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Let $X$ be an abelian variety over an algebraically closed field $k$. I have read that one has for an arbitrary closed point $x$ on $X$ a canonical identification

$$T_x(X)\simeq \operatorname{Ext}^1(k(x),k(x))$$

where on the left we have the tangent space in the point $x$ and on the right I take the first $\operatorname{Ext}$-group of the skyscraper sheaf in $x$ with itself.

Can someone explain in detail how you get this identification? Does it hold for arbitrary varieties?

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up vote 9 down vote accepted

$\def\Ext{\operatorname{Ext}}$You can compute it explicitly!

Let $X$ be an arbitrary variety and let $x$ be a closed point.

Since $k(x)$ is supported at $x$, we can do the computation in an affine open neighborhood of $x$ (this is more or less obvious for $\Ext^1$: its elements are extensions and there are very few options for the support of the middle term of a short exact sequence starting and ending in $k(x)$...)

So suppose $X=\operatorname{Spec}A$ and that $x$ corresponds to the maximal ideal $\def\m{\mathfrak{m}}\m$. Then $\Ext^1_X(k(x),k(x))=\Ext^1_A(A/\m,A/\m)$ (this is exercise III.6.7 in Hartshorne) There is a short exact sequence of modules $$0\to\m\to A\to A/\m\to0$$ Applying $\Ext(\mathord-,A/\m)$ we get the exact sequence of $A$-modules \begin{multline} 0\to\hom_A(A/\m,A/\m)\to\hom_A(A,A/\m)\to\hom_A(\m,A/\m)\to\\\to\Ext^1_A(A/\m,A/\m)\to\Ext^1_A(A,A/\m)=0 \end{multline} The last zero comes from the fact that $A$ is a projective $A$-module, of course. We have $\hom_A(A/\m,A/\m)\cong A/\m$ and $\hom_A(A,A/\m)\cong A/\m$ as $A$-modules, and the map $\hom_A(A/\m,A/\m)\to\hom_A(A,A/\m)$ appearing in the exact sequence is the identity up to these isomorphisms. It follows that the connecting homomorphism $\hom_A(\m,A/\m)\to\Ext^1_A(A/\m,A/\m)$ appearing in the sequence is an isomorphism, so $$\Ext^1_A(A/\m,A/\m)\cong\hom_A(\m,A/\m)$$ An $A$-homomorphism $\m\to A/\m$ must vanish on $\m^2$, so it induces an $A$-homomorphism $\m/\m^2\to A/\m$ which is $A/\m$-linear. This correspondence is easily seen to be bijective,so we get an isomorphism $$\Ext^1_A(A/\m,A/\m)\cong\hom_{A/\m}(\m/\m^2,A/\m).$$ The right hand side is the dual of the cotangent space $\m/\m^2$ at $x$, so it is the Zariski tangent space.

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I would venture the claim that all the tags on the question are irrelevant. This is just homological algebra :) – Mariano Suárez-Alvarez Oct 25 '11 at 15:40
This is great, Mariano! But the alg-geo tag is justified: you had to take an open affine... ;) I will accept your answer in any case, but still I would like to see if someone has an idea about the second question. – Veen Oct 25 '11 at 18:20
This is wonderful. – Piotr Pstrągowski Jul 21 '13 at 22:13

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