Torsion free and divisible Z-modules

**robeuler** · September 28th 2009, 02:21 AM

If A is a torsion free Z module (abelian group) and D is a divisble Z module, then Hom(A,D) is divisible.

I'm also wondering if it is necessary to assume that A is torsion free. (I think so but I can't think of a counter example)

**NonCommAlg** · September 28th 2009, 03:45 AM

Originally Posted by robeuler

If A is a torsion free Z module (abelian group) and D is a divisble Z module, then Hom(A,D) is divisible.

this is a strange claim! are you sure it's not Hom(D,A) instead?

**robeuler** · September 28th 2009, 06:22 AM

Originally Posted by NonCommAlg

this is a strange claim! are you sure it's not Hom(D,A) instead?

I am fairly certain it is Hom(A,D)

**aliceinwonderland** · September 28th 2009, 01:04 PM

Originally Posted by robeuler

If A is a torsion free Z module (abelian group) and D is a divisble Z module, then Hom(A,D) is divisible.

I'm also wondering if it is necessary to assume that A is torsion free. (I think so but I can't think of a counter example)

Below is my attempt and I'll assume A is a finitely generated torsion free Z module.

Lemma 1. Let A, B, $\{A_i | i \in I \}$ and $\{B_j | j \in J \}$ be modules over $\mathbb{Z}$ . Then there are isomorphisms of abelian groups:
$Hom(\sum_{i \in I}A_i, B) \cong \prod_{i \in I}Hom(A_i, B)$
Proof. (Hungerford, "Algebra" p201).

Lemma 2. If M is a module over $\mathbb{Z}$ , then there is an isomorphism of $\mathbb{Z}$ -modules $M \cong Hom(\mathbb{Z}, M)$
Proof. (Hungerford, "Algebra" p202).

Lemma 3. Any product of (even infinitely many) injective modules is injective (link).

Lemma 4. An abelian group (Z-module) is injective if and only if it is divisible.

By the structure theorem for finitely generated modules over PID, a finitely generated torsion free $\mathbb{Z}$ -module A can be denoted as:
$A \cong \mathbb{Z} \times \mathbb{Z} ... \times \mathbb{Z} \cong \mathbb{Z}^n$ , where n is a rank of A.

Thus, by lemma 1 and 2, $Hom(A,D) \cong Hom(\sum_{i \in I}A_i, D)\cong \prod_{i \in I}Hom(A_i, D) \cong D^n$ , where $A_i \cong \mathbb{Z}$ , $I=\{1,2, .. , n\}$ , and n is a rank of A. Thus, Hom(A,D) is divisible by lemma 3 and 4.

There are certain issues if A is an infinite product of torsion-free Z modules. Then, A is not isomorphic to a direct sum of Z's, so lemma 1 cannot be applied. This issue is discussed in (Hungerford, p201) but there is no explanation about it in the book. I don't have an answer for this issue yet.

**NonCommAlg** · September 28th 2009, 06:45 PM

Originally Posted by aliceinwonderland

Below is my attempt and I'll assume A is a finitely generated torsion free Z module.

Lemma 1. Let A, B, $\{A_i | i \in I \}$ and $\{B_j | j \in J \}$ be modules over $\mathbb{Z}$ . Then there are isomorphisms of abelian groups:
$Hom(\sum_{i \in I}A_i, B) \cong \prod_{i \in I}Hom(A_i, B)$
Proof. (Hungerford, "Algebra" p201).

Lemma 2. If M is a module over $\mathbb{Z}$ , then there is an isomorphism of $\mathbb{Z}$ -modules $M \cong Hom(\mathbb{Z}, M)$
Proof. (Hungerford, "Algebra" p202).

Lemma 3. Any product of (even infinitely many) injective modules is injective (link).

Lemma 4. An abelian group (Z-module) is injective if and only if it is divisible.

By the structure theorem for finitely generated modules over PID, a finitely generated torsion free $\mathbb{Z}$ -module A can be denoted as:
$A \cong \mathbb{Z} \times \mathbb{Z} ... \times \mathbb{Z} \cong \mathbb{Z}^n$ , where n is a rank of A.

Thus, by lemma 1 and 2, $Hom(A,D) \cong Hom(\sum_{i \in I}A_i, D)\cong \prod_{i \in I}Hom(A_i, D) \cong D^n$ , where $A_i \cong \mathbb{Z}$ , $I=\{1,2, .. , n\}$ , and n is a rank of A. Thus, Hom(A,D) is divisible by lemma 3 and 4.

There are certain issues if A is an infinite product of torsion-free Z modules. Then, A is not isomorphic to a direct sum of Z's, so lemma 1 cannot be applied. This issue is discussed in (Hungerford, p201) but there is no explanation about it in the book. I don't have an answer for this issue yet.

if $A$ is finitely generated, the problem can be solved much easier because we know that any finitely generated torsion free $\mathbb{Z}$ module is free. now let $\{a_1, \cdots , a_k \}$ be a basis for $A$ and suppose

that $0 \neq n \in \mathbb{Z}$ and $f \in \text{Hom}(A,D)$ are given. then for every $j \leq k$ there exists $d_j \in D$ such that $f(a_j)=nd_j,$ because $D$ is divisible. define $g: A \longrightarrow D$ by $g \left(\sum_{j=1}^k m_ja_j \right)=\sum_{j=1}^k m_jd_j.$

note that $g$ is well-defined because $\{a_1, \cdots , a_k \}$ is a basis for $A.$ it's clear that $g \in \text{Hom}(A,D)$ and $f=ng.$ therefore $\text{Hom}(A,D)$ is divisible. Q.E.D.

for the general case, i.e. if $A$ is not finitely generated, i'm not convinced that the claim remains true, although i have no counter-example yet.

**robeuler** · September 29th 2009, 02:02 AM

Thank you for the tips. I think I might be able to prove it using the theorem that

A Z-module is injective if and only if it is a divisible group. I am working with some diagrams but I'm not sure.

**robeuler** · September 30th 2009, 01:26 AM

I think I have it!

Prove a lemma that a Z module is injective if and only if it is a divisible group.

Then look at the sequence
0 -> A -> A
Where the map from A to A is multiplication by n. It is injective because A is torsion free.

Lift the middle A to D by a homomorphism f and see that there exists a map from the right A to D, call it g such that g(nx)=f(x). Since we are working with Z modules we can pull the n out. And since our choice of f was arbitrary and f is divisible we see that Hom(A,D) is divisible.

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