# Brauer group of a curve over non-algebraically closed field

It is a famous consequence of Tsen's theorem that a smooth curve over an algebraically closed field has trivial Brauer group. But what about curves over non algebraically closed fields?

Let us fix a smooth, projective curve $$X$$ over some field $$k$$. If $$X$$ has a rational point $$x\in X(k)$$, then the natural map $$\text{Br}(k)\to\text{Br}(X)$$ has a retraction $$\text{Br}(X)\to\text{Br}(k)$$, thus it is injective. Moreover, it is widely known that $$\text{Br}(X)$$ injects in $$\text{Br}(k(X))$$. But what about closed points? A Brauer class that is trivial on each closed point is globally trivial?

Precise question: Is the map $$\text{Br}(X)\to\prod_{x\in X^1}\text{Br}(k(x))$$ injective? If not, can we describe its kernel?

Observe that, for $$k$$ algebraically closed, the injectivity above is precisely $$\text{Br}(X)=0$$.

• It is also trivially injective for $k$ finite (Albert-Brauer-Hasse-Noether). Why does $\mathrm{Br}(X)$ map to the direct sum? – user19475 Oct 8 '18 at 17:08
• It obviously doesn't, I meant the product. I corrected, thanks. – Giulio Bresciani Oct 8 '18 at 17:11
• You can BTW replace the Brauer groups of the residue fields by the Brauer groups of the Henselisations or completions of the local rings. – user19475 Oct 8 '18 at 17:43
• If $X$ is a conic over the real numbers without a real point then the map is clearly not injective. – naf Oct 9 '18 at 4:27

I don't think your map is injective. Here is an attempt at a recipe for constructing a counterexample.

The ingredients are a $$C_1$$-field $$F$$ of characteristic zero and a smooth projective curve $$X_0$$ over $$F$$ having non-trivial Brauer group. For a concrete example take $$F=\mathbf{C}(t)$$ and $$X_0$$ to be an elliptic curve with non-trivial Brauer group; a calculation of such a curve can be found in O. Wittenberg, Transcendental Brauer–Manin obstruction on a pencil of elliptic curves, PDF file here.

Edit: What's below is unnecessary. $$X_0$$ is already a counterexample, since every finite extension of $$F$$ has trivial Brauer group.

Given these ingredients, let $$k$$ be the field $$F((t))$$, let $$\mathcal{X}$$ be a smooth proper curve over $$F[[t]]$$ whose special fibre is $$X_0$$, and let $$X$$ be the generic fibre of $$\mathcal{X}$$.

There is a natural injective map $$\mathrm{Br}(\mathcal{X}) \to \mathrm{Br}(X)$$. Let $$\alpha$$ lie in the image of this map, and let $$P$$ be a closed point of $$X$$. If $$R$$ is the integral closure of $$F[[t]]$$ in $$k(P)$$, then $$P$$ extends to an $$R$$-point of $$\mathcal{X}$$, and $$\alpha(P) \in \mathrm{Br}(k)$$ lies in the image of $$\mathrm{Br}(R)$$, which is trivial ($$R$$ is a Henselian local ring whose residue field is $$C_1$$). Therefore $$\alpha(P)=0$$. This holds for all closed points $$P$$, so $$\alpha$$ lies in the kernel of your map.

It remains to show that $$\mathrm{Br}(\mathcal{X})$$ is non-trivial. For any $$n$$ coprime to the characteristic of $$F$$, proper base change gives $$\mathrm{H}^2(\mathcal{X},\mu_n) \cong \mathrm{H}^2(X_0, \mu_n)$$. Then the Kummer sequence shows that $$\mathrm{Br}(\mathcal{X})[n] \to \mathrm{Br}(X_0)[n]$$ is surjective. In particular, $$\mathrm{Br}(\mathcal{X})$$ is non-trivial whenever $$\mathrm{Br}(X_0)$$ is non-trivial.

Remark: if you don't insist that $$X$$ is a curve, then things are much easier: take a surface over a finite field having non-trivial Brauer group.

Just for completeness: The "correct" way to understand the Brauer group of $$X$$ using its codimension $$1$$ points is via residue maps.

Specifically: Let $$X$$ be a regular integral noetherian scheme. Then for each codimension $$1$$ point $$x$$ of $$X$$, there is residue map $$\mathrm{Br}(\kappa(X)) \to H^1(\kappa(x), \mathbb{Q}/\mathbb{Z})$$, where $$\kappa(X)$$ denotes the function field of $$X$$ and $$\kappa(x)$$ the residue field of $$x$$. Then the sequence $$0 \to \mathrm{Br}(X) \to \mathrm{Br}(\kappa(X)) \to \bigoplus_{x \in X^{(1)}} H^1(\kappa(x), \mathbb{Q}/\mathbb{Z})$$ is exact, with the caveat that one exclude $$p$$-primary parts if $$\kappa(x)$$ has characteristic $$p$$ for some $$x$$. Moreover, if $$X$$ is a curve over a perfect field, then the sequence is in fact exact, i.e. one does not need to exclude any $$p$$-primary parts.

This is all a consequence of Grothendieck's purity theorem, and can be found in Section 6.8. of "Poonen - Rational points on varieties".

Here's an explicit example (joint with A. Landesman). Let $$k$$ be an algebraically closed field of characteristic not $$2$$ or $$3$$, and let $$X/k$$ be a non-supersingular K3 with Neron-Severi rank $$\geq 5$$. Then $$X$$ admits an elliptic fibration $$f: X \to \mathbf{P}^1$$.

Observation: Let $$\ell$$ be a prime not equal to $$\operatorname{char} k$$. Then there is an exact sequence $$0 \to \operatorname{Pic}(X) \otimes \mathbf{Z}_\ell \to H^2(X,\mathbf{Z}_{\ell}(1)) \to T_{\ell}\operatorname{Br}(X) \to 0.$$ By Hodge theory, the middle term has rank $$22$$ and therefore by the assumption that $$X$$ is not supersingular, $$\operatorname{Br}(X) \neq 0$$.

Let $$E$$ be the generic fiber of $$f : X \to \mathbf{P}^1$$, it is a smooth elliptic curve over $$K := k(\mathbf{P}^1)$$. We claim that $$\operatorname{Br}(E) \neq 0$$. Indeed, it is enough to show that $$\operatorname{Br}(X) \subseteq \operatorname{Br}(E)$$.

Let $$\alpha$$ be a Brauer class on $$X$$. If $$\alpha|_E = 0$$, then $$\alpha$$ must die on some open $$U \subseteq X$$. But $$X$$ is a regular integral scheme and so $$\operatorname{Br}(X) \hookrightarrow \operatorname{Br}(U)$$. Hence $$\alpha = 0$$ and we are done.