# Primitive element theorem for division rings extension

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 $$A$$?

• does a division ring 'automatically' have the invariant basis number property? Yes. The proof is the same as the fact that finite dimensional vector spaces have a unique dimension. – rschwieb Sep 12 '17 at 18:52
• Thanks, that is what I suspected. Please, do you know if a division ring is staby finite? en.wikipedia.org/wiki/Stably_finite_ring – user237522 Sep 12 '17 at 18:57
• At the article you cite, it says: noetherian rings and artinian rings are stably finite, and that does include division rings. Did you read it? In fact it could say one-sided Noetherian or one-sided Artinian. – rschwieb Sep 12 '17 at 19:14
• The answer is yes, since a division ring is Noetherian (it has no one-sided ideals other than 0 and itself). – user237522 Sep 12 '17 at 19:15

I don't think there's a theorem of this sort. In particular, consider $k=A=\mathbb{R}$ and $B=\mathbb{H}$. Then $B$ is a separable algebra over $k=A$ (there is no question of what this means since $A$ is commutative), but it does not have a primitive element. More generally, if $k$ is any field and $B$ is a finite dimensional noncommutative division algebra whose center is $k$, then $B$ is a separable $k$-algebra but cannot have a primitive element over $k$ (otherwise it would be commutative!).
• Thanks! I like your answer, but I still wonder what happens if $A$ is non-commutative. Anyway, in the special case of rank two, is there a primitive element? – user237522 Sep 13 '17 at 11:23
• Yes. just take any element of $B\setminus A$ and it obviously must be primitive. – Eric Wofsey Sep 13 '17 at 14:54
• More generally, in the special case of rank prime number, also $B=A[b]$, for every $b \in B-A$; this follows from math.stackexchange.com/questions/2428065/…. Am I right? – user237522 Sep 13 '17 at 19:42