Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am reading Lang's Algebra and trying to fill in the gaps in my mathematical background while I train for the quals. So I came across the following exercise (chapter 20, ex. 26 in the third edition).

$A$ is a commutative ring, $E$ an $A$-module and its dual is $E^*=\mathrm{Hom}_A(E; \mathbb Q/\mathbb Z)$.

It is asked to prove that a short sequence \begin{equation} 0 \rightarrow N \rightarrow M \rightarrow E \rightarrow 0 \end{equation} is exact iff the dual sequence \begin{equation} 0 \rightarrow E^* \rightarrow M^* \rightarrow N^* \rightarrow 0 \end{equation} is exact.

It is not hard to prove that the dual sequence is exact if the first one is, using the arguments Lang shows, in particular the injectivity of the module $\mathrm{Hom}_A(A; \mathbb Q/\mathbb Z)$. The vice versa turns out to be harder (at least, I can't find a clear way for that), besides the easy proof that $ N \rightarrow M$ is injective, thanks to the canonical embedding of a module $M$ in its bidual $M^{**}$. Any hint or reference? Also any interesting reading on the topics of injective and projective module, resolutions, etc, will be appreciated.

Thanks a lot.

share|improve this question
    
I guess you meant to say $N \to M$ is injective? –  Tunococ Aug 16 '12 at 18:32
    
Ok, I edited that. You're right, I meant injective. –  Niccolò Aug 16 '12 at 18:34
    
You are also assuming $A = \mathbb Q / \mathbb Z$ too, right? I have the feeling that you need divisibility. I'm not sure yet though. –  Tunococ Aug 16 '12 at 20:56
    
Well, the exercise needs another edit. The dual is defined as $Hom_Z(E;Q/Z)$ and not as $A$-maps, sorry again. I should re-check my posts more than once apparently. –  Niccolò Aug 16 '12 at 22:23
    
@Tunococ Your guess is correct: $\textrm{Hom}_\mathbb{Z}(-; I)$ is exact if and only if $I$ is an injective abelian group, and injective abelian groups are precisely the divisible abelian groups. –  Zhen Lin Aug 17 '12 at 1:42
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.