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I am reading Illusie's lecture notes "topics in algebraic geometry", and I have difficulty in following his proof of unobstructedness of deformation of curves. Here is the statement of the proposition:

Prop Let $f_0: X_0 \to Y_0$ be a smooth proper morphism with relative dimension 1, and $i : Y_0 \to Y$ a first-order thickening with ideal $I$. If moreover $Y$ is affine, then there always exists a lifting of $X_0$ over $Y$ .

Proof. First, since $Y_0$ is affine, we note that $H^q(X_0, T_{X_0/Y_0} \otimes f_{0}^{*}I)= \Gamma (Y_0,R^qf_{0*}(T_{X_0/Y_0}\otimes f_{0}^{*}I))$
By Zariski’s main theorem, for any $q > 1$,
$R^qf_{0*}(T_{X_0/Y_0}\otimes f_{0}^{*}I))=0$ (*)
Hence the obstruction $o(f_0, i)\in H^2(X_0, T_{X_0/Y_0}\otimes. f_0^*I)$ vanishes. #

And I wonder how to derive (*) by Zariski's main theorem.

Thank you!

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up vote 5 down vote accepted

Probably Illusie wrote "Zariski's Main Theorem", but he intended the Theorem of Formal Functions (which is the key result needed in the modern proof of Zariski's Theorem).

In fact, the Theorem of Formal Functions implies the following result, see [Hartshorne, Algebraic Geometry, Corollary 11.2 page 279].

Proposition. Let $f \colon X \to Y$ be a projective morphism of noetherian schemes, and let $r= \textrm{max} \{\dim X_y | y \in Y \}$. Then $R^qf_* (\mathscr{F})=0$ for all $q >r$ and for all coherent sheaves $\mathscr{F}$ on $X$.

Now by projection formula we have $$R^qf_{0*}(T_{X_0/Y_0}\otimes f_{0}^{*}I))=R^qf_{0*}(T_{X_0/Y_0}) \otimes I,$$ so you can apply the previous proposition with $r=1$ and $\mathscr{F}=T_{X_0/Y_0}$ in order to get the desired vanishing.

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Thank you very much! – Xiaobo Zhuang Oct 3 '12 at 11:09

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