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Let $K$ be a field and $D$ be a central division algebra over $K$ of degree $n$. Suppose that $L\subset D$ is a maximal subfield, so that $[L:K]=n$. Then we know that $L$ is a splitting field, so there exists an isomorphism $f:D\otimes_KL\to M_n(L)$ of $L$-algebras.

My question is: does it always exists an $f$ as above such that $f(L\otimes _KL)$ is the set of diagonal matrices in $M_n(L)$?

(asked in https://math.stackexchange.com/q/1619116/217671 but I think it might be more appropriate here).

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    $\begingroup$ The answer to your question is "no" in general. Indeed to have $L\otimes_K L\simeq L^n$ as $L$-algebras you need the extension $L/K$ to be Galois. Of course your affirmation holds under the assumption $L/K$ Galois : Anton's argument work in this case. $\endgroup$
    – Paul Broussous
    Commented Jan 20, 2016 at 10:26

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The answer is "yes" if the field extension $L/K$ is Galois : Anton's proof works in that case.

Classical examples where there exists a Galois degree $n$ extension $L/K$ embedded in $D$ are those of global fields (e.g. numbers fields) and local fields (completions of global fields). In this case the division algebras are moreover cyclic : one may choose $L/K$ cyclic.

In the case of a cyclic division algebra, you may even choose an explicit isomorphism $f$. Indeed take $L/K$ cyclic with generator $\sigma\in {\rm Gal}(L/K)$. By Skölem-Noether theorem, there exists $x\in D$, well defined up to a right factor in $L^\times$ such that $\sigma (l)=xlx^{-1}$, $l\in L$.

First considering $D$ as a right $L$-vector space, you have a canonical isomorphism of $L$-algebras:

$$ D\otimes_K L \simeq {\rm End}_L (D), \ d\otimes l \mapsto \{ u\mapsto dul\} $$

The family $(1,x,x^2 ,...,x^{n-1})$ is a basis of $D/L$. In this basis, the image of $L\otimes_K L$ takes a diagonal form.

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The invertible elements $L^\times$ are the rational points of a maximal torus, which is mapped to a torus, therefore each element is diagonalizable. Take a regular element $a\in L$, i.e., such that $L$ equals the cetralizer of $a$ in $D$. Then $d=f(a\otimes 1)$ is diagonalizable over the algebraic closure of $L$, hence over $L$. So diagonalize it and assume that $f(a\otimes 1)$ is diagonal. But it still is regular, so $f(L\otimes L)$ sits in its centralizer, which is the set of diagonal matrices. By dimension reasons the set $f(L\otimes L)$ equals the set of diagonal matrices.

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    $\begingroup$ To make your argument work you need to assume that $L/K$ is Galois. $\endgroup$
    – Paul Broussous
    Commented Jan 20, 2016 at 10:18
  • $\begingroup$ Thanks! So now I will state a related question would be: can you always find L such that L/K is Galois? I think these are called crossed product division algebras, is that correct? We can assume in this context that K is a number field, or a completion thereof. $\endgroup$
    – dbluesk
    Commented Jan 20, 2016 at 11:27

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