Show that xp - x + a is irreducible over GF(p) if a is different from 0.
I assumed for contradiction that xp - x + a is reducible. Then it must split into irreducible factors (none of them linear) over GF(p) or just Zp. Say one of the irreducible factors has degree m < p. Then the size of its splitting field must be pm. The splitting field of xp - x + a consists of all the roots of that polynomial. Let b be a root of that polynomial. Then since the multiplicative group of the splitting field is cyclic, all roots of that polynomial can be expressed as some power of b. So I need to find the minimum polynomial of b. If the minimum polynomial of b is just xp - x + a then I am done. So I assume the minimum polynomial of b is the irreducible factor with degree m < p that I mentioned above. Then the splitting field of the irreducible factor of degree m must be the same as the splitting field of xp - x + a since they both contain all the roots of xp - x + a.
I am having trouble finding a contradiction in this argument. Can someone help me with this proof? Thanks.