TITLE: Obstruction to rationality of del Pezzo surfaces of degree 4
QUESTION [7 upvotes]: 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?

REPLY [5 votes]: I will supplement Dan's nice answer by claiming that the answer is almost always no.  Specifically, almost every degree 4 del Pezzo surface over the rational numbers with a rational point is non-rational but has trivial Brauer group.
To support this, I wrote a little Magma program that picked two random quadrics in $\mathbb{P}^4$ containing the point $(1:0:0:0:0:0)$ and with integer coefficients in $[-5,5]$.  Let $X$ be the intersection of these two quadrics.  I computed the Fano scheme of lines on $X$.  If the Fano scheme is irreducible over $\mathbb{Q}$, that is, the lines are all Galois-conjugate, then a calculation shows that the Brauer group is trivial and the surface is not rational.  I ran the program 100 times and, apart from 1 time when $X$ was singular, this was always the case.  I'm sure it would be straightforward to show that the Fano scheme is always irreducible outside a thin set of the space parametrising such pairs of quadrics.