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There are $p^{p - 1} - 1$ monic irreducible polynomials of prime degree $p$ over $\mathbb{F}_p$ by this post.

The chance of picking one of them randomly is $\cfrac{p^{p - 1} - 1}{p^p} = \cfrac{1}{p} - \cfrac{1}{p^p}$ .

Is there a systematic way to list all of them?

Here are $p - 1$:

$f(x) = x^p - x + a \in \mathbb{F}_p[x]$, where $a \in \mathbb{F}_p^*$

And here are another $p - 1$:

$\cfrac{1}{a}x^pf(\cfrac{1}{x}) = x^p - \cfrac{1}{a}x^{p - 1} + \cfrac{1}{a} = x^p - bx^{p - 1} + b \in \mathbb{F}_p[x]$, where $b \in \mathbb{F}_p^*$

Many more to go...

EDIT: I'm thinking there can't be a clean, systematic way to list these particular irreducible polynomials. But I'm asking just in case because an exercise in a book on Galois theory (and I'm just working through the exercises for myself) asks to factorize $x^{p^p} - x$ over $\mathbb{F}_p$, which almost makes it seem like a constructive factorization is expected. But I now came across this post for only the case $p = 3$, and the process there seems to be enumerate and check.

Oscar
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    The naive algorithm is to sieve for them. If $f(x)$ is reducible, then it has an irreducible factor of degree at most $p/2$. So if you remove all such multiples from monic polynomials of degree $p$, what remains are the irreducible polynomials you seek. Compare Eratosthenes sieve to find prime numbers. – hardmath Jun 02 '20 at 18:44
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    In theory you can generate them algorithmically. Use for example the known irreducible polynomial $f(x)=x^p-x-1$ to construct the field $\Bbb{F}_{p^p}$. Then go through its elements and generate their minimal polynomials using the Frobenius action. But I really would not want to do that :-) – Jyrki Lahtonen Jun 02 '20 at 20:19
  • @JyrkiLahtonen That would take lots of work! But let me make sure I understand. For $\alpha \in \mathbb{F}_{p^p}$, collect all the conjugate roots $\alpha_i$ using the Frobenius automorphism repeatedly. Then multiply out the factorization $\prod(x - \alpha_i)$ to get the minimimal polynomial of $\alpha$ over $\mathbb{F}_p$. – Oscar Jun 02 '20 at 20:59
  • That was the idea. Factoring is likely faster. – Jyrki Lahtonen Jun 02 '20 at 21:02

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