Central division algebras and splitting fields

Question

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 $M_n(L)$?

Answer 1

A sketch suggesting that the answer is affirmative, if we make the extra assumption that $L/K$ is Galois (with Galois group $G$ of order $n$). If you knew this beforehand, and were only looking for a general argument, then I apologize.

Namely, in that case it is known that $D$ can be written as a crossed product. IOW there is a factor set $k$ such that $D\cong(L,G,k)$. This means that $D$ has a basis $\{u_s\mid s\in G\}$ such that the products of those basis elements correspond (modulo the factor set, i.e. bunch of non-zero constants from $L$) to the product in $G$. Furthermore, $L\subset D$ is the $L$-span of $u_1$. See section 8.4. of Jacobson's Basic Algebra II for the relevant theory.

All this allows us to view $D$ as a $n^2$ dimensional $K$-space $V(D)$ of $n\times n$ matrices over $L$ in such a way that $L=Lu_1$ is exactly the set of diagonal matrices.

I would be surprised if it turned out that the obvious $K$-bilinear mapping $D\times L\to M_{n\times n}(L)$ would not give the kind of mapping from $D\otimes_KL\to M_{n\times n}(L)$ that you want. Observe that in $V(D)$ the elements of $L$ of appear along the diagonal as their various Galois conjugates, so only the elements of $K$ are scalar matrices.

Answer 2

@JyrkiLahtonen has already argued that the answer is 'yes' if $L/K$ is Galois. In general, the answer is 'no'. Indeed, suppose that $L \ne K$ and $L/K$ is purely inseparable. (@Skip at MO gives an explicit construction of an algebra for which this happens.) Suppose that $f(L \otimes_K L)$ consists of semisimple elements. (This may be automatic—I don't know—but, if it fails, then certainly $f(L \otimes_K L)$ doesn't equal the ring of diagonal matrices!) If $x \in L$, then $f(x \otimes_K 1)$ satisfies the polynomial $X^{p^n} - x^{p^n} = (X - x)^{p^n}$ for some non-negative integer $n$, hence, by semisimplicity, must be scalar.