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Let us work over a ground field of characteristic zero. As is well-known, a K3 surface is a smooth projective geometrically integral surface $X$ whose canonical class $\omega_X$ is trivial and for which $\operatorname{H}^1(X,\mathscr{O}_X)$ vanishes.

A bit of folklore (proven e.g. in Beauville's Complex Algebraic Surfaces) is that if $X$ is a smooth complete intersection of a quadric and a cubic in $\mathbb{P}^4$, or a smooth complete intersection of three quadrics in $\mathbb{P}^5$, then $X$ is K3.

The question

Suppose I have a (non-smooth) complete intersection $X_{2,3}$ of a quadric and a cubic in $\mathbb{P}^4$ all of whose singularities are rational double points, and a (non-smooth) complete intersection $X_{2,2,2}$ of three quadrics in $\mathbb{P}^5$, again with at most rational double points.

Do these two surfaces have K3 surfaces as their minimal regular models?

I suspect the answer is yes; for example, since the singularities are assumed to be rational double points, they should admit a crepant resolution. However, I can't really locate this fact in the literature. As for the vanishing of the appropriate $\operatorname{H}^1$, I really haven't a clue.

R.P.
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2 Answers2

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Yes, this is true: the smooth minimal model is a $K3$ surface.

In fact, let $\bar{X}$ be the resolution of the singularities of $X$. Then the following holds.

(1) Rational double points impose no adjunction conditions to canonical forms, hence $\omega_{\bar{X}}$ is trivial.

(2) Rational double points have simultaneous resolution, hence $\bar{X}$ is deformation equivalent (and so diffeomorphic by Ehresmann's Theorem) to a smooth complete intersection $X^{sm}$ of the same type as $X$. This implies $b_1(\bar{X})=b_1(X^{sm})=0$, hence $H^1(\bar{X}, \, \mathcal{O}_{\bar{X}})=0$.

For part (1) you can look at M. Reid's Young Person Guide to Canonical Singularities, whereas part (2) can be found in Kollar-Mori's book Birational Geometry of Algebraic Varieties.

Francesco Polizzi
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    Alternative to (2), valid in any characteristic: since $X$ has rational singularities, a resolution $f:\tilde{X}\rightarrow X $ satisfies $R^if_*\mathcal{O}_{\tilde{X} }=0$ for $i>0$. – abx Mar 16 '15 at 13:05
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For what it's worth, I wrote up a proof (pretty detailed) following the hints in Francesco Polizzi's answer. It's in an unpublished preprint found here (p. 38 onwards). I am not a geometer, so the exposition may be a little clunky, but at least I think all the crucial steps are there.

R.P.
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