Polynomial with integer coefficients which is irreducible over all prime fields

Question

I have been thinking about this for a bit and I feel like the answer is no, but I cannot prove it. Formally, can there be a polynomial $f(x)$ which is irreducible over $\mathbb{F}_{p} \forall p$ where $f(x) \in \mathbb{Z}[x]$?

At first I thought that a polynomial like $f(x) = x^2 - p$ for some prime $p$ might work, but for instance in $\mathbb{F}_{7}$ you have $4^{2} \equiv 2$ $(mod \, 7)$.

Then I though maybe considering Galois groups over $\mathbb{F}_{p}$ might lead somewhere, but I couldn't figure anything out there either. I have the suspicion that there is a very obvious answer that I'm missing, any feedback would be much appreciated!

Answer 1

I don't think it's as complicated:

Let $P(x)$ be a (nonconstant) polynomial with integer coefficients. $P(x)$ (Considered as a polynomial $\mathbb{C} \to \mathbb{C}$) can only take the values $1, -1$ finitely many times, or else it would be either constant or not a polynomial.

So take some integer $n$, such that $P(n) \neq 1, -1$

If $P(n) = 0$, then it will be reducible for any prime ($n$ is a root). If $P(n) \neq 0$, then factor $P(n)$ into primes, its prime factorization has at least $1$ prime, call such a prime $p$, as $P(n) \neq 1, 0, -1$.

$n \mod p$ is then a root in $\mathbb{F}_p$

Answer 2

Any non constant polynomial $f \in \mathbf Z[X] $ has a root mod $p$ for infinitely many primes $p$. This is exercise 30 of chapter 3 in Marcus' "Number Fields" (hint : prove this under the assumption $f(0)=1$ by considering prime divisors of the numbers $f(n!)$ ; then reduce to this case by setting $g(X)=X(f(0)X))/f(0))$.

If you want to use more powerful machinery, consider the splitting field $K$ of $f$ over $\mathbf Q$ . Tchebotarev tells us that infinitely many primes $p$ are completely decomposed in $K$, so that any root of $f$ belongs to infinitely many $p$-adic rings $\mathbf Z_p$, which implies that any root modulo $p$ belongs to infinitely many prime fields $\mathbf F_p$ .