Back

The first two terms of the sequence factor as $Out(M_n(A)$, $H^1(A,{\mathbb{G}}_m)$ is just a cute nickname for the class goup of $A$, and the pointed set $H^1(A,G{L}_n)$ classifies locally-free $A$-modules of rank $n$.
The map $Cl(A)\to H^1(A,G{L}_n)$ comes from the map sending a unit of $A$ to the corresponding scalar matrix. Hence, it sends the class of a fractional ideal $I$ to the isomorphism class of $I\oplus I\dots \oplus I$.

3/n

By the theory of projective modules over a Dedekind domain, this sum is simply isomorphic to $A\oplus A\dots \oplus A\oplus I^n$, and it is free if and only if $I^n$ is principal.
Hence, by exactness of the sequence, the $n$-torsion of the class group is isomorphic to the image of $PG{L}_n(A)$, which is in turn isomorphic to $Out(M_n(A)$.

3/n

Even though this all looks like nasty homological algebra, the maps involved are quite explicit:

An outer automorphism of $M_n(A)$ is simply conjugation by a matrix $P$ in $G{L}_n(k)$, where $k$ and $P{M}_n(A)P^{-1}=M_n(A)$.
The ideal class of $A$ associated to $P$ is the class of the fractional ideal generated by the coefficients of $P$.

It is even possible to construct inverse images of ideal classes, see https://doi.org/10.1515/CRELLE.2006.098 (sorry, I don't have an open access url).
4/n

I really like this result, but always had the impression that it is not so well known. I guess I just thought that people should know...

Please let me know if you knew about it, if you like it, if you know (or can think of) an application of it, or if you have anything to say about this.
5/5

@montessiel - Thanks, I really like this stuff! I didn't know it, but that doesn't mean much since I'm just a raw beginner in class field theory.

I don't quite understand why the image of ${\mathrm{P}\mathrm{G}\mathrm{L}}_n(A)$ in the ideal class group $H^1(A,{\mathbb{G}}_m)$ is the $n$- torsion in the ideal class group. Let's see, it must be because this image is the kernel of the next map, to $H^1(A,{\mathrm{G}\mathrm{L}}_n)$. (1/n)

@montessiel - So it seems you're claiming that the $n$-fold direct sum $I\oplus \cdots \oplus I$ is a free $A$-module of rank $n$ iff $I$ represents an $n$-torsion element of the ideal class group. And I guess that's right. Okay.

Nice!

(2/n, n = 2)

@johncarlosbaez Yes, this follows from the definition and properties of the Steinitz class of a projective module over a Dedekind domain.
Essentially, writing things in rank 2 for simplicity, we have that
$I\oplus J\simeq A\oplus IJ$ for ideals $I$ and $J$, and also that $A\oplus I\simeq A^2$ if and only if $I\simeq A$.

@montessiel @johncarlosbaez What I don't catch is why ${\mathrm{P}\mathrm{G}\mathrm{L}}_n(A)/{\mathrm{G}\mathrm{L}}_n(A)$ corresponds to outer automorphisms of $M_n(A)$. I know that all automorphisms are inner if $A$ is a field, but I have no idea about what those automorphisms could be in general.

@antoinechambertloir @johncarlosbaez If $A$ is an integral domain with field of fractions $K$, an automorphism of $M_n(A)$ extends to one of of $M_n(K)$, and is therefore conjugation by some $P\in G{L}_n(K)$. Now, if $P$ is, up to multiplication by a scalar in $K^{\times }$, in $G{L}_n(A)$, this is an inner automorphism. However, it is possible that $\lambda P\notin G{L}_n(A)$ for any $\lambda \in K^{\times }$. In this case, conjugation by $P$ is in fact an outer automorphism.
1/n

@antoinechambertloir @johncarlosbaez If $A$ is a local ring, the Skolem-Theorem applies and every automorphism of $M_n(A)$ is inner. If $A$ is not local, this might not be the case. But then we can see $PG{L}_n$ as a group sheaf for the Zariski/Etale/fppf topology.
Concretely, you can have a matrix $P\in G{L}_n(K)$ that is not colinear to any matrix of $G{L}_n(A)$, but such that for any prime $\mathfrak{p}$, it is colinear to some matrix of $G{L}_n(A_{\mathfrak{p}})$.
2/n

@antoinechambertloir @johncarlosbaez In this case, you still have $P{M}_n(A)P^{-1}=M_n(A)$ and you get a non-inner automorphism.
If you think of it this way, you can really see the cohomology map $Aut(M_n(A))\to Cl(A)$: take $(f_i{)}_{1\le i\le n}$ such that, as ideals, $(f_1,...,f_n)=A$, and such that there exists ${\lambda }_i$ such that ${\lambda }_{i}P\in G{L}_n(A(f_i^{-1}))$.
3/n

@antoinechambertloir @johncarlosbaez
Now, the data of these ${\lambda }_i$ forms a cocycle and yield a class in $H_{zar}^1(A,A^{\times })$ which gives an ideal class $[I]\in Cl(A)$. In fact, this is the class of the ideal generated by the coefficients of $P$.
If you put things this way, what I was originally talking about is the fact that the ideal $I^n$ is principal.
Another neat consequence is that the $n$-th power of any automorphism of $M_n(A)$ is inner.
4/4

@antoinechambertloir @johncarlosbaez One last correction about an unfortunate choice of notation in toot 3/n:
I talked about a family $(f_i{)}_{1\le i\le n}$. The size $n$ of the family is completely unrelated to the degree of the matrix algebra $M_n(A)$ on which we have been working.

@lievenlebruyn I was happily surprised when I saw you interact with this post, because this note is part of the litterature I read when I first discovered this fact.
I remember that you mentioned non-commutative algebraic geometry as a motivation in this note (and in your PhD manuscript, I believe?)
I care about maximal orders and Azumaya algebras in my research and I was wondering if learning some non-commutative geometry would help. Do you know if this approach is still relevant today?

@montessiel Sure, maximal orders and non-commutative *Algebraic* geometry are very relevant today, see the work of Michel Van den Bergh (and others) on non-commutative desingularisations (Google for 'non-commutative crepant resolutions'). Michel's approach is more homological than mine (using representation varieties). A readable (imo) account (though also almost 20 years old) are my 3 talks on noncommutative geometry@n :
http://matrix.uantwerpen.be/lieven.lebruyn/LeBruyn2003e.pdf
Please ask if you want more references.

@lievenlebruyn Thank you for the references.
I will certainly need some time to digest this, but I will know to ask again if I have more questions.