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Topic:
about the KroneckerWeber theorem
Replies:
6
Last Post:
Feb 4, 2013 7:45 PM




Re: about the KroneckerWeber theorem
Posted:
Feb 4, 2013 3:01 AM


On 02/03/2013 08:03 PM, quasi wrote: > 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).
I have a further question about conjugate roots ...
The nontrivial 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.



