Let $k$ be a field and $A$ a commutative $k$-algebra. The enveloping algebra of $A$, denoted by $A^e$ is defined as $A\otimes_kA^o$ where $A^o$ is the opposite algebra. Here the opposite algebra does not make a real difference. But I'm looking to understand the proof of Witherspoon of the Hochschild-Konstant-Rosenberg theorem for cohomology.
Recall that for a prime(in particular a maximal ideal) $P$ the localization at $P$, is the localization with respect $S=A\setminus P$.The localization induces an exact functor on $A$-Mod. Moreover, is $B$ is an $A$-algebra, then $S^{-1}A$ is an $A$-algebra. Also, a $S^{-1}A$-algebra.
So my question is about the last paragraph, we get an Ext over $(A^e)_M$ but for the proof we need it over $(A_M)^e$.
So for a maximal ideal $M$ of $A$, it happens that $(A^e)_M\cong(A_M)^e$?