finite place of $k$ that splits in $k(\sqrt{D})$, then $C^f(k_v)$ is the union of two projective lines over $k_v$. Thus $C^f({\mathbb{A}}_k)=C^f({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$ is infinite, and hence so is $X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$. This construction gives a surface of simpler geometric structure than the surface in Proposition \[aa\], but it does not possess the stronger property of the previous example: here no element of $X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$ is contained in the Zariski open set $X\setminus X(k)$. For this argument we assume that the Jacobian of $E$ has rank $0$ and a finite Shafarevich–Tate group, e.g. $E$ is an elliptic curve over ${\mathbb{Q}}$ of analytic rank $0$. Let $(M_v)$ be an adelic point in $X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$. Its image $(N_v)$ in $E$ is contained in $E({\mathbb{A}}_k)^{\operatorname{Br}}$, but this set is just the connected component of $0=E(k)$ in $E({\mathbb{A}}_k)$, see [@Sk Prop. 6.2.4]. Thus for all finite places $v$ we have $N_v=0$. For any place $v$ that does not split in $k(\sqrt{D})$, this implies $M_v\in X(k)$.

The Hasse principle {#HP}
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In this section we construct a smooth projective surface $X$ over an arbitrary number field $k$ such that $X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$ is infinite and $X(k)$ is empty. This means that $X$ does not satisfy the Hasse principle and this failure is not 