有限域中$x^p-x-a$为本原多项式的充分条件及a的集合刻画

Alright, let's break down these two questions about primitive polynomials of the form $f(x) = x^p - x -a$ over $\mathbb{F}_p$:

First, let's recall that a polynomial is primitive over $\mathbb{F}_p$ if its root generates the multiplicative group of the finite field extension it defines. For $f(x) = x^p -x -a$, here's a straightforward sufficient condition:

Suppose $p$ is a prime and $a \in \mathbb{F}_p^*$ (so $a \neq 0$). If:

• $f(x)$ is irreducible over $\mathbb{F}_p$ (which, as we'll see below, is automatically true when $a \neq 0$), AND
• For every prime divisor $q$ of $N = p^p - 1$, the element $x^{N/q}$ is not equal to 1 in the quotient ring $\mathbb{F}_p[x]/(f(x))$

Then $f(x)$ is a primitive polynomial.

Why does this work? Since $a \neq 0$, $f(x)$ is irreducible, so $\mathbb{F}p[x]/(f(x)) \cong \mathbb{F}{p^p}$, whose multiplicative group is cyclic of order $N$. For $x$ to generate this group (making $f(x)$ primitive), its order must be exactly $N$. Checking that $x^{N/q} \neq 1$ for all prime divisors $q$ of $N$ ensures that the order of $x$ can't be a proper divisor of $N$, hence it must be $N$ itself.

Let's start with easy cases and build up:

• Case $a = 0$: $f(x) = x^p -x = x(x^{p-1} - 1)$, which factors completely into linear terms over $\mathbb{F}_p$. This polynomial is reducible, so it can't be primitive (primitive polynomials must be irreducible, since they define a finite field extension).

• Case $a \neq 0$: First, we note $f(x)$ is irreducible over $\mathbb{F}_p$. Why? If $\alpha$ is a root of $f(x)$, then all roots are $\alpha + k$ for $k = 0,1,...,p-1$ (since $(\alpha +k)^p - (\alpha +k) -a = \alpha^p +k^p - \alpha -k -a = (\alpha^p - \alpha -a) + (k -k) = 0$). The orbit of $\alpha$ under the Frobenius automorphism has size $p$ (since $a \neq 0$, adding $k a$ gives distinct elements), so the minimal polynomial of $\alpha$ has degree $p$—exactly $f(x)$. Thus, $\mathbb{F}p[x]/(f(x)) \cong \mathbb{F}{p^p}$.

Now, $f(x)$ is primitive if and only if $x$ (as an element of $\mathbb{F}{p^p}$) generates the multiplicative group $\mathbb{F}{p^p}*$ (which is cyclic of order $N = p^p -1$).

To characterize the set of such $a$:
The set is exactly all elements of the form $\alpha^p - \alpha$, where $\alpha$ is a primitive element of $\mathbb{F}_{p^p}*$.

Alternatively, we can describe this set using an order condition:
An element $a \in \mathbb{F}_p^*$ is in the set if and only if for every prime divisor $q$ of $N = p^p -1$, $x^{N/q} \not\equiv 1 \pmod{f(x)}$.

We can also note the size of this set: there are $\phi(N)/p$ such elements, where $\phi$ is Euler's totient function. This is because each $a \neq 0$ is the image of exactly $p$ elements of $\mathbb{F}{p^p}$ (all roots of $f(x)$), and there are $\phi(N)$ primitive elements in $\mathbb{F}{p^p}*$.

内容的提问来源于stack exchange，提问作者Reyx_0