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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 $L$).

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You can find this in either Scharlau's book Quadratic and hermitian forms' or in The book of involutions'. –  Matthew Stover May 12 '10 at 14:56
    
Is the point of the question to allow arbitrary $k$? (For arithmetically interesting fields -- local, global -- it seems clear from structure of Brauer groups without even requiring mention of $\sigma$.) –  BCnrd May 12 '10 at 15:06
    
@Matthew: The book of involutions is a big book. Do you recall roughly where in the book this is discussed? –  BCnrd May 12 '10 at 15:07
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@BCnrd: Exactly, the point is to allow an arbitrary field $k$ of any characteristic. –  Mikhail Borovoi May 12 '10 at 17:09
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Now that I have the references more handy, I can be more specific. I would recommend Chapter 8 of Scharlau for the basic question of the field splitting the algebra. He works in full generality. The book of involutions doesn't deal with this directly. However, for the functorial properties of the unitary group, their section on group schemes and Hopf algebras is quite good. –  Matthew Stover May 13 '10 at 15:00
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3 Answers 3

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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.

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Very nice. Since you wish to work with arbitrary fields (e.g., allowing char. > 0), perhaps better to say $K_s$ rather than $\overline{K}$, and to say $L$ is a commutative etale $K$-subalgebra of $D$ (not just a commutative $K$-algebra) so that you even get $F/k$ to be separable. –  BCnrd May 14 '10 at 4:14
    
@BCnrd: Edited! –  Mikhail Borovoi May 14 '10 at 7:54
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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.

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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.)

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