Splitting of a division algebra with an involution of second kind 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 $L$).
 A: 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.
A: You find the answer to the question in "The Book of Involutions". More precisely, in the first  5 lines of the proof of Lemma 10.27.
A: This is related to the splitting of the unitary group $SU(B, \sigma)$, a topic which is addressed in "Generic Splitting of Reductive Groups," by Kersten and Rehmann. Specifically, Corollary 6.3 produces a field which seems to satisfy your desired properties.
I would think that you could take the function field of $R_{E/k}(SB(B))$, the Weil restriction of the Severi-Brauer variety of $B$. That would definitely split $B$, but I am not sure why you would still have a field when you take the tensor product with $E$. (This may be what they do in the paper I cited, but I couldn't follow the notation.)
