Splitting of $2$ in a cubic extension Let $L$ be the splitting field of $X^3-3X+1=0$. How does the prime $2$ split in $L$? I have figured out that either $2=\mathfrak{P}$ or $2=\mathfrak{P}\mathfrak{Q}\mathfrak{R}$. I guess it is the second form, but I do not know how to find the prime divisors.
 A: Use this proposition from p. 47 of Algebraic Number Theory by Neukirch: Let $L/K$ be a separable extension of number fields and let $\mathcal{o}$ and $\mathcal{O}$ be the rings of integers of $K$ and $L$, respectively. Assume that $L=K(\theta)$ for some $\theta\in\mathcal{O}$ with minimal polynomial over $K$ being $p(X)\in\mathcal{o}[X]$, then we have:

(8.3) Proposition. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{o}$ which is relatively prime to the conductor $\mathfrak{F}$ of $\mathcal{o}[\theta]$, and let
  $$
  \overline{p}(X)=\overline{p}_1(X)^{e_1}\cdots\overline{p}_r(X)^{e_r}
$$
  be the factorization of the polynomial $\overline{p}(X)=p(X)\pmod{\mathfrak{p}}$ into irreducibles $\overline{p}_i(X)=p_i(X)\pmod{\mathfrak{p}}$ over the residue class field $\mathcal{o}/\mathfrak{p}$, with all $p_i(X)\in\mathcal{o}[X]$ monic. Then
  $$
  \mathfrak{P}_i=\mathfrak{p}\mathcal{O}+p_i(\theta)\mathcal{O},\quad i=1,\ldots,r,
$$
  are the different prime ideals of $\mathcal{O}$ above $\mathfrak{p}$. The inertia degree $f_i$ of $\mathfrak{
P}_i$ is the degree of $\overline{p}_i(X)$, and one has
  $$
  \mathfrak{p}=\mathfrak{P}^{e_1}_1\cdots\mathfrak{P}^{e_r}_r.
$$

Now we apply this to our problem.
In our problem $K=\mathbb{Q}$ and $L=\mathbb{Q}(\theta)$, where $\theta$ is a root of $p(X)=X^3-3X+1$. $\mathcal{o}=\mathbb{Z}$ and $\mathcal{O}=\mathbb{Z}[\theta]$. The minimal polynomial of $\theta\in\mathcal{O}$ over $\mathbb{Q}$ is $p(X)\in\mathbb{Z}[X]$. Let $\mathfrak{p}=2\mathbb{Z}$. Since the conductor of $\mathbb{Z}[\theta]$ is $\mathbb{Z}[\theta]$ itself, $2\mathbb{Z}$ is relatively prime to it. Now $\overline{p}(X)=p(X)\pmod{2}=X^3-3X+1\pmod{2}$. $\overline{p}(X)$ is irreducible over $\mathbb{Z}/2\mathbb{Z}$ because it is of degree $3$ and has no roots in $\mathbb{Z}/2\mathbb{Z}$. Whence the factorization of $\overline{p}(X)$ into irreducibles over $\mathbb{Z}/2\mathbb{Z}$ is just itself, namely
$$
  \overline{p}(X)=X^3-3X+1\pmod{2}.
$$
Notice that $X^3-3X+1\in\mathbb{Z}[X]$ is monic. Then
$$
\mathfrak{P}=\mathfrak{p}\mathcal{O}+p(\theta)\mathcal{O}=2\mathbb{Z}\mathbb{Z}[\theta]+0\mathbb{Z}[\theta]=2\mathbb{Z}[\theta]
$$
is the only prime ideal of $\mathbb{Z}[\theta]$ above $2\mathbb{Z}$. The inertia degree of $2\mathbb{Z}[\theta]$ is the degree of $X^3-3X+1\pmod{2}$, i.e. $3$, and one has
$$
  2\mathbb{Z}=2\mathbb{Z}[\theta].
$$
In particular, $2\mathbb{Z}$ is inert in $\mathbb{Q}(\theta)$, i.e. $2$ is inert in $\mathbb{Q}(\theta)$.
