noncommutative algebra - Linear disjointness of subfields of a centrally finite division algbera

If K and L are maximal subfields, then most of the time one would expect the map g to be injective, but not always. I will sketch how to construct examples of this form for which g is not injective.

We will take D to be a cyclic division algebra of index greater than two. Explicitly, let E/F be a cyclic Galois extension of degree n, with Galois group generated by $sigma$. Let $E[x]_{sigma}$ be the twisted polynomial ring, with multiplication defined by $xa=(sigma a)x$.

Pick $tin F^times$ and let D be the quotient of $E[x]_{sigma}$ by the central element xⁿ-t. Let K be obvious copy of E in D, and let L be the conjugate of K by an element of the form $l_0+l_2x^2+cdots+l_{n-1}x^{n-1}$ with all $l_i$ non-zero (yes, there is deliberately no linear term. This is the crux of the construction). I claim that this gives the desired counterexample.

To show this, it suffices to tensor with E, and show that the product map from $(Kotimes E)otimes(Lotimes E)$ to $Dotimes E=M_n(F)$ is not injective. K embeds via D into M_n(F) via $amapsto diag(a,sigma a,ldots,sigma^{n-1}a)$ and hence $Kotimes E$ is the space of diagonal matrices. Pick the obvious basis k₁,...,k_n of this space, and let l₁,...,l_n be the conjugate basis of $Lotimes E$.

Then (exercise), one can find indices i and j such that $k_il_j=0$. This yields the desired non-injectivity.