# Dimension of division rings extension

Let $A \subseteq B$ be two division $k$-algebras, where $k$ is a field of characteristic zero. I am not sure if I wish to further assume that $B$ is affine over $A$, namely, if $B$ is finitely generated as an $A$-algebra.

If $A$ and $B$ happen to be commutative, namely fields, then the dimension of the field extension $A \subseteq B$ is naturally defined as the dimension of $B$ as a vectore space over $A$. (Actually, it is enough that $A$ is a field, and $B$ can be a non-commutative division ring).

I wonder if there is a 'natural' way to define the dimension of $A \subseteq B$, when both $A$ and $B$ are non-commutative?

Perhaps this depends on the specific division rings $A$ and $B$? What about Gelfand-Kirillov dimension? What about global dimension? See also this question.

Any ideas are welcome. Thanks!

• The closest analogue to the commutative case you describe would be the rank of B over A in the case B is a free A-module and the invariant basis number property holds. – Kimball Sep 11 '17 at 0:54
• @Kimball, thanks for your comment. I like it and I guess my question has no 'unique' answer, so I would also like to hear additional ideas (if we do not know that $B$ is a free $A$-module etc.). – user237522 Sep 11 '17 at 11:32
• A module over a division algebra is always free. Assuming either a set theory with the axiom of choice or that the module is finitely generated. The proof from the field case works. Differences come from the fact you need to be careful about the difference between left and right modules. For example $\{(1,i),(j,k)\}$ are linearly independent vectors in $\Bbb{H}^2$ when that is viewed as a left $\Bbb{H}$-module, but when viewed as a right $\Bbb{H}$-module they become linearly dependent: $(j,k)=(1,i)j$. See this answer by Matt Emerton for more. – Jyrki Lahtonen Sep 11 '17 at 15:30
• @JyrkiLahtonen, thank you very much for your comment. So if we wish to talk about the 'dimension'/rank of $B$ over $A$ (when $A$ and $B$ are division rings), then we only need to know that $A$ has the IBN property? – user237522 Sep 11 '17 at 15:55
• Hmm. I have always thought that all division algebras automatically have IBN. If I'm wrong please educate me! – Jyrki Lahtonen Sep 11 '17 at 16:01