TITLE: How do primes split in $\mathbb{Z}[\sqrt[3]{2}]$?
QUESTION [12 upvotes]: I don't know much about the ring $\mathbb{Z}[\sqrt[3]{2}]$.  According to this note it is a Euclidean domain.
I wanted to know how primes split in $\mathbb{Q}[\sqrt[3]{2}]$.  There's an obvious case: $2 = (\sqrt[3]{2})^3$ but this already tells us that splitting occurs.  Apparently $p = 3$ also splits as:
\begin{eqnarray*} 3 &=& \big(\sqrt[3]{2} + 1\big)\big(\sqrt[3]{4} - \sqrt[3]{2} +1\big)  \\
5 &=& \big(\sqrt[3]{4} + 1\big)\big(1 +2 \sqrt[3]{2} - \sqrt[3]{4}\big) 
\end{eqnarray*}
There's a small typo in the note.
If we have a prime $p \in \mathbb{Z}$ do we know when it factors in the ring $\mathbb{Z}[\sqrt[3]{2}]$ ? Can we get more formula like the two listed above?

REPLY [4 votes]: Note $\mathcal{O} = \mathbf{Z}\left[\sqrt[3]{2}\right]$ and $k = \textrm{Frac}(\mathcal{O}) = \mathbf{Q}\left(\sqrt[3]{2}\right)$. How a prime $p\in\mathbf{Z}$ factors (or not) in $\mathcal{O}$ is given by how the polynomial $T^3 - 2$ factors in $\mathbf{F}_p [T]$. (This is a result of Dedekind.) To know the latter we need to know when $2$ is cube in $\mathbf{F}_p$ and when it is and when $p\not=2$ whether or not there is a primitive cubic root of $1$ in $\mathbf{F}_p$. Previous condition for $p\not=3$ is the same as $-3$ being a square in $\mathbf{F}_p$ which is the same (by the law of quadratic reciprocity) as $p \equiv 1[3]$. Take now a prime $p\geq 5$. If $p \equiv 2[3]$ then $2$ is a cube in $\mathbf{F}_p$ and there's no primitive cubic root of $1$ in $\mathbf{F}_p$ so that $(p) = (x)(y)$ where the ideals $(x)$ and $(y)$ have respective norms $p$ and $p^2$. As $-1$ is a norm in $k$, modulo action from $\mathcal{O}^{\times}$, you write $p=xy$ with $N_{k/\mathbf{Q}}(x)=p$ and $N_{k/\mathbf{Q}}(y)=p^2$. If If $p \equiv 1[3]$ then if $2^{\frac{p-1}{3}}\equiv 1[p]$ the polynomial $T^3-2$ splits completely in $\mathbf{F}_p[T]$ and $p$ will be (in $\mathcal{O}$) the product of three elements of norm over $\mathbf{Q}$ equal to $p$. If $2^{\frac{p-1}{3}}\not\equiv 1[p]$ then the polynomial $T^3-2$ is irreducible in $\mathbf{F}_p[T]$ and $p$ is prime in $\mathcal{O}$.