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About LinkBacksIf 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)
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.Originally Posted by robeuler
Lemma 1. Let A, B, $\displaystyle \{A_i | i \in I \}$ and $\displaystyle \{B_j | j \in J \}$ be modules over $\displaystyle \mathbb{Z}$. Then there are isomorphisms of abelian groups:
$\displaystyle Hom(\sum_{i \in I}A_i, B) \cong \prod_{i \in I}Hom(A_i, B)$
Proof. (Hungerford, "Algebra" p201).
Lemma 2. If M is a module over $\displaystyle \mathbb{Z}$, then there is an isomorphism of $\displaystyle \mathbb{Z}$-modules $\displaystyle M \cong Hom(\mathbb{Z}, M)$
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 $\displaystyle \mathbb{Z}$-module A can be denoted as:
$\displaystyle A \cong \mathbb{Z} \times \mathbb{Z} ... \times \mathbb{Z} \cong \mathbb{Z}^n$, where n is a rank of A.
Thus, by lemma 1 and 2, $\displaystyle Hom(A,D) \cong Hom(\sum_{i \in I}A_i, D)\cong \prod_{i \in I}Hom(A_i, D) \cong D^n$, where $\displaystyle A_i \cong \mathbb{Z}$, $\displaystyle I=\{1,2, .. , n\}$, 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 $\displaystyle A$ is finitely generated, the problem can be solved much easier because we know that any finitely generated torsion free $\displaystyle \mathbb{Z}$ module is free. now let $\displaystyle \{a_1, \cdots , a_k \}$ be a basis for $\displaystyle A$ and suppose
that $\displaystyle 0 \neq n \in \mathbb{Z}$ and $\displaystyle f \in \text{Hom}(A,D)$ are given. then for every $\displaystyle j \leq k$ there exists $\displaystyle d_j \in D$ such that $\displaystyle f(a_j)=nd_j,$ because $\displaystyle D$ is divisible. define $\displaystyle g: A \longrightarrow D$ by $\displaystyle g \left(\sum_{j=1}^k m_ja_j \right)=\sum_{j=1}^k m_jd_j.$
note that $\displaystyle g$ is well-defined because $\displaystyle \{a_1, \cdots , a_k \}$ is a basis for $\displaystyle A.$ it's clear that $\displaystyle g \in \text{Hom}(A,D)$ and $\displaystyle f=ng.$ therefore $\displaystyle \text{Hom}(A,D)$ is divisible. Q.E.D.
for the general case, i.e. if $\displaystyle A$ is not finitely generated, i'm not convinced that the claim remains true, although i have no counter-example yet.
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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