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Math Help - Torsion free and divisible Z-modules

  1. #1
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    Torsion free and divisible Z-modules

    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)
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  2. #2
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    Quote Originally Posted by robeuler View Post
    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?
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    Quote Originally Posted by NonCommAlg View Post
    this is a strange claim! are you sure it's not Hom(D,A) instead?
    I am fairly certain it is Hom(A,D)
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    Quote Originally Posted by robeuler View Post
    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.
    Last edited by aliceinwonderland; September 28th 2009 at 04:07 PM. Reason: Included several words and notations for clarity
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  5. #5
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    Quote Originally Posted by aliceinwonderland View Post
    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.
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    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.
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  7. #7
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    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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