Splitting of a division algebra with an involution of second kind

Question

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 $L$).

Answer 1

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.

Answer 2

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.

Answer 3

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.)