Re: Galois theory: how to fix this proof? Does it need fixing?


Please excuse the slow reply! I haven't slept properly for the
last three weeks, I keep dozing off at odd moments, and I haven't
found much time (or, to be honest, much will) for thinking about
mathematics today - but I'd better make some effort to reply!
Maybe tomorrow I'll actually understand what I'm doing ... 8-P

On Fri, 19 Sep 2008 01:03:22 +0200, Jannick Asmus
<jannick.newsREMOVE_ME@xxxxxx> wrote:

----- Original Message -----
From: Angus Rodgers <twirlip@xxxxxxxxxxx>
Sent: Sep/18/2008 8:09:24 PM +0200

The proof also divides into two cases, depending on whether F does
or does not contain a primitive $p$th root of unity, for a certain
prime p: whose definition, to my mind, is part of the problem with
the proof.

[...]

Because G = Gal(E/F) is soluble, it has a normal subgroup H of
prime index, p. (This is proved in the appendix on group theory.)

[...]

I must at least say briefly that it seems to be necessary to
subdivide case 2 further, into:

(2a) E contains a primitive $p$th root of unity: call it omega.

(2b) E does not contain a primitive $p$th root of unity.

[...]

The problem arises in case (2), where he defines F^* = F(omega)
and E^* = E(omega), notes that E^*/F^* is a Galois extension, and
defines G^* = Gal(E^*/F^*).

By the way, I find it simplest to think of w as any primitive $p$th
root of unity adjoined to E, and then F(w)is simply the subfield of
E(w) generated by F and w.

[...] establishes an injection G^* -> G; and he
then notes that G^*, being isomorphic to a subgroup of a soluble
group, is soluble.

If we knew that G^* had a normal subgroup of index p, we would
be fine.

Here the benefits of the different cases you have introduced are quite
evident:

In case (2a) we have F(w) is contained in E=E(w) such that |G*| < |G|.
Now apply the induction hypothesis to G*.

OK, no problem there.

Case (2b) can be approached with the following lemma:

Let K|k and L|k be two finite extensions such that [K:k] and [L:k] are
coprime, then [LK:k]=[K:k].[L_k] - where LK denotes the compositum of L
and K in some algebraically closed field containing K and L.

.... or in any common extension of K and L. I didn't know the lemma,
and it's not in Rotman's book, but I looked it up in Dummit & Foote,
and its proof is elementary and short. So far, so good.

This lemma applies to the situation above with k = E^H, K = E, L =
E^H(w), since [E:E^H] = [G:H] = p and [E^H(w):E^H]<p are coprime. In
particular, the extension E(w)|E^H(w) has degree p.

I'm having a lot of trouble making sense of this. (Apologies if it's
just because I'm having a bad day - which I definitely am!)

Let M = E^H. Then [M:F] = [G:H] = p. But you're writing [E:M] = p.
Help!

I expect there's a good (even "obvious") reason why [M(w):M] < p, but
I'm not seeing it. (Apologies again - bad day, etc. etc.)

Ignoring these difficulties, I see we would have:

E(w)
/ \
/ \
M(w) E
/ \ /
/ \ /
F(w) M
\ /
\ /
F

where [E:M] = p and [M(w):M] = m, say, where p does not divide m.

Then certainly, by the lemma you mention, [E(w):M] = mp, therefore
[E(w):M(w)] = p. But as I say, I cannot yet accept either of the
premises of this argument (although I expect to be kicking myself
over at least one of them, as soon as I post this).

It is normal and
separable (here you need that p is prime to the characteristic of the
fields involved), hence Galois. This gives you a normal subgroup H* in
G* of index p.

(Excuse me if I don't try to verify this yet, until I've overcome
the earlier difficulties.)

[...] The proof is definitely flawed in at least one minor
detail: he says that a certain polynomial is "necessarily separable",

Hmm, everything is separable and the only polynomial in sight is the
minimal polynomial of w (over some of the fields involved) which is
separable due to the assumption that p is prime to the characteristic.

(You mention this again in your other post; I'll deal with it there.)

--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.

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