7
$\begingroup$

Given a regular local ring $R$ and an $R$-algebras $S$, which is torsion free and finitely generated (even free if needed) as an $R$-module.

Assume we have a nontrivial surjective map $f: M \rightarrow T$, where $M$ is a projective $S$-module, finitely generated and torsion free, and $T$ is a torsion module over $S$. If $N$ denotes $ker(f)$, we get an exact sequence: $0\rightarrow N\rightarrow M\rightarrow T\rightarrow 0$.

Given another torsion module $Q$, when is the induced map $f^{\*}: Hom_S(M,Q)\rightarrow Hom_S(N,Q)$ non trivial, when is it trivial?

My first idea was to use the long exact $Ext$-sequence: Since $M$ is projective we have $Ext^1_S(M,Q)=0$, thus if $f^{\*}=0$, the sequence gives an isomorphism $Hom_S(N,Q)\cong Ext^1_S(T,Q)$.

So what can be said about the groups $Ext^1_S(T,Q)$? Are they always/sometimes/never trivial? Can we compute them if we assume that one of these moudles is a simple $S$-module? Are there other approaches to this question?

$\endgroup$
1
  • $\begingroup$ I retagged because there's no reason to create a "homological-algbera" tag when we already have the perfectly good (and correctly spelled) "homological-algebra" ;) $\endgroup$ Jun 28, 2011 at 13:53

1 Answer 1

4
$\begingroup$

Actually, it would be easier to look at the other end of the exact sequence. Namely, your map $f^*$ is trivial implies the map $g^*: Hom_S(T,Q) \to Hom_S(M,Q)$ is an isomorphism.

Now, since $M$ is projective, the support of $Hom_S(M,Q)$ is equal to the support of $Q$. Thus we have $Supp(Hom_S(T,Q)) = Supp(Q)$, which implies $$Supp(T) \supseteq Supp(Q) \ \ (1)$$

When $Q$ is simple, (so $Q=S/m$ where $m$ is a maximal ideal) as you alluded to in the last paragraph, then $(1)$ is also sufficient, provided that the surjection $M \to T$ is minimal when localizing at $m$, as you can easily check for yourself.

Another situation when $(1)$ also suffices is when $T,Q$ are both cyclic $S$ module and $M=S$ (you always need the map $M\to T$ to be minimal, may be that what you meant by "non-trival" surjection?)

Other than what described above, I think what you want will fail most of the times, even with $(1)$.

$\endgroup$
1
  • $\begingroup$ How could i not see this. I was to obsessed with the $Ext$-groups. But in my example i can actually compute the $Hom$-groups and see if they are isomorphic or not, and if they are not, then $f^{\*}$ cannot be trivial. Thanks a lot for pointing that out! $\endgroup$
    – TonyS
    Jun 19, 2011 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.