2
$\begingroup$

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!

$\endgroup$
  • 5
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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!).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.