# Family of Enriques surfaces and Grothendieck-Riemann-Roch

Currently I'm studying the article Moduli of Enriques surfaces and Grothendieck-Riemann-Roch by Pappas.

I am particularly interested in how he applies the GRR.

Q1. What is meant by a "family of Enriques surfaces"? I was guessing a flat morphism $f:Y\longrightarrow T$ of smooth projective varieties such that each fibre is an Enriques surface, but maybe this is too general?

Q2. In the article it says that the higher direct images $R^i f_\ast O_Y$ are zero ($i>0$). Is this an application of Grauert's theorem (III.12, Cor. 12.9, Hartshorne)?

Q3. How to see easily that $R^0f_\ast O_Y = O_T$ ? I am guessing Grauert's theorem again.

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Welcome Ariyan :) –  Wanderer Mar 2 '10 at 11:31

This is a somewhat technical remark, related to Andrea's answer, which is a bit too big to fit into the comment box.

If $f: Y \rightarrow T$ has connected fibres, to conclude that $R^0f_*\mathcal O_Y = \mathcal O_T$, one needs some assumptions beyond just that $f$ is a projective morphism of Noetherian schemes. (Consider these examples: a closed embedding will have connected fibres. To give such an example in which all fibres non-empty, consider a non-reduced $T$, and let $Y$ be the underlying reduced subscheme. Or one could take $T$ to be a cuspidal cubic curve and $Y$ to be its normalization.)

What the theorem on formal functions shows (assuming that $f$ is projective, and that $Y$ and $T$ are Noetherian, so we can apply the result as it is proved in Hartshorne) is that for any point $P$ in $T$, the $\mathfrak m_P$-adic completion $(R^0f_*\mathcal O_Y\hat{)}_P$ is equal to $H^0(\hat{Y}_P,\mathcal O)$, the global sections of the structure sheaf on the formal fibre $\hat{Y}_P$ over $P$.

So if $f$ has connected fibres, and hence connected formal fibres, so that $H^0(\hat{Y}\_P,\mathcal O)$ is a local ring, we see that $(R^0f_*\mathcal O_Y\hat{)}_P$ is a finite local $\hat{\mathcal O}\_{T,P}$-algebra. In general, one can't do better than this.

But, if $f$ is flat with geometrially connected and reduced fibres (e.g $f$ is smooth with geometrically connected fibres), then base-change for flat maps (Hartshorne III.9.3) shows that the fibre mod $\mathfrak m_P$ of $R^0f_*\mathcal O_Y$ is equal to $H^0(Y_P,\mathcal O_P)$ (the actual fibre over $P$, now, not the formal fibre), which equals $k(P)$ (the residue field at $P$), since $Y_P$ is projective, geometrically reduced, and geometrically connected over $k(P)$.

So, maintaining these assumptions on $f$, we see that for each point $P$ of $T$, the stalk $(R^0f_*\mathcal O_Y)_P$ is a finite $\mathcal O\_{T,P}$-algebra with the property that its reduction modulo $\mathfrak m_P$ is isomorphic to the residue field $k(P)$ of ${\mathcal O}\_{T,P}$. This implies (by Nakayama) that the natural map ${\mathcal O}\_P \rightarrow (R^0f_*\mathcal O_Y)\_P$ is surjective. This is true at every $P$, and so we see that $\mathcal O_T \rightarrow R^0f_*\mathcal O_Y$ is surjective.

Now one can combine this with the Grauert result to conclude (since a surjection of invertible sheaves is necessarily an isomorphism) that the natural map $\mathcal O_T \rightarrow R^0f_*\mathcal O_Y$ is an isomorphism. (We probably don't need to use the full force of Grauert here; for example, suppose that $T$ is connected; a flat map is open, and a projective map is closed, so $f$ is surjective, hence faithfully flat, and this implies that the map $\mathcal O_T \rightarrow R^0f_*\mathcal O_Y$ is injective, I think.)

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There is not much to say, because you answer yourself. :-)

A family of Enriques surface is as you define it. The equality in 2) follows from Grauert's theorem.

The only thing is that Grauert is not enough to prove that $R^0f_\ast O_Y = O_T$, since it would only imply that $R^0f_\ast O_Y$ is a line bundle. But in characteristic 0 the equality $R^0f_\ast O_Y = O_T$ (for a projective morphism of Noetherian shcemes) is equivalent to the fact that the fibers are connected. One implication is [Har III.11.3]. I cannot find a reference for the other implication now, but it follows in your case from Stein factorization [Har III.11.5].

EDIT: I misquoted the result. One implication is always true, but the requires stronger assumptions, in particular $f$ should be flat. See Emerton's answer for a correction.

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The equality $R^0f_\ast\mathcal O_Y=\mathcal O_T$ is in fact quite easy: We have a map $O_T \to R^0f_\ast\mathcal O_Y$ and if we know that the right hand side commutes with base change then by Nakayama's lemma the map is surjective and as both sides are line bundle it is an isomorphism. –  Torsten Ekedahl Aug 25 '10 at 13:21

I couldn't figure out how to comment on Emerton's post, so I apologize for this non sequitur nitpick of an answer.

"But, if $f$ is flat with geometrially connected and reduced fibres (e.g $f$ is smooth with geometrically connected fibres), then base-change for flat maps (Hartshorne III.9.3) shows that the fibre mod $P$ of $R^0fY$ is equal to $H^0(Y_P,\mathcal{O}_P)$ (the actual fibre over $P$, now, not the formal fibre), which equals $k(P)$ (the residue field at $P$), since $Y_P$ is projective, geometrically reduced, and geometrically connected over $k(P)$."

I am not sure that result from Hartshorne quite does that. It needs the map along which the base change is being made to be flat (since we need the Cech complex to remain exact under base change). In this case, we are pulling back along the decidedly unflat inclusion of the point $P$ into $Y$, and so would need the further technology of semi-continuity and base change from Hartshorne III.12 (which is unfortunately a piece of bad exposition, IMO).

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Thank you for this. –  Emerton Sep 1 '10 at 5:53