Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


quasi
Posts:
11,911
Registered:
7/15/05


Re: about the KroneckerWeber theorem
Posted:
Feb 3, 2013 8:03 PM


David Bernier <david250@videotron.ca> wrote: > >The KroneckerWeber 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 nontrivial 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 KroneckerWeber theorem, L isn't an abelian extension.
Right, L isn't an abelian extension of Q.
>But we have a nonabelian 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



