Re: Three practice algebra qual problems


Snis Pilbor wrote:

2. Let p be prime and consider g(x) in GF(p)[x] given by g(x)=x^p-x+a,
where a in GF(p) is nonzero. Prove g(x) is irreducible over GF(p).

What I figured out: g(x) obviously has no root in GF(p). It has
distinct roots over a splitting field, and any field containing one
root of g(x) contains all of them-- as a matter of fact if r is any
root of g(x) in any field then r+1, r+2, ..., r+(p-1) are a full set of
roots of g(x).


What you can do is assume that g(x) has a nontrivial factorization over
GF(p), say g(x)=f(x)h(x). If f and h have coefficients in GF(p) then
they will be fixed by the frobenius automorphism x -> x^p. But you can
also observe that the frobenius automorphism acts transitively on the
roots of g(x) in the splitting field, so if f or h has one root of g in
the splitting field, then it must have all the roots, which clearly
implies that one of f or g must be a constant.


3. Give an example of a Unique Factorization Domain S and a nonzero
prime ideal P in S such that S/P is not a field.

What I figured out: if S is a PID and P is a nonzero prime ideal in S
then S/P is a field. So we can restrict our attention to UFDs which
arent PIDs. I seem to recall that these are extremely exotic, and I
seem to remember that the "easiest" example of such a thing is
Q(sqrt(-19)) but, even if this memory is accurate, this fact was highly
nontrivial to prove, and anyway even if Q(sqrt(-19)) is a non-PID UFD,
I have no idea how to pick P...


If you want a number ring that is a non-PID UFD, then yes, you'll be
stuck with something kind of exotic and I'm not sure how you would
proceed. But it's a pretty standard fact that Z[x] (the ring of
integer polynomials) is UFD, and you should be able to find the needed
prime ideal in there fairly easily.

.

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