# Ext groups of a point on a scheme

Given a scheme $X$ over a field $k$ and a closed point $x$ with residue field $k(x)$ and inclusion $i:{x}\rightarrow X$ one can consider the following abelian groups

(1)$Ext^1_{\mathcal O_X}(i_*k(x),i_*k(x))$

and

(2)$Ext^1_{\mathcal O_{X,x}}(k(x),k(x))$.

The second one is just seen as Ext-group in the sense of modules over a ring.

Are they isomorphic?

I would define a map from (1) to (2) by just taking stalk in $x$, but I dont really see how one would come from (2) to (1).

And is there a structure of a $k-$ vector space on (2) such that the iso is also one of $k-$spaces? (1) surely has a $k-$vector space structure as the scheme is defined over $k$.

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The second Ext-group is an $\mathcal O_{X,x}$-module. Since $\mathcal O_{X,x}$ is a $k$-algebra, it is in particular a $k$-vector space (and this is compatible with the $k$-v.s. structure on the first Ext-group and the map from the first to the second).
Furthermore, the two Ext-groups are isomorphic (i.e. the map you defined is an isomorphism). To see this, choose an affine open, say $U =$ Spec $A$, around $x$, and let $\mathfrak p$ be the prime ideal of $A$ corresponding to $x$.
Firstly, restriction to $U$ induces an isomorphism $Ext^1_{\mathcal O_X}(k(x),k(x)) \cong Ext^1_{\mathcal O_U}(k(x),k(x))$ (since any extension of the two skyscrapers at $x$ is again supported on $x$, and hence on $U$). The second Ext group in this isomorphism can then be computed in terms of modules, and the isomorphism we want reduces to the isomorphism $Ext^1_A(A/\mathfrak p, A/\mathfrak p) = Ext^1_{A_{\mathfrak p}}(A/\mathfrak p,A/\mathfrak p),$ which is straightforward. (Any $A$-module which is an extension of $A_{\mathfrak p}$-modules is again an $A_{\mathfrak p}$-module.)
How do you get the structure of $A_p$-module you are talking about in the last bracket? I see this was actually my problem. – Cyril Nov 7 '11 at 20:07
@Cyril: Dear Cyril, An $A_{\mathfrak p}$-module is the same as an $A$ module on which every element of $A\setminus \mathfrak p$ acts invertibly. This latter condition is preserved under extensions. Regards, – Matt E Nov 7 '11 at 21:19