The Brauer group is a set

Question

When one presents the Brauer group of a field $F$, it is usually said that the group's elements are "equivalence classes of finite dimensional central simple algberas over $F$ under the Brauer equivalence relation".

Now, in this statement it is implicitly said that this object is indeed a set - but usually no explanation for this statement is given.

Notice that by the Wedderburn-Artin Theorem and the Brauer equivalence, it is enough to show that the object "all division algebras over a field" forms a set. So my question is, is this true for any field and why?

If the question is not set theoretic, don't tag it with set theory tags. — Asaf Karagila, Jul 12 '15 at 13:19
This is very interesting, can you link or cite some source where I can learn more about these concepts of Brauer groups? — mathreadler, Jul 12 '15 at 13:24

Martin Brandenburg · Accepted Answer · 2015-07-12T13:13:07.810

9

The algebras in that context are all finite-dimensional vector spaces, so that these vector spaces may be assumed to be in the set $\{K^n : n \in \mathbb{N}\}$. There is only a set of algebra structures on the vector space $K^n$, because it corresponds to a subset of $\hom(K^n \otimes K^n,K^n)$.

edited Jul 12 '15 at 13:13

answered Jul 12 '15 at 12:27

Martin Brandenburg

163,620

But, each vector space can be given with different multiplicaion. – Mike Jul 12 '15 at 12:31
3

Sure. But there are not so many of them. – Zhen Lin Jul 12 '15 at 12:34

The Brauer group is a set

1 Answers1