Assume that $A \subseteq B$ are division $k$-algebras, where $k$ is a field of characteristic zero. Further assume that $A$ has the IBN property (does a division ring 'automatically' have the invariant basis number property?).
One version of the primitive element theorem says that a finite separable field extension $F \subseteq E$ has a primitive element, namely, there exists an element $e_0 \in E$ such that $E=F(e_0)$.
Now, according to the comments in this question, the rank of $B$ as an $A$-module is well-defined.
Is there an analogue theorem to the primitive element theorem for such division algebras?
Namely, if $A \subseteq B$ has finite rank and if it is 'separable', then there exists $b \in B$ such that $B=A[b]$ (the finiteness of rank implies that $b$ is 'left' algebraic over $A$).
I am not sure if I know what 'separable' should mean. There is the notion of a separable algebra over a commutative algebra. It is easy to see that a division algebra $D$ over a field $k$ is separable, since the kernel of the canonical map $D \otimes_k D \to D$, $d_1 \otimes_k d_2 \mapsto d_1d_2$, is zero. Therefore, each of $A$ and $B$ is separable over $k$. What should be the meaning for $B$ separable over $A$?
Any comments are welcome!