It is section $4.4$, exercise number $5$. It says the following:

Let $F$ be a field of characteristic $p$. Show that $f(x)=x^p-x-a$ has no multiple roots and that $f(x)$ is irreducible in $F[x]$ if and only if $a\neq c^p-c$ for any $c\in F$.

$\bf{Proof:}$ First of all we note that $f'(x)=-1$, hence $\gcd(f,f')=1$, and we see that $f$ contains no multiple roots.

Secondly, assume that $a=c^p-c$ for some $c\in F$. Then we have that $$x^p-x-a=x^p-x-(c^p-c)=x^p-c^p-x+c=(x-c)^p-(x-c)=(x-c)((x-c)^{p-1}-1)$$Hence $f(x)$ is reducible. (That is, we have proven that if $f$ is irreducible then $a\neq c^p-c$ for any $c\in F$).

For the converse, I was trying to do something along the lines consider $E$ to be a splitting field for $f(x)$. Then, let $b$ be a root of $f(x)$ in $E$. This $b$ satisfies the identity $b^p-b=a$, hence I would like to show the existence of some root $b\in E$ to be actually in our field $F$. This is were I got stuck.

  • $\begingroup$ Is $a \in F$ an arbitrary element? $\endgroup$ – m_l Apr 24 '12 at 9:22
  • $\begingroup$ yes.${}{}{}{}{}{}$ $\endgroup$ – Daniel Montealegre Apr 24 '12 at 9:25
  • $\begingroup$ The polynomial can be reducible over $F$ even, if it has no zeros in $F$. For example, if $p=5$ you have to show that $f(x)$ does not split into a product of quadratic and cubic. The answers to this question. explain, how you can deduce irreducibility of $f(x)$ over $F$. There were several approaches. Mine used a trick in showing that the coefficient of the second highest degree term of an eventual factor of $f$ is in the same field as the root. $\endgroup$ – Jyrki Lahtonen Apr 24 '12 at 10:14
  • $\begingroup$ Of those answers, my favorite is the one given by zyx. However, that depends on the mapping $x\mapsto x^p$ being an automorphism of $F$. This is not the case for all fields of characteristic $p$, so ... $\endgroup$ – Jyrki Lahtonen Apr 24 '12 at 10:20
  • $\begingroup$ Possible duplicate of How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$? $\endgroup$ – Bach Feb 5 '19 at 16:52

Let $E$ be a splitting field of $f(x)=x^p-x-a$ over $F$. Let $b\in E $ be a root of $f$. Then for any $n=1,2,\ldots,(p-1)$, the element $b+n$ is also a root of $F$, because $$f(b+n)=(b+n)^p-(b+n)-a=(b^p-b-a)+(n^p-n)=0+0=0$$ (to be precise, $b+n$ is just shorthand for $b+(n\cdot 1_F)$). Because $f$ is of degree $p$, we've now found all the roots. Clearly, we have that $E=F(b)$.

Let $g(x)\in F[x]$ be the minimal polynomial for $b$ over $F$. Note that $g\mid f$ because $f(b)=0$. Since $E=F(b)$, we have that $[E:F]=\deg(g)$.

Now note that $g(x-n)$ is the minimal polynomial for $b+n$ over $F$; if any monic irreducible polynomial $h(x)\in F[x]$ with $\deg(h)<\deg(g)$ had $h(b+n)=0$, then $h(x+n)$ would have $b$ as a root and still have smaller degree than $g$, contradicting the fact that $g$ is $b$'s minimal polynomial.

Let the factorization of $f$ into monic irreducibles be $f=q_1 q_2\cdots q_r$ (there can't be any repeated factors, because that would lead to repeated roots). Each of the $q_i$ has as a root at least one of the $b+k$, and therefore (being monic and irreducible) is its minimal polynomial. This implies that $\deg(q_i)=\deg(g)$ for all $i$, and therefore $\deg(g)\mid \deg(f)=p$. Thus, either

  • $\deg(g)=p$, hence $f=g$ is irreducible, or,

  • $\deg(g)=1$, hence $[E:F]=1$, hence $b\in F$.

  • $\begingroup$ Why is $n^p-n=0$? $\endgroup$ – Daniel Montealegre Apr 24 '12 at 10:37
  • $\begingroup$ That's because $n$, or to be precise $n\cdot 1_F$, lies in the prime field of $F$, which is a field of $p$ elements since $F$ is of characteristic $p$. $\endgroup$ – Zev Chonoles Apr 24 '12 at 10:40
  • $\begingroup$ Got it. Very good answer. Thanks. $\endgroup$ – Daniel Montealegre Apr 24 '12 at 10:43
  • $\begingroup$ Very nice argument, +1. Though, the other answer is better in delivering the gist. $\endgroup$ – Cauchy Aug 8 '17 at 8:21

Conversely, suppose $f$ is reducible. Then we can write $f=\prod_{i=1}^k f_i$ for some irreducible $f_i$ and some $k\in\mathbb N$. We know that the $f_i$ are distinct factors. Consider any two of the $f_i$; without loss of generality, we consider $f_1,f_2$. Now consider $F[t]/(f_1)$, which contains a root $b$ of $f_1$ which is consequently a root of $f$. Any field which contains a root of $f$ contains all roots of $f$. Thus $F[t]/(f_1)$ is a splitting field of $f$ over $F$, with $[F[t]/(f_1):F]=\deg f_1$. We argue analogously for $f_2$. As splitting fields are isomorphic, we conclude that $\deg(f_1)=\deg(f_2)$.

Hence $f=\prod_{i=1}^k f_i$ where all of the $f_i$ have the same degree, say $\deg(f_i)=n$. Therefore, $p=nk$ and so we have either $n=1$ and $k=p$, or $n=p$ and $k=1$. In the former case, we see that $F$ contains a root of $f$, contradicting our hypothesis. In the latter case, $f$ is irreducible.


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