# Finiteness of X(k)/B for Rational Surfaces

Recall our setup. We start with a projective surface ${X/k}$ that becomes rational after some finite extension of scalars ${k'/k}$. Let ${C_0(X)}$ be the group of ${0}$-cycles of degree ${0}$. Last time we defined the Manin pairing ${(-,-): C_0(X)\times (Br(X)/Br(k))\rightarrow Br(k)}$ using the corestriction map ${(\sum n_ix_i, a)=\prod_i cor_{k(x_i)/k}(a(x_i))^{n_i}}$. Two rational points are called Brauer equivalent if ${(x-y, a)=1}$ for all ${a\in Br(X)}$, and denote the set of rational points up to Brauer equivalence by ${X(k)/B}$.

Now let ${N=NS(X_{k'})}$ be the Néron-Severi group of ${X_{k'}}$. It turns out we can factor the Manin pairing as follows:

$\displaystyle \begin{matrix} C_0(X)\times (Br(X)/Br(k)) & \longrightarrow & Br(k) \\ \downarrow & & \uparrow \\ A_0(X)\times H^1(G, N) & \longrightarrow & H^1(G, N\otimes \overline{k}^\times)\times H^1(G, N)\end{matrix}$

The goal of today is to say something about this factoring. Last time we wrote down the Hochschild-Serre spectral sequence and said the map ${Br(k)\rightarrow Br(X)}$ was just the quotient map ${E_2^{2,0}\rightarrow E_\infty^{2,0}}$ followed by the inclusion. Note that since all differentials are ${0}$ for all ${E_n^{1,1}}$ we get that it equals ${H^1(G, N)}$ and sits inside ${Br(X)}$. Thus we have a sequence ${Br(k)\rightarrow Br(X)\rightarrow H^1(G,N)}$ whose composition is ${0}$ and hence gives a map

$\displaystyle Br(X)/Br(k)\rightarrow H^1(G, N).$

This defines for us the left vertical map, since the left factor is just projection from all ${0}$-cycles of degree ${0}$ to ${0}$-cycles modulo rational equivalence of degree ${0}$. The right vertical map is just the one induced on group cohomology via the standard intersection pairing on the surface ${N\otimes \overline{k}^\times \times N\rightarrow \overline{k}^\times}$.

This leaves us with the bottom map. Call it ${\Phi \times id}$ where ${\Phi:A_0(X)\rightarrow H^1(G, N\otimes \overline{k}^\times )}$. It turns out that the majority of Bloch’s paper is merely defining this map and checking that the above diagram commutes, so we won’t get into that. It involves lots of K-theory which I’m not going to get into.

Supposing the above, the main theorem of the paper is that ${Im \Phi}$ is finite in the case of our hypotheses. We can check the nice corollary that ${X(k)/B}$ is finite. If ${X(k)}$ is empty we’re done, so fix some ${x_0\in X(k)}$. The proof is that we can make ${\Psi: X(k)\rightarrow H^1(G, N\otimes \overline{k}^\times)}$ by ${\Psi(x)=\Phi([x]-[x_0])}$. Since ${Im \Psi\subset Im \Phi}$, it must have finite cardinality. We need only check that distinct Brauer classes stay distinct to get the result, but this follows from commutativity of the diagram and the fact that Brauer classes are by definition distinguished under the Manin pairing.

It turns out that Manin had already proved that ${X(k)/B}$ is finite for cubic surfaces, so Bloch’s result extends this to any rational surface. As a consequence of the construction of ${\Phi}$, Bloch also gets the strange result that if ${X}$ is a conic bundle, i.e. ${X\rightarrow \mathbf{P}^1}$ has generic fiber a conic, and ${k}$ is local, then if ${X}$ has good reduction then ${|Im\Phi |=1}$. Thus at places of good reduction ${A_0(X)}$ is trivial.

Note how useful this is. For example, take some conic bundle over ${\mathbf{Q}_p}$. Good reduction means that there exists some proper, regular model ${\frak{X}/\mathbf{Z}_p}$ whose generic fiber is ${X}$ and whose special fiber is non-singular. It is hard to tell whether or not ${X}$ has good reduction, because you might accidentally be picking the wrong model. With this condition of Bloch, one can sometimes explicitly calculate some non-trivial element of ${A_0(X)}$ (Manin actually does this using the defining equation of a class of Châtalet surfaces!) which tells you ${X}$ has bad reduction.

To phrase this a different way, to test for honest bad reduction without some criterion requires a choice of model over ${\mathbf{Z}_p}$. There could be infinitely many distinct choices here, so it could be hard to tell if you’ve exhausted all possibilities. This criterion of Bloch says that no choice needs to be made. Bad reduction can be tested inherently from the variety over ${\mathbf{Q}_p}$.