$x^3+ax+b$ is reducible over infinitely many $\Bbb{F}_p$?

Question

Let $a,b\in \Bbb{Z}$ be a arbitrary fixed integer.

Does there exists infinitely many prime numbers $p$ such that $x^3+ax+b$ is reducible over $\Bbb{F}_p$ ?

I know there is a $p$ such that $f(x)=x^3+ax+b$ is reducible over $\Bbb{F}_p$. Indeed, take $c\in \Bbb{Z}$ such that $f(c)$ is not $±1$. Then $f(x)$ is reducible over every $\Bbb{F}_p$, $p\mid f(c)$. But I'm stuck with how can I gain infinitely many such $p$.

If $x^3+ax+b$ has a root in $\Bbb{F}_p$, then it's reducible over $\Bbb{F}_p$.

So if I could prove $\{x^3+ax+b\mid x\in \Bbb{Z} \}$ contains infinitely prime factors, we are done. How can I prove that ?

Answer 1

Let $f \in \Bbb Z[x]$ of monic of degree $\geq 2$. Without loss of generality, assume that $f$ is irreducible.

Let $\alpha$ be a root of $f$ and consider $\Bbb Z[\alpha]$. This is an order in $\Bbb Q(\alpha)$, so up to finitely many primes, the splitting behaviour of primes $p$ in $\Bbb Q(\alpha)$ is determined by the reduction of $f$ modulo $p$.

Now in an extension of number fields, such as $\Bbb Q(\alpha)/\Bbb Q$, infinitely many primes split completely. This follows for instance from the Chebotarev density theorem. (But I think there are analyis-free proofs available as well). But if a prime $p$ that is coprime to the conductor of $\Bbb Z[\alpha]$ (this condition only rules out finitely many primes) splits completely in $\Bbb Q(\alpha)$, then we get that $f$ factors into linear factors modulo $p$. In particular, $f$ is reducible mod $p$.

Answer 2

I add an argument using the theory of elliptic curves. Let $E:y^2=f(x)$ be an elliptic curve defined over $\mathbb{Q}$.　Here, $f(x)$ is a polynomial in my question.

Suppose $p$ is a prime that splits completely in $\mathbb{Q}(E[2])/\mathbb{Q}$ and is a good prime for the elliptic curve $E/\mathbb{Q}$.

We can choose infinitely many such primes $p$ thanks to the Chebotarev density theorem. In particular, Chebotarev's theorem implies that $1/2$ percent of the primes split completely in $\mathbb{Q}(E[2])/\mathbb{Q}$, meaning there are infinitely many such primes. And we can avoid primes of bad reduction and can choose prime $p$ such that it splits complete in $\Bbb{Q}(E[2])/\Bbb{Q}$ and it has good prime for $E/\Bbb{Q}$.

Then $\mathbb{Q}_p(E[2]) = \mathbb{Q}_p$, and thus $E(\Bbb{Q}_p(E[2])) = E(\mathbb{Q}_p)$ injects into $\tilde{E}(\mathbb{F}_p)$.

Therefore, for a prime $p$ that splits completely in $\mathbb{Q}(E[2])/\mathbb{Q}$, we have $\#E(\mathbb{F}_p)[2] = 4$, which implies that $f(x)$ is reducible over $\mathbb{F}_p$.