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 1Let 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.

First, let’s see that it commutes with products. Let be a collection of -algebras. The claim is that

The problem is that the rings and might not be the same; there is a map

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.

April 17, 2012 at 4:23 pm

Very enjoyable!

It seems that in the post you define an \’etale ring map as a finitely presented ring map which is formally \’etale. Right?

Suppose on the other hand you define an \’etale map A —> B as a smooth ring map of relative dimension zero, i.e., as a map such that the naive cotangent complex is zero. Then it isn’t hard to show that such a ring map has the form B = A[x_1, …, x_n]/(f_1, …, f_n) with the Jacobian matrix invertible in B. It is trivial to lift such algebras directly (even when the ideal isn’t nilpotent).

I don’t think it requires ZMT to show that the two definitions agree… but I could be wrong.

April 17, 2012 at 10:09 pm

Thanks! I was a bit confused about this for a while, but I do agree that one can show this much of the local structure theorem by bare hands, and lifting such algebras is indeed straightforward. I’m not totally sure why I thought ZMT was necessary to make the lift — it’s surprising that Raynaud’s book doesn’t give this simpler argument.

April 17, 2012 at 6:14 pm

[…] over to read Akhil Mathew’s very nice blog post before reading this one. Then read my comment on his post (which was somehow a bit off topic). My […]

April 18, 2012 at 11:46 am

“so we need to check that it commutes with filtered colimits.” -> “so we need to check that it commutes with filtered limits.”

April 19, 2012 at 1:41 pm

Or: “that it cocomutes with filtered colimits”?

April 20, 2012 at 12:23 am

I do mean filtered colimits here: any corepresentable functor (covariant) will commute with sufficiently filtered colimits. If is a ring of cardinality , then homming out of commutes with -filtered colimits.