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?

  • $\begingroup$ 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


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 $(1)$.

  • $\begingroup$ 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

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