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Let $0 \rightarrow E_1 \stackrel{\sigma}{\rightarrow} E_2$ be an exact sequence of free $A$-modules, $A$ being a commutative ring. Let $F$ be a free $A$ module. Then is it true that the sequence $Hom_A(E_2,F) \rightarrow Hom_A(E_1,F) \rightarrow 0$ is exact? If yes, how can i show it?

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Look up the definition of injective module. What you wrote is essentially the same thing. Take all three modules to be $\mathbb{Z}$ in the category of $\mathbb{Z}$-modules and $\sigma$ is $1\mapsto 2$. –  Matt Dec 10 '11 at 17:36
    
@Matt: good point. So as i understand, the answer to my question is no. Now, what if $F=A$, so that the hom sequence is in fact the dual sequence? How does this change the situation? –  Manos Dec 10 '11 at 18:09
    
My example takes $F=A$. Unless I'm misunderstanding something, the only way you can guarantee surjectivity there is if $F$ is an injective $A$-module. The point is that if $\sigma$ is something like "multiplication by 2", then you need to know that there is a map $E_2\to A$ that acts like "division by 2" so that when you compose you get the identity. For instance, if you work with $\mathbb{Z}$-modules then you need $F$ to be at least as "big" as $\mathbb{Q}$. –  Matt Dec 10 '11 at 18:37
    
@Matt: I understand your comments, thank you. However, do you have Lang's Algebra? Can you take a look at theorem 6.3 p. 143? I am trying to see where does the exactness of the dual sequence come from the right. –  Manos Dec 10 '11 at 19:24
    
Ah. I see the confusion. I'll answer. –  Matt Dec 10 '11 at 19:56

1 Answer 1

up vote 2 down vote accepted

In Lang there is actually a stronger hypothesis. The theorem is that when you have $0\to U\to W\to V\to 0$ where all terms are finite rank free $A$-modules then $Hom_A(-,A)$ is exact. Note that the way it is written in the question allowed me to pick $\sigma$ as multiplication by $2$, but the cokernel is then $\mathbb{Z}/2\mathbb{Z}$ which is not a free $\mathbb{Z}$-module.

The key here is that any such short exact sequence with the given hypothesis actually splits and so you can reduce to the case of checking it works for the sequence $0\to U\to U\oplus V\to V\to 0$ where the first is just an inclusion of the free module and the second is the projection. (Note again that $\mathbb{Z}$ is not isomorphic to $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$, so the sequence I constructed for the counterexample does not split).

In that case what we want to check is that $Hom_A(U\oplus V, A)\to Hom_A(U,A)\to 0$ is exact. I.e. if you have a map $f: U\to A$ can you find some map $g: U\oplus V\to A$ such that composing with the inclusion $U\to U\oplus V\to A$ is the original? Sure. Just take $g=f\oplus 0$.

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Excellent answer. Thanks. –  Manos Dec 10 '11 at 21:33

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