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

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)?

• I assume you want conditions on $K$ and $L$ that make this work for all $G$? For your example, $L$ need only be separably, not necessarily algebraically, closed. Another case where this can be done is when $K$ is a Henselian field, and $L$ is its strict Henselisation. This is in Bruhat–Tits 2. I'm almost positive it's not true in general, but don't have an example off the top of my head. – LSpice Jun 9 '17 at 18:53
• A silly example is when $K$ is finite (or, more generally, of cohomological dimension $\le 1$), at least if $L/K$ is algebraic; for then, if $S$ is a maximal $K$-split torus in $G$, then $C_G(S)_{/L}$ is a torus, hence contains a unique maximal split sub-torus, which is therefore $\mathrm{Gal}(L/K)$-stable, hence defined over $K$. – LSpice Jun 9 '17 at 19:00

• $K = \mathbb Q_5$, 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.