Why are the first-order deformations of a scheme $X$ over a field $k$ given by $Ext^1_{\mathcal O_X}(\Omega_X,\mathcal O_X)$, where here I mean the Ext group?

Furthermore, for an integral affine scheme $Spec B$ of finite type over algebraically closed $k$, why is $T^1(B/k,M)\cong Ext^1(\Omega_{B/k},M)$ for any torsion-free $B$-module $M$?

Writing $B=A/I$, where $A=k[x_1,...,x_n]$, we get an exact sequence, $0\rightarrow Hom(\Omega_{B/k},M)\rightarrow Hom(\Omega_{A/k}\otimes B,M)\rightarrow Hom(I/I^2,M)\rightarrow T^1(B/k,M)\rightarrow 0$. Since the $T^i$ functors commute with localization, we can localize to a principal open set, say $D(f)$, contained inside the dense open nonsingular locus of $Spec B$. But there the conormal exact sequence $I/I^2\rightarrow \Omega_{A/k}\otimes B\rightarrow \Omega_{B/k}\rightarrow 0$ is left exact as well, and the corresponding exact sequence for the $T^i$ functors is precisely the same long exact sequence obtained by applying $Hom(--,M)$ to the now left exact conormal sequence. Thus $T^1(B/k,M)= Ext^1(\Omega_{B/k},M)$ over $D(f)$. So my real question is why does this imply the result before localization? I suspect it has something to do with $M$ being torsion-free, but I'm not sure how to finish the argument.

These seem to be basic facts in Deformation theory, but I can't find a proof of them.


  • $\begingroup$ This is explained in Hartshorne's book on deformation theory. $\endgroup$
    – J.C. Ottem
    Jun 28, 2012 at 15:36
  • $\begingroup$ @J.C. Ottem: The second part is in fact an exercise there, but I'm asking how to finish it. I just don't know how to get the unlocalized claim from what I've explained so far. As for the first question, do you know on what page it's explained? From another exercise there, I know that the global deformation functor T^1 fits in the same spot in a 4-term exact sequence (see p. 42, exercise 5.8) as Ext^1(\Omega,O_X) does in the s.e.s. coming from the local-to-global spectral sequence. But does this necessarily mean they're isomorphic? A priori the maps might be different. Let me know, thanks $\endgroup$
    – HNuer
    Jun 28, 2012 at 16:14

1 Answer 1


The second question you asked is Theorem 1.1.10 in Sernesi "Deformations of Algebraic Schemes". In particular the theorem proves the (canonical) identification $$Ex(X/S, I) = Ext^1_{{\mathcal O}_X} (\Omega^1_{X/S}, I) $$ where the left hand side is the set of isomorphism classes of extension of a scheme $X$ over $S$ with kernel $I$.

Now in the affine case with $X = \text{Spec}\, B$ and $S = \text{Spec}\, A$ the sheaf of $\mathcal O_X$-modules $I$ verifies $I = \tilde{M}$; this gives the identification $$Ex_A (B,M) = Ex(X/S,I)$$

Finally using the definition of the $T^1$ functor for $A$-algebra extensions one gets

$$T^1(B/A, M) \cong Ext^1_{\mathcal O_X} (\Omega^1_{X/S}, I)$$

Your question on the localization and torsion freeness is addressed in Sernesi's proof: in particular the relative conormal sequence one can build up from an extension of $X/S$ is left exact because the kernel of the map $I \to (\Omega^1_{X'/S})_{\vert X}$ is torsion and $I$ is locally free being a coherent locally free sheaf.

As for the first question, as pointed out in the comments, the answer is in Hartshorne "Deformation Theory" as well as in Sernesi. In particular Theorem 5.3 (Hartshorne) and Proposition 1.2.9 (Sernesi) establish the one-to-one correspondence between (isomorphism classes of) deformations of a nonsingular variety $X$ over the dual numbers (first order deformations) and the group $H^1(X,\mathcal T_X)$ via the Kodaira Spencer map $\kappa$; you get your answer by the following easy computation $$H^1(X, \mathcal T_X) \cong Ext^1(\mathcal O_X, \mathcal T_X) \cong Ext^1 (\mathcal O_X, {\Omega^1_X}^\vee) \cong Ext^1 (\Omega^1_X, \mathcal O_X)$$


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