For each prime \(p\), give an explicit polynomial of degree \(p\) that is irreducible over the finite field \(\mathbb{F}_p\), constructed in a uniform way across all primes \(p\). Prove that your polynomial is irreducible over \(\mathbb{F}_p\).
Solution
For every prime \(p\), take \[ f(x)=x^p-x-1\in\mathbb F_p[x]. \] Let \(\alpha\) be a root in an algebraic closure. Then \[ \alpha^p=\alpha+1. \] Iterating Frobenius gives \[ \alpha^{p^m}=\alpha+m \] for every integer \(m\ge0\), where \(m\) is read in \(\mathbb F_p\).
The degree of the minimal polynomial of \(\alpha\) over \(\mathbb F_p\) is the size of its Frobenius orbit. The first positive \(m\) for which \[ \alpha^{p^m}=\alpha \] is therefore the first positive \(m\) with \(m=0\) in \(\mathbb F_p\), namely \(m=p\). Hence \[ [\mathbb F_p(\alpha):\mathbb F_p]=p. \] Since \(f\) has degree \(p\) and vanishes at \(\alpha\), it is the minimal polynomial of \(\alpha\). Thus \[ \boxed{x^p-x-1\text{ is irreducible over }\mathbb F_p} \] for every prime \(p\).