Let $k$ be a field, $K/k$ a separable quadratic extension,
and $D/K$ a central division algebra of dimension $r^2$ over $K$
with an involution $\sigma$ of second kind
(i.e. $\sigma$ acts non-trivially on $K$ and trivially on $k$).
Does there exist a field extension $F/k$ such that $L:=K\otimes\_k F$
is a field, and $D\otimes\_K L$ splits (i.e. is isomorphic to the matrix algebra $M\_r(L)$ over $L$)?


Motivation: Let $h\in D$ be a Hermitian element ($h^\sigma =h$), and let
 $G$ be the $k$-group with
$G(k)=${$g\in D^\times\ | \ ghg^\sigma=h$}.
I want to find a field extension $F/k$ such that $G\times\_k F$
is a unitary group over a field $L$ (and not over a division algebra over $L$).



Splitting of a division algebra with an involution of second kind
I answer my own question. The answer is *yes*.


Since there are no non-trivial division algebras over finite fields,
we may assume that $k$ and $K$ are infinite.
Let $H=${$h\in D\ |\ h^\sigma=h$} denote the $k$-space of Hermitian elements of $D$.
Consider the embedding $D\hookrightarrow M\_r(\bar K)$ induced
by an isomorphism $D\otimes\_K \bar K\simeq M\_r(\bar K)$.
An element x of $D$ is called semisimple regular,
if its image in $D\otimes\_K \bar K\simeq M\_r(\bar K)$
is a semisimple matrix that has $r$ different eigenvalues.
A standard argument using an isomorphism
$D\otimes\_k \bar K\simeq M\_r(\bar K)\times M\_r(\bar K)$
shows that there is a dense Zariski open subset
$H\_{reg}$ consisting of semisimple regular elements in $H$.
Clearly $H\_{reg}$ contains $k$-points.


Let $h\in H\_{reg}$ be a semisimple regular Hermitian element.
Let $L$ be the centralizer of $h$ in $D$.
Since $h$ is Hermitian ($\sigma$-invariant), the $k$-algebra $L$ is $\sigma$-invariant.
Since $h$ is semisimple and regular, the algebra $L$ 
is a commutative étale $K$-subalgebra of $D$
of dimension $r$ over $K$ (we calculate in $D\otimes\_K K\_s$).
Clearly $L$ is a field, $[L:K]=r$.
Since $L\subset D$ and $[L:K]=r$, the field $L$ is a splitting field for $D$,
see e.g. Scharlau, Quadratic and Hermitian Forms, Ch. 8, Thm. 5.4.


Since $L\supset K$, we see that $\sigma$ acts non-trivially on $L$.
Let $F$ denote the subfield of fixed points of $\sigma$ in $L$,
then $[L:F]=2$ and $[F:k]=r$.
Clearly $F\cap K=k$ and $FK=L$, hence $L=K\otimes\_k F$.
The extension $F/k$ is separable.


Another version of the proof vas proposed by Uzi Vishne.


