Suppose $X$ is a projective smooth variety over a non-algebraically closed field , do we still have $Ext^i(F,\omega)\to H^{n-i}(X,F)^{\vee}$? (Hartshorne's proof Thm III 7.6 requires $k$ to be algebraically closed)
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1$\begingroup$ See Grothendieck duality, then you get statements over much more general bases. $\endgroup$– Karl SchwedeCommented Aug 23, 2015 at 18:25
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1$\begingroup$ @KarlSchwede In case $k$ is not algebraically closed, is the morphism still an isomorphism? (I am not familiar with derived language, I am not sure how the duality concretely writes out?) $\endgroup$– user39380Commented Aug 23, 2015 at 18:41
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2$\begingroup$ Need $X$ geometrically connected. In this case, I think you can prove it using flat base change, since passing to the algebraic closure is flat. $\endgroup$– Will SawinCommented Aug 23, 2015 at 19:07
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11$\begingroup$ The proof works verbatim for any projective Cohen-Macaulay scheme of pure dimension $n$ over any field. Look at the proof! The only place $k$ being algebraically closed is "used" is that the local ring at a closed point of $\mathbf{P}^N_k$ is regular, but every local ring at a closed point on an affine space over any field is regular (since $R[t]$ is a regular ring for any regular noetherian ring $R$, as a standard application of Serre's homological criterion for regularity: see Theorem 19.5 in Matsumura's book "Commutative Ring Theory"). $\endgroup$– grghxyCommented Aug 23, 2015 at 22:31
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1 Answer
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Ok, here's a proof via specializing Grothendieck duality. It is probably useful for people to see this worked out. Say $f: X \to \text{Spec }k$ is the structural map and its proper and $X$ is just a scheme of finite type over $k$, say $F$ is a coherent sheaf on $X$ (for simplicity). Then Grothendieck duality says that $$R f_* R \mathcal{H}\text{om}_{O_X}(F, f^! k) \simeq R\mathcal{H}\text{om}_{k}(R f_* F, k).$$ In particular, this is an isomorphism in the derived category.
Analyzing the Left side
If $X$ is Cohen-Macaulay, then $f^! k$ is a canonical module on $X$ with a shift (locally by the dimension, note that Cohen-Macaulay means it is locally equidimensional). So lets work with one connected component at a time, where it is of some dimension $d$. Then $f^! k \simeq \omega_X[d]$ (take it as a definition if you aren't comfortable with it). We also notice that $\mathcal{H}\text{om}(F, \bullet)$ takes injectives to flasques and so we get the composition of derived functors: $$ Rf_* R\mathcal{H}\text{om}(F, \bullet) = R\text{Hom}(F, \bullet). $$ (ie, global sections of sheafy hom are non-sheafy-hom) We take the $(i-d)$th cohomology of the left side and get $$ \begin{array}{rl} & \mathcal{H}^{i-d}(R f_* R \mathcal{H}\text{om}_{O_X}(F, f^! k) ) \\ = & \mathcal{H}^{i-d} R \text{Hom}_{\mathcal{O}_X}(F, \omega_X[d]) \\ = & \mathcal{H}^{i} R \text{Hom}_{\mathcal{O}_X}(F, \omega_X) \\ = & \text{Ext}^i(F, \omega_X). \end{array} $$
Analyzing the right side
This is even easier, $R\mathcal{H}\text{om}_{k}(\bullet, k)$ is just the derived functor of $k$-vector space duality, which we identify with itself (its already exact). So I'll just write it as $(\bullet)^{\vee}$. Again we take the $(i-d)$th cohomology and get $$ \mathcal{H}^{i-d} R\mathcal{H}\text{om}_{k}(R f_* F, k) = \mathcal{H}^{i-d}(( R f_* F)^{\vee}) = (\mathcal{H}^{d-i} Rf_* F)^{\vee} = (H^{d-i}(X, F))^{\vee}. $$
In conclusion
Plugging these two things in we recover the desired Serre duality. Notice we never used the fact that $k$ was algebraically closed. We only used the fact that it was a field when noticing that $\mathcal{H}\text{om}_k(\bullet, k)$ was exact already. You can still say similar things (that look a bit more complicated) when $k$ is a Gorenstein ring instead of a field.
answered Aug 24, 2015 at 3:41
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2$\begingroup$ Strictly speaking, one has to know how the Grothendieck duality isomorphism is defined in order to affirm that the concrete isomorphism ${\rm{Ext}}^i(F,\omega) \simeq {\rm{H}}^{d-i}(X,F)^{\vee}$ that you obtain in the end really coincides (up to a universal sign depending only on $i$ and $d$) with the map built from Ext-pairings and a (suitable) trace map ${\rm{H}}^d(X,\omega) \rightarrow k$. So it is necessary to "look under the hood" a bit. $\endgroup$– grghxyCommented Aug 24, 2015 at 4:02
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$\begingroup$ True. I guess it depends on exactly what statement you are looking for. $\endgroup$ Commented Aug 24, 2015 at 15:26