Let $K$ be a field, let $L$ be a field containing $K$, and let $G$ be a reductive group over $K$. Does there always exist a torus $T$ of $G$ so that $T\_{/L}$ is a maximal split torus of $G\_{/L}$? If such a torus does not exist in general, are there assumptions on $K$ and $L$ that guarantee that such a torus does exist (aside from the well-known case of $L$ algebraically closed)?



Does a reductive group over $K$ always have a torus that becomes maximal split over $L$?
Suppose that


* $K = \mathbb Q\_5$, say;
* $L$ is a cubic, totally ramified extension of $K$; and
* $G$ is the algebraic group underlying the multiplicative group of a cubic division algebra over $K$.


Then $G$ is $L$-split. On the other hand, because the splitting field of a torus in $G$ is a Galois extension of $K$, and the only intermediate field $L/L'/K$ with $L'/K$ Galois is $L' = K$, any subtorus of $G$ that becomes split over $L$ is already split.


