Twists of cubic threefolds Let k be a field of characteristic $0$.


Let $X$ be a variety over k which is isomorphic to a smooth cubic threefold over $\bar{k}$. Then is $X$ isomorphic to a smooth cubic threefold over $k$?


For motivation, let's consider some other cases. 


*

*For cubic curves, the analogous question has a a negative answer in general, as elliptic curves can have quite complicated twists. 

*For cubic surfaces however the answer is yes. This is because cubic surfaces are embedded by the anticanonical divisor, which is already defined over $k$.


This latter argument does not work however for cubic threefolds, as for a cubic threefold, the hyperplane section is half the anticanonical divisor. So the question is whether the anticanonical divisor is always divisible by $2$ in the Picard group of $X$.
Note that the Hochschild-Serre spectral sequence yields the exact sequence
$$0 \to \mathrm{Pic}(X) \to \mathrm{Pic}(X_\bar{k})^{\mathrm{Gal}(\bar{k}/k)}=\mathbb{Z} \to \mathrm{Br}(k),$$
so the obstruction to $X$ being isomorphic to a cubic threefold over $k$ is given by an element of $\mathrm{Br}(k)[2]$.
 A: As pointed out by Noam Elkies, my comment above can be turned into a positive answer to the question.
In the exact sequence

$\DeclareMathOperator{\Pic}{Pic}\DeclareMathOperator{\Br}{Br}0 \to \Pic X \to (\Pic \bar{X})^{G_k} \to \Br k$,

the arrow $\delta \colon (\Pic \bar{X})^{G_k} \to \Br k$ sends the class of an effective Cartier divisor $D$ to the class of the Severi--Brauer variety whose $\bar{k}$-points are isomorphic to the projectivisation of $\mathrm{H}^0(\bar{X},\mathcal{O}(-D))$.  See Serre, Local Fields, X.6 for a passing mention of this fact, and Grothendieck, Le groupe de Brauer, III, 5.4 for more details (though he doesn't actually prove it, saying "on (plus précisément, J. Giraud) vérifie que la classe de cet élément dans $\Br(Y)$ est bien $\delta(\xi)$.")
In the particular example here, $\Pic X$ is of index $2$ in $(\Pic \bar{X})^{G_k}$, so the image of $\delta$ is killed by $2$.  On the other hand, taking $D=-K_X/2$, the vector space $\mathrm{H}^0(\bar{X},\mathcal{O}(-D))$ has dimension $5$, since we are assuming that $\bar{X}$ is isomorphic to a cubic threefold, on which $D$ is the class of a hyperplane section.  It follows (for example Serre, loc. cit.) that $\delta(D) \in \Br k$ is killed by $5$.  So $\delta(D)=0$, and $D$ lies in $\Pic X$.
A: As noted above the obstruction is given by some $A \in Br(k)[2]$, and this obstruction vanishes if $X$ has a $k$-rational point. Since $X$ has a rational point over some odd degree extension, $A$ must become trivial over an odd degree extension. This means that $A$ is already trivial over $k$.
