Date: Feb 4, 2013 3:01 AM Author: David Bernier Subject: Re: about the Kronecker-Weber theorem On 02/03/2013 08:03 PM, quasi wrote:

> David Bernier<david250@videotron.ca> wrote:

>>

>> The Kronecker-Weber theorem characterizes abelian extensions

>> of Q.

>>

>> If we look at p(X) = X^3 - 2 over Q, then according to

>> Wikipedia, the splitting field L of p over Q is

>>

>> Q(cuberoot(3), -1/2 +i*srqrt(3)/2)

>>

>> where -1/2 +i*srqrt(3)/2 is a non-trivial third root of unity.

>>

>> By Artin, because L is a splitting field,

>

> Yes.

>

>> L is a Galois extension of Q.

>

> Yes.

>

>> So L is an abelian extension of Q.

>

> No, L is not an abelian extension of Q.

>

>> <http://en.wikipedia.org/wiki/Splitting_field#Cubic_example>.

>>

>> Then L is an extension of degree 6 (as a vector field over Q)

>> of Q.

>

> Yes.

>

>> By Galois theory, the automorphisms of L fixing Q form a group

>> of order 6.

>

> Right.

>

>> By the Kronecker-Weber theorem, L isn't an abelian extension.

>

> Right, L isn't an abelian extension of Q.

>

>> But we have a non-abelian group of order 6 ...

>

> Right.

>

>> So I guess the automorphism group of L (which fix Q) is

>> isomorphic to S_3, the symmetric group on three objects.

>>

>> So, is this right?

>

> Yes.

>

>> Some automorphisms:

>> (a) identity

>> (b) complex conjugation

>>

>> Supposedly, there should be 4 more automorphisms of L leaving

>> Q invariant.

>>

>> Perhaps cuberoot(3) can be sent to either of

>> cuberoot(3)*(-1/2 +i*srqrt(3)/2),

>> cuberoot(3)*(-1/2 -i*srqrt(3)/2)?

>

> Yes.

>

>> Anyway, finding and constructing these automorphisms of L

>> doesn't look too easy.

>

> It's not hard.

>

> For each element of L\Q, an automorphis of L over Q must

> send that element of L to one of its conjugates (over Q).

I have a further question about conjugate roots ...

The non-trivial third roots of unity

-1/2 +i*srqrt(3)/2 and -1/2 -i*srqrt(3)/2 are complex conjugates.

I don't know of a definition where, for example, in the setting

above,

2^(1/3) is said to be conjugate to

2^(1/3) * (-1/2 +i*srqrt(3)/2).

I looked at Conjugate (group theory) at Wikipedia here:

http://en.wikipedia.org/wiki/Conjugate_%28group_theory%29 .

David ?ernie?

> To get an automorphism of L over Q,

>

> send 2^(1/3)

> to any of the 3 cube roots of 2

>

> and send -1/2 +i*srqrt(3)/2

> to any of the 2 nontrivial cube roots of 1

>

> That yields 6 choices (the above choices are independent),

> and for each of those 6 choices there is a uniquely determined

> automorphism of L over Q.

>

>> But I guess people require an extension to be Galois so that the

>> fundamental theorem of Galois theory applies ...

>

> Right, Galois extensions are needed for certain theorems.

>

> quasi

--

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