PART 2 :)
INTEGRAL
DOMAINS
Basic definitions ………………………………………………………..03
Unique factorization domains (UFD) ……..….………………………06
Principal ideal domains (PID) ……….………..……………………….12
GCD and LCM …………………………………………………………..15
Euclidean domain ……………………………………………………16
The field of fractions
Factorization in polynomial rings
BASIC DEFINITIONS
O NO TAN BASIC
1. "A" is an integral domain All
ab =elements
0 implies
a∈ a =A0are
or non-zero
b=0 divisors
2. "a" is a zero divisor a≠0 ∃ b≠0 such the ab = 0
3. "a" is an invertible element a ≠ 0 ∃ non zero b∈A such that ab = 1
4. "a" and "b" are associates a~b if a = ub with u∈Aˣ
5. "p" is irreducible p ∉ Aˣ ∪ {0}
p = ab implies a ∈ Aˣ or b ∈ Aˣ
Implies
IntegralUFD
domain implication
6. "p" is prime p ∉ Aˣ ∪ {0}
p | ab implies p | a or p | b
Z is an integral domain. The only invertible elements of Z are +1,-1 .
Clearly all elements in Z are non zero divisors.
Any field is an integral domain. For any field K the ring of polynomials K[x] is an
integral domain.
E
X Assume A is an integral domain. Elements a,b ∈ A are associates iff (a) = (b)
A (meaning a | b and b | a).
M "Associates => (a)=(b)"
P If a = ub then a ∈ (b) and (a) ⊆ (b).
L Since u∈ Ax u-1 a = b implying b ∈ (a) and (b) ⊆ (a) => (a) = (b)
E "(a)=(b) => Associates"
If (a) = (b) then a = ub and b=va for some u,v ∈ A.
Thus a = ub = uva = a => uv = 1, u and v are units => associates
L
E
M
M
A
UNIQUE FACTORIZATION DOMAIN
UFD NOT UFO!!!
UFD
COMMUTATIVE RING
DONEC<< QUIS
INTEGRAL
NUNCDOMAIN << UFD
An integral domain is called a UFD if:
1.
1. Any a ∈ A\{0} can be written as a product of a unit and irreducible
elements, that is a = u (p₁ …pᵢ) where u ∈ Aˣ and pᵣ are irreducible
2. The above factorization is unique up to units and a permutation of
factors.
An integral domain satisfying just (1) is called a factorization domain (FD).
Z is a UFD.
Any field is a UFD.
For any field K the ring of polynomials K[x] is a UFD.
If A is a UFD so is A[x]
E
X
A Assume A is an integral domain. Elements a,b ∈ A are associates iff (a) = (b)
M (meaning a | b and b | a).
P "Associates => (a)=(b)"
L If a = ub then a ∈ (b) and (a) ⊆ (b).
E Since u∈ Ax u-1 a = b implying b ∈ (a) and (b) ⊆ (a) => (a) = (b)
"(a)=(b) => Associates"
If (a) = (b) then a = ub and b=va for some u,v ∈ A.
Thus a = ub = uva = a => uv = 1, u and v are units => associates
L
E
M
M
A
EXAM QUESTION
Consider a ring (integral domain) which is not a UFD.
Answer:
A = Z[√-5] = {a + bi√5 | a,b ∈ Z } ⊆ C
To prove this claim that the element 6 has two different factorisations.
6 = 2 · 3 = (1+ i√5) · (1- i√5)
You should reach the conclusions:
There exist two different factorisations (to show that they are not equivalent
prove that they are not associate), meaning not a UFD.
Elements are irreducible, meaning it is an FD.
Although elements are irreducible, they are not prime.
THEORE
M
Let A be a factorization domain. Then A is a UFD iff every irreducible element is prime.
"=>"
Assume UFD. Take "p" irreducible then p=ab implies a ∈ A or b ∈ A.
Consider p | ab, that is pc = ab with a, b, c ∈ A. We want to show p | a or p | b.
Let
a = ∏aᵢ b = ∏bᵢ c = ∏cᵢ
be factorizations into irreducible element. Then the element ab = pc has two
factorizations….
∏aᵢ ∏bᵢ = p ∏cᵢ
Since we assumed UFD then both factorizations are equal and p equals (up to a unit) to
one of aᵢ or bᵢ meaning that p divides a or b.
Hence, "p" is prime
Let A be a factorization domain. Then A is a UFD iff every irreducible element is prime.
"<="
Assume FD. Assume that an element a has two decompositions into irreducible
elements a = ∏ᵐpᵢ = ∏ⁿqᵢ. By assumption, all pᵢ and qᵢ are prime.
Let p = pᵣ. Then p should divide one of q (assume it is qᵣ).
But qᵣ is irreducible, therefore qᵣ = u·p for some u ∈ Aˣ.
Dividing by p we get ∏ᵐ⁻¹pᵢ = u∏ⁿ⁻¹qᵢ. By induction on m, these factorizations are
the same up to the permutation of factors and multiplication of units => UFD.
This completes the proof of our theorem. Both directions have been shown.
PRINCIPAL IDEAL DOMAIN
BUT NOT THE SCHOOL’S PRINCIPAL!!
PRINCIPAL IDEAL DOMAIN
COMMUTATIVE RING
DONEC
< INTEGRAL
QUIS NUNC
DOMAIN < UFD < PID
An integral domain is called a PID if every ideal of A is a principal ideal.
An ideal I ⊆ A is called principal if it can be generated by one element, that is
∃ a ∈ A such that I = (a) = aA
Every ideal of Z is a principal ideal meaning Z is a PID.
!!!! This is different for Z[x] which is NOT a PID (this does not work as an example
for exam questions asking a case of UFD but not PID).
Any polynomial ring K[x] over a field K is a PID (hence a UFD).
THEORE
M
Let A be a PID. Then every irreducible element is prime.
Ley p ∈ A be irreducible and let p | ab. Assume p ∤ a. The ideal (p,a) is a principal
ideal (PID condition), therefore (p,a) = (d) for some d ∈ A.
In particular p ∈ (d) implying p = cd. Irreducibility means that either c ∈ Aˣ or d ∈
Aˣ hold. If c ∈ Aˣ then (p) = (d) ∋ a and p | a. CONTRADICTION.
Our assumption is wrong and p | a, meaning p is prime.
Every PID is a UFD (study proof from notes)
GCD AND LCM
NOT LSD …
GREATEST COMMON DIVISOR
EL MÁSDONEC
GRANDE
QUIS
DENUNC
LOS PEQUES
Let A be an integral domain. The gcd of two elements a l b ∈ A is an element
d ∈ A (denoted by gcd(a,b) ) such that…
1.
1. d | a and d | b
2. If c | a and c | b then c | d
Elements a, b are called coprime if gcd(a,b) = 1
Not all rings have gcd but it always exists in PID and UFD
If d is a gcd then any of its associates is also a gcd. If c, d are two gcd of a
and b then both c | d and d | c must hold!!!!
LEAST COMMON MULTIPLE
EL MÁSDONEC
PEQUEQUIS
DE LOS
NUNC
GRANDES
Let A be an integral domain. The lcm of two elements a l b ∈ A is an element
m ∈ A (denoted by lcm(a,b) ) such that…
1.
1. a | m and b | m
2. If a | c and b | c then m | c
Not all rings have lcm but it always exists in PID and UFD
THEORE
M
Let A be a PID. Then a, b ∈ A are coprime (gcd=1) if there exist x, y ∈ A such that ax + by
=1
"=>"
We consider the previous lemma:
"d = gcd(a,b) if and only if (a,b) = (d)"
Now, condition ax + by = 1 with x, y ∈ A implies that (a,b) = A = (1).
And by the lemma above 1 = gcd(a,b), and elements a and b are coprime
THEORE
M
GCD and LCM exist in UFDs.
Any two elements in a UFD can be written in terms of irreducible elements and a unit
(invertible element)… ∀a, b ∈ A …
a = u ∏ pᵢᵏ b = v ∏ pᵢᴸ
where u, v ∈ Aˣ; p₁… pᵢ are distinct prime elements (not associate) with k, l ≥ 0.
Define…
d = ∏pˢ m = ∏pᵀ
s= min{k,l} t= max{k,l}
We claim that d = gcd(a,b) and m = lcm(a,b). Let us show this for d. Assume that c | a and
c | b. Up to a unit we can write c = ∏ pʳ where r ≤ k and r ≤ l. This implies that r ≤ l
meaning c | d as required.
EUCLIDIAN DOMAINS
EL DE EL EUCLIDES
An integral domain A is called a Euclidean domain if there exists a
function δ: R\{0}→ ℕ = {0, 1, 2, …} (degree function or Euclidean
function) such that ∀a,b ∈A\{0} there exist elements q, r ∈A with a =
bq + r and either r = 0 or 𝛿 (r) < 𝛿 (b).
The ring ℤ with degree function 𝜹(n) = |n| is a Euclidean domain.
For any field K, the algebra of polynomials K[x] with degree function 𝜹(f) = deg(f)
is a Euclidean domain
THEORE
M
Let A be a commutative ring and let f,g ∈ A be polynomials such that g is monic of degree
d (i.e g = xᵈ + gᵣxᵈ⁻¹+ … + g₁x + g₀). Then there exist unique polynomials q, r ∈ A[x] such
that f = gq + r and deg(r)< d.
Let us choose a polynomial q ∈ A[x] such that the polynomial r:= f - gq has the minimal
possible degree. Assume that r = rᵤxᵁ + … + r₀ has degree u ≥ d. Then
F = (g · q) + r => F - (g · q) = r
F - (g · q) - (g · rᵤxᵁ⁻ᵈ) = r - (g · rᵤxᵁ⁻ᵈ )
F - g · (q + rᵤxᵁ⁻ᵈ) = r - (g · rᵤxᵁ⁻ᵈ ) = (∑ᵁ⁻¹(rᵦxᵝ)) - (∑ᵈ⁻¹ (rᵤgᵦxᵁ⁻ᵈ⁺ᵝ))
Which has degree < d, a contradiction. To prove the uniqueness, assume that f = gq’ + r’
with deg (r’) < d. Then g(q - q’) = r - r’. If q ≠ q’ then deg(g(q − q’)) ≥ deg g = d and
deg(r − r′) < d, a contradiction. Therefore q = q’, hence also r = r’.
THEORE
M
An element a ∈ A is a root of a polynomial f ∈ A[x] (i.e. f(a) = 0) iff (x - a) divides f.
"=>"
We can divide f with a remainder (r) as follows:
f = g · (x - a) + r
where deg(r) < deg(x-a) = 1. Hence r ∈ A (a constant). We know that "a" is a root so at
f(a) we have:
f(a) = g (a - a) + r = g · 0 + r = 0 => r = 0
Hence f = g (x-a) and (x - a) divides f.
"<="
If (x-a) divides f then it is clear that f = (x-a)g and a is a root
If A is an integral domain, then a polynomial f ∈ A[x] has at most deg(f)
roots. If "a" is a root of f then we write… f = (x-a)q with deg(q) = deg(f) - 1.
Any root of f different from "a" is a root of q.
If A is NOT an integral domain, then f ∈ A[x] can have more than deg(f)
roots.
The polynomial x² - 1 in ℤ₈[x] has four roots
If K is an infinite field and f, g ∈ K[x] are polynomials such that f(a) = g(a)
for all a ∈ K then f - g have infinitely many roots and f-g = 0.
The ring Z[i] ={x + iy| x,y ∈ Z}⊆ℂ is called the ring of Gaussian integers. It is
a Euclidean domain with degree function 𝜹 (x + iy) = |x + iy|²
The ring of Gaussian integers is a Euclidean domain.
Let a, b ∈ Z[i]\{0}. Consider ab⁻¹ = x +iy ∈ C. There exist m, n ∈ ℤ such that |x -
m| ≤ 1/2 and |y - n| ≤ 1/2.
Let q = m + ni and a = bq + r => r = a - bq. Since a = abb⁻¹ we get…
r = a - bq = abb⁻¹ - bq = b · (ab⁻¹ - q)
Replacing q and ab⁻¹ in the last equation…
r = b · (ab⁻¹ - m - ni) = b · (x +iy - m - ni) = b(x - m) + ib(y - n)
𝛅(r) = |r|² = |b|² (|x-m|²+|y-n|²) ≤ |b|²/2< |b|² = 𝛅(b)
EUCLIDEAN ALGORTIHM
DONEC QUIS NUNC
THE FIELD OF FRACTIONS
I
FACTORIZATION IN POLYNOMIAL RINGS
I
I
