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Let $f$ be the polynomial $X^3 -rX -s$, where $r>0$ and $s>0$ and let $\Delta$ be its discriminant. Suppose that $\Delta$ is a square, and that $p$ and $q$ are primes such that $\Delta$ divides $p-q$. Under those conditions, is it always the case that $f$ is irreducible over $F_{p}$ iff f is irreducible over $F_{q}$? If $\Delta$ is not a square, it's easy to find counterexamples. For quadratic polynomials $X^2 -a$, one has periodicity in $n$ of the Jacobi symbol $(n/a)$ and a quadratic reciprocity law for $(n/a)$ and $(a/n)$ when $a$ and $n$ are odd.

Jyrki Lahtonen
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David Bernier
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  • Yes, $r$ and $s$ are positive integers. The discriminant $\Delta$ is for the polynomial over $Z$. I could share code in PARI/gp that does random tests concerning this question. – David Bernier Nov 08 '24 at 04:41

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This basically uses the ideas in Jyrki's comments. If $p,q\mid \Delta$ the result is clear because then $f$ has a square factor mod both $p,q$. Otherwise $f$ is separable mod $p,q$. Then the splitting of $f$ mod $p$ determines the splitting of $p$ in $K=\Bbb Q[t]/(f)$ and conversely. Since $\Delta$ is a square, $K$ is an abelian extension of $\Bbb Q$. Let $N$ be the conductor of $K$. Then $N$ divides the discriminant of $K$, in fact $\Delta_K=N^2$ (see this, or this, maybe there is a way to avoid this? I don't know enough about the relation between conductor/discriminant). Also $\Delta_K\mid \Delta$. So if $p\equiv q\pmod\Delta$, then $p\equiv q\pmod N$, so they have the same image under the Artin map $(\Bbb Z/N\Bbb Z)^\times\to\operatorname{Gal}(K/\Bbb Q)$ and from this we get that $p,q$ have the same splitting behaviour in $K$. In particular $f$ is irreducible mod $p$ iff $f$ is irreducible mod $q$.

In a way this is a generalization of the proof of the quadratic case you mention.

leoli1
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