# Polynomial $x^p-x+a$ is Irreducible and Separable Over a Finite Field

## Problem 229

Let $p\in \Z$ be a prime number and let $\F_p$ be the field of $p$ elements.
For any nonzero element $a\in \F_p$, prove that the polynomial
$f(x)=x^p-x+a$ is irreducible and separable over $F_p$.

(Dummit and Foote “Abstract Algebra” Section 13.5 Exercise #5 on p.551)

## Proof.

### Separability

The separability can be checked by noting that the derivative of $f(x)$ is
$f'(x)=px^{p-1}-1=-1$ and is relatively prime to $f(x)$, and thus $f(x)$ is separable.

We give two proofs for the irreducibility of $f(x)=x^p-x+a$. Both proofs use the following lemmas.

### Lemma 1

Lemma 1. If $\alpha$ is a root of $f(x)=x^p-x+a$, then $\alpha+j$ is a root of $f(x)$ for any $j\in \F_p$.

#### Proof of Lemma 1

We have
\begin{align*}
f(\alpha +j)&=(\alpha+j)^p-(\alpha+j)+a\\
&=\alpha^p+j^p-\alpha-j+a\\
&=f(\alpha)=0.
\end{align*}
Here, we used Fermat’s little theorem that $j^a=j$ in $\F_p$.
Thus, $\alpha+j$ is also a root of $f(x)$ for any $j\in \F_p$.

### Lemma 2

Lemma 2. The polynomial $f(x)=x^p-x+a$ does not have a root in $\F_p$.

#### Proof of Lemma 2

If $\alpha$ is a root of $f(x)$ in $\F_p$, then we have
\begin{align*}
0=f(\alpha)=\alpha^p-\alpha+a=a
\end{align*}
since $\alpha^p=\alpha$ in $\F_p$ by Fermat’s little theorem.
However, this contradicts that $a$ is a nonzero element of $\F_p$.

### Proof 1 of the irreducibility

Suppose that we have
$f(x)=g(x)h(x)$ for $g(x), h(x) \in \F_p[x]$.

By Lemma 1, if $\alpha$ is a root of $f(x)$, then $\alpha+j$ is also a root of $f(x)$ for any $j\in \F_p$. Thus, $\alpha + j$ is a root of either $g(x)$ or $h(x)$. From this, we obtain that
$g(x)=\prod_{j\in I}(x-(\alpha+j)),$ where $I$ is a subset of $\F_p$.

Expanding the product, we see that
$g(x)=x^n-\sum_{j\in I} (\alpha +j)x^{n-1}+\text{ (lower terms)},$ where $n=|I|$.
Since $g(x) \in \F_p[x]$, the coefficient of $x^{n-1}$ is in $\F_p$. Thus we have
\begin{align*}
\F_p \ni \sum_{j\in I} (\alpha +j)=n\alpha +\sum_{j\in I} j.
\end{align*}

Since $\sum_{j\in I} j\in \F_p$, we deduce that $n \alpha \in \F_p$.
Since by Lemma 2, $\alpha \not \in \F_p$, we must have $n=0$ in $\F_p$.
(Otherwise $n$ is invertible in $\F_p$ and we would get $\alpha \in \F_p$.)
Therefore $n=|I|$ is either $0$ or $p$.
If $n=0$, then $g(x)=1$. If $n=p$, then $h(x)=1$. This implies that $f(x)$ is irreducible over $\F_p$.

### Proof 2 of the irreducibility

We give another proof that $f(x)=x^p-x+a$ is irreducible over $\F_p$.

Let $m(x)\in \F_p[x]$ be the minimal polynomial of the root $\alpha$ of $f(x)$.
Then we have $f(x)=m(x)f_1(x)$ for some $f_1(x) \in \F_p[x]$.

If $f_1(x)=1$, then $f(x)=m(x)$ is irreducible as the minimal polynomial is irreducible.
If not, then $f_1(x)$ has some root, and it must be of the form $\alpha+j$ for some $j\in \F_p$ by Lemma 1.
It is straightforward to check that the minimal polynomial of $\alpha+j$ is $m(x-j)$. Thus we can write $f_1(x)=m(x-j)f_2(x)$.

If $f_2(x)=1$, we stop here. If not we iterate the same procedure with $f_2$ as above. Eventually we obtain
$f(x)=\prod_{j\in I}m(x-j)$ for some subset $I$ of $\F_p$.

Let $n$ be the degree of $m(x)$. Then comparing the degree of both sides, we have
$p=n|I|.$ Since $p$ is prime, we have either $n=1, |I|=p$, or $n=p$, $|I|=1$.
The former case implies that the minimal polynomial $m(x)$ of $\alpha$ is degree $1$, and hence $\alpha \in \F_p$. However, this cannot happen by Lemma 2.
So we must have $n=p$ and $|I|=1$, which deduce that $f(x)=m(x)$ and $f(x)$ is irreducible over $\F_p$.

## Comment.

The polynomial $f(x)=x^p-x+a$ is studied in Artin–Schreier theory.

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### 2 Responses

1. Hansi says:

minor issues:
In the line with f'(x) you should imo differentiate, as if you were in the Real numbers
so f'(x)=p*x^(p-^)-1

In the Proof of Lemma1, second line, there’s a minor error with the sign of the second j, should be -j

• Yu says:

Dear Hansi,
Thank you for pointing out the typos. I edited the post. Thanks!

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##### Show that Two Fields are Equal: $\Q(\sqrt{2}, \sqrt{3})= \Q(\sqrt{2}+\sqrt{3})$

Show that fields $\Q(\sqrt{2}+\sqrt{3})$ and $\Q(\sqrt{2}, \sqrt{3})$ are equal.

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