TITLE: Brauer group of a curve over non-algebraically closed field
QUESTION [12 upvotes]: 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 $\text{Br}(X)=0$.

REPLY [3 votes]: 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.