GF(q); x^p-x+a has no root => irreducible [PROOF]

[UD1]

Homework Statement

Let q=p^e, where p is a prime and e is a positive integer. Let a be in GF(q). Show that f(x)=x^p-x+a is irreducible over GF(q) if and only if f(x) has no root in GF(q)

Homework Equations

The Attempt at a Solution

One of the directions seems obvious. Namely, if f is irreducible, the f has no roots: if f had a root, say b in GF(q), then f(x)=(x-b)*g(x), where g is a poly with coefficients from GF(q).

The other direction: if f has no root, then f is irreducible. Or equivalently, f is reducible, then f has a root in GF(q). Frankly, I don't have any ideas. The only "observation" I made is that if the power of f were q rather than p and it had a root, say b, then b+1, b+2,...,b+p-1 would be roots, too. I think the key point here is that I need to "guess" what the linear factorization of f in its splitting field looks like.

 

[Hurkyl]

The only "observation" I made is that if the power of f were q rather than p and it had a root, say b, then b+1, b+2,...,b+p-1 would be roots, too.

I could understand this if you were adding arbitrary elements of GF(q) -- but since you're just adding elements of GF(p), I don't understand why you are insisting on q instead of p.

 

[UD1]

I could understand this if you were adding arbitrary elements of GF(q) -- but since you're just adding elements of GF(p), I don't understand why you are insisting on q instead of p.

OK. It looks that there is some notation misunderstanding. I'm not that well familiar with finite fields so I might have used the type of notation not usually used.

By GF(q) I mean a finite field of order q (with q elements) where q is a power of prime p, and that p is a characteristic of the field.

 

[Hurkyl]

Right, that's what I thought you meant. The numbers you're adding to b: 0,1,2,...,p-1, are all elements of GF(p).

 

[UD1]

Right, that's what I thought you meant. The numbers you're adding to b: 0,1,2,...,p-1, are all elements of GF(p).

So, how does that help?

 

[Hurkyl]

Well, if I'm right, then your observation applies. Surely that's better than your observation not applying?

 

[UD1]

Well, if I'm right, then your observation applies. Surely that's better than your observation not applying?

So, we know that, assuming that b is a root in some extension of F, then

$$f(x)=\prod_{j=0}^{p-1}(x-(b+j))$$.

Now suppose that f is reducible, that is f=gh for some polynomials g and h whose coefficients are in GF(q). There must exists a proper subset I of {0,...,p-1} such that

$$g(x)=\prod_{j\in I}(x-(b+j))$$.

How does it follow that f has a root in F. Namely, that b is in F?