Re: Another Galois theory problem


Doug B wrote:
Hi, I am trying to show that a field of characteristic p>0 is perfect
iff it has the property that every element has a pth root (the
definition of perfect field I have is, "F is perfect if every algebraic
extension is separable").

I don't really know how to proceed either direction. I've had the most
luck proving if it is perfect, it has the pth root property. I do this
by supposing it does not. Then there's an a such that x^p-a has no
roots. I take a splitting field and, enlarging it if necessary, assume
it is an algebraic extension.

That last statement makes no sense whatsoever. A splitting field of a
(set of) polynomials over F is ->necessarily<- an algebraic extension!
It is generated by adjoining roots of polynomials, and therefore it is
generated over F by algebraic elements; hence the extension must be
algebraic. So why are you saying that you are "enlarging it if
necessary" to make it algebraic? Methinks you are missing some
very important and basic facts about field extensions!

There, x^p-a splits, so pick r with
r^p=a. Then what I *WANT* to write is (x-r)^p = x^p-r^p = x^p - a,

This follows trivially from the fact that the characteristic is p. The
coefficient of x^i in the expansion of (x-r)^p is (p choose i), which
is a multiple of p for all i, 0<i<p.



so
that r is a multiple root of x^p-a, contradiction.

Indeed. See below.

But this is not
rigorous, since in fields of characteristic >0, just because two
polynomials agree everywhere does not imply they are the same
polynomials.

Just multiply out (x-r)^p. Don't know why you are trying to evaluate
anything.

(By the way: it ->is<- true for polynomials of degree less than
char(F), though that is neither here nor there in this situation)

So even though the polynomials x^p-a and (x-r)^p agree
everywhere, I don't know how to show they are the same polynomials, if
indeed they even are.

Yes,. they are. Just write out (x-r)^p and remember the characteristic
is p.

Also, I am implicitly assuming here that x^p-a
is irreducible, otherwise even if I could show it has multiple roots in
a splitting field, that would prove nothing.

Think minimal polynomial: the minimal polynomial of r over F must
divide x^p-a. Since x^p-a is equal to (x-r)^p, then the minimal
polynomial of r over F must be of the form (x-r)^k for some k, 0<k<=p.
If k>1, then this gives you the inseparabilit, and if k=1, that tells
you that a already had a p-th root in F, which contradicts your choice
of a.



But I have no idea how to
show that x^p-a is even irreducible in the original field.

You don't have to (although it will be, since it will turn out that for
an irreducible polynomial to be inseparable, it must be of the form
g(x^p) for some polynomial g).

And all
this is just one direction of the statement, the other direction I have
even less of a clue how to proceed...

If every algebraic extension is separable, let a in F and consider the
polynomial x^p-a. Let r be a root, and consider F(r). Now argue as
above about what the minimal polynomial of r must look like, and use
the fact that the extension is separable to deduce that r is already in
F.

Arturo Magidin. sans .sig

.

Relevant Pages