Talk:Galois theory

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Quadratic example[edit]

I have a minor beef with the "Example of a quadratic equation" section. Roots should be thought of algebraicly and the discussion of symmetries in a graph is misleading. For instance, we can permute the three roots of x^3-2 but it doesn't correspond to any reflections or rotations of a graph. Maybe this should be rewritten? dunkstr 01:13, 25 May 2004 (UTC)

done --Dmharvey 14:04, 27 May 2005 (UTC)

The quadratic example (x - 2)(x - 1) is not clear, at least to me. The commentary says that there is only one permutation; to me it looks like there are two, (1,2)-> (1,2) and (1,2) -> (2, 1). Am I missing something? 2001:558:6011:1:38F3:171E:261:EBEA (talk) 19:56, 29 June 2017 (UTC)

One of the roots satisfies the equation

x-1=0,

which does not remain true if the root are exchanged. Thus the permutation that exchanges the roots does not belong to the Galois group, which, here, is reduced to the identity permutation. D.Lazard (talk) 20:14, 29 June 2017 (UTC)

Algebraic Equation[edit]

The first example says "Furthermore, it is true, but far less obvious, that this holds for every possible algebraic equation satisfied by A and B." And I thought, this clearly isn't true for the equation "A * A + B = c". But then I realised that that equation isn't algebraic because c contains sqrt(3). Anyway, my point is that the example is confusing to the non-mathematician who doesn't know the technical definition of algebraic equation. Algebraic equation currently redirects to Algebraic Geometry, which also isn't very helpful because it talks about Algebraic equations without defining them. I think wikipedia needs a simple article defining algebraic equations with links to and from this article and algebraic geometry. But I'm not qualified to do it. Reilly 15:02, 13 October 2005 (UTC)

On a related subject the piece currently reads:

'One might raise the objection that

A

and

B

are related by yet another algebraic equation,

A-B-2{\sqrt {3}}=0

,

which does not remain true when A and B are exchanged. However, this equation does not concern us, because it does not have rational coefficients; in particular, √3 is not rational.'

But this last sentence cannot be correct can it?

a) Can galois theory really not handle complex coefficients?

b) isn't the real reason that that equation isn't part of the Galois group since it cannot be permuted?

WolfKeeper 19:05, 2 May 2006 (UTC)

I've had a stab at this: Algebraic equation

Reilly 17:55, 5 July 2006 (UTC)

Consider reading the article more than once.

"An important proviso is that we restrict ourselves to algebraic equations whose coefficients are rational numbers. (One might instead specify a certain field in which the coefficients should lie, but for the simple examples below, we will restrict ourselves to the field of rational numbers.)"

Consider F=Z[x]/(x^2+1). There are two automorphisms identity, and a+bx->a-bx. Here we are working over the field of integers. The equation x^2-1=0 has two solutions A=i and B=-i. However the equation A-B=2i implies that there is only the identity automorphism (since A and B cannot be interchanged). This is because the equation A-B=2i has coefficents in F which are not in the integers.

-kaz — Preceding unsigned comment added by 70.73.85.118 (talk) 04:34, 2 November 2006 (UTC)

Kaz, I am not satisfied with your answer. The link by Reilly also does not help. I support the claim by WolfKeeper. In which sense it "does not remain true when

A

and

B

are exchanged". What does in mean, "to remain true", while the equation has no solution among pairs of rational numbers? It cannot "remain true" because, for any rational

A

and

B

, it was not true. This example is not clear, we should either rewrite or remove it. dima (talk) 11:41, 11 October 2008 (UTC)

P.S. Is it possible to suggest some example, where

A-B

appear (disabeling the swapping), but still remain in the set of rational numbers? dima (talk) 11:50, 11 October 2008 (UTC)

Guys, I think it's simpler than that.

The initial constraints within the polynomial that define A and B are:

A+B=4

and

AB=1

.

This has two solutions for [A,B].

For a _specific_ solution of [A,B], it happens to also be true that

A-B-2{\sqrt {3}}=0

. For that same _specific_ solution, it happens to also be true that

A=2-{\sqrt {3}}

.

But neither of those relations are not true for _all_ solutions of [A,B]. That is why these relations are not to be considered. The fact that this particular polynomial happens to have irrational roots is coincidental.

As your discussion is over a decade old, I think I'll go ahead and make this change shortly unless I hear back. The point was very confusing to me when reading this article and it hurts readability. Jurgen Hissen (talk) 20:00, 7 May 2020 (UTC)

Looks awful on a Macintosh computer[edit]

This article really looks awful on a Mac, Mac OS X, Version 10.5.5

Lots failed to parse. But looks great on a Sun Ray (at Sun Microsystems as I write)!

dino (talk) 21:30, 14 January 2009 (UTC)

Now looks fine on a Mac[edit]

Now looks OK on a Mac. What it was I don't know -- maybe a temporary local weirdness.

dino (talk) 22:21, 15 January 2009 (UTC)

Solvable[edit]

The definition of solvable given in the article is wrong. Cyclic should be replaced with abelian. 159.115.238.165 (talk) 00:15, 8 December 2009 (UTC)

There's nothing wrong with it. You are presumably referring to the definition of solvable group as one having a subnormal series with abelian factors. Now, (for finite groups) it does not make a difference whether you use "abelian" or "cyclic" in this definition, as every finite abelian group is a direct sum of cyclic groups. Note that the (equivalent) definition in this article is still different, it asks for a composition series with cyclic factors, not just subnormal series; since factors of a composition series are simple by definition, it cannot have noncyclic abelian factors, thus it would be pointless to refer to general abelian groups here. Moreover, in the context of Galois theory it makes much more sense to refer to cyclic instead of abelian even for subnormal series, as radical field extensions correspond to cyclic factors. — Emil J. 12:41, 8 December 2009 (UTC)

Discriminant is not a symmetric function in the coefficients[edit]

But only in the roots, or am I missing something? —Preceding unsigned comment added by 87.4.176.2 (talk) 21:39, 1 January 2010 (UTC)

You're not missing anything, it was indeed wrong. — Emil J. 14:06, 4 January 2010 (UTC)

Solvable groups and solution by radicals not clear[edit]

thanks for writing this great article

the last sentence of the first paragraph in the section "solvable groups and solution by radicals" is very very long and includes two "if"

It's much too confusing for me! can you please rewrite it. —Preceding unsigned comment added by 84.229.239.105 (talk) 20:28, 19 March 2011 (UTC)

The great triumph[edit]

The article says "One of the great triumphs of Galois Theory was the proof that for every n > 4, there exist polynomials of degree n which are not solvable by radicals—the Abel–Ruffini theorem. This is due to the fact that for n > 4 the symmetric group Sn contains a simple, non-cyclic, normal subgroup, namely An."

But the criterion for solvability is that the factor groups of the normal composition chain are cyclic, not that the normal subgroup in the chains themselves are cyclic.

According to me the real reason why Sn is unsolvable, is because its subgroup An is unsolvable, because that one has no normal subgroups anymore which could lead to cyclic factor groups and the chain ends right there without resolving to {1}.

It's been too long since I've studied group theory so perhaps there is equivalence here, but for the Galois layman with some undergraduate math skills this argument is not intelligible I'm afraid. — Preceding unsigned comment added by 194.78.35.195 (talk) 10:49, 27 February 2013 (UTC)

The fact that An is simple and normal implies, by the definition of simple that (1) ⊂ An ⊂ Sn is a composition series. The fact that it is simple and non-cyclic implies that it is not solvable. Thus the formulation of the article and yours are both correct.

However, there is something else that is wrong in this paragraph: The "great triumph" of Galois is not that one, which was not a surprise after Abel-Ruffini result. It is that Galois gave an algorithmic criterion to decide which equations are solvable or not. This is better explained at the end of section "History". D.Lazard (talk) 12:17, 27 February 2013 (UTC)

Vague phrase[edit]

Under the heading "Inverse Galois problem", the vague phrase "For that" is used. The word "that" might refer to the easy construction, the ground field or something else. — Preceding unsigned comment added by 31.53.53.179 (talk) 06:25, 6 September 2018 (UTC)

Fixed. This led me to further edit the section. D.Lazard (talk) 07:29, 6 September 2018 (UTC)

up to sign?[edit]

Pre-history says coefficients are symmetric functions of the roots "up to sign"; looks to me they are functions of the roots either "up to a common factor", or, assuming the highest-order coefficient is fixed to 1, uniquely defined by the roots. Am I missing something? 12.104.156.31 (talk) 00:28, 22 October 2019 (UTC)

The assertion in the article is "the coefficients of a monic polynomial are (up to sign) the elementary symmetric polynomials in the roots". This is cocorrect, as "monic" means that the highest-degree coefficient is 1 (you can follow the link). D.Lazard (talk) 08:15, 22 October 2019 (UTC)

ANHDUC - WIKI 2

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