24
$\begingroup$

Given an ring extension of a (commutative with unit) ring, Is it possible to give a "good" notion of "degree of the extension"?. By "good", I am thinking in a degree which allow us, for instance, to define finite ring extensions and generalize in some way the Galois' correspondence between field extensions and subgroups of Galois' group.

I suppose one can call a ring extension $A\subset B\ $ finite if $B$ is finitely generated as an $A$-module, and the degree would be the minimal number of generators, but is that notion enough to state a correspondence theorem?

Thanks in advance!

$\endgroup$
2
  • 21
    $\begingroup$ There's even a Galois theory of schemes, namely, the fundamental group of a scheme classifies the finite etale coverings of the scheme. When the scheme is affine, this becomes a Galois theory of rings. When the scheme is the spec of a field, it becomes classical Galois theory. The theory goes back to Grothendieck's seminar SGA1 from the early 1960s. $\endgroup$
    – mephisto
    May 3, 2011 at 0:09
  • 3
    $\begingroup$ Jacobson (1956) discusses Galois theory of rings of linear transformations. See www-history.mcs.st-and.ac.uk/Extras/… $\endgroup$ May 3, 2011 at 8:07

4 Answers 4

23
$\begingroup$

There is indeed a theory of Galois extension of rings. See, for example, the very nice paper [Chase, S. U.; Harrison, D. K.; Rosenberg, Alex. Galois theory and Galois cohomology of commutative rings. Mem. Amer. Math. Soc. No. 52 1965 15--33. MR0195922 (33 #4118)] The theory developed there does include a Galois correspondence.

There is even a Hopf-Galois theory, where the Galois group is replaced by a Hopf algebra (co)acting on the big ring, for extra fun---the correspondence in this case, though, is quite more delicate/complicated.

$\endgroup$
10
$\begingroup$

In addition to the above references, I would like to mention some non-commutative extensions of the Galois theory. See

P. M. Cohn, Skew Fields, Cambridge University Press, 1995

for the Galois theory of skew fields. Extensions to some classes of noncommutative rings are given in the book

V. K. Kharchenko, Noncommutative Galois theory, Novosibirsk, 1996,

available only in Russian, and many papers of its author, some of which exist also in the English translation.

$\endgroup$
4
  • 1
    $\begingroup$ The Hopf extension (and non-comm.) is discussed in S. Montgomery's Hopf algebras and their actions on rings. $\endgroup$ May 3, 2011 at 17:12
  • $\begingroup$ Since I do not read russian, this might be the right place where to ask: are you aware of a generalization of the norm in a Galois extension of noncommutative algebras? Of course, if one treats central simple algebras, there is a reduced norm, but I am in a more general setting: $k$ a field, $B/A$ a finite (Galois?) extension of noncommutative $k$-algebras; and would like something like $\mathrm{Norm}_{B/A}\colon B^\times\to A^\times$. $\endgroup$ Jan 23, 2015 at 8:36
  • 1
    $\begingroup$ @FilippoAlbertoEdoardo, re, the Dieudonné determinant suggests that maybe one should expect such a norm actually to be a map $B^\times_\text{ab} \to A^\times_\text{ab}$. $\endgroup$
    – LSpice
    Aug 5, 2023 at 4:32
  • 1
    $\begingroup$ @LSpice Thanks, I was not aware of the notion. I agree that this sounds much more reasonable. $\endgroup$ Aug 24, 2023 at 15:58
7
$\begingroup$

For a "survey" of Galois theory of commutative rings, there is one book:

The Separable Galois Theory Commutative Rings by Andy R. Magid (1974).

which has a nice section summarizing the state of the development up to 1974.

There is also a more general book aiming at a topos-theory style general Galois theory (although I haven't read it) including also a nice survey:

Galois Theories by Francis Borceux and George Janelidze (2001).

$\endgroup$
4
$\begingroup$

see related question

and SGA1 as well as Lenstra's notes

and Manjul Bhargava and Matt Satriano's paper

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.