## If F is a finite field, F[x] contains infinitely many primes

Let $F$ be a finite field. Prove that $F[x]$ contains infinitely many prime polynomials.

Suppose to the contrary that $F[x]$ contains only finitely many primes, and label these $p_1,p_2,\ldots,p_n$. Let $q = \prod p_i$, and consider $q+1$. Because $F[x]$ is a Euclidean domain, it is a unique factorization domain. Note that each $p_i$ has degree at least 1, since the constant polynomials in $F[x]$ are units. Thus $q+1$ has degree at least 1, and in particular is not a unit or zero. Thus $q+1$ can be written as a product of irreducibles in $F[x]$, which (again because $F[x]$ is a UFD) are precisely the $p_i$. Suppose $q+1 = p_kd$. Then $1 = p_k(d - q^\prime)$, and in particular $p_k$ is a unit- a contradiction.

Thus $F[x]$ contains infinitely many primes.