Math 323: Solutions to homework 9

Problem 1. (a) Let R be an integral domain, and let M be an R-module. Prove

that if Tor(M ) 6= {0}, then M cannot be free on any set of generators.
Note that as proved in the last homework, if R has zero divisors, then any R-
module has torsion elements, so it makes sense to discuss this condition only for
integral domains.
Remark. Let R be an integral domain that is not a PID, and let I be a non-
principal finitely generated ideal in R. Then we proved in class that I considered
as an R-module is torsion-free (i.e. Tor(I) = {0}), but not free. As a reminder,
here is the proof: Since R is an integral domain, the equation ra = 0, where r ∈ R,
a ∈ I implies either r = 0 or a = 0. Thus, Tor(I) = 0. Thus, I considered as
an R-module is torsion-free. Assume, to the contrary, that I is free on some finite
subset A. Since A is not empty, |A| ≥ 1. But given {x, y} ⊂ A, x, y 6= 0, we observe
that (y)x + (−x)y = 0. The non-trivial linear dependence forces |A| ≤ 1. Hence
|A| = 1. Assume A = {a}. Then any non-zero element x ∈ I can be represented as
x = ra for unique r ∈ R. We conclude that I = Ra as a R-module, which forces I
to be a principal ideal, contradiction.
Thus we see that over a general integral domain, being torsion-free is necessary
but not sufficient for being free. By now we have proved that over a PID this
condition is also sufficient.
Solution: Assume Tor M 6= {0}. Assume, to the contrary, that M is free on
some subset A ⊂ M of generators. Choose x ∈ Tor(M ) with x 6= 0. By definition,
there exists r ∈ R with r 6= 0 such that rx = 0. Then, by definition of freeness,
there exists unique nonzero elements r1 , r2 , ..., rn of R and unique a1 , a2 , ..., an in
A such that
x = r1 a1 + r2 a2 + · · · + rn an
for some n ∈ N. Acting on both sides by r, we have
0 = rx = r(r1 a1 ) + r(r2 a2 ) + · · · + r(rn an )
This is equivalent to
0 = (rr1 )a1 + (rr2 )a2 + · · · + (rrn )an
Since r 6= 0 and ri 6= 0 for each 1 ≤ i ≤ n, we get rri 6= 0 for each 1 ≤ i ≤ n. (Here
we are using the fact that R is an integral domain). But the above equation gives
a non-trivial relation between the generators, which contradicts the fact M is free
on the subset A.

2. Section 10.2, Problem 12.

We first solve Problem 11 in Section 10.2 in order to solve Problem 12.

Section 10.2, Problem 11. Let A1 , A2 , ..., An be R-modules and let Bi be a

submodule of Ai for each i = 1, 2, ..., n. Prove that
(A1 × · · · × An )/(B1 × · · · × Bn ) ∼
= (A1 /B1 ) × · · · × (An /Bn )

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Solution: We consider the map ψ : A1 × · · · × An → (A1 /B1 ) × · · · × (An /Bn )

defined by
ψ : (a1 , a2 , ..., an ) 7→ (a1 + B1 , a2 + B2 , ..., an + Bn )
It is clear that ψ is a surjective homomorphism. Its kernel is given by
ker(ψ) = B1 × B2 × · · · × Bn
By the First Isomorphism Theorem for modules, we obtain that
(A1 × · · · × An )/(B1 × · · · × Bn ) ∼
= (A1 /B1 ) × · · · × (An /Bn )
as desired.

Section 10.2 Problem 12. Let I be a left ideal of R and let n be a positive
integer. Prove
Rn /IRn ∼
= R/IR × · · · × R/IR
Solution: We have the following R-module isomorphisms:
Rn ∼
= R × R × ···× R
and
IRn ∼
= IR × IR × · · · × IR
both of which are easily checked by constructing explicit maps. By invoking the
result of previous problem, we obtain that
Rn /IRn ∼= (R × R × · · · × R)/(IR × IR × · · · × IR) ∼
= R/IR × · · · × R/IR
as desired.
Section 10.3 Problem 2. Assume R is commutative. Prove that Rn ∼ = Rm if
and only if n = m, i.e. two free R-modules of finite rank are isomorphic if and only
if they have the same rank.
Solution: If n = m, it is immediate that Rn ∼ = Rm . Conversely, assume
n ∼ m
R = R . Let I be a maximal ideal in R (Proposition 11 of Section 7.4 shows
that in a ring with identity maximal ideals exist). Suppose φ : Rn → Rm is the
corresponding R-module isomorphism. Let π : Rm → Rm /IRm be the natural
projection map. Then, π ◦ φ : Rn → Rm /IRm is an R-module homomorphism. We
claim that ker(π ◦ φ) = IRn . Indeed,
ker(π◦φ) = {x ∈ Rn : (π◦φ)(x) = 0} = {x ∈ Rn : φ(x) ∈ IRm } = φ−1 (IRm ) = IRn
where the last equality follows from the fact φ is an isomorphism. From First
Isomorphism Theorem, we obtain that
Rn /IRn ∼
= Rm /IRm
From the result of previous problem Rn /IRn = (R/IR) × (R/IR) × · · · × (R/IR).
Identifying R/IR by R/I (since I is an ideal), we have Rn /IRn = (R/I)n . So the
above equation now reads:
(R/I)n ∼= (R/I)m
But R/I is a field, as I is a maximal ideal. But two finite dimensional vector spaces
over a field F = R/I is isomorphic if and only if n = m. This completes the proof.

Section 10.3 Problem 5. Let R be an integral domain. Prove that every

finitely generated torsion R-module has a nonzero annihilator, i.e. there is a nonzero
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element r ∈ R such that rm = 0 for all m ∈ M . Give an example of a torsion R-

module whose annihilator is the zero ideal.
Solution: Assume M is a finitely generated torsion R-module. By definition,
there exists a finite subset A ⊂ M , say A = {a1 , a2 , ..., am } such that
M = {s1 a1 + s2 a2 + · · · + sm am | s1 , s2 , ..., sm ∈ R, a1 , a2 , ..., am ∈ A, m ∈ Z+ }
Since M is a torsion R-module, for each m ∈ M there is a non-zero r ∈ R such
that rm = 0. In particular, {a1 , a2 , ..., am } = A ⊂ M , and so there exist non-zero
elements r1 , r2 , ..., rm such that ri ai = 0 for each 1 ≤ i ≤ m. Let
r = r1 r2 · · · rm
Since R is an integral domain, and ri 6= 0 for each 1 ≤ i ≤ m, it follows that r 6= 0.
Let m ∈ M . We will show that rm = 0. Since m ∈ M , there exists s1 , s2 , ..., sm ∈ R
such that
m = s1 a1 + s2 a2 + · · · + sm am
But then,
rm = r(s1 a1 + s2 a2 + · · · + sm am )
= (r1 r2 · · · rm )(s1 a1 + s2 a2 + · · · + sm am )
= (s1 r2 r3 · · · rm )(r1 a1 ) + (s2 r1 r3 · · · rm )(r1 a2 ) + · · · + (sm r1 r2 · · · rm−1 )(rm am )
= 0 + 0 + ···+ 0 = 0
where we have used commutativity of R (which comes from the assumption that R
is an integral domain). We have shown that rm = 0. Since m ∈ M was arbitrary,
it follows that r is a non-zero annihilator of M , as desired.
Here is an example of torsion R-module, whose annihilator is the zero ideal.
Since Q/Z is an abelian group, we can consider Q/Z as a Z-module. Assume, to
the contrary, that there exists a non-zero annihilator of Q/Z; i.e. there exists z ∈ Z
such that zq = 0 for each q ∈ Q/Z. Take k to be any integer that does not divide
z. Then, using q = 1/k, we have
0 = z1/k = 1/k + 1/k + · · · + 1/k = z/k
| {z }
z times
z
So z/k = 0, implying that ∈ Z. But then z = bk for some b ∈ Z, which means
k
that k divides z, contradiction.

Section 10.3 Problem 6. Prove that if M is a finitely generated R-module

that is generated by n elements then every quotient of M may be generated by n
(or fewer) elements. Deduce that quotients of cyclic modules are cyclic.
Solution: Suppose M is finitely generated by n elements. By definition, there
exists a finite subset A ⊂ M , say A = {a1 , a2 , · · · an } such that
M = {r1 a1 + r2 a2 + · · · + rn an | r1 , r2 , ..., rn ∈ R, a1 , a2 , ..., an ∈ A, m ∈ Z+ }
Suppose N is a submodule of M . Then, we claim that the set
A = {a1 + N, a2 + N, ..., an + N }
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generates M/N . Indeed, every element of M/N can be written as m + N for some
m ∈ M . Now, m = r1 a1 + r2 a2 + · · · + rn an for some r1 , r2 , ...., rn ∈ R. Thus,

m + N = (r1 a1 + r2 a2 + · · · + rn an ) + N
= (r1 a1 + N ) + (r2 a2 + N ) + · · · + (rn an + N )
= r1 (a1 + N ) + r2 (a2 + N ) + · · · + rn (an + N )

So m + N is generated by elements of A. Since m + N was arbitrary element of

M/N , it follows that M/N is generated by A. Therefore, M/N is generated by n
or possibly fewer elements.

Section 10.3 Problem 9. An R-module M is called irreducible if M 6= 0

and if 0 and M are the only submodules of M . Show that M is irreducible if and
only if M 6= 0 and M is a cyclic module with any nonzero element as a generator.
Determine all the irreducible Z-modules.

Solution: First assume M is irreducible. Then, by definition M 6= 0, and 0 and

M are the only submodules of M . Choose any nonzero element x ∈ M . Consider
the cyclic submodule N = Rx. Since N 6= 0 as 0 6= x ∈ N , it follows that N = M ,
that is Rx = M . So M is a cyclic module with any nonzero element as a generator.
Conversely, assume M 6= 0, and M is a cyclic module with any nonzero element
as a generator. We want to show that M is a irreducible. Let N be any submodule
of M . We need to show N is either 0 submodule or N = M . Assuming N 6= 0,
choose 0 6= x ∈ N , then Rx ⊂ N . But by hypothesis, M = Rx, and so we get
M ⊂ N . We conclude that M = N . As a result M is an irreducible R-module.

We claim that irreducible Z-modules are precisely cyclic groups of prime order.
First, Z-modules can be precisely identified with abelian groups. Irreducibility of
Z-module is equivalent to the abelian group having no nontrivial normal subgroups,
i.e. it must be a simple group. But the only simple abelian groups are cyclic groups
of prime order. So Z-module being irreducible implies that it must be cyclic group
of prime order. Conversely, it is easily checked that every cyclic group of prime
order is an irreducible Z-module.

Section 10.2 Problem 13. Let R be a commutative ring, and let F be a

free R-module of a finite rank. Prove the following isomorphism of R-modules:
HomR (F, R) ∼
= F.

Solution: We will construct an explicit R-module isomorphism. Assume F

is free on some subset A ⊂ F . Since F has a finite rank, A is finite, say A =
{a1 , a2 , ..., an }. Then, for every x ∈ F , there exists unique elements r1 , r2 , ..., rn ∈
R, such that
x = r1 a1 + r2 a2 + · · · + rn an
Define Ψ : HomR (F, R) → F by

Ψ(φ) = φ(a1 )a1 + φ(a2 )a2 + · · · + φ(an )an

for every φ ∈ HomR (F, R). Note that φ(aj ) ∈ R for each 1 ≤ j ≤ n, and so in the
above equation, each φ(aj ) acts on aj ∈ F , as F is an R-module. We claim that Ψ
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is a R-module isomorphism. First, for any φ, σ ∈ HomR (F, R), we have

Ψ(φ + σ) = (φ(a1 ) + σ(a1 ))a1 + (φ(a2 ) + σ(a2 ))a2 + · · · + (φ(an ) + σ(an ))an
= (φ(a1 ) + φ(a2 ) + · · · + φ(an )) + (σ(a1 ) + σ(a2 ) + · · · + σ(an ))
= Ψ(φ) + Ψ(σ)
and for each r ∈ R
Ψ(rφ) = rφ(a1 )a1 + rφ(a2 )a2 + · · · + rφ(an )an
= r(φ(a1 )a1 + φ(a2 )a2 + · · · + φ(an )an ) = rΨ(φ)
Thus Ψ is an R-module homomorphism. To prove that Ψ is injective, assume
Ψ(φ) = Ψ(σ) for some φ, σ ∈ HomR (F, R). Then, by definition of Ψ, we have
φ(a1 )a1 + φ(a2 )a2 + · · · + φ(an )an = σ(a1 )a1 + σ(a2 )a2 + · · · + σ(an )an
or equivalently,
(φ(a1 ) − σ(a1 ))a1 + · · · + (φ(an ) − σ(an ))an = 0
Since F is a free R-module on A = {a1 , a2 , ..., an }, there cannot exist any non-trivial
linear relation between a1 , a2 , ..., an . Thus, φ(aj ) − σ(aj ) = 0 for each 1 ≤ j ≤ n,
that is φ(aj ) = σ(aj ) for each 1 ≤ j ≤ n. Now, take x ∈ F be any element. Then,
x has a unique representation of the form
x = r1 a1 + r2 a2 + · · · + rn an
Applying φ we get
φ(x) = r1 φ(a1 ) + · · · + rn φ(an ) = r1 σ(a1 ) + · · · + rn σ(an ) = σ(x)
Thus, φ(x) = σ(x). Since x was arbitrary, it follows that φ = σ, and so Ψ is
injective. To prove that Ψ is surjective, take any y ∈ F . Then, y has a unique
representation of the form
y = s1 a1 + s2 a2 + · · · + sn an
where si ∈ F . By a consequence of universal property, there exists a R-module
homomorphism ϕ ∈ HomR (F, R) such that ϕ(aj ) = sj for each 1 ≤ j ≤ n. Then,
Ψ(ϕ) = ϕ(a1 )a1 + ϕ(a2 )a2 + · · · + ϕ(an )an
= s1 a1 + s2 a2 + · · · + sn an = y
So Ψ(ϕ) = y. Since y ∈ F was arbitrary, it follows that Ψ is surjective. We
conclude that Ψ is a bijective R-module homomorphism, that is, Ψ is R-module
isomorphism. Therefore, HomR (F, R) ∼ = F as R-modules.
A shorter way to state essentially the same solution: We want to prove
that HomR (F, R) ∼ = F . Let {a1 , . . . , an } be a basis of F , as above. We make a map
from F to HomR (F, R) in the following way: let a ∈ F . By definition, there exist
unique elements r1 , . . . , rn such that a = r1 a1 + · · · + rn an . Then, consider the map
φa sending each generator ai to ri . By the universal property of the free module,
the map φa extends to a unique module homomorphism Φa from F to R. Thus, we
constructed a map from F to HomR (F, R), by a 7→ Φa . It is clearly injective, and
a module homomorphism (again, by the universal property, we just need to check
that on the generators, φa+b = φa + φb , and φra = rφa (and both are obvious). It
remains toPcheck that this map is surjective. Let Φ : F → R be a homomorphism.
Let a = Φ(ai )ai (as in the above solution). Then clearly Φ = Φa (again, it
suffices to check that they coincide on the generators). So the point is that this is
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the same solution, but the fact that we have the universal property allows us to
reduce the number of things to check.

Section 10.3 Problem 23. Show that a direct sum of free R-modules is free.
L
Solution: Suppose M = i∈I Ni is a direct sum of free R-modules. So Ni is a
free R-module for each i ∈ I. Here I is some index set. By definition, Ni is free on
some subset Ai ⊂ Ni . We claim that M is free on the subset
[
A= Ai ⊂ M
i∈I

Indeed, take any x ∈ M . Then, by definition of a direct sum, x has a unique

representation of the form

x = x1 + x2 + x3 + · · · + xn

for some positive integer r. Here xi ∈ Ni . But freeness of Ni tells us that xi has a
representation of the form

xi = ri,1 ai,1 + ri,2 ai,2 + · · · + ri,k ai,k

where ri,j and ai,j are unique. And this is true for each 1 ≤ i ≤ r. Substituting
the above expressions as i runs from 1 to r into the equation for x, we obtain

x = r1,1 a1,1 + r1,2 a1,2 + · · · + r1,k a1,k

+ r2,1 a2,1 + r2,2 a2,2 + · · · + r2,k a2,k
+ ······························+
+ rn,1 an,1 + rn,2 an,2 + · · · + rn,k an,k

Furthermore, the representation above is unique (by following the above steps). We
conclude that M is a free module on its subset A, where
[
A= Ai
i∈I

One more exercise about a universal property (which was not-to-hand-

in): Let A, B be arbitrary sets. Let C = A × B be their Cartesian product

C = {(a, b) | a ∈ A, b ∈ B},

and let π1 : C → A, π2 : C → B be the projections onto the first and second factor:
π1 (a, b) = a, π2 (a, b) = b. Prove that the Cartesian product C = A × B with the
projections π1 and π2 satisfies the following universal property: for any set M and
any two maps f1 : M → A and f2 : M → B, there exists a unique map Φ : M → C,
such that f1 = π1 ◦ Φ, f2 = π2 ◦ Φ. Prove that if D is any other set (equipped
with maps p1 : D → A and p2 : D → B) satisfying the same universal property,
then there exists a bijective map from C to D, such that it commutes with the
projections π1 , p1 and π2 , p2 .
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f2
M
Φ

π2
f1
C =A×B B
π1

Solution: We define Φ : M → C by
Φ(m) = (f1 (m), f2 (m))
Then, it is clear that π1 ◦ Φ = f1 and π2 ◦ Φ = f2 . To prove uniqueness, we observe
(a, b) = (π1 ((a, b)), π2 ((a, b)))
And so if Ψ : M → C is any map with the same universal property, then for any
m ∈ M , we have
Ψ(m) = (π1 (Ψ(m)), π2 (Ψ(m)))
The conditions π1 ◦ Ψ = f1 and π2 ◦ Ψ = f2 force
Ψ(m) = (f1 (m), f2 (m)) = Φ(m)
| {z }
by definition

So Ψ(m) = Φ(m) for each m ∈ M . It follows that Φ : M → C is unique map

enjoying this particular universal property.
Now let’s suppose that D is any other set (equipped with maps p1 : D → A,
and p2 : D → B) satisfying the same universal property. We need to prove that a
bijective map g : C → D exists such that the following diagrams commute:
g g
C D C D
π1 π2
p1 p2

A B
Now, from the universal property of C = A × B, the maps p1 : D → A and
p2 : D → B induce unique map h : D → C such that π1 ◦ h = p1 and π2 ◦ h = p2 .
Similarly, from universal property of D, the maps π1 : C → A and π2 : C → B
induce unique map g : C → D such that p1 ◦ g = π1 and p2 ◦ g = π2 . We claim
that g is a bijection, whose inverse is h.
Consider the map g ◦ h : D → D. We will use the universal property of D. In
the definition of universal property, we can let M = D. Indeed, we have the maps
p1 : D → A and p2 : D → B. We observe that
p1 ◦ (g ◦ h) = (p1 ◦ g) ◦ h = π1 ◦ h = p1
p2 ◦ (g ◦ h) = (p2 ◦ g) ◦ h = π2 ◦ h = p2
But the identity map idD : D → D also satisfies:
p1 ◦ idD = p1
p2 ◦ idD = p2
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By the uniqueness guaranteed by the universal property, we get g ◦ h = idD .

Similarly, we consider the map h ◦ g : C → C. We will use the universal property
of C. In the definition of universal property, we can let M = C. Indeed, we have
the maps π1 : C → A and π2 : C → B. We observe that
π1 ◦ (h ◦ g) = (π1 ◦ h) ◦ g = p1 ◦ g = π1
π2 ◦ (h ◦ g) = (π2 ◦ h) ◦ g = p2 ◦ g = π2
But the identity map idC : C → C also satisfies:
π1 ◦ idC = π1
π2 ◦ idC = π2
By the uniqueness guaranteed by the universal property, we get h ◦ g = idC .
So we have shown that maps g : C → D and h : D → C satisfy h ◦ g = idC and
g ◦ h = idD . It follows that g is a bijection with h as its inverse. It is also clear
from the above discussion that g commutes with the projections π1 , p1 and π2 , p2 .