I said we’d apply our pure algebra theory from the past few days to a concrete example today. I’m pretty sure this must be quite well known to experts, but I’ve never seen it written down somewhere. I’ve certainly found use of this fact in my own research.

Let be a smooth projective variety with . This just says the variety has no infinitesimal automorphisms and hence we are in a very general situation. This applies to curves of genus , or K3 surfaces, or Calabi-Yau varieties of higher dimension, etc. It is just one cohomological criterion that should be relatively easy to check.

The lemma we’ll prove is that given some other variety , these varieties are isomorphic over if and only if the are isomorphic over , where the perfect closure of inside some fixed algebraic closure (the smallest perfect field containing inside of ). What this lemma allows you to do in practice is sometimes quickly reduce to the case of working over a perfect field which can greatly simplify things.

Here is the proof. Consider the Isom scheme whose functor of points is given by (the -valued points are just the isomorphisms over ). In our nice situation this is well-known to be representable by a quasi-projective scheme over . We will simply check now that the functor is formally unramified and hence étale. To simplify notation, call the Isom scheme .

Let be a -algebra and a square zero ideal. We must check that the natural map given by restricting an isomorphism to is injective.

Suppose and are isomorphisms over that agree over . Then is an infinitesimal automorphism of . But infinitesimal automorphisms are parametrized by by assumption. Thus is the identity isomorphism over all of and hence .

This shows that is injective and hence is étale. We now prove that the canonical map is a bijection. The fact that is étale over tells us that it is a finite product of separable field extensions of . Let be the étale algebra representing the Isom functor.

We must show that given by composing with the embedding , i.e. is a bijection. Since just composes with an inclusion, it is injective. Since is a product of separable field extensions, any homomorphism must have separable image. The field extension is purely inseparable, so the image must land inside . This shows that is surjective. Thus our map is a bijection.

What we proved is just a restatement of the lemma. Of course any isomorphism of and over base changes to one over , but using the algebra we’ve developed the past few posts we see that an isomorphism over uniquely descends to one over .

July 22, 2012 at 6:11 pm

Very cool! This was quite a nice argument.

July 23, 2012 at 8:28 am

I’m pretty sure this is well known. My advisor seemed to know it, and I’ve seen it “referenced” without proof for example by Brian Conrad at this mathoverflow question. It is weird I haven’t seen people really spell it out anywhere though.