Let be a ring. An -algebra is said to be étale if it is finitely presented and for every -algebra and every nilpotent ideal (or ideal consisting of nilpotents), we have an isomorphism
In other words, given a homomorphism of -algebras , we can lift it uniquely to the “nilpotent thickening” .
The algebras étale over form a category, ; this is the étale site of . For example, for a field, it consists of the category of all finite separable extensions. \’Etaleness is preserved under base-change, so for any morphism , there is a functor
A basic property of étale morphisms is the following:
Theorem 1 Let be a ring and a square zero (or nilpotent) ideal. Then there is an equivalence of categories
given by tensoring with .
This result is often proved using the local structure theory for étale morphisms, but this is fairly difficult: as far as I know, the local structure theory requires Zariski’s Main Theorem for its proof. (Correction: as observed below, one only needs a portion of the local structure theory to make the argument, and that portion does not require ZMT.) Here is a more elementary argument.
First, observe that the functor is fully faithful by the definition of an étale morphism. Given étale -algebras , we observe that -maps are in bijection with -maps , which in turn are in bijection with -maps . In other words, we have an isomorphism
This is exactly the statement of full faithfulness.
The more difficult part consists of showing that the functor is essentially surjective: that is, that every algebra étale over is obtained via reduction of an algebra étale over .
So, let be an étale -algebra, which we want to “lift” to . In other words, we want to find an such that . The idea of producing is to write down the functor that it corepresents and then appeal to the adjoint functor theorem.
In fact, we know that if exists, it must satisfy the property
for all -algebras . This is, again, by the nilpotent lifting condition. So, we need to prove that the functor
is actually corepresentable. This will give us the desired .
How can we show that the above functor is corepresentable? Since rings form a presentable category, we have to check that it commutes with limits and sufficiently filtered colimits, in view of the adjoint functor theorem. Well, it certainly commutes with all filtered colimits (as is finitely presented over ), so we need to check that it commutes with filtered colimits.
This is certainly surjective, and the kernel is surjected upon by the kernel of
which is a square-zero ideal. Consequently, the two rings in question in (1)differ by a square-zero ideal, and so maps out of into them are the same.
Next, we need to show that our functor commutes with equalizers. So, let’s say we have an equalizer diagram of -algebras
Then we want an equalizer diagram of sets
If we had an equalizer diagram of rings
then we’d be done. But we don’t. But by the same logic, the map from to the honest equalizer is a surjection whose kernel consists of square-zero elements. In fact, if an element in has image in , then its square in must have image zero, and so must be zero.
It follows that we have defined a corepresentable functor on the category of -algebras; let be a corepresenting object. So we have a functorial isomorphism:
valid for every -algebra .
Now, finally, we need to show that is actually étale over and that . The second property follows from taking to be an -algebra in the above isomorphism. The first follows from the fact that can’t see nilpotent thickenings because doesn’t. This completes the proof.