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!

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    $\begingroup$ 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. $\endgroup$ – rschwieb Sep 12 '17 at 18:52
  • $\begingroup$ Thanks, that is what I suspected. Please, do you know if a division ring is staby finite? en.wikipedia.org/wiki/Stably_finite_ring $\endgroup$ – user237522 Sep 12 '17 at 18:57
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    $\begingroup$ 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. $\endgroup$ – rschwieb Sep 12 '17 at 19:14
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    $\begingroup$ The answer is yes, since a division ring is Noetherian (it has no one-sided ideals other than 0 and itself). $\endgroup$ – 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!).

  • $\begingroup$ 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? $\endgroup$ – user237522 Sep 13 '17 at 11:23
  • $\begingroup$ Yes. just take any element of $B\setminus A$ and it obviously must be primitive. $\endgroup$ – Eric Wofsey Sep 13 '17 at 14:54
  • $\begingroup$ 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? $\endgroup$ – user237522 Sep 13 '17 at 19:42
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Eric Wofsey Sep 13 '17 at 19:45
  • $\begingroup$ Great. Thank you very much. $\endgroup$ – user237522 Sep 13 '17 at 19:48

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