Talk:Galois group

Typo?

Hello,

I'm far from being certain of what I'm going to say but isn't there a problem when saying: "automorphisms of E/F - that is, isomorphisms α from E to itself"

Shouldn't it be "automorphisms of E - that is, isomorphisms α from E to itself" or "automorphisms of E/F - that is, isomorphisms α from E/F to itself"

Randomblue 22:34, 24 July 2007 (UTC)

The punctuation was wrong (I've fixed it now). By an automorphism of E/F we mean an automorphism of E that fixes F pointwise. --Zundark 07:24, 25 July 2007 (UTC)

I think this is wrong because at Field extension#Normal, separable and Galois extensions the term "automorphisms" refers to "field automorphisms" and then Aut(L/K) is formed by those which share the property of fixing the elements of F:

Given any field extension L/K, we can consider its automorphism group Aut(L/K), consisting of all field automorphisms α: L → L with α(x) = x for all x in K.

The article Automorphism#Examples also says

In the case of a Galois extension L/K the subgroup of all automorphisms of L fixing K pointwise is called the Galois group of the extension.

So, again, the automorphisms are just field automorphisms and those which fixes K are the elements of the Galois group. Helder (talk) 22:51, 16 June 2010 (UTC)

The group of automorphisms of the extension E/F (denoted Aut(E/F)) consist of the automorphisms of E that fix the elements of F pointwise. If E/F is Galois, then you call this the Galois group. And when E/F is not Galois it's simply called the automorphism group of E/F. The group of automorphisms of E is denoted Aut(E) and consists of all field automorphisms of E. RobHar (talk) 03:41, 17 June 2010 (UTC)

Gal and Aut

I note that the author has been careful to use Aut instead of Gal in the case of R/Q. However, are we sure that Gal is appropriate in the line below ("Gal(C/Q)"). I am only knowledgeable on Galois theory of finite extensions and may be overlooking some aspect of the theory that develops infinite extensions, but C is not the splitting field for any polynomial of any degree (even countably infinite) over Q. Is it possible that Gal here is a typo and should be Aut as in the case of R? —Preceding unsigned comment added by Gtg207u (talk • contribs) 05:42, 7 December 2007 (UTC)

I think C/Q is a Galois extension according to the definition given in that article. --Zundark (talk) 15:41, 7 December 2007 (UTC)

It fails even the basic precondition of being an algebraic extension. -- EJ (talk) 16:46, 7 December 2007 (UTC)

True - I hadn't noticed that someone (you) had added that restriction to the first paragraph. The condition is still not present in the definition itself - perhaps you should add it there too. --Zundark (talk) 17:55, 7 December 2007 (UTC)

Yes, I added the restriction. I thought it was simply omitted: the definition without the restriction was at odds with the standard usage of the term (to the best of my knowledge), as well as its usage in other WP articles (e.g., profinite group, glossary of field theory, or this very article). In particular, the fundamental theorem of Galois theory breaks badly under the unrestricted definition, which makes the definition kind of pointless (IMHO), and directly contradicts statements to the contrary in several WP articles.

I've reformulated the definition now to make it more explicit. -- EJ (talk) 15:07, 10 December 2007 (UTC)

"It can be shown that E is algebraic over F if and only if the Galois group is pro-finite." is a nice fact from this article, so one might want to decide what a galois extension is like when there are transcendental elements. Obviously we cannot replace galois group with field automorphism group, so something in the fact depends on the extension being galois. Perhaps K=k[x] is galois over k, and F is galois over k if F is galois over K and K is galois over k?

At any rate, the Aut/Gal confusion is widespread, so be careful in what might seem like minor changes. JackSchmidt (talk) 18:00, 7 December 2007 (UTC)

The statement you mention is certainly puzzling. As it is written, it actually requires every field extension to have a Galois group. That's probably unintended, but even in the reading "a Galois extension is algebraic iff bla bla" it's unclear what is meant by a Galois extension. I suspect it's actually possible the statement is in error, so at any rate I'd like to see a clear reference for that. -- EJ (talk) 15:07, 10 December 2007 (UTC)

I checked the page history, and the fact was in the original version of the page. I think for any field k, Aut(k(x)/k)=PGL(2,k). For k=Z/2Z it is finite and so pro-finite. If there are *any* non-algebraic extensions that are allowed to be "galois", then I would think k(x)/k would be. For infinite fields, PGL(2,k) is infinite and rarely pro-finite, so perhaps that is what was meant. Making such a statement (true and) precise would be fairly difficult, I think. JackSchmidt (talk) 17:34, 10 December 2007 (UTC)

Oh and to be clear, I am not trying to say the definition of Galois extension needs to achieve the maximum level of abstruse generality to apply equally and meaninglessly to all cases, just that when changing (or clarifying) its generality to be careful of all the places it might break. Honestly, I think the fact from Galois group should have just indicated more clearly what sort of context it meant, and that Galois extension should primarily apply to finite dimensional extensions, with infinite dimensional algebraic extensions as a generalization, and whatever the fact intends as yet another example of how to extend the definition. JackSchmidt (talk) 18:18, 7 December 2007 (UTC)

I agree, though it's worth pointing out that the generalization to infinite algebraic extensions works literally the same way as for finite extensions in many cases (including the definition itself), so one can as well state such stuff directly for all algebraic extensions, without separating the finite-dimensional case first. -- EJ (talk) 15:07, 10 December 2007 (UTC)

Beneficial to mention a certain theorem?

Does anyone think it might benefit to mention (I believe it's called) the Conjugation Isomorphism Theorem? It basically says that if two numbers a and b algebraic over a field, then a mapping defined between the field extensions obtained by ajoining a and b which sends $\Sigma _{i}c_{i}a^{i}$ to $\Sigma _{i}c_{i}b^{i}$ is an isomorphism if and only if a and b are conjugate over the original field. This theorem has some useful corollaries.LkNsngth (talk) 05:11, 6 April 2008 (UTC)

Examples

How about getting a lot more examples into this page ? —Preceding unsigned comment added by 188.26.49.57 (talk) 23:53, 30 March 2010 (UTC)

Adding examples to this article is definitely a good idea, and by all means do so if you feel that you can. However, remember that Wikipedia is not a textbook, and thus it is important to include only the most significant examples of Galois group computations in Galois theory. For instance, while it may not be advisable to add examples of the Galois group of arbitrary polynomials (with rational coefficients), it is useful to add an example of the Galois group of an extension being the symmetric group on n letters, for instance, where n is a natural number.

That said, the most important task at this stage is to write a strong opening for the article, add a good set of references, and include a more comphrensive treatment of the theoretical aspects of Galois theory. (As I said, if you know this material well, feel free to improve the article; I do not think anyone will object as long as your changes are encyclopaedic and accurate.) PST 04:21, 16 April 2010 (UTC)

Automorphism in definition

I find the definition slightly confusing. The second sentence reads: "An automorphism of E/F is defined to be an automorphism of E that fixes F pointwise." However, the general definition of automorphisms says nothing about any "pointwise fixing". Thus it would be perfectly possible (I think) to define an automorphism of E/F such that no such pointwise fixing occurs. Shouldn't then the sentence in question read: "Let us define an automorphism of E/F such that F is fixed pointwise"? The following third example would then explain what it means, and then the fourth sentence would read "The set of all SUCH automorphisms..." etc. Regards, Borys (27.03.2011) —Preceding unsigned comment added by 80.213.206.177 (talk) 12:53, 27 March 2011 (UTC)

We need to make it clear that this is the definition of an automorphism of a field extension, not just something that we are defining. An automorphism of an algebraic object is required to fix the distinguished elements (constants / nullary operations), so this definition is the expected one anyway - for what else is E/F, if not E with the elements of F considered as distinguished elements? --Zundark (talk) 13:39, 27 March 2011 (UTC)

Abandoned user draft

In 2014 a new user made substantial changes to a copy of this page at User:Wjxzhang/sandbox.

Please would an interested editor assess the usefulness of that work, incorporate what is appropriate into this article, and leave a note here when done? – Fayenatic London 14:11, 25 June 2020 (UTC)