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Suppose the base field is $\mathbb{C}$ and the Clifford algebra is the classical one (i.e. associated to a quadratic form in $n$ variables). It seems that there are relations between Clifford algebras and Azumaya algebras?

I guess this from an indirect source, that is, the results of Kuznetsov on derived categories of quadrics and Caldararu's result on twisted derived categories . A quick search shows that when $n=3$, Clifford algebra is also an Azumaya algebra (see the paper "On the Clifford algebra of a binary cubic form" for example). However, is there any result about the general case?

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    $\begingroup$ Over $\mathbb R$, Clifford algebras give super Azumaya algebras and make up the super Brauer group. I learned this here: math.ucr.edu/home/baez/week212.html I'm not sure about the situation over $\mathbb C$. $\endgroup$
    – Will Sawin
    Sep 9, 2015 at 2:23
  • $\begingroup$ Over any field, the structure of Clifford algebras is described in terms of central simple algebras (i.e., Azumaya algebras) in Chapter 9 of Bourbaki's Algebra (Sect 9, no 4). $\endgroup$
    – anon
    Sep 9, 2015 at 13:10

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