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?
 A: 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.
