Thread: Polynomial in GF(p^n) irreducible or...

I'd appreciate some hints or advice for this assignment:

Let F be a field of characteristic p and f(x) = x^p - x - c be a polynomial in F[x]. Show that f is either irreducible in F[x] or that all its roots lie in F.

So F=GF(p^n), n>0. If n=1 we have x^p - x - c = x - c, but what properties could I use for general n?

Let E be a splitting field for f(x) = over F of char p. Then for be a root for f(x), then r + 0, r + 1, ..., r + p - 1 are also roots.
I have to make a quick point, that f has no repeated roots.for if, assuming root a in E, f(x) = h(x)(x-a) , then f'(x) = h(x) + h'(x)(x-a), and f'(a) = h(a).
Since f'(x) = f'(a) = h(a) , so no repeated roots.


Since = = 0 . for (Since
so
Take g(x) to be the minimal polynomial such that g(r) = 0.

Then [E : F ] = deg(g).

If f = is the decomposition of f into unique factorization, since g is the min. polynomial s.t g(r) = 0, g divides f. since we made f into a unique factorization g divides one of the , since are monic and irreducible, this means g(x)*1 = , thus deg( ) = deg(g) for all .

So deg(g) divides deg(f).
Since deg(f) = p, either f = g and f is irreducible
or deg(g) = 1, and [E: F] = 1, and so

Thank you for your reply. I don't understand this however: "this means g(x)*1 = f_i" Why does this follow from the f_i being monic and irreducible?

Also, is it true in general in GF(p^n)[x] that if r is a root then so is r+i, i=0,1,...,p-1? If so, how? Suppose \sum_{k=0}^m b_k x^k has r for a root. Then \sum b_k (a+i)^k = \sum b_k (a^k + i^k) = \sum b_k i^k. How is that zero?

Originally Posted by spudwish

Thank you for your reply. I don't understand this however: "this means g(x)*1 = f_i" Why does this follow from the f_i being monic and irreducible?

Also, is it true in general in GF(p^n)[x] that if r is a root then so is r+i, i=0,1,...,p-1? If so, how? Suppose \sum_{k=0}^m b_k x^k has r for a root. Then \sum b_k (a+i)^k = \sum b_k (a^k + i^k) = \sum b_k i^k. How is that zero?

if g(x) is the minimal polynomial such that g(r) = 0, then by def it is the polynomial of lowest degree such that if any other polynomial p(r) = 0, g divides p.
if g(x) divides f(x). Then since f(x) is factorized as the product of irreducible elements, then the minimal polynomial dividing f(x) is one of those irreducible elements means it is one of those irreducible elements,
for irreducible means that it cannot be factored anymore, so anything that divides it means if q is a factor that is not a unit then f_i wouldn't really be irreducble in the first place right?

As for your second question. It is not true in general. Take in it has a root 1. But 1+1=0, if not a root of it.

jakncoke, i'm a little unclear on your argument as to why deg(g)|deg(f). i understand that g = f_i for some i, but this only shows deg(g) ≤ deg(f) = p. the degrees of the factors sum, not multiply. something is missing here, perhaps you should also show that if r + a (for a in Z[SUB]p[/SUP]) is another root of f, with minimal polynomial h, that deg(g) = deg(h), hmm?

Originally Posted by Deveno

jakncoke, i'm a little unclear on your argument as to why deg(g)|deg(f). i understand that g = f_i for some i, but this only shows deg(g) ≤ deg(f) = p. the degrees of the factors sum, not multiply. something is missing here, perhaps you should also show that if r + a (for a in Z[SUB]p[/SUP]) is another root of f, with minimal polynomial h, that deg(g) = deg(h), hmm?

I said deg(g) = deg( ) for all i which implies that either f factorizes into linear factors(which is showed earlier must be true if it has a root in E) or is irreducible. Thus deg(g) = p or 1 which divides deg(f)=p