A Mind for Madness

Musings on art, philosophy, mathematics, and physics

Brauer Equivalence of Rational Points

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Today I’m going to start a series on the arithmetic of rational surfaces. I feel like this theory is fairly unknown, but it provides such a wonderful source of examples that don’t exist in the curve case. Everything is so explicit with equations and concrete calculations.

The theorem I want to get to is in Bloch’s paper “On the Chow Groups of Certain Rational Surfaces.” It says that for a certain class of rational surfaces, good/bad reduction can be detected on the Chow group of {0}-cycles of degree {0}. We will also pull a lot from Manin’s book Cubic Surfaces and possibly from some papers of Colliet-Thélène.

Rather than prove this amazing result, I want to describe some of the constructions and ideas that go into it. Let’s get some terminology out of the way. For our purposes, a projective surface {X/k} will be called rational if there exists some {k'/k} for which the extension of scalars {X_{k'}} is birational with {\mathbf{P}^2_{k'}}.

Fix an algebraic closure {\overline{k}} and let {G=Gal(\overline{k}/k)}. Nothing in the immediate theory should require this, but since the goal is a theorem about reduction type, without loss of generality {k} is either assumed global or complete, local, arising as the fraction field of some DVR of mixed characteristic. If you want to push the theory through for positive characteristic, you will need to at least alter all the {\overline{k}} to {k^{sep}}.

Bloch’s proof involves playing around with Brauer groups and in particular realizing a certain pairing of Manin in a new way. Our goal for today will just be to work out some of the basic Brauer group theory. Recall that {Br(X)=H^2(X, \mathbf{G}_m)}. Unless otherwise stated, all cohomology will be étale or Galois (the distinction should be obvious from context and so subscripts will be omitted). This means that {Br(k)=H^2(Spec(k), \mathbb{G}_m)=H^2(G, \overline{k}^\times)}.

First, we’ll check that there is a map {Br(k)\rightarrow Br(X)} in a sort of ridiculous way, because we’ll need to use the map involved at a later point. For étale cohomology, we have the convergent first-quadrant Hochschild-Serre spectral sequence {E_2^{p,q}=H^p(G, H^q(X_{\overline{k}}, \mathbf{G}_m))\Rightarrow H^{p+q}(X, \mathbf{G}_m)}. Thus writing {H^2(X, \mathbf{G}_m)=Br(X)\simeq E_{\infty}^{0,2}\oplus E_{\infty}^{1,1}\oplus E_{\infty}^{2,0}} and using that {E_{\infty}^{2,0}} is a quotient of {E^{2,0}_2=Br(k)} we get a map via inclusion then projection.

Given any point {x_i: Spec(k(x_i))\rightarrow X}, we have a finite index subgroup {G'=Gal(\overline{k}/k(x_i))} of {G}. By functoriality of Galois cohomology, this gives us two group homomorphisms. The restriction, {res: Br(k)\rightarrow Br(k(x_i))} and corestriction (aka the transfer), {cor_{k(x_i)/k}: Br(k(x_i))\rightarrow Br(k)}. A well-known property of these two maps is that {(cor)\circ (res)=[G: G']} is multiplication by the index (or by Galois theory in this case, {[k(x_i):k]}).

The last map we need before we can piece together Manin’s pairing is that we can pullback on cohomology given a point {x_i^*: Br(X)\rightarrow Br(k(x_i))}. We denote this by {a(x_i)} (thought of as the Brauer class restricted to {x_i}). Define {C_0(X)} to be the free abelian group generated by the points of {X} (i.e. {0}-cycles) of degree {0}. Caution: This is not the Chow group or anything because we aren’t moding out by any sort of rational or algebraic equivalence yet. We only require degree {0}.

For those not used to the degree map when over non-algebraically closed fields, recall that {deg(\sum n_ix_i)=\sum [k(x_i): k]n_i}, so it is possible that {2x_1-3x_2} is a degree {0} cycle if the residue field extensions are {3} and {2} respectively.

The Manin pairing is {(-,-): C_0(X)\times (Br(X)/Br(k))\rightarrow Br(k)} given by {\displaystyle \left(\sum n_ix_i, a\right)=\prod_i cor_{k(x_i)/k}(a(x_i))^{n_i}}. Certainly at the level of {C_0(X)\times Br(X)} everything is well-defined by the above maps. There is some work in checking that we can pass to the quotient.

Here is why it works. If we take a class {a\in Br(X)} that is in already in {Br(k)}, then {\displaystyle Br(k)\rightarrow Br(X)\stackrel{x_i^*}{\rightarrow} Br(k(x_i))} is just {res(a)} (this is somewhat non-obvious from our definition). Now given any element {\sum n_ix_i\in C_0(X)} we need to check that it pairs with {a} to {1}. We crucially use our arithmetic definition of degree {0} here.

\displaystyle \begin{matrix} (\sum n_ix_i, a) & = & \displaystyle \prod_i cor_{k(x_i)/k}(a(x_i))^{n_i} \\ & = & \displaystyle\prod_i (cor_{k(x_i)/k}\circ res)(a^{n_i}) \\ & = & \displaystyle\prod_i a^{[k(x_i):k]n_i} \\ & = & \displaystyle a^{\sum [k(x_i):k]n_i} \\ & = & a^0 =1 \end{matrix}

This gives us a well-defined pairing which I’ll refer to as the Manin pairing (non-standard terminology to my knowledge). We say that two rational points {x,y\in X(k)} are Brauer equivalent if {(x-y, a)=1} for all {a\in Br(X)}. The key idea of the next post will be to rewrite this pairing in a way that allows us to check that for a smooth rational surface over a global field the set of rational points up to Brauer equivalence: {X(k)/B} is finite. But more importantly, the set has only one element (i.e. all points are Brauer equivalent) at places of good reduction.

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Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

One thought on “Brauer Equivalence of Rational Points

  1. Pingback: Brauer-Manin Obstruction « A Mind for Madness

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