12
$\begingroup$

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 $\text{Br}(X)=0$.

$\endgroup$
  • $\begingroup$ It is also trivially injective for $k$ finite (Albert-Brauer-Hasse-Noether). Why does $\mathrm{Br}(X)$ map to the direct sum? $\endgroup$ – user19475 Oct 8 '18 at 17:08
  • 1
    $\begingroup$ It obviously doesn't, I meant the product. I corrected, thanks. $\endgroup$ – Giulio Bresciani Oct 8 '18 at 17:11
  • 2
    $\begingroup$ You can BTW replace the Brauer groups of the residue fields by the Brauer groups of the Henselisations or completions of the local rings. $\endgroup$ – user19475 Oct 8 '18 at 17:43
  • 4
    $\begingroup$ If $X$ is a conic over the real numbers without a real point then the map is clearly not injective. $\endgroup$ – naf Oct 9 '18 at 4:27
8
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

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".

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.