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I want to know the number of non-isomorphic quaternion algebras over a non-Archimedean local field $K$. What is the number of non-isomorphic central simple algebras of dimension $n^2$ over a non-Archimedean local field $K$?

I know the Brauer group of $K$ is isomorphic to $\dfrac{\mathbb{Q}}{\mathbb{Z}}$. I know the structure of the group $\dfrac{\mathbb{Q}}{\mathbb{Z}}$ very well, and it has only one element of order $2$.

Let $n \in \mathbb{N}$ be arbitrary. Is there any relation between the elements of order $n$ (or elements of order dividing $n$) in the group $\dfrac{\mathbb{Q}}{\mathbb{Z}}$, and the central simple algebras of dimension $n^2$?

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  • $\begingroup$ I think the order (squared) of a CSA in the Brauer group divides the dimension (unless I have things backwards) but they aren’t always equal. Perhaps look in Serre first since if there is a simple relationship it would very likely be there $\endgroup$
    – user208649
    Commented Oct 2, 2020 at 22:32

1 Answer 1

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The elements of order $n$ in $\frac{\mathbb{Q}}{\mathbb{Z}}$ correspond bijectively to the isomorphism classes of central simple algebras over $K$ of dimension $n^2$. In particular there is a unique isomorphism class of (non-split) quaternion algebra. See Remark 4.4 on p. 110 here for an explicit construction.

https://www.jmilne.org/math/CourseNotes/CFT310.pdf

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    $\begingroup$ So can we conclude that there are exactly two equivalence classes of non-split algebras of dimension $9$? Am I right? $\endgroup$ Commented Oct 3, 2020 at 7:40
  • $\begingroup$ @Tirelessandhardworking this is probably not helpful given that it's been two years and a week, but yes, you are right. $\endgroup$
    – hunter
    Commented Nov 10, 2022 at 16:51

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