MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello, I asked this question already on StackExchange with no answer, maybe it will be better suited here. It is well known that $\operatorname{Ext}^1_\mathbb{Z}(\mathbb{Z}/p\mathbb{Z},A) \simeq A/pA$. Is there a generalization of this formula outside PIDs? I mean $\operatorname{Ext}^1_R(R/I,M) \simeq M/IM$ for maximal or prime ideals? Does it hold for Dedekind domains? The classic proof I know for abelian groups seem to depend on principality of the ideal and I was unable to either think of different way make it work or to find an counterexample. Thanks for any input.

share|cite|improve this question
up vote 6 down vote accepted

In general $Ext^1_R(R/I,M) \not\cong M/IM$.

As an example take a finite abelian group $G$ and set $R := \mathbb{Z}G$ and let $I:= I_G = \ker(\mathbb{Z}G \to \mathbb{Z},\; g \mapsto 1)$ be the augmentation ideal. $I_G$ is a prime ideal since $\mathbb{Z}G/I_G \cong \mathbb{Z}$. Then, with trivial coefficients $$Ext_R^1(R/I,\mathbb{Z})=Ext_{ZG}^1(\mathbb{Z},\mathbb{Z})=H^1(G;\mathbb{Z})=Hom(G,\mathbb{Z})=0$$ while $\mathbb{Z}/I_G\mathbb{Z}=\mathbb{Z}$. The latter holds because $I_G$ is generated by $g-1$ $(g \in G)$ and $(g-1) \cdot 1 = 0$.

In general, the following holds: $$Ext^1_R(R/I,M)=\dfrac{Hom_R(I,M)}{i(M)}$$ where $i: M \to Hom_R(I,M),\; m \mapsto (x \mapsto xm)$.

If $I$ is finitely generated by $a_1,...,a_n$ we can choose a presentation $R^n \to I \to 0$ that induces an embedding $0 \to Hom_R(I,M) \to Hom_R(R^n,M) \cong M^n$. Thus $$Ext_R^1(R/I,M) \le \dfrac{M^n}{\lbrace (a_1m,...,a_nm) \mid m \in M \rbrace}.$$

share|cite|improve this answer
$I$ being locally free is good enough, because then there is a dual module $I^{\vee}$ such that $Hom_R(I,M)=I^{\vee} \otimes M$. This occurs, for instane, in Dedekind domains, which is one case OP was interested in. – Will Sawin Aug 8 '12 at 20:29
Clarifying the comment in light of the edit: Will was helping Ralph discover when you can get a closed form expression for $Hom_R(I,M)$. – David White Aug 8 '12 at 20:47
In a Dedkenind domain, $(I^\vee \otimes M)/M=I^\vee \otimes (M/IM)=M/IM$, with the last identification non-canonical. – Will Sawin Aug 8 '12 at 21:04
@Will: Thanks. I was thinking about the following: Suppose $I$ is locally free and assume the localization $R_P^n \to I_P$ is an isomorphism for each max. P (). Then the localization of the inclusion $j: Ext_R^1(R/I,M) \to M^n/(...)$ is an isomorphism for each max. P and hence $j$ is an isomorphism. Do you know if () can be expected for Dedekind domains or other interesting categories of rings ? – Ralph Aug 8 '12 at 21:33
@Ralph: I'm not sure what you mean. (*) sounds like either the definition of locallly free or else the statement $R^n \cong I$, depending on whether the local maps could be any map or are localizations of the same map. Note than an ideal, to be locally free, must be locally free of rank one. – Will Sawin Aug 9 '12 at 0:45

Interesting question! Here is a generalization which includes the Dedekind domain case:

Proposition: Let $R$ be a Noetherian regular domain with $n=\dim R$ and $I\subset R$ an ideal such that $R/I$ is artinian and Gorenstein. Then for a finitely generated $R$-module $M$, we have $Ext^n(R/I,M)\cong M/IM$.

(when $R$ is Dedekind $R/I$ is locally a quotient of a DVR, so it is locally a hypersurface, thus Gorenstein).

Proof. It is enough to prove for $M=R$ because of the following

Claim: if $N$ is an artinian $R$-module then the natural map $Ext^n_R(N,R)\otimes M \to Ext^n_R(N,M)$ is an isomorphism.

Here's why: We have $pd_RN = n$. Let $P_{\bullet}: 0\to P_n \to \cdots \to P_0$ be a projective resolution of $N$. Then one compute the LHS by taking the cohomology at the end of $Hom(P_{\bullet},R)$ and tensor with $M$, and the RHS by taking cohomology at the end of $Hom(P_{\bullet},M)$. But tensor product is right-exact, proving that we get the same answer.

Assuming we can prove for $M=R$, then apply the claim with $N=R/I$ completes the proof.

Now we prove it for $M=R$. Since $R/I$ is artinian, we only need to prove it after localizing at each maximal ideal containing $I$. Thus we can assume $(R,m)$ is local. But by Local Duality $Ext^n(R/I,R)$ is isomorphic to the Matlis dual of $H^0_m(R/I) =R/I$. But since $R/I$ is Gorenstein this dual is actually isomorphic to $R/I$. QED

share|cite|improve this answer
Thanks very much, this helped a lot ! – Fred.Fred Aug 8 '12 at 23:05

Let $(R,\mathfrak m, k)$ be an arbitrary (say Noetherian) local ring of dimension at least $2$ and $M$ an arbitrary non-zero (say finitely generated) $R$-module of $\mathrm {depth}_R M\geq 2$. Then $\mathrm{Ext}^1_R(R/\mathfrak m,M)=0$, but $M/\mathfrak m M\neq 0$ by Nakayama.

share|cite|improve this answer
In general this fails even when $R=k$. – Hailong Dao Aug 8 '12 at 20:34
@Hailong, could you be more specific about what you mean? I believe $k$ does not have dimension at least $2$. – Sándor Kovács Aug 8 '12 at 20:57
Dear Sándor, I just meant that there is a simple counter example when the dimension is not $1$. – Hailong Dao Aug 8 '12 at 20:58
I see. You probably meant "it fails" and not "this fails"... Also, this is a series of not-entirely-obvious examples for any dimension larger than $1$. I figured that it might be more interesting to give an example where not every module is projective in case someone is thinking about trying to find the most general conditions under which the original statement may hold... – Sándor Kovács Aug 8 '12 at 21:07
Dear Sándor, sorry for my poor use of language. Of course, it is useful to have a counter-example in the general case, I thought I would just point out the most obvious one. +1 to your nice examples! Cheers! – Hailong Dao Aug 8 '12 at 21:12

Your Answer


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

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