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.

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

Let $A$ be a local noetherian ring, $M$ an $A$-module finitely generated. Let $f$ be an $A$-regular and $M$-regular element (i.e. $f$ is not a zero divisors on $A$ nor on $M$). Then inside the category of $A/fA$-modules (I think we can suppose finitely generated), we have the following isomorphisms of functors:

Ext$^n_{A/fA}(M/fM,\_)\cong$ Ext$^n_A(M,\_)$ for every $n\geq0$

Ext$^n_{A/fA}(\_,M/fM)\cong$ Ext$^{n+1}_A(\_,M)$ for every $n\geq0$

I found this theorem into some notes but I don't know how to prove it. I'm even not sure to understand what it means, it seems to imply that the functor Ext$^n_A(M,\_)$ is defined on the category of $A/fA$-modules. Could you tell me how to prove it and could you help me to understand better the statement, please?

share|cite|improve this question
I guess, it is meant like, for all $A$-modules $M$ and $N$: $$\mathrm{Ext}^n_{A/fA}(M/fM,\ N/fN) \cong \mathrm{Ext}^n_A(M, N)$$ I tried to prove it for $n=0$, i.e. for the case of $\ \mathrm{Hom}\ $ functor, but couldn't finish. Probably some property of local rings is needed to be considered. – Berci Sep 20 '12 at 14:31
$A/f$-modules can be considered as $A$-modules by change of rings using the map $A \twoheadrightarrow A/f$. That's why, with some abuse of notation, we can talk about $\operatorname{Ext}_A^n (M,N)$ for $N$ an $A/f$-module. Which notes are you reading? – m_t_ Sep 20 '12 at 16:14
@mt_ this one:… – Chris Sep 23 '12 at 0:58
up vote 2 down vote accepted

Let $R,S$ be rings, let $\phi:R \to S$ be a ring map. Then we have functors $\uparrow = \uparrow_R^S : \operatorname{mod}_R \to \operatorname{mod}_S$ ("induction") and $\downarrow = \downarrow^S_R : \operatorname{mod}_S \to \operatorname{mod}_ r$ ("restriction") defined as follows: $M \downarrow $ is the $R$-module with the same underlying set as $M$ and with $R$-action $r \cdot m := \phi(r)m$ and $N \uparrow : S \otimes _R M$, where the right-action of $R$ on $S$ is $s\cdot r := s\phi(r)$.

Example: $\phi$ is a quotient map $A \to A/f$. Then $N \uparrow = A/f \otimes _A N \cong N/fN$.

In general we have $\hom_S(M\uparrow, X ) \cong \hom_R( M, X\downarrow)$ - see Cartan and Eilenberg p.29. This is "change of rings" and says that induction and restriction are mutually adjoint functors. We always have a map $\phi^*: \operatorname{Ext}^n_S(M\uparrow, X) \to \operatorname{Ext}^n_R(M, X\downarrow)$ (use the Yoneda definition of Ext for example). It is not an isomorphism in general, but if $\operatorname{Tor}^R_n(S, X) = 0$ for $n>0$ it is an isomorphism (CE page 118).

In your case, a projective resolution of $A/f$ over $A$ is given by

$$ 0 \to Af \to A \to A/f \to 0 $$

To compute the Tor groups, tensor with $M$:

$$ 0 \to Af \otimes _A M \to M \to M/f \to 0$$

(making some identifications). The only way this could fail to be exact is if $xf \otimes _A m \mapsto xfm$ were not injective, but $f$ is not a zero divisor on $M$. Thus all higher Tor groups vanish.

share|cite|improve this answer
To type the up/down arrow you need to say \mathord\uparrow; otherwise, being a relation, it gets the spacing wrong: compare M\uparrow=N $M\uparrow=N$ vs. M\mathord\uparrow=N $M\mathord\uparrow=N$ (inthe first case \uparrow= gets interpreted as a single relation, and spaced accordingly) – Mariano Suárez-Alvarez Sep 24 '12 at 9:03
@Mariano thanks, I'd always hacked that with backslash exclamation mark but felt vaguely guilty about it. – m_t_ Sep 24 '12 at 9:22

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.