exact sequence of modules

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January 20th 2009, 12:54 PM
Banach

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?
January 21st 2009, 12:02 AM
NonCommAlg

Quote:

Originally Posted by Banach

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:

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

Quote:

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 $k=n-m.$
January 21st 2009, 08:27 AM
Banach

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?
January 21st 2009, 08:14 PM
NonCommAlg

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: $\frac{M}{T(M)} \simeq R^{n-m}.$
January 30th 2009, 05:49 AM
Banach

Thanks once more! Very nice solution indeed!