Irreducibly of $x^5-x-1$ over $\mathbb{F}_3$

Question

Using the affine linearized polynomial $f(x)=x^p-x-b$ is irreducible over $\mathbb{F}_q$ if and only if $\operatorname{Tr}_{\mathbb{F}_q/\mathbb{F}_p} \neq 0$. But here $\mathbb{F}_3$ is not a field extension of $\mathbb{F}_5$.

Also we have that: Let $a \in \mathbb{F}_q$ and let $p$ be the characteristic of $\mathbb{F}_q$. Then the trinomial $x^P - x - a$ is irreducible in $\mathbb{F}_q[x]$ if and only if it has no root in $\mathbb{F}_q$.

Answer 1

$x^5-x-1$ is indeed irreducible over $\mathbb{F}_3$, because it has no root and no quadratic factor.

Concerning primitivity, a counting shows that there are $\frac{\phi(3^5-1)}{5}=22$ monic primitive polynomials of degree $5$ over $\mathbb{F}_3$, and $\frac{3^5-3}{5}=48$ monic irreducible polynomials of degree $5$ over $\mathbb{F}_3$. Taking out the obvious non-primitive ones, the chances are good for $x^5-x-1$ to be primitive.

Answer 2

Easy to see that our polynomial has no roots from $\{1,-1,0\}$ and since $x^2+1,$ $x^2+x-1$ and $x^2-x-1$ they are unique irreducible polynomials with degree $2$, it remains an easy checking.

Answer 3

The quickest way of proving irreducibility of $p(x)=x^5-x-1$ in $\Bbb{F}_3[x]$ that I can think of is the following.

1. Everybody started out by checking that $p(x)$ has no roots in $\Bbb{F}_3$ and therefore no linear factors.
2. Similarly quadratic factors can be ruled by checking that $p(x)$ has no roots in $\Bbb{F}_9$. All the non-zero elements of the field of nine elements are zeros of $x^8-1=(x^4-1)(x^4+1)$. If $\beta\in \Bbb{F}_9$ is a zero of $x^4+1$, then $\beta^4=-1$. Hence $\beta^5=-\beta$ and $$p(\beta)=\beta^5-\beta-1=-2\beta-1.$$ For this to be equal to zero $\beta$ would have to an element of the prime field. But that possibility was ruled out in step 1 already. OTOH, if $\beta$ is a zero of $x^4-1$ then $\beta^4=1$. Consequently $\beta^5=\beta$ and $$p(\beta)=\beta^5-\beta-1=-1\neq0.$$
3. If a degree five polynomial is not irreducible, then it has at least one at most quadratic factor. We saw that $p(x)$ does not have such factors so it must be irreducible.