Show that $f(x) = x^p -x -1 \in \Bbb{F}_p[x]$ is irreducible over $\Bbb{F}_p$ for every $p$.

Question

Let $p$ be a prime.

a) Show that $f$ has no roots in $\Bbb{F}_p$.

Let $F^*$ be the multiplicative group of $\Bbb{F}_p$. Then, by lagrange's thoerem for all nonzero $\alpha \in \Bbb{F}_p$, $\alpha^{p-1} = 1 \implies \alpha^p=\alpha \implies \alpha^p-\alpha=0$. Of course $0^p=0$, so this is true for all elements of $F$ and not just the nonzero ones. But then $\alpha^p - \alpha - 1 = -1$ for all $\alpha \in \Bbb{F}_p$ and so it must have no roots in $\Bbb{F}_p$. I could have also done this using the Frobenius automorphism, right?

b) Let $\alpha$ be a root of $f$ (in some algebraic closure of $\Bbb{F}_p$). Show that $\alpha + s$ is also a root for all $s \in \Bbb{F}_p$.

Let $\alpha^p - \alpha -1 =0$. Let $E$ be an algebraic closure of $\Bbb{F}_p$. Since $E$ has characteristic $p$, $(\alpha + s)^p = \alpha^p + s^p$. So we have,

$$(\alpha + s)^p - (\alpha + s) -1 = \alpha^p + s^p - \alpha -s -1 = s^p - s = 0.$$

c) Conclude that $f$ is irreducible over $\Bbb{F}_p$, for every $p$.

By b) and the fact that $\Bbb{F}_p$ has $p$ distinct elements, we know that the roots of $f$ are $\alpha, \alpha+1, ... , \alpha + p-1$. So if $K$ is a splitting field, we have $$x^p - x -1 = (x-\alpha)(x-(\alpha+1))...(x-(\alpha+p-1)).$$

Now let's assume that $f$ is reducible over $\Bbb{F}_p$. Then $f=gh$ for some $g$ and $h$ with degrees less than that of $f$. So $g$ and $h$ must be of the form $(x-(\alpha+s_1))...(x-(\alpha+s_k))$ where k is less than n. Let's say that g has degree 2, because the other cases are similar.

So $g =(x-(\alpha + s_i))(x-(\alpha + s_j))$ and the constant term for $g$ is,

$$(\alpha + s_i)(\alpha + s_j) = \alpha^2 + s_is_j\alpha + s_is_j.$$

Since $\Bbb{F}_p$ is a field, if $\alpha s_is_j$ is an element of $\Bbb{F}_p$ , then so is $((\frac{1}{s_is_j})(\alpha s_is_j) = \alpha$, a contradiction.

We can show by induction that if we multiply $(x-(\alpha+s_1))...(x-(\alpha+s_k)$, we get a term that looks like $s_1s_2...s_k\alpha$. So this is also true for $k>2$.

Do you think that my answer is correct?

Thank you in advance

Answer 1

I like your idea very much, but I didn't understand how you managed to ignore the $\alpha^2$ term from your constant term.

I have written up an answer to this question with a similar idea, where I look at the next to highest degree term of $g(x)$ (=the term of degree $k-1$). In your degree two example, this would be the linear term. Its coefficient is $2\alpha+(s_i+s_j).$ As $s_i,s_j$ are in the prime field, and $2$ is invertible, we can, as in your argument, conclude that $\alpha\in\mathbb{F}_p$, which is a contradiction.

By studying the lowest degree term, you get a lot of clutter from powers of $\alpha$. The degree $k-1$ term is IMHO easier to manage.

Answer 2

a) and b) are fine. I haven't checked c) but you could also note that if the polynomial splits into $h \cdot g$ with $h,g \in \mathbb{F}_p[x]$ then, say $$ h(x) = \prod_{i \in S} (x - (\alpha + i)) \in \mathbb{F}_p[x] $$

with $S$ proper subset of $\{0,1,\ldots,p-1\}$. Therefore $$ \sum_{i \in S} (\alpha + i) \in \mathbb{F}_p $$ hence $|S| \alpha \in \mathbb{F}_p$ but since $0 < |S| < p$ this implies that $\alpha \in \mathbb{F}_p$, but then $$ f(x) = \prod_{i} (x - i) = x^p - x $$ which is not true.

Answer 3

(This is an alternate approach that doesn't use the outlined argument.)

In general, if $h(x)|x^{p^n}-x$ in $\mathbb F_p[x]$ then $h(x)$ is the product of distinct prime polynomials, each of degree equal to a factor of $n$.

If you know this, a fun way to prove this theorem (but skipping the outlined steps in the question) is to prove that $x^p-x-1$ is a divisor of $x^{p^p}-x$. Since $x^p-x-1$ has no roots in $\mathbb F_p$, this would mean that it has to be a prime polynomial.

You show that it is a factor by noting that:

$$x^{p^p}-x = \sum_{k=0}^{p-1} \left(x^{p^{k+1}}-x^{p^k}\right) = \sum_{k=0}^{p-1} (x^p-x)^{p^k} = \\\sum_{k=0}^{p-1}((x^p-x-1)+1)^{p^k} \equiv 0\pmod{x^p-x-1}$$