# Why are these two functors isomorphic?

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?

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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? –  mt_ Sep 20 '12 at 16:14
@mt_ this one: archive.numdam.org/ARCHIVE/SAC/SAC_1966-1967__1_/… –  Chris Sep 23 '12 at 0:58

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.

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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. –  mt_ Sep 24 '12 at 9:22