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

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 quasi Posts: 12,067 Registered: 7/15/05
Posted: Feb 3, 2013 8:03 PM

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).

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

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