# $x^p -x-c$ is irreducible over a field of characteristic $p$ if it has no root in the field

Let $c$ be an element of a field $F$ of characteristic $p$ (prime). Then how to show that $x^p -x-c$ is irreducible over $F$ if it has no root in $F$?

I was trying using contradiction and by considering the corresponding splitting field, but could not conclude.

Let $L$ be the splitting field and $\alpha \in L$ a root. Consider the map $Gal(L/F) \to \mathbb Z/p\mathbb Z$ given by $\sigma \mapsto \sigma(\alpha)-\alpha$ (You should verify that this is well defined). Since $\mathbb Z/p\mathbb Z$ has no non-trivial subgroups, the map is either trivial or surjective. If the map is surjective, we deduce that the galois group acts transitively on the roots, hence the polynomial is irreducible. If the map is trivial, we deduce $\alpha \in F$, which is a contradiction to the assumption.
Without Galois theory (but somehow the same proof): Consider the isomorphism $\varphi: F[x]\to F[x], x \mapsto x+1$. The polynomial $f = x^p-x-c$ is fixed by $\varphi$, hence the irreducible factors are permuted by $\varphi$. Let the irreducible factors be $M := \{f_1, \dotsc, f_n \}$. Since there is no root in $F$, we have $n < p$.
$\langle \varphi \rangle \cong \mathbb Z/p\mathbb Z$ acts on $M$, hence we get a homomorphism $\langle \varphi \rangle \cong \mathbb Z/p\mathbb Z \to S_n$. From $n<p$ we deduce that this homomorphism is trivial, hence the action is trivial, so all irreducible factors are fixed by $\varphi$. You should finish the proof by showing that $\varphi$ does not fix any polynomials of degree $1, 2 ,\dotsc, p-1$. Hence the only irreducible factor can be $f$ itself.
• It is an isomorphism, since we have an obvious inverse given by $x \mapsto x-1$. If you do not understand this isomorphism, the whole proof might be out of reach for you. But to be honest, the proof of this will not be any easier. This is somehow the standard way of showing this. – MooS Apr 29 '15 at 6:28