The complete problem appears in Hungerford's Algebra.

Let $c\in F$, where $F$ is a field of characteristic $p$ ($p$ prime). Then $x^p−x−c$ is irreducible in $F[x]$ if and only if $x^p−x−c$ has no root in $F$.

One direction has already been answered:

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

I wish to prove the other direction:

If $x^p−x−c$ is irreducible in $F[x]$, then $x^p−x−c$ has no root in $F$.

I tried to prove this using the contrapositive:

Assume $\alpha\in F$ such that $\alpha^p−\alpha−c=0$. We show that we can factor $x^p−x−c$ into simpler polynomials, $f,g$ each of degree $<p$. I am stumped because I cannot show such a polynomial factorization.

I was thinking that maybe I could say that, hence, $x-\alpha$ is a linear factor of $x^p−x−c$, and thus, $x^p−x−c = (x-\alpha)f(x)$ where $\deg f(x) \le p-1$.

Therefore, $x^p−x−c$ is a reducible polynomial.

When $\mathrm{char} F = 0$, then this statement is supposedly false. However, I cannot think of a reducible polynomial over $\mathbb Q$ of degree $5$ whose roots are not rational, or an irreducible polynomial over $\mathbb Q$ of degree $5$ but has roots in $\mathbb Q$.

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    $\begingroup$ I do not understand the question. No irreducible polynomial of degree at least has a root. This is also true in char 0. The proof is as you say. Maybe you want to show something else, say an "if and only if." $\endgroup$ – quid Apr 13 '16 at 14:08
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    $\begingroup$ The other direction is already proven: math.stackexchange.com/questions/1256088/… $\endgroup$ – Harry Evans Apr 13 '16 at 14:10
  • $\begingroup$ I see. As said the part you ask about is a consequence of a result that holds for every polynomial over every feel, and you gave the proof. (Just the final paragraph is not correct.) $\endgroup$ – quid Apr 13 '16 at 14:13
  • $\begingroup$ @Quid: $X^4+2X^2+1$ is reducible but has no roots in $\mathbb R$. $\endgroup$ – Surb Apr 13 '16 at 14:13
  • $\begingroup$ @Surb The question is about "if P is irreducible then it has no roots" not "if it has no roots then it is irreducible" $\endgroup$ – quid Apr 13 '16 at 14:15

The contrapositive of your statement is: If $f(x)$ has a root in $F$, then $f(x)$ is reducible over $F$. This is true over any field.

Over any field you have a factorization theorem: If $f\in F[x]$ and $\alpha\in F$, then $f(x)=q(x)(x-\alpha)+f(\alpha)$. In particular, if $\alpha$ is a root of $f$, then $(x-\alpha)$ is a factor of $f$.

The way this is proved is to compute $f(x)-f(\alpha)$ and use the factorization $$x^k-\alpha^k=(x-\alpha)(x^{k-1}+x^{k-2}\alpha+\cdots+x\alpha^{k-2}+\alpha^{k-1}).$$

It is the other direction that fails in characteristic 0.


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