(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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)

2. Relevant equations

3. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**