# Do K3-surfaces have Weierstrass equations

I've been wondering a bit about K3-surfaces and their analogy to elliptic curves. I've just started so this might be a very silly question.

Do all K3-surfaces have a Weierstrass equation (up to birational equivalence)?

I'm thinking about an equation of the form $y^2= x^3+A(t) x + B(t)$ in $\mathbf{A}^3$ for a variety birational to $X$. Of course, this would imply a K3-surface is a two-cover of $\mathbf{P}^2$ (up to birational equivalence).

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For a K3, "Up to birational equivalence" is the same as "up to isomorphism" since they are minimal surfaces and even though some K3's are double covers of $\mathbf{P}^2$ branched over a sextic (not a cubic, by adjunction formula degree 6 is needed to make canonical trivial), not all are of this form. For example, consider a Kummer K3. – Matt Nov 8 '12 at 3:01
I don't understand your comment on "birational equivalence" being the same as "isomorphism". I'm trying to say that we need to restrict to an open, or maybe to an open of some blow-up. Certainly this is necessary if we would want a Weierstrass equation in that form. This is already the case for elliptic curves. – Harry Nov 8 '12 at 12:39
I guess I have no idea why you are picking such a strange definition then. Of course you have to work birationally if you choose some affine model of a projective thing. Why not eliminate "birational" in the elliptic curve case by defining the Weierstrass equation as the projectivized equation? Every elliptic curve is isomorphic to a closed subvariety of $\mathbb{P}^2$ defined by a (homogeneous) Weierstrass equation. Also, the question should be altered to be $w^2=f(x,y,z)$ where $f$ is degree six if there is any hope for a K3. – Matt Nov 8 '12 at 19:40