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Galois Theory and Cyclic Extensions

Date: 04/28/2005 at 19:01:04
From: Jessica
Subject: Galois Theory and cyclic extensions

Let a be an element of the algebraic closure of the rational numbers 
(Q{bar}), but not in the rational numbers, and let F be a subfield 
of Q{bar} that is maximal for the property that a is not in F.  Prove
that every finite extension of F is cyclic.

I think I have a start on this problem, but I keep getting stuck.  One
of the only facts I know about cyclic extensions is that if F(a) > F
is finite and a^n is in F for some n, then the extension is cyclic. 
So, I'm trying to use that.  If K is any finite extension of F, then 
K contains F(a).  If I can show a^n is in F (which I can't), 
then F(a) > F is cyclic.  Does that imply that K > F is cyclic?

Am I on the right track?  Thank you!



Date: 04/28/2005 at 23:25:12
From: Doctor Wilkinson
Subject: Re: Galois Theory and cyclic extensions

Hi, Jessica. 

This problem seems to me to be quite difficult, but I can show that
F(a) is cyclic.  Maybe that will help you, and you will see something
that I don't.  I claim first that F(a) is a normal extension of F. 
This is so because if a' is any conjugate of a, F(a') contains a and
therefore contains F(a), but has the same degree over F as F(a) does,
and is therefore equal to F(a).  But now the Galois group of F(a) has
no proper subroups because there are no fields between F and F(a).  So
the Galois group of F(a) over a must be cyclic of prime order.  But I
don't know how to get from that to the conclusion in the problem.

- Doctor Wilkinson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/05/2005 at 04:37:40
From: Doctor Jacques
Subject: Re: Galois Theory and cyclic extensions

Hi Jessica,

Unless otherwise specified, we will only use finite extensions and 
finite groups (this will save some typing).

Note first that it is enough to prove the theorem for (finite) Galois 
extensions.  If E is any extension, and H is the Galois closure of E, 
and if we know that G(H/F) is cyclic, and therefore Abelian, all 
subgoups are normal, and all intermediate extensions are Galois, so
E = H after all--all finite extensions are Galois.

Note first that Doctor Wilkinson's argument can be used to prove a 
little more: if p = [F(a) : F] (p is prime), F(a) is the only 
extension of degree p of F, since any such extension contains F(a), 
and the degrees are the same.

This implies that the degree of any extension is a multiple of p.  In 
fact, we can say a little more.

Assume that there is a Galois extension E of degree m*p^k, with
gcd(m,p) = 1.  A Sylow p-subgroup of G(E/K) has index m, and the 
corresponding fixed field is an extension of degree m.  That extension 
has to contain F(a), by maximality, and its degree must therefore be 
divisible by p--this is a contradiction.  The conclusion is that the 
degree of any Galois extension is a power of p (and, of course, this 
is therefore true for any extension).  In particular, the Galois 
groups are p-groups.

If E is any Galois extension, G(E/F) is a p-group with only one 
subgroup of index p, since F(a) is the only extension of degree p.

We now move the focus to group theory.  It is enough to prove that a 
finite p-group with only one subgroup of index p is cyclic, or, 
equivalently, that a non-cyclic finite p-group has more than one 
subgroup of index p.  I'll let you try to prove that--here are a few 
hints:

* Induction on the order of the group is always a good thing to try.

* If G is an Abelian, non-cyclic p-group, it is the direct product of
  at least two cyclic p-groups; in that case, you can extract a
  subgroup of index p from either of those subgroups, and that will
  give you different subgroups of G.

* If G is any non-Abelian group, its center Z(G) is a proper (normal)
  subgroup, and the quotient group G/Z(G) is not cyclic.

* If G is a finite p-group, its center Z(G) is not the trivial
  subgroup.

(Please write back if you require further assistance on this).

A few additional remarks:

F is a very interseting field--all irreducible polynomials have degree 
p^k for some k.

F contains the primitive p-th roots of unity, since they satisfy an 
equation of degree at most p-1.  (This is required to use the theorem 
you mention about cyclic and radical extensions.)  By induction, F 
contains all primitive roots of order p^k.

I don't know (although I have a hunch that it is true) if there are 
extensions of degree p^k for all k (the theorem is still true if this 
is not the case); in particular, it could  (?) be possible that
F(a) = Qbar.  This is certainly not always the case; for example, if
a = sqrt(2), you should be able to show that there is a maximal 
extension F such that F does not contain a and F(a) does not contain 
2^(1/4).

Here is a trick question: if a = 2^(1/6), what is the degree of
[F(a) : F] ? (Obvious candidates are 2 and 3.)

Does this help?  Write back if you'd like to talk about this some 
more, or if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Modern Algebra

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