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 $\mathrm{Br}(k)[2]$.

Local Fields, X.6). Looking at the case of a smooth cubic threefold, it seems to have dimension 4, since then it's just the dual of the ambient $(\mathbf{P}^4)$. $\endgroup$ – Martin Bright Sep 11 '15 at 15:09