Is there an irreducible polynomial of degree $3$, which is reducible modulo every prime?

Is there an irreducible polynomial of degree $$3$$, which is reducible modulo every prime?

Motivation:

In this question (Irreducible polynomial which is reducible modulo every prime) it is simply proved that $$x^4+1$$ is reducible modulo every prime number.

I am curious about the least possible $$2\leq d$$, such that there is an irreducible polynomial of degree $$d$$, which is reducible modulo every prime.

If $$f(x)$$ is an irreducible polynomial of degree $$2$$, then it is easy to show that there exists a prime such that it is irreducible modulo $$p$$.

• – Grigory M Oct 25 '20 at 22:06
• @GrigoryM +1 for sharing such a nice answer. But I can not find anything relevant about degree $3$ polynomials there. – NeoTheComputer Oct 25 '20 at 22:15
• One has to think through that answer in detail, but "It follows that $f$ is reducible mod $p$ for all $p$ if and only if $G$ does not contain an $n$-cycle. The smallest value of $n$ for which this is possible is $n=4$" is indirectly telling you exactly what you wanted to know. – Greg Martin Oct 25 '20 at 22:20
• @GregMartin Thanks for clarifying. – NeoTheComputer Oct 25 '20 at 22:22

As explained in the answer Grigory M linked to in the comments, an irreducible polynomial $$f(x)$$ is reducible $$\bmod p$$ for every $$p$$ iff the Galois group of (the splitting field of) $$f$$ does not contain an $$n$$-cycle, where $$n = \deg f$$. When $$n = 3$$ the only possible Galois groups are $$S_3 \cong D_3, A_3 \cong C_3$$ both of which contain a $$3$$-cycle so this can't happen for irreducible cubics. In fact we have the following dichotomy:
• If $$\text{Gal}(f) \cong S_3 \cong D_3$$ then $$f$$ is irreducible $$\bmod p$$ for a set of primes $$p$$ with natural density $$\frac{1}{3}$$.
• If $$\text{Gal}(f) \cong A_3 \cong C_3$$ then $$f$$ is irreducible $$\bmod p$$ for a set of primes $$p$$ with natural density $$\frac{2}{3}$$.
This reflects the density of $$3$$-cycles in the Galois group. The latter case can be completely understood using the Kronecker-Weber theorem.
More generally, if $$n = p$$ is prime then a transitive subgroup of $$S_p$$ contains a $$p$$-cycle (exercise), so this can't happen for irreducible polynomials of prime degree. The smallest $$n$$ for which a transitive subgroup of $$S_n$$ need not contain an $$n$$-cycle is $$n = 4$$ and we can take $$C_2 \times C_2$$ to be the Galois group, again as explained in the linked answer.