Symmetric Galois group / irred poly
Fernando Maranel
Just working through an exercise in Steven Roman's "Field Theory" (p.282)
and have become somewhat stuck. We are asked to prove that if p(x) is a
degree n polynomial over Q [rationals] with Galois group G isomorphic to S_n
[Symmetric group] then f(x) is irreducible.
I've googled and come across a number of proofs which use the transitive
property (that G acts transitively on the roots of the polynomial or some
such) but I suspect there is a simpler way of showing the above result (and
would prefer there to be because I'm not particularly fond of group theory
and didn't really understand the transitive proof). I might be way off but
can we not assume p is not irreducible and deduce a contradiction on the
order of G (=n!) or some such? Any ideas how to proceed or other
suggestions if is this not possible?
Many thanks.
Arturo Magidin
It depends a bit on what you already know; do you know that the Galois
group of the polynomial is the same as the Galois group of the
splitting field of the polynomial over Q, and that the order of the
Galois group is equal to the degree of this extension?
If so, then basically, the degree of the splitting field over Q is
equal to the order of the Galois group. If g(x) is irreducible of
degree k, then the splitting field of g has degree at most k! over Q
(prove it). So if f(x) = f_1(x)*...*f_r(x) is a factorization into
irreducibles of degrees d1,...,dr (with d1+...+dr = n, of course),
then the splitting field of f(x) would have degree at most d1!
d2!...dr! over Q. Show that for this to equal (d1+...+dr)! you must
have r=1.
--
Arturo Magidin
Fernando Maranel
group of the polynomial is the same as the Galois group of the
splitting field of the polynomial over Q, and that the order of the
Galois group is equal to the degree of this extension?
# Firstly, thank you for your reply and help.
# Yes, I know these two facts.
If so, then basically, the degree of the splitting field over Q is
equal to the order of the Galois group. If g(x) is irreducible of
degree k, then the splitting field of g has degree at most k! over Q
(prove it).
# I think this just follows from the fact G is isomorphic to a subgroup of
the symmetric group S_n and since the order of any subgroup of S_n divides
|S_n| = n!, the splitting field of g has a degree which divides n! and
therefore is no more than n!.
So if f(x) = f_1(x)*...*f_r(x) is a factorization into
irreducibles of degrees d1,...,dr (with d1+...+dr = n, of course),
then the splitting field of f(x) would have degree at most d1!
d2!...dr! over Q. Show that for this to equal (d1+...+dr)! you must
have r=1.
# To show this can we just assume that d1=d2=...=dr=1 and then if r is not 1
we have 1=x! where x is bigger than 1 and therefore RHS is clearly not equal
to 1. I couldn't prove this without fixing the d's. Would that be okay?
Thanks again.
Arturo Magidin
Me:
> So if f(x) = f_1(x)*...*f_r(x) is a factorization into
> irreducibles of degrees d1,...,dr (with d1+...+dr = n, of course),
> then the splitting field of f(x) would have degree at most d1!
> d2!...dr! over Q.
You'll need Dedekind's Theorem here, that if K<L<M are fields, then
[M:K]=[M:L][L:K], by the way.
> Show that for this to equal (d1+...+dr)! you must
> have r=1.
>
> # To show this can we just assume that d1=d2=...=dr=1 and then if r is not 1
> we have 1=x! where x is bigger than 1 and therefore RHS is clearly not equal
> to 1.
I do not see why you can "just assume that d1=d2=...=dr=1". How would
this prove the general case?
>I couldn't prove this without fixing the d's. Would that be okay?
I don't think so, nor do I see why you are having trouble proving it
in general. Here's an easy example; suppose r=2. You need to compare
d1!d2! with (d1+d2)!.
d1!d2! = 1 * 2 * 3 * ... * d1 * 1 * 2 * 3 * ... * d2
(d1+d2)!= 1 * 2 * 3 * ... * d1*(d1+1)*(d1+2)*...*(d1+d2)
And if you know it for r=2, can you prove it in general?
--
Arturo Magidin
hagman
Simply note that the function f: (d1, ..., dr) |-> d1! * ... * dr!
defined on r-tuples of non-negative integers of sum d1+...+dr=n
assumes its maximum.
If i != j and di >= dj>0 then increasing di by one and decreasing dj
by 1 increases f.
Hence at most one di is !=0 at maximum