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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 $p$.

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    $\begingroup$ No. See math.stackexchange.com/a/160855/152 $\endgroup$ – Grigory M Oct 25 '20 at 22:06
  • $\begingroup$ @GrigoryM +1 for sharing such a nice answer. But I can not find anything relevant about degree $3$ polynomials there. $\endgroup$ – NeoTheComputer Oct 25 '20 at 22:15
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    $\begingroup$ 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. $\endgroup$ – Greg Martin Oct 25 '20 at 22:20
  • $\begingroup$ @GregMartin Thanks for clarifying. $\endgroup$ – NeoTheComputer Oct 25 '20 at 22:22
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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.

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