Splitting of $X^{3} - 2$ modulo $p$

Question

I am trying to prove the existence of infinitely many primes $p$ modulo which the polynomial $X^{3} - 2$ does not have any root. In particular, I want to find the exact value of the density of the set of such primes. I think somewhere I have to use the Chebotarev Density Theorem but I am unable to do so. Can someone tell how to find the density with/without using the Chebotarev Density Theorem?

Answer 1

Let $f$ be an irreducible polynomial over $\Bbb Z$ with degree $n$. Let $G$ be its Galois group considered as a subgroup of $S_n$ via its action on the zeros of $f$.

Let $a_1,\ldots,a_n$ be nonnegative integers with $a_1+2a_2+\cdots+na_n=n$. Let $A$ be the set of primes $p$ such that the modulo $p$ factorization of $f$ has $a_j$ factors of degree $j$, for each $j$ Chebotarev density means the Dirichlet density of $P$ is $c/|G|$ where $c$ is the number of elements of $G$ with cycle structure $1^{a_1}2^{a_2}\cdots n^{a_n}$. In particular, the set of primes $p$ with $f$ irreducible modulo $p$ has Dirichlet density $c/|G|$ where $c$ is the number of $n$-cycles in $G$. This density is $1/n$ when $G=S_n$.

In your example where $f=X^2-2$, $G=S_3$ and the Dirichlet density you seek is $1/3$.

Answer 2

In his non-analytic proof of the main theorems on class field theory, Chevalley had to prove the existence of prime ideals that remain inert in cyclic extensions without using any density theorem. You should find this result in a few books on idelic class field theory. In your case, look at the polynomial $X^3 - 2$ over ${\mathbb Q}(\sqrt{-3})$.

Answer 3

Let us assume $p>3$. If $p\equiv 2\pmod{3}$, the map $x\mapsto x^3$ is bijective over $\mathbb{Z}/(p\mathbb{Z})$, hence $x^3-2$ has for sure a root in $\mathbb{F}_p$. If $p\equiv 1\pmod{3}$, the map $x\mapsto x^3$ sends $\mathbb{Z}/(p\mathbb{Z})^*$ into a subgroup $H$ with $\frac{p-1}{3}$ elements, and $x^3-2$ factors over $\mathbb{F}_p$ iff $2\in H$. If $p\equiv 1\pmod{3}$, this is expected to happen with a probability $\approx\frac{1}{3}$, since the size of $H$ is roughly one third of the size of $\mathbb{Z}/(p\mathbb{Z})^*$.

It follows that a for a random prime $p$ in the interval $[4,M]$, the polynomial $x^3-2$ has no root in $\mathbb{F}_p$ with a probability given by

$$\mathbb{P}[p\equiv 1\!\!\!\!\pmod{3}]\cdot \mathbb{P}[2\not\in H] \approx \frac{1}{2}\cdot\frac{2}{3}=\color{red}{\frac{1}{3}}.$$ You should not meet any issue in turning this argument into a rigorous argument through Chebotarev Density Theorem. Cubic reciprocity can be used together, or in place of the previous result.