# an isomorphism of extension functors

Let $(R,m)$ be a Noetherian *local ring and suppose that $m$ is maximal in the ordinary sense. Then why is it true that $\operatorname{Ext}^i_R(R/m^j,M) \cong \operatorname{Ext}^i_{R_m}(R_m/m^jR_m,M_m)$ for every graded $R$-module $M$?

Reference: Bruns and Herzog, Cohen-Macaulay Rings, Remark 3.6.18

• Is there any other sense to $m$ being maximal? You could've just written "Let $(R,m)$ be a Noetherian local ring". I thought that $R$ and $R_m$ would be isomorphic since localizing at $m$ only adds units. Similarly for $M$ and $M_m$. – Patrick Da Silva Mar 26 '14 at 22:44
• Sorry but you are wrong: *local has a meaning of its own, i.e. $m$ is a homogeneous ideal that is unique with respect to the property that if any other homogeneous ideal $I$ contains properly $m$, then $I = R$. But this doesn't imply that there do not exist other non-homogeneous maximal ideals. – Manos Mar 26 '14 at 22:52
• ahhhh, so it's a *local ring, not a local ring. Sorry, I didn't notice that the * had a meaning! So it's like a ''locally homogeneous'' ring, if we would want to give it a fancy name. (Although I don't like it.) – Patrick Da Silva Mar 27 '14 at 2:15

$\newcommand{\Ext}{\text{Ext}}$As you know, $\Ext^i_{R_m}(R_{m}/m^{j}R_m, M_m) \cong \Ext^i_R(R/m^j, M) \otimes_R R_m$ (since $R$ is Noetherian, $R \to R_m$ is flat, and $R/m^j$ is a finite $R$-module). Thus, it suffices to see that every element of $R \setminus m$ acts as a unit on $\Ext^i_R(R/m^j, M)$. But since $m$ is maximal, any element $x \in R \setminus m$ acts as a unit on $R/m^j$ for any $j$, hence also on $\Ext^i_R(R/m^j, M)$ (e.g. by applying the functor $\Ext^i_R(\_, M)$ to the exact sequence $0 \to R/m^j \xrightarrow{x} R/m^j \to 0$)