Re: Another Galois theory problem


On 11-Jul-2006, "Doug B" <doug_protocols@xxxxxxxxx>
wrote in message <1152667122.492475.163730@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:

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. There, x^p-a splits, so pick r with
r^p=a.

You don't need a splitting field. Let irr(F,r) be the irreducible
polynomial of r over F. In F(r), x^p-a = x^p-r^p = (x-r)^p, so
irr(F,r) divides this and so irr(F,r) = (x-r)^n for some n <= p.
If n > 1 then r is separable over F which contradicts that F is
perfect, so n = 1 and r is in F.

Then what I *WANT* to write is (x-r)^p = x^p-r^p = x^p - a, so
that r is a multiple root of x^p-a, contradiction. 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. 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. 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. But I have no idea how to
show that x^p-a is even irreducible in the original field.

In fact it isn't, but that doesn't matter. See above.

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

Assume there's an r inseparable over F, with f(x) = irr(F,r). Then
f'(r) = 0 since r is inseparable, so f(x) must divide f'(x), so
f'(x) = 0 since the degree of f'(x) is less than that of f(x). So
f(x) is in F[x^p], so f(x) = a_n*x^{pn} + a _{n-1}*x^{p(n-1)}... +
a_0 and since each a_i has a pth root in F, say b_i^p = a_i, then
f(x) = (b_n*x^n)^p + (b_{n-1}*x^{n-1})^p + ... + b_0 = (b_n*x^n +
b_{n-1}*x^{n-1} + ... b_0)^p, contradicting that f(x) is
irreducible over F.

--
Jim Heckman
.

Relevant Pages
Quantcast