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Replies: 6   Last Post: Feb 4, 2013 7:45 PM

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 David Bernier Posts: 3,887 Registered: 12/13/04
Posted: Feb 4, 2013 3:01 AM

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

--
dracut:/# lvm vgcfgrestore
File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID
993: sh
Please specify a *single* volume group to restore.

Date Subject Author
2/3/13 David Bernier
2/3/13 quasi
2/4/13 David Bernier
2/4/13 quasi
2/4/13 David Bernier
2/4/13 quasi
2/4/13 Leon Aigret