# exact sequence of modules

• Jan 20th 2009, 12:54 PM
Banach
exact sequence of modules
Hi once again! I have two questions:
a) Let $\displaystyle R$ a PID and $\displaystyle M$ a finitely generated $\displaystyle R$-module. Show that the $\displaystyle R$-module $\displaystyle Hom_R(M,R)$ is free and has finite rank.

b) We have an exact sequence $\displaystyle 0 \longrightarrow R^m \longrightarrow R^n \longrightarrow M \longrightarrow 0$ of modules over a PID $\displaystyle R$. Show that $\displaystyle M/T(M)$ is a free $\displaystyle R$-module of rank n-m ($\displaystyle T(M)$ 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?
• Jan 21st 2009, 12:02 AM
NonCommAlg
Quote:

Originally Posted by Banach

a) Let $\displaystyle R$ a PID and $\displaystyle M$ a finitely generated $\displaystyle R$-module. Show that the $\displaystyle R$-module $\displaystyle Hom_R(M,R)$ is free and has finite rank.

let $\displaystyle T(M)$ be the torsion submodule of $\displaystyle M.$ we have $\displaystyle M=T(M) \oplus R^k,$ for some $\displaystyle k.$ it's very easy to see that $\displaystyle \text{Hom}_R(T(M),R)=\{0\}$ and $\displaystyle \text{Hom}_R(R,R) \simeq R.$ thus:

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

Quote:

b) We have an exact sequence $\displaystyle 0 \longrightarrow R^m \longrightarrow R^n \longrightarrow M \longrightarrow 0$ of modules over a PID $\displaystyle R$. Show that $\displaystyle M/T(M)$ is a free $\displaystyle R$-module of rank n-m ($\displaystyle T(M)$ denoting the submodule of torsionelements).
the exact sequence tells us that $\displaystyle M$ is finitely generated and thus $\displaystyle M=T(M) \oplus R^k,$ for some $\displaystyle k.$ so $\displaystyle M/T(M) \simeq R^k.$ the claim is that $\displaystyle k=n-m.$ 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 $\displaystyle 0 \longrightarrow R^m \longrightarrow R^n \longrightarrow R^k \longrightarrow 0,$ which obviously implies that $\displaystyle k=n-m.$
• Jan 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 $\displaystyle M=T(M) \oplus R^k$ onto $\displaystyle R^k$. Let $\displaystyle f:R^m \rightarrow R^n$ and $\displaystyle g:R^n \rightarrow M$ be the maps of the exact sequence. The composition $\displaystyle p \circ g$ is then a surjective map $\displaystyle R^n \rightarrow R^k$, but we do not have $\displaystyle Ker(p \circ g)=Im(f)$. 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 $\displaystyle M$ to $\displaystyle R^{n-m}$ with kernel $\displaystyle T(M)$ but how find it?
• Jan 21st 2009, 08:14 PM
NonCommAlg
well, the problem is even easier than what i thought! so we have this exact sequence: $\displaystyle 0 \longrightarrow R^m \overset{f}{\longrightarrow} R^n \longrightarrow M \longrightarrow 0.$ then $\displaystyle f(R^m) \simeq R^m$ is a submodule of $\displaystyle R^n.$ since $\displaystyle R$ is a PID, there exists

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

$\displaystyle M \simeq \frac{\bigoplus_{j=1}^nRe_j}{\bigoplus_{j=1}^m Ra_je_j} \simeq \bigoplus_{j=1}^m \frac{R}{Ra_j} \oplus Re_{m+1} \oplus \cdots \oplus Re_n \simeq \bigoplus_{j=1}^m \frac{R}{Ra_j} \oplus R^{n-m}.$ therefore: $\displaystyle T(M) \simeq T \left(\bigoplus_{j=1}^m \frac{R}{Ra_j} \oplus R^{n-m} \right)=\bigoplus_{j=1}^m \frac{R}{Ra_j},$ and hence: $\displaystyle \frac{M}{T(M)} \simeq R^{n-m}.$
• Jan 30th 2009, 05:49 AM
Banach
Thanks once more! Very nice solution indeed!