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.



Singular models of K3 surfaces
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*.


