No finite field is algebraically closed.Suggest a slogan
A finite field \(F\) is not algebraically closed.
Write \(F = \theset{a_1, \ldots, a_q}\) and let \(f(x) \coloneqq 1 + \prod_{i=1}^q (x - a_i) \in F[x]\). Then \(f\) is nonconstant, and \(f(a_j) = 1 \neq 0\) for every \(j\), so \(f\) has no root in \(F\).