Date: Tue, 28 Nov 1995 09:14:59 -0800 (PST)
From: [Permission pending]
To: Dave Rusin
Subject: Re: Galois group question
> In some appropriate sense (sorry, I forget the exact wording) almost
> every polynomial of degree 2n over Q has Galois group isomorphic to
> S_(2n). So take say almost any polynomial close to (x^2+1)(x^2+2)...(x^2+n)
> and you get what you want.
So, is it true that these polynomials
are dense in C^n, i.e. given (z_1, ..., z_n) in C^n you can find an
irreducible polynomial over Q with maximal Galois group, with roots
arbitrarily close to z_1, conj(z_1), ..., z_n, conj(z_n)? This is
again a consequence of the Hilbert Irreducibility theorem?
Interesting -- one might think that since complex conjugates
satisfy the same polynomials over the reals, that that would impose
some conditions on the Galois group.
==============================================================================
Date: Wed, 29 Nov 1995 11:49:37 -0800 (PST)
From: Gene Ward Smith
Subject: Re: Galois group question
To: Dave Rusin
On Wed, 29 Nov 1995, Dave Rusin wrote:
> I had mailed the same thing to Halloway, but I couldn't remember the
> argument or reference for this last point: how do you know that most
> choices of the a_i which yield irreducible polynomials do in fact give
> S_n as the group?
This is an application of Hilbert irreducibility, I think Fried and
Jarden has the general stiutation if I recall. Anyway, Hilbert applies
to more than just curves. Also, it is even a *generic* polynomial
for S_n. The condiftion for not giving S_n creates a variety, and
points not on the variety will give S_n, which is another way of
looking at it.
I seem to recall that the reason is that for each
> (maximal) subgroup of S_n the condition that the Galois group be
> contained in that subgroup imposes as Diophantine condition on the a_i
> (e.g., group <= A_n iff discriminant=square), but I don't really know
> the proof of such a statement except from seat-of-the-pants experience
> with small n. Do you have a reference?
Ooops--this is what I just said. The reference is, this is just
Galois theory! Another point of departure is the theory of invariants
of ordinary representations of finite groups--the algebra of invariants
characterizes the group (which is a permutation subgroup of S_n, and
hence also a linear rep, and hence has invariants=resolvents.) There
was a nice survey in the Bullitin back in '79 or thereabouts on
invariants of groups.
==============================================================================
Date: Wed, 29 Nov 1995 13:36:55 -0800 (PST)
From: [Permission pending]
To: Dave Rusin
Subject: Re: Galois group question
> the way I always remembered it was that getting the galois group to be
> less than s_n required some condition on the coefficients, e.g. the
> galois group is contained in a_n iff the discriminant is a square.
I was a tad puzzled about this A_n stuff -- how about cyclic Galois
groups etc. that aren't contained in A_n. Perhaps the transitive
proper subgroups of S_n that aren't in A_n are rather limited.
> >Interesting -- one might think that since complex conjugates
> >satisfy the same polynomials over the reals, that that would impose
> >some conditions on the Galois group.
>
> It just ensures that the galois group has a nontrivial element of order 2.
> I suppose that is something.
>
& various subgroups which are normalized by conjugation are contained
in subgroups of index 2 over them.
One thing you DO know is that if the minimal polynomial is of
degree 2*n, n odd, and it has no real roots, then the Galois group
isn't in A_(2n) because complex conjugation is an odd permutation.
Possibly the discriminant is imaginary then.
One other thing I was thinking about, is if you wanted to see
whether two roots of a polynomial generate the same extension
field, possibly this could be found out from the minimal
polynomial. If two roots a and b generate the same extension field
then s: a --> b is defined by s(a) = p(a), p(a) a polynomial in a,
so p(p(...(p(a))...)) = a, so either p is linear or the minimal
polynomial divides p composed with itself however many times. And
perhaps you could figure out if this is possible from the minimal
polynomial. And that would give you information on the Galois
group.
Doug Zare sent me this:
" Have you encountered
the Cebotarev density theorem? If you look it up, it might be listed as
Tsch-, Ch-, or Tch- instead of C-."
He seemed a bit vague about Slavic spelling, but maybe that is the
name of the "almost all Galois groups are maximal" theorem.