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

  • $\begingroup$ 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, 2018 at 17:08
  • 1
    $\begingroup$ It obviously doesn't, I meant the product. I corrected, thanks. $\endgroup$ Oct 8, 2018 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, 2018 at 17:43
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    $\begingroup$ If $X$ is a conic over the real numbers without a real point then the map is clearly not injective. $\endgroup$
    – naf
    Oct 9, 2018 at 4:27

3 Answers 3


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.


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