Recall that an finite-dimensional algebra $A$ over a field $k$ is central simple iff there is an iso

$A \otimes_k A^{op} \cong M_n(k)$

where $A^{op}$ is the opposite ring and $M_n(k)$ is the matrix ring.

On the other hand, a finite field extension $K / k$ is Galois iff there is an iso

$K \otimes_k K \cong K^{Aut_k(K)}.$

These two statements strike me as similar and I am wondering:

  1. Is there a Galois theory for non-commutative extensions in which the central simple algebras take the role of the Galois extensions in the usual Galois theory? Or, is there a theory which subsumes these two facts as some special cases?
  2. Is there some sort of absolute "non-commutative" Galois group which features automorphisms of all non-commutative extensions of $k$ at a time? I could imagine this group $Gal_{nc}(k)$ to map (on?)to the usual group $Gal(k)$.

Thank you!

  • $\begingroup$ Your second statement is not correct: take $k=\mathbb{R}$ and $K=\mathbb{C}$ $\endgroup$ – Ofra Mar 20 '16 at 20:11
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    $\begingroup$ @Ofra: yes it is. The superscript denotes taking a product, not taking fixed points. $\endgroup$ – Qiaochu Yuan Mar 20 '16 at 20:54

Let $k$ be a field. Say that a $k$-algebra $A$ is separable if any of the following equivalent conditions holds (it is not obvious that they are equivalent):

  1. $A$ is projective as an $(A, A)$-bimodule.
  2. $A$ is geometrically semisimple in the sense that $A \otimes_k L$ is semisimple for any field extension $L$ of $k$.
  3. $A \otimes_k A^{op}$ is semisimple.

Theorem: The separable $k$-algebras are precisely the finite products of matrix algebras over finite-dimensional division algebras $D$ over $k$ whose centers $Z(D)$ are separable (in the usual sense) field extensions of $k$.

Hence the classification of separable algebras up to Morita equivalence is a mix of usual Galois theory and the classification of central simple algebras: the ones which are indecomposable with respect to product are classified by pairs of a finite separable extension $k \to L$ and a class in the Brauer group $\text{Br}(L)$ of $L$.

Over an arbitrary base $k$ the correct generalization requires an additional finiteness hypothesis: an algebra $A$ is 2-dualizable if $A$ is finitely generated projective both as an $(A, A)$-bimodule and as a $k$-module. I think the classification of these is a mix of the classification of finite etale $k$-algebras and the classification of Azumaya algebras. The terminology comes from topological field theory: a 2-dualizable $k$-algebra defines a 2-dimensional topological field theory taking values in the Morita 2-category of $k$-algebras, bimodules, and bimodule homomorphisms.

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  • $\begingroup$ As for the question about getting a bigger Galois group, the relevant group in this story is still the usual Galois group, but it's acting on more interesting things (categories instead of sets). $\endgroup$ – Qiaochu Yuan Mar 20 '16 at 6:30
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    $\begingroup$ Could you extend a bit this comment ? (or pointing to a reference ?) Do we know a classification for those algebras ? $\endgroup$ – Simon Henry Mar 20 '16 at 9:12
  • $\begingroup$ Is it possible that in 1. above it should be projective as a $k,k$-bimodule? Or otherwise how does it depend on $k$? $\endgroup$ – Jakob Mar 20 '16 at 13:09
  • $\begingroup$ Dear Quiaochu, thank you for your answer! Can you please provide a reference for the statements above? Also, can you elaborate more on your comment that this "non-commutative" Galois group is the same as the usual one? $\endgroup$ – Jakob Mar 20 '16 at 13:10
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    $\begingroup$ @Jakob: I have a reference now: qchu.wordpress.com/2016/03/27/separable-algebras $\endgroup$ – Qiaochu Yuan Mar 31 '16 at 3:32

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