Let $X$ be a del Pezzo surface over a number field $k$. (A *del Pezzo surface* over $k$ is a smooth, projective, geometrically connected surface whose anti-canonical class $K_X$ is ample.) Let $d := K_X^2$ be the *degree* of $X$. It is well-known that $d$ satisfies $1 \leq d \leq 9$.

If $d \geq 5$, and $X(k) \neq \emptyset$, then $X$ is birationally equivalent to $\mathbb{P}^2_k$. Moreover, if $d=5$ or $d=7$, the condition on $X(k)$ is automatically satisfied. (This is Theorem 9.4.7 in Bjorn Poonen's Rational Points on Varieties.) In other words, for del Pezzo surfaces of degree $5$ and higher, the only obstruction to $k$-rationality is the possible lack of $k$-rational points.

My question concerns the "first" (i.e. highest-degree) non-trivial case, as far as rationality is concerned, namely the case where $d=4$. In this case, there are examples of $X$ with $X(k)\neq \emptyset$ but where $X$ is non-rational. An obstruction to rationality is given by the (non-trivial part of the) *Brauer group* $\operatorname{Br}(X)$ of $X$: that is, if $\operatorname{Br}(X)/\operatorname{Br}(k) \neq 0$, then $X$ is not $k$-rational. (To be sure, such $X$ exist, even among those $X$ that have rational points. This is where the $d=4$ case differs from the higher degree cases.) Indeed, this simply follows from the fact that $\operatorname{Br}$ is a birational invariant of smooth, projective, geometrically connected varieties, and from the fact that $\operatorname{Br}(\mathbb{P}^2_k) = \operatorname{Br}(k)$. My question is:

Is the converse true? That is, if $X$ is a del Pezzo surface of degree $4$ over a number field $k$ with $X(k)\neq\emptyset$, and $\operatorname{Br}(X) = \operatorname{Br}(k)$, does it follow that $X$ is a $k$-rational surface?

no. But if you want at least a partial result in this direction, see theorem 3.36 in Wittenberg's bookIntersections de deux quadriques et pinceaux de courbes de genre 1. $\endgroup$2more comments