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Given an irreducible polynomial $f(x)$ with integer coefficients, I want to characterize all the primes $p$ based on how $f(x)$ factors $\mod p$. For quadratic polynomials, it follows directly from the Law of Quadratic Reciprocity. Specifically, if $f(x)$ has discriminant $D$, then $f(x)$ splits completely $\mod p$ ($p>2$) precisely when $D$ is a square $\mod p$ and remains irreducible otherwise. I want to generalize this as much as I can for the case where $f(x)$ is a cubic polynomial. Specifically, I want to determine for which primes, $f$ splits, remains irreducible, or factors as a product of a linear and quadratic polynomial. Any hints, full or partial solutions, or references to papers or textbooks with ideas that could help me would be greatly appreciated! Here's what I've done so far:

Let $f(x)$ be an irreducible cubic polynomial with integer coefficients. Let $\alpha, \beta, \gamma$ be the roots and $D = (\alpha-\beta)^2(\beta-\gamma)^2(\alpha-\gamma)^2$ its discriminant. Let $p$ be a prime. For simplicity, I'm assuming $p>3$ and that $p \nmid D$. Let $K/\mathbb{Q}$ be its splitting field and $G$ its Galois group. Let $K_p/\mathbb{F}_p$ be the splitting field of $f$ over the finite field with $p$ elements.

Case 1: $D = d^2$ for some integer $d$. In this case, $G$ is cyclic of order 3. This implies that $\beta$ and $\gamma$ can be written as polynomials in $\alpha$, so we have either $f$ splits completely $\mod p$, or $f$ remains irreducible. I'm conjecturing that $K \subset \mathbb{Q}(\zeta_d)$ ($\zeta_d$ is a primitive $d$-th root of unity) and from there I can obtain congruence conditions for which the polynomial splits. I imagine that this can be proven easily using Class Field Theory, but I'm still in the process of learning it. My main idea so far has been to use Gaussian periods corresponding to index 3 subgroups of $(\mathbb{Z}/d\mathbb{Z})^\times$. It worked for all the explicit examples I've checked, but I have no idea how to prove it works in general.

Case 2: $D$ is not a square. If $\left(\frac{D}{p}\right) = -1$, then I'm able to show that $f$ has exactly one root in $\mathbb{F}_p$ as then $\mathbb{F}_p(\sqrt{D})$ is a quadratic subextension of $K_p$, which implies $K_p$ is itself a quadratic extension. I have no idea how to show the converse statement or if it is even true.

If $\left(\frac{D}{p}\right) = 1$, things get a lot more complicated and this where I'm having the most trouble. In the case $f(x) = x^3 - 2$, we have that $D = -3\cdot6^2$. Thus, $$\left(\frac{D}{p}\right) = \left(\frac{-3\cdot6^2}{p}\right) = \left(\frac{-3}{p}\right) = 1 \iff p \equiv 1\mod 3.$$ In this case, I know that there exists integers $A,B$ such that $p = A^2 + 3B^2$ and that $f$ splits $\mod p$ iff $3 | B$. These congruence conditions were derived from the Law of Cubic Reciprocity, which heavily relies on the Eisenstein integers and $p = A^2 + 3B^2$ being possible because the Eisenstein integers are a UFD.

Back to the general case, my first guess has been to write $$kp = A^2 - DB^2$$ for some appropriate $k$, and then hopefully be able to derive some congruence conditions for $A$ and $B$. I imagine that this might involve deriving some kind of analogue for Cubic Reciprocity, but the ring of integers of $\mathbb{Q}(\sqrt{D})$ would replace the Eisenstein integers.

My next idea (which might be more fruitful), is to use modular (or automorphic) forms. Admittedly, I don't know much about them, just a few basic facts like how they form a finite dimensional $\mathbb{C}$-vector space etc. If I try this, my approach would be to first try to use modular forms (after learning more about them) to try solving the case $f(x) = x^3 - 2$, replacing $x^2 + 3y^2$ with $$\theta = \sum_{i,j \in \mathbb{Z}} q^{x^2+3y^2}$$ and see if this sheds any light on the general problem. I've heard tidbits from people much smarter than me that makes me think this could maybe work out, but I'm mostly in the dark here.

If you have any input on what would be the best approach, or if you're thinking, "From what you've written, it doesn't seem like you know enough math. You should probably know X,Y, and Z before you can even hope to solve this," let me know! I'd be glad to hear any solution, even if it goes way over my head! I've been trying to solve this problem for over a year now, and I feel as if I've made no significant progress. It's VERY disheartening.

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  • $\begingroup$ First and foremost: Welcome to Math.SE! A nice first post! $\endgroup$ – Jyrki Lahtonen Jun 9 at 5:16
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    $\begingroup$ Indeed, class field theory has a lot to say about this. Hopefully you soon get someone well versed in it to answer. Below I will say what I can. Doesn't amount to much :-( $\endgroup$ – Jyrki Lahtonen Jun 9 at 5:18
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    $\begingroup$ In Case 1 Kronecker-Weber says that $K$ is contained in some cyclotomic field. I think your plan is fine. For sure you can describe the cubic subfields with the aid of Gaussian periods. But aren't the cubic subfields the fixed fields of subgroups of index three (rather than of order three)? $\endgroup$ – Jyrki Lahtonen Jun 9 at 5:23
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    $\begingroup$ Case 2 is probably more complicated. IIRC (wait for people who really know) the splitting pattern of a cubic $f(x)\in\Bbb{Z}[x]$ modulo primes can be described using a congruence if and only if the Galois group $Gal(K/\Bbb{Q})$ is abelian. Here (assuming $f(x)$ is irreducible over the rationals) the Galois group is $S_3$, and there cannot be such a pattern. More precisely, there cannot exist an iinteger $F$ such that the splitting behavior of $f(x)$ modulo $p$ would only depend on the residue class of $p$ modulo $F$. $\endgroup$ – Jyrki Lahtonen Jun 9 at 5:27
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    $\begingroup$ Yes, congruence describes splitting iff abelian Galois group. It is this MO question. $\endgroup$ – user10354138 Jun 9 at 5:44

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