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Let K and L be two subfields of a non-commutative division algebra D with the center Z. Suppose that K and L contain Z and that D is finite dimensional over Z. Let V be the tensor product of K and L over Z, viewed as a vector space over Z. Consider the linear map g from V to D defined by g(x \otimes y) = xy, for every x in K and y in L. My question is this:

Under what condition(s) on K and L the map g is injective?

What is clear is that the intersection of K and L has to be Z. But that is not sufficient in general. Would this necessary condition be also sufficient if K and L were maximal subfields?

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2 Answers 2

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 $t\in F^\times$ 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 $(K\otimes E)\otimes(L\otimes E)$ to $D\otimes E=M_n(F)$ is not injective. K embeds via D into Mn(F) via $a\mapsto diag(a,\sigma a,\ldots,\sigma^{n-1}a)$ and hence $K\otimes E$ is the space of diagonal matrices. Pick the obvious basis k1,...,kn of this space, and let l1,...,ln be the conjugate basis of $L\otimes E$.

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

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Do you assume that elements of $K$ and $L$ commute ?

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