# An Application to Isomorphisms of Varieties

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 ${X/k}$ be a smooth projective variety with ${H^0(X, \mathcal{T})=0}$. This just says the variety has no infinitesimal automorphisms and hence we are in a very general situation. This applies to curves of genus ${g\geq 2}$, 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 ${Y/k}$, these varieties are isomorphic over ${k}$ if and only if the are isomorphic over ${k^{perf}}$, where ${k^{perf}}$ the perfect closure of ${k}$ inside some fixed algebraic closure ${\overline{k}}$ (the smallest perfect field containing ${k}$ inside of ${\overline{k}}$). 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 ${\text{Isom} (A)=\{X\otimes A \stackrel{\sim}{\rightarrow} Y\otimes A\}}$ (the ${A}$-valued points are just the isomorphisms over ${A}$). In our nice situation this is well-known to be representable by a quasi-projective scheme over ${k}$. We will simply check now that the functor is formally unramified and hence étale. To simplify notation, call the Isom scheme ${Z}$.

Let ${A}$ be a ${k}$-algebra and ${I}$ a square zero ideal. We must check that the natural map ${Z(A)\rightarrow Z(A/I)}$ given by restricting an isomorphism ${X\otimes A\rightarrow Y\otimes A}$ to ${X\otimes (A/I)\rightarrow Y\otimes (A/I)}$ is injective.

Suppose ${\phi_1}$ and ${\phi_2}$ are isomorphisms over ${A}$ that agree over ${A/I}$. Then ${\phi_2^{-1}\circ \phi_1}$ is an infinitesimal automorphism of ${X}$. But infinitesimal automorphisms are parametrized by ${H^0(X, \mathcal{T})\otimes I=0}$ by assumption. Thus ${\phi_2^{-1}\circ \phi_1}$ is the identity isomorphism over all of ${A}$ and hence ${\phi_1=\phi_2}$.

This shows that ${Z(A)\rightarrow Z(A/I)}$ is injective and hence ${Z}$ is étale. We now prove that the canonical map ${Z(k)\rightarrow Z(k^{perf})}$ is a bijection. The fact that ${Z}$ is étale over ${k}$ tells us that it is a finite product of separable field extensions of ${k}$. Let ${E}$ be the étale algebra representing the Isom functor.

We must show that ${T: Hom_k(E, k)\rightarrow Hom_k(E, k^{perf})}$ given by composing with the embedding ${i:k\hookrightarrow k^{perf}}$, i.e. ${(E\rightarrow k)\mapsto (E\rightarrow k\stackrel{i}{\rightarrow} k^{perf})}$ is a bijection. Since ${T}$ just composes with an inclusion, it is injective. Since ${E}$ is a product of separable field extensions, any homomorphism ${E\rightarrow k^{perf}}$ must have separable image. The field extension ${k^{perf}/k}$ is purely inseparable, so the image must land inside ${k}$. This shows that ${T}$ is surjective. Thus our map ${\text{Isom}(k)\rightarrow \text{Isom}(k^{perf})}$ is a bijection.

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