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Math Help - exact sequence of modules

  1. #1
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    exact sequence of modules

    Hi once again! I have two questions:
    a) Let R a PID and M a finitely generated R-module. Show that the R-module Hom_R(M,R) is free and has finite rank.

    b) We have an exact sequence 0 \longrightarrow R^m \longrightarrow R^n \longrightarrow M \longrightarrow 0 of modules over a PID R. Show that M/T(M) is a free R-module of rank n-m ( T(M) denoting the submodule of torsionelements).

    To solve a) i tried to create an exact sequence involving Hom(M,R) but i got stuck. And for b) i do not have any ideas. Can anyone please help me?
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  2. #2
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    Quote Originally Posted by Banach View Post

    a) Let R a PID and M a finitely generated R-module. Show that the R-module Hom_R(M,R) is free and has finite rank.
    let T(M) be the torsion submodule of M. we have M=T(M) \oplus R^k, for some k. it's very easy to see that \text{Hom}_R(T(M),R)=\{0\} and \text{Hom}_R(R,R) \simeq R. thus:

    \text{Hom}_R(M,R)=\text{Hom}_R(T(M) \oplus R^k,R) \simeq \text{Hom}_R(T(M),R) \oplus \text{Hom}_R(R^k,R) \simeq R^k.



    b) We have an exact sequence 0 \longrightarrow R^m \longrightarrow R^n \longrightarrow M \longrightarrow 0 of modules over a PID R. Show that M/T(M) is a free R-module of rank n-m ( T(M) denoting the submodule of torsionelements).
    the exact sequence tells us that M is finitely generated and thus M=T(M) \oplus R^k, for some k. so M/T(M) \simeq R^k. the claim is that k=n-m. right now i don't know a quick way to prove

    the claim but i'm sure it's not hard. one idea is to use the given exact sequence to find an exact sequence 0 \longrightarrow R^m \longrightarrow R^n \longrightarrow R^k \longrightarrow 0, which obviously implies that k=n-m.
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  3. #3
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    Hi! Thank you very much for your answer! I think a) is clear to me now.

    b)
    I tried to follow your idea with the new exact sequence and projected M=T(M) \oplus R^k onto R^k. Let f:R^m \rightarrow R^n and g:R^n \rightarrow M be the maps of the exact sequence. The composition p \circ g is then a surjective map R^n \rightarrow R^k, but we do not have Ker(p \circ g)=Im(f). Do you have an idea how to alter this map so that the condition is satisfied?

    Another idea would be to find a surjective map from M to R^{n-m} with kernel T(M) but how find it?
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  4. #4
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    well, the problem is even easier than what i thought! so we have this exact sequence: 0 \longrightarrow R^m \overset{f}{\longrightarrow} R^n \longrightarrow M \longrightarrow 0. then f(R^m) \simeq R^m is a submodule of R^n. since R is a PID, there exists

    a basis \{e_1, \cdots , e_n \} for R^n and non-zero elements a_j \in R, \ 1 \leq j \leq m, such that \{a_1e_1, \cdots , a_me_m \} is a basis for f(R^m). from the exact sequence it's clear that M \simeq \frac{R^n}{f(R^m)}. therefore:

    M \simeq \frac{\bigoplus_{j=1}^nRe_j}{\bigoplus_{j=1}^m Ra_je_j} \simeq \bigoplus_{j=1}^m \frac{R}{Ra_j} \oplus Re_{m+1} \oplus \cdots \oplus Re_n \simeq \bigoplus_{j=1}^m \frac{R}{Ra_j} \oplus R^{n-m}. therefore: T(M) \simeq T \left(\bigoplus_{j=1}^m \frac{R}{Ra_j} \oplus R^{n-m} \right)=\bigoplus_{j=1}^m \frac{R}{Ra_j}, and hence: \frac{M}{T(M)} \simeq R^{n-m}.
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  5. #5
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    Thanks once more! Very nice solution indeed!
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