Tag: etale algebra

Intro to Brauer Groups

I want to do a series on the basics of Brauer groups since they came up in the past few posts. Since I haven’t really talked about Galois cohomology anywhere, we’ll take a slightly nonstandard approach and view everything “geometrically” in terms of étale cohomology. Everything should be equivalent to the Galois cohomology approach, but this way will allow us to use the theory that is already developed elsewhere on the blog.

I apologize in advance for the sporadic nature of this post. I just need to get a few random things out there before really starting the series. There will be one or two posts on the Brauer group of a “point” which will just mean the usual Brauer group of a field (to be defined shortly). Then we’ll move on to the Brauer group of a curve, and maybe if I still feel like continuing the series of a surface.

Let {K} be a field and {K^s} a fixed separable closure. We will define {Br(K)=H^2_{et}(Spec(K), \mathbb{G}_m)=H^2(Gal(K^s/K), (K^s)^\times)}. This isn’t the usual definition and is often called the cohomological Brauer group. The usual definition is as follows. Let {R} be a commutative, local, (unital) ring. An algebra {A} over {R} is called an Azumaya algebra if it is a free of finite rank {R}-module and {A\otimes_R A^{op}\rightarrow End_{R-mod}(A)} sending {a\otimes a'} to {(x\mapsto axa')} is an isomorphism.

Define an equivalence relation on the collection of Azumaya algebras over {R} by saying {A} and {A'} are similar if {A\otimes_R M_n(R)\simeq A'\otimes_R M_{n'}(R)} for some {n} and {n'}. The set of Azumaya algebras over {R} modulo similarity form a group with multiplication given by tensor product. This is called the Brauer group of {R} denoted {Br(R)}. Often times, when an author is being careful to distinguish, the cohomological Brauer group will be denoted with a prime: {Br'(R)}. It turns out that there is always an injection {Br(R)\hookrightarrow Br'(R)}.

One way to see this is that on the étale site of {Spec(R)}, the sequence of sheaves {1\rightarrow \mathbb{G}_m\rightarrow GL_n\rightarrow PGL_n\rightarrow 1} is exact. It is a little tedious to check, but using a Čech cocycle argument (caution: a priori the cohomology “groups” are merely pointed sets) one can check that the injection from the associated long exact sequence {H^1(Spec(R), PGL_n)/H^1(Spec(R), GL_n)\hookrightarrow Br'(R)} is the desired injection.

If we make the extra assumption that {R} has dimension {0} or {1}, then the natural map {Br(R)\rightarrow Br'(R)} is an isomorphism. I’ll probably regret this later, but I’ll only prove the case of dimension {0}, since the point is to get to facts about Brauer groups of fields. If {R} has dimension {0}, then it is a local Artin ring and hence Henselian.

One standard lemma to prove is that for local rings a cohomological Brauer class {\gamma\in Br'(R)} comes from an Azumaya algebra if and only if there is a finite étale surjective map {Y\rightarrow Spec(R)} such that {\gamma} pulls back to {0} in {Br'(Y)}. The easy direction is that if it comes from an Azumaya algebra, then any maximal étale subalgebra splits it (becomes the zero class after tensoring), so that is our finite étale surjective map. The other direction is harder.

Going back to the proof, since {R} is Henselian, given any class {\gamma\in H^2(Spec(R), \mathbb{G}_m)} a standard Čech cocycle argument shows that there is an étale covering {(U_i\rightarrow Spec(R))} such that {\gamma|_{U_i}=0}. Choosing any {U_i\rightarrow Spec(R)} we have a finite étale surjection that kills the class and hence it lifts by the previous lemma.

It is a major open question to find conditions to make {Br(X)\rightarrow Br'(X)} surjective, so don’t jump to the conclusion that we only did the easy case, but it is always true. Now that we have that the Brauer group is the cohomological Brauer group we can convert the computation of {Br(R)} for a Henselian local ring to a cohomological computation using the specialization map (pulling back to the closed point) {Br(R)\rightarrow Br(k)} where {k=R/m}.

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Étale algebras

One interesting thing we could do at this point is work out the much more interesting structure theory of separable algebras if we relax some of the standing assumptions such as commutativity. Instead, we’re going to start working in a more general situation, but the idea is to put stronger conditions into our definition to keep the structure close to the same.

Note that since we were working over a field all of our algebras were automatically flat. For this reason, they were all something called étale algebras. Geometrically speaking this means they are smooth and the fibers all have dimension {0}. Another equivalent way to say this is that the structure map is flat and unramified. This condition of an algebra being unramified is what we’ll discuss today.

If you look in Milne’s book Étale Cohomology, you’ll find that there’s a mind-boggling large number of equivalent ways to check a map is étale. The way that algebraic geometers (at least those working with functors) tend to check this condition is almost entirely passed over in the book. There is just one quick mention that this method can be done, so today we’ll write down this condition and next time we’ll work out a neat example using it.

The idea of being unramified can maybe be stated more easily in terms of spaces. A variety {X/k} is formally unramified if for any {Y\rightarrow X} and any infinitesimal thickening, say {\overline{Y}} of {Y}, there is at most one way to extend {Y\rightarrow X} to the thickened scheme {\overline{Y}\rightarrow X}. This is intentionally vague, but now if we think about everything being affine, by the equivalence of categories with algebras we could work out what the exact condition should be. One should be careful about the word “formally” existing everywhere, but since we’ll assume our algebras are again unital, associative, commutative, and finite dimensional this formal condition is completely equivalent to being unramified/étale in the usual sense.

An {A}-algebra {B} is called (formally) unramified if for any sequence of {A}-algebras {0\rightarrow I\rightarrow C'\rightarrow C\rightarrow 0} with {I} nilpotent we have that any {A}-map {B\rightarrow C} has at most one extension through {C'}, i.e. there is at most one composition {B\rightarrow C'\rightarrow C} which yields {B\rightarrow C}. If in addition one can always find such an extension, {B} is called an étale {A}-algebra. One might have learned in Hartshorne that this latter condition is called the infinitesimal lifting criterion and implies the map is smooth. We already talked about this here.

To wrap up today I’ll point out why this is the condition that comes up most for me. Suppose you are given the points of some scheme {h_X: \text{k-alg} \rightarrow Set}, but you don’t know much else about it. Maybe you want to check that {X} is étale. If you try to use one of the standard methods, you may get stuck since those other methods tend to assume you know more about {X} already. Merely from knowing the functor of points you only need to check that it is formally unramified by checking that given the situation above {0\rightarrow I\rightarrow C'\rightarrow C\rightarrow 0} the natural map coming from the functor {h_X(C')\rightarrow h_X(C)} is injective.

Next time we’ll work out one common place where this happens, and then see that checking this map is injective (which will be easy in the example) implies some incredibly strong things about the variety {X}.

Separable Algebras 2

Before continuing with some basics of separable algebras, I want to give some motivation for why we should care. First consider the correspondence between the existence of nilpotents and not being reduced. Separable algebras are related to the notion of being geometrically reduced, i.e. reduced after any base change. This isn’t great motivation, but I think the next one is.

Étale morphisms of varieties are pervasive throughout algebraic geometry. Since they are smooth maps of relative dimension {0}, you could think of them as the replacement for covering spaces from topology in algebraic geometry. The local structure on rings of these maps will give you separable algebras, so if you want to understand the covering space theory and related notions of fundamental groups it would be good to know something about separable algebras.

Lastly, if you care about (affine) group schemes then you’ve probably seen that such a {G} has an exact sequence associated to it called the connected-étale sequence and breaks the group scheme into a semi-direct product of its connected component of the identity and the spectrum of the maximal separable subalgebra. Thus separable subalgebras tell us something about the number of connected components of a variety.

There’s even more motivation, but we’ll move on to the post now. That was just meant to show that this isn’t some vacuous definition that algebraists made up to play around with. Recall we are making standard assumptions that when we use the term algebra it means unital, associative, commutative, and finite dimensional.

Let {k} be any field and fix an algebraic closure {\overline{k}} and separable closure {k_s}. Today we’ll check that a {k}-algebra, {A}, is separable if and only if it satisfies one of the following equivalent conditions:

1) {A\otimes_k \overline{k}} is reduced

2) {A\otimes_k \overline{k}\simeq \overline{k}\times \cdots \times \overline{k}}

3) {A} is a finite product of separable field extensions

4) {A\otimes_k k_s\simeq k_s\times \cdots \times k_s}

Note how strong this structure theorem is. In algebraic geometry terms, suppose you find out somehow that the structure map is étale (maybe by checking a condition we’ll discuss next time). Then your variety is just {X=Spec(A)} where {A} is just a finite product of separable field extensions of the base field {k}. Thus {X} is just a finite collection of points with no weird structure going on at all.

To prove these equivalences we need the fact that any finite dimensional {k}-algebra is a finite product of algebras each of which contain a unique maximal ideal consisting entirely of nilpotents. Since this is a standard first-year algebra fact, we’ll assume it (recall the series is on what “ought” to be taught in first-year algebra). We’ll start by noting that {A} separable is equivalent to 1. The previous fact gives us the equivalence of 1 and 2, since having no nilpotents implies all the maximal ideals must be {0} and hence the algebra must be a product of field extensions.

By definition of tensor product we get 3 implies 4 and 4 implies 2, so all we need to show is 2 implies 3. Now 2 implies that the number of {k}-algebra maps {A\rightarrow \overline{k}} is the dimension of {A}. Also, {\dim A=\sum \dim A_i} where {A_i} are the algebras with nilpotent maximal ideals. Since {A} is separable, these maximal ideals are {0} and hence the {A_i} are field extensions of {k}. If they are separable, then our dimension count matches up, but if even one of them was a non-separable extension, then the dimension will be less due to the multiple root. Thus they all must be separable extensions of {k}. This proves the equivalence of all definitions.

Now that we know the general form of any separable {k}-algebra you might think they are uninteresting, but next time we’ll show there is something interesting going on and it is related to covering spaces/deck transformation/fundamental groups. Then we’ll move on to étale maps in more generality than just really nice algebras over fields.