 (V (a1 , . . . , ad )) = V (f (a1 , . . . , ad )).
V (m)
By Krull’s height theorem III.4.8, we deduce
b ≤ d.
ht(m)
□
Remark 3.8: Many more properties of rings “descend” along faithfully flat ring morphisms. More
precisely, let f : A → B be a faithfully flat ring map, then the following hold [Sta25, Tag 033D]:
(1) If B is Noetherian, so is A.
(2) If B is reduced, so is A.
(3) If A and B are integral domains and B is normal, so is A.

5Such a sequence of a is called system of parameters for A.

i
 4. REGULAR LOCAL RINGS 99

4. Regular local rings

Recall that a Noetherian local ring A with maximal ideal m satisfies
dim(A) ≤ edim(A) := dimA/m m/m2 .
The regular local rings are those where equality holds:
Definition 4.1: Let A be a Noetherian local ring with maximal ideal m. Then, A is called regular
if
dim(A) = edim(A).
Remark 4.2: If A = k[x1 , . . . , xn ]/(f1 , . . . , fr ) and P ∈ Spec A is a closed point corresponding
to the maximal ideal m and with residue field k, then one can show that m/m2 is isomorphic, as a
k-vector space, to the cokernel of the Jacobian matrix
 
∂x1 f1 . . . ∂x1 fr
 .. .. .

 . .
∂xn f1 . . . ∂xn fr

Motivated by differential geometry, we should therefore think of m/m2 as the dual of the tangent
space of Spec A at the point P . This is why m/m2 is also called Zariski cotangent space of A at m.
The condition that A is regular is then the condition that its tangent space has the same dimension
as A itself and is an analogue of smoothness in the algebraic setting.
Example 4.3: The following local rings are regular:
(1) Fields.
(2) DVRs.
6
(3) The localization of the polynomial ring k[x1 , . . . , xn ] over a field k at any prime ideal.
(4) The power series ring k[[x1 , . . . , xn ]] over a field k. 7
Power series rings are the universal examples of complete regular local rings, by the following
result of Cohen (see e.g. [Sta25, Tag 0323]):
Theorem 4.4 (Cohen’s structure theorem): Let A be a Noetherian complete local ring. Then, there
exists a field k or a complete DVR Λ with a prime p ∈ Z ⊆ Λ as uniformizer, an integer n ≥ 0, and
a surjective ring homomorphism
k[[x1 , . . . , xn ]] → A
or
Λ[[x2 , . . . , xn ]] → A.
If A is regular and contains a field, then there is an isomorphism k[[x1 , . . . , xn ]] ∼
= A.
We can prove the following simple version of the above result:

6For maximal ideals, this is clear. The case of non-maximal ideals is an exercise (it follows, for example, from
Noether normalizatin).
7In fact, one can show that if A is regular, then so is A[[x]].
100 VI. REGULAR LOCAL RINGS

Corollary 4.5: Let A be a Noetherian complete regular local ring containing a field k mapping
isomorphically to the residue field of A. Then there is an isomorphism
k[[x1 , . . . , xn ]] ∼
= A.
P ROOF. Choose a minimal set of generators a1 , . . . , an ∈ m of the maximal ideal of A. Since
A is regular, we have n = dim(A) by Nakayama’s lemma. Since A is complete, one checks as in
Proposition VI.3.2 the map
φ : k[[x1 . . . , xn ]] → A
xi 7→ ai
is well-defined and surjective. If φ is not injective, then it contains a non-zero divisor g, because
k[[x1 , . . . , xn ]] is an integral domain. But then dim(A) < dim k[[x1 , . . . , xn ]] = n by the principal
ideal theorem, contradicting n = dim(A). □
Thus, the real content of Cohen’s structure theorem for complete regular local rings containing
a field is that they contain a field mapping isomorphically to the residue field. We have many
characterizations of regular local rings.
Proposition 4.6: Let A be a Noetherian local ring with maximal ideal m and residue field k. Let
d = dim(A). Then, the following are equivalent:
(1) A is regular.
(2) A
b is regular.
(3) The ideal m is generated by d elements.
(4) There is an isomorphism of rings
gr(A) ∼
= k[x1 , . . . , xd ].

P ROOF. (1) ⇔ (2) : By Lemma VI.3.1(5) and Corollary VI.3.7, we have dim(A) = dim(A) b
and edim(A) = edim(A), b so this equivalence is clear.
(1) ⇔ (3) : This follows immediately from Nakayama’s lemma III.4.4 applied to m.
(4) ⇒ (1): This follows from gr(A)1 = m/m2 .
(2) ⇒ (4) : Assume for simplicity that A contains a field mapping isomorphically to its residue
field 8. By Corollary VI.4.5, we have A ∼ = k[[x1 , . . . , xn ]]. Then, the isomorphism gr(A) ∼
=
k[x1 , . . . , xd ] is clear. □
Since regular local rings correspond to smooth points in the setting of algebraic geometry, we
expected them to have many nice properties. A simple one is that they are all integral, as follows
from the following more general result (note that the assumptions are satisfied for regular local rings
by Corollary VI.2.9 and Proposition VI.4.6):
Lemma 4.7: Let A be a ring and I ⊆ A an ideal. Assume that i≥0 I i = 0 and that grI (A) is an
T
integral domain. Then, A is an integral domain.
P ROOF. Let a, b ∈ A be non-zero. Since i≥0 I i = 0, there exist r, s ≥ 0 such that a ∈ I r , a ̸∈
T

I r+1 , b ∈ I s , b ̸∈ I s+1 . Let a and b be the images of a and b in grI (A)r and grI (A)s , respectively.
Then, a, b ̸= 0 and, since grI (A) is an integral domain, 0 ̸= ab = ab. Thus, ab ̸= 0, as desired. □
8The classical proof of the general case uses Hilbert polynomials, see for example [Kem11, Theorem 13.4]
 4. REGULAR LOCAL RINGS 101

Recall the definition of global dimension of an Abelian category with enough projectives from
Sheet 13. For a ring A, we let gldim(A) be the global dimension of ModA . The following charac-
terization of regular local rings is due to Serre (see [Sta25, Tag 065U]):
Theorem 4.8 (Serre’s characterization of regular local rings): Let A be a Noetherian local ring.
Then, the following are equivalent:
(1) A is regular.
(2) gldim(A) < ∞.
If these conditions hold, then gldim(A) = dim(A).
Note that the finiteness of gldim(A) is stable under localization at primes, since Ext commutes
with localization. One of the fundamental consequences of Theorem VI.4.8 is the following, which
was a long-standing open problem before Serre’s characterization:
Corollary 4.9: Let A be a Noetherian local ring and p ∈ Spec A. If A is regular, then so is Ap .
Remark 4.10: In the geometric interpretation of regularity, this means that the locus of points in
Spec A that are nonsingular is stable under generalization.
Another algebraically interesting property is the following, which is an application of homolog-
ical machinery developed by Auslander and Buchsbaum (also look up the “Auslander–Buchsbaum
formula” in this context):
Theorem 4.11 (Auslander–Buchsbaum theorem): Let A be a Noetherian local ring. Then, A is a
UFD.
Remark 4.12: Since UFDs are normal by Proposition IV.1.10 and normal 1-dimensional Noether-
ian local rings are regular by Proposition IV.3.4, being a UFD characterizes regular local rings in
small dimension.
If k is a field, then the ring A = k[[x, y, z]]/(z 2 + x3 + y 5 ) is a UFD (see e.g. [Lip69]), but it
is not regular: Set m = (x, y, z). Then,
A/m2 = k[[x, y, z]](z 2 + x3 + y 5 , x2 , xy, xz, y 2 , yz, z 2 ) ∼
= k[[x, y, z]]/(x, y, z)2 ∼
= k3 .
In fact, if k is algebraically closed and char(k) ̸= 2, 3, 5, then A is the only complete Noetherian
local k-algebra with residue field k that is a UFD but not regular by [Lip69, Theorem 25.1].
CHAPTER VII

Appendix

102
 1. CATEGORIES 103

1. Categories
For the sake of completeness, we include the formal definitions of categories, functors, and
natural transformations here. In the following, a class is a collection of sets with a common property,
e.g. the collection of all sets is a class (but not a set!).
Definition 1.1 (Category): A category C consists of the following data:

(1) A class ob(C). Elements of ob(C) are called objects.

(2) A class Mor(C). Elements of Mor(C) are called morphisms (or arrows).
(3) For every morphism f , a source s(f ) ∈ ob(C) and a target t(f ) ∈ ob(C). We write
f : A → B for an arrow with source A and target B.
(4) For any two morphisms f, g with t(f ) = s(g)1, an arrow g ◦ f called composition of g and
f.
(5) For every object A ∈ ob(C), a morphism 1A := idA called identity of A.

Satisfying the following conditions:

(1) s(g ◦ f ) = s(f ) and t(g ◦ f ) = t(g) for all composable arrows f, g.
(2) s(1A ) = t(1A ) = A for all objects A.
(3) h ◦ (g ◦ f ) = (h ◦ g) ◦ f for all composable arrows f, g, h.
(4) 1B ◦ f = f = f ◦ 1A for f : A → B.
Remark 1.2: In any category, one can define isomorphisms as those arrows that have a two-sided
inverse.
Example 1.3: The collection of sets, groups, Abelian groups, vector spaces over a field, rings,
topological spaces, etc. are all categories. The most important category for this lecture course is the
category ModA of modules over a ring A.
Example 1.4: If C is a category, the opposite category C op of C is the category with ob(C op ) =
ob(C) and Mor(C op ) = Mor(C) but with source and target functions interchanged.
Definition 1.5 (Functor): Let C and D be categories. A (covariant) functor F : C → D consists of

(1) A map F : ob(C) → ob(D).

(2) A map F : Mor(C) → Mor(D).

Satisfying the following conditions:

(1) s(F (f )) = F (s(f )), t(F (f )) = F (t(f )).

(2) F (g ◦ f ) = F (g) ◦ F (f ) for composable arrows f, g.
(3) F (1A ) = 1F (A) for all A ∈ ob(C).

A contravariant functor from C to D is a covariant functor F : C op → D.

1We call g and f composable if this condition holds.

104 VII. APPENDIX

Example 1.6: Given a ring A and a module N , the collections of maps

HomA (−, N ) : M 7 → HomA (M, N )
f : M → M′ 7→ (− ◦ f ) : HomA (M ′ , N ) → HomA (M, N )
HomA (N, −) : M 7→ HomA (N, −)
f : M → M′ 7→ (f ◦ −) : HomA (N, M ) → HomA (N, M ′ )
(−) ⊗A N : M 7→ M ⊗N
f : M → M′ 7→ (f ⊗A id) : M ⊗A N → M ′ ⊗A N
are functors ModA → ModA (the first one is contravariant, the other two covariant).
Definition 1.7: Let C and D be categories and F, G : C → D functors. A natural transformation
α : F ⇒ G is a collection of arrows α(A) : F (A) → G(A), one for every object A of C, such that
for all arrows f : A → B the following diagram commutes:
F (f )
F (A) / F (B)

α(A) α(B)
 G(f ) 
G(A) / G(B)

A natural transformation is called natural isomorphism (or natural equivalence) if all the α(A) are
isomorphisms.
L Q
Example 1.8: If A is a ring, consider the functors HomA ( i∈I A, −), i∈I (−) : ModA →
ModA 2. Let ιi : AL
L
→ i∈I A be the Q inclusion of the i-th summand. We define a natural transfor-
mation α : HomA ( i∈I A, −) ⇒ i∈I (−) by setting
M Y
α(M ) : HomA ( A, M ) → M
i∈I i∈I
g 7→ ((g ◦ ιi )(1))i∈I .
If f : M → M′ is an A-linear map, then
Y
( f )(((g ◦ ιi )(1))i∈I ) = (f ◦ g ◦ ιi )i∈I = α(M ′ )(f ◦ g),
i∈I
so α is indeed a natural transformation. In fact, each α(M ) is an isomorphism by the universal
property of direct sums, hence α is a natural isomorphism.
Remark 1.9: If F, G : ModA → ModA are two functors that are naturally isomorphic, then F is
exact if and only if G is. Using this and the above example, we proved Proposition II.2.9.

2Here, Q is a functor by setting

Q
f to be the map that is f on every component
i∈I i∈I
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