xᵖ-x+a is irreducible and separable over ZZ/(p)

Prove that for any prime p and any nonzero a \in \mathbb{F}_p, q(x) = x^p-x+a is irreducible and separable over \mathbb{F}_p.

Note that D_x(q) = px^{p-1}-1 = -1, so that q and D(q) are relatively prime. So q is separable.

Now let \alpha be a root of q. Using the Frobenius endomorphism, (\alpha+1)^p-(\alpha+1)+a = \alpha^p - \alpha + a = 0, so that \alpha+1 is also a root. By induction, \alpha_k is a root for all k \in \mathbb{F}_p, and since q has degree p, these are all of the roots.

Now \mathbb{F}_p(\alpha+k) = \mathbb{F}_p(\alpha), and in particular the minimal polynomials of \alpha and \alpha+k have the same degree over \mathbb{F}_p – say d. Since q is the product of the minimal polynomials of its roots, we have p = dt for some t. Since p is prime, we have either d=1 (so that \alpha \in \mathbb{F}_p, a contradiction) or d=p, so that q itself is the minimal polynommial of \alpha, hence is irreducible.

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