Three practice algebra qual problems
Snis Pilbor
1. Let S be a nonzero vector subspace of Q^n (Q the rational
numbers). Let p be a prime number. Suppose that whenever
(x_1,...,x_n) in S, then (p x_n,x_1,...,x_{n-1}) in S. In otherwords,
S is invariant under the operation of multiplying the nth entry by p
and then cycling entries.
What I figured out: The map sending (x_1,...,x_n) to (p
x_n,x_1,...,x_{n-1}) is a linear transformation whose matrix has row 1
equal to pe_n and row i equal to e_{i-1} for i=2,...,n, where e_i is
the usual element of the standard basis. This matrix is obviously
invertible and (less obviously) has min polynomial x^n-p. Maybe this
would be useful for solving the problem...
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).
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...
Can anyone help out, these are not homework, they're practice qual
problems I'm doing on my own so you guys are pretty much my only hope
=)
Snis Pilbor
karlfre...@gmail.com
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.
Snis Pilbor
Snis Pilbor wrote:
> Here are 3 practice algebra qual problems which stumped me.
>
> 1. Let S be a nonzero vector subspace of Q^n (Q the rational
> numbers). Let p be a prime number. Suppose that whenever
> (x_1,...,x_n) in S, then (p x_n,x_1,...,x_{n-1}) in S. In otherwords,
> S is invariant under the operation of multiplying the nth entry by p
> and then cycling entries.
>
> What I figured out: The map sending (x_1,...,x_n) to (p
> x_n,x_1,...,x_{n-1}) is a linear transformation whose matrix has row 1
> equal to pe_n and row i equal to e_{i-1} for i=2,...,n, where e_i is
> the usual element of the standard basis. This matrix is obviously
> invertible and (less obviously) has min polynomial x^n-p. Maybe this
> would be useful for solving the problem...
>
Argh, I was so wrapped up making sure the map in question was clear, I
forgot to type the actual problem. The problem is, given these
hypotheses, show S=Q^n.
Achava Nakhash, the Loving Snake
For problem 3, first you are mistaken about Q(sqrt(-19). In fact any
algebraic number field is something called a Dedekind domain (unique
factorization of ideals into prime ideals or equivalent definitions to
be found in any algebraic number theory book.) Therefore any algebraic
number field that is a UFD is also a PID, I am not sure how what I
said proves this, but it is a highly interesting fact, anyway, and it
is true that UFD implies PID for algebraic number fields. Also ideals
in any algebraic number field can be generated by either one or two
elements.
Back to your problem. Since polynomials in a UFD form a UFD, K[x, y]
for any field K is a UFD that is not a PID. The ideal generated by x
is prime but not maximal, so this should answer your question.
For problem 2, I believe that you have done the hard part. I need to
find some air-conditioning before heat stroke carries me off. I will
return with the rest of this later today, I hope.
Regards,
Achava
Robert Israel
Hint: if a subspace S of a vector space V is invariant under linear
transformation T, then T induces a linear transformation T^ of V/S
by T^ (v + S) = T v + S. What can you say about the minimal polynomial
of T^?
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada