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Let $p$ be a prime and $b$ is a non-zero element of the prime field $\mathbb{F}_p$. Then what are the finite fields $\mathbb{F}_q$ of characteristic $p$ over which the trinomial $x^p-x-b$ is irreducible ?

I am completely stuck at this problem, please help.

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Please accept an answer or explain why you don't accept it - accepting a solution helps our community :) – Ofir Dec 31 '12 at 7:55
up vote 3 down vote accepted

If $\alpha$ is a root of $f(x)=x^p-x-b$ in the algebraic closure of $\mathbb{F}_{p}$, then it can be seen (by using the Frobenius automorphism $x \to x^p$) that the rest of the roots of $f$ are given by $\alpha +i, 1 \le i < p$.

If $f(x)$ factors as $\prod_{j=1}^{r} g_j(x)$ over $\mathbb{F}_{p^n}$, it means that each $g_j$ is the minimal polynomial of some $\alpha + i$. Those minimal polynomials must be of the same degree, since if $h(x)$ is irreducible and vanishes on $\alpha+i$ then $h(x-i+j)$ is also irreducible and vanishes on $\alpha+j$.

Let $d=\deg g_1$. Then by comparing degrees, $p = rd$. There are 2 cases -

  1. $d=1$ - in this case, $f$ splits into linear factors, so it must have a root in $\mathbb{F}_{p^n}$. But if $\alpha^{p}-\alpha-b=0$ then $$\alpha^{p^n} = \alpha + nb$$ By induction (apply the Frobenius automorphism $x \to x^p$ on both sides). $y \in \mathbb{F}_{p^n}$ iff $y^{p^n} = y$, so we must have $nb=0$, i.e. $p|n$.

  2. $d=p$ - in this case, $f$ is irreducible.

In conclusion, in $\mathbb{F}_{p^p}$ $f$ splits into linear factors (and also in any finite field containing it). In the rest of cases ($p \nmid n$), $f$ is irreducible. $\blacksquare$

Fun related fact: $x^{p^n}-x$ is the product of all irreducible polynomials of degree dividing $n$.

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