3
$\begingroup$

Let $A$ be a commutative ring with $1$ and $\mathcal m$ be a maximal ideal. One knows that then there is a canonical isomorphism

$A_{\mathcal m}/{{\mathcal m}A_{\mathcal m}} \simeq A/{\mathcal m}$.

Does one have an isomorphism of extension groups

$\operatorname{Ext}^1_A(A/{\mathcal m}, A/{\mathcal m}) \simeq \operatorname{Ext}^1_{A_{\mathcal m}}(A_{\mathcal m}/{{\mathcal m}A_{\mathcal m}},A_{\mathcal m}/{{\mathcal m}A_{\mathcal m}})$

via $A\rightarrow A_{\mathcal m}?$

  • 1
    Can you compute both things in an example, say with $A=k[x,y]$ and $m$ the maximal ideal at the origin?2011-10-29

1 Answers 1

4

Let $A$ be a commutative ring with $1$ and $\mathfrak m$ a maximal ideal. The natural morphism from $A$ to $A_{\mathfrak m}$ inducing an isomorphism of $A/\mathfrak m$ onto $A_{\mathfrak m}/\mathfrak m_{\mathfrak m}$, we may denote these two fields by the single letter $K$. Then $K$ is also an $A$-module and an $A_{\mathfrak m}$-module, and we have $ \text{Hom}_A(A,K)=\text{Hom}_{A_{\mathfrak m}}(A_{\mathfrak m},K)=K.\tag1 $ Let $n$ be a nonnegative integer. We claim the existence of a canonical isomorphism $ \text{Ext}^n_A(K,K)\simeq\text{Ext}^n_{A_{\mathfrak m}} (K,K). $ To prove this, apply the localization functor, which is exact and commutes with direct sums, to a free resolution of $K$ in the category of $A$-modules, and use (1).


Thank you to Mariano (see his comment) for having pointed out a mistake: in the previous version I incorrectly called "injection" the natural morphism from $A$ to $A_{\mathfrak m}$.

  • 0
    Cher Mariano: It's not by masochism. It's just to let people understand the comments.2011-10-30