TITLE: How does 11 split in the ring $\mathbb{Z}[\sqrt[3]{2}]$
QUESTION [10 upvotes]: I learned that the splitting of primes in a number field $K = \mathbb{Q}(x)/p(x)$ depends on the factorization of $p(x) \pmod p$.  While this is not at all obvious to me, let's use it:
$$x^3 - 2 \equiv (x-7)(x^2 + 7x + 6) \pmod {11}$$
and I do not think that second factor splits.  What does this say about the factorization of $p = 11 \in \mathbb{Z}[\sqrt[3]{2}]$ ?

My best guess is that $p = 11$ splits as a linear and a quadratic.  For some integers: $a,b,c,d,e \in \mathbb{Z}$ we have:
\begin{eqnarray*} 11 &=& (a + b\sqrt[3]{2}) (c + d \sqrt[3]{2}+ e\sqrt[3]{4}) \\ 
&=& (ac + 2be) + (bc + ad)\sqrt[3]{2} + (bd+ae)\sqrt[3]{4}\end{eqnarray*}
Most by tautology, we get 3 equations in 5 unknown integers and we can try to solve it:
\begin{eqnarray*}
ac + 2be &=& 11 \\
bc + \,\,ad &=& 0 \\
bd + \,\,ae &=& 0
\end{eqnarray*}
This is looking pretty bad, however I noticed the three determinants.  And maybe that could be useful.

REPLY [3 votes]: I'm not a bounty sniper, I swear. I'm just thinking out loud. I'm wondering how much of the naïve methods that work for quadratic rings can be carried over to cubic rings.
For example, we find that 11 splits in $\mathbb Z[\sqrt{-2}]$ since $-2$ is a square modulo 11, e.g., $3^2 = 11 - 2$. Then $3 + \sqrt{-2}$ is a divisor of 11, and we find that $$\frac{11}{3 + \sqrt{-2}} = 3 - \sqrt{-2}.$$ Since those are both conjugates with norms of 11 each, we're done.
Similarly in $\mathbb Z[\root 3 \of 2]$ with find that 2 is a cube modulo 11, e.g., $7^3 \equiv 2 \pmod{11}$. But $N(7 + \root 3 \of 2) = 345$... damn it. Gotta do $N(7 - \root 3 \of 2) = 341$ instead, but then we must figure out how to find $x \in \mathbb Z[\root 3 \of 2]$ such that $\langle 11, 7 - \root 3 \of 2 \rangle = \langle x \rangle$. I guess that can't be done by naïve methods.