How does 11 split in the ring $\mathbb{Z}[\sqrt[3]{2}]$

The factorization $x^3 - 2$ mod $11$ tells us about the factorization of the ideal $(11)$ in $\mathbb{Z}[\sqrt[3]{2}]$. Let $K = \mathbb{Q}(\sqrt[3]{2})$ and let $\mathcal{O}_K = \mathbb{Z}[\sqrt[3]{2}]$ be its ring of integers. Then the factorization of the ideal $(11)$ can be detected from the structure of $\mathcal{O}_K/(11)$. By the Third Isomorphism Theorem, we have $$ \frac{\mathbb{Z}[\sqrt[3]{2}]}{(11)} \cong \frac{\mathbb{Z}[x]}{(11, x^3 - 2)} \cong \frac{(\mathbb{Z}/11\mathbb{Z})[x]}{(x^3 - 2)} $$ and the structure of this last ring is determined by the factorization of $x^3 - 2$ mod $11$. For more on this, I recommend Keith Conrad's blurb Factoring After Dedekind.

So factoring mod $11$, we find $x^3 - 2 = (x - 7)(x^2 + 7x + 5)$. By Theorem 8 in the linked PDF, then we have the factorization of ideals $$ (11) = (11, \alpha - 7)(11, \alpha^2 + 7 \alpha + 5) = \mathfrak{p}_1 \mathfrak{p}_2 $$ where $\alpha = \sqrt[3]{2}$.

However, this hasn't yet given us a factorization of the element $11$. It turns out that $\mathbb{Z}[\sqrt[3]{2}]$ is a PID, so the prime ideals above are principal. Since $\mathfrak{p}_1, \mathfrak{p}_2$ have norms $11$ and $121$, then they must be generated by elements of norm $\pm 11$ and $\pm 121$. Finding such elements amounts to solving the Diophantine equations $$ a^3 - 6 abc + 2 b^3 + 4 c^3 = N(a + b \sqrt[3]{2} + c \sqrt[3]{2}^2) = \pm 11, \pm 121 \, . $$ I don't know of a good way to do this in general, but for this case it turns out that $\mathfrak{p}_1 = (1 - \sqrt[3]{2} - \sqrt[3]{2}^2)$ and $\mathfrak{p}_2 = (1 - 3 \sqrt[3]{2} - 2 \sqrt[3]{2}^2)$ and $$ 11 = (1 - \sqrt[3]{2} - \sqrt[3]{2}^2)(1 - 3 \sqrt[3]{2} - 2 \sqrt[3]{2}^2) \, . $$

Addendum: Here’s how we can use lattice reduction algorithms to find generators for $\mathfrak{p}_1$ and $\mathfrak{p}_2$. Using the Minkowski embedding we embed $\mathfrak{p}_1$ and $\mathfrak{p}_2$, obtaining lattices $L_1$ and $L_2$ in $\mathbb{R} \times \mathbb{C} \cong \mathbb{R}^3$. The LLL algorithm allows us to find short vectors \begin{align*} v_1 &= (1.8473221018630726395, -3.4275743288228073350, -0.40107945302261334667)\\ v_2 &= (5.9545652536210184438, -6.3318338134128589259, 0.74092293817182559026) \end{align*} for $L_1$ and $L_2$, respectively, which should correspond to generators for $\mathfrak{p}_1$ and $\mathfrak{p}_2$. The first entries $\alpha_1$ and $\alpha_2$ of each of these vectors corresponds to the real embedding. To recognize these real numbers as algebraic numbers, we embed $1, \sqrt[3]{2}, \sqrt[3]{2}^2$ in $\mathbb{R}$, and then use numerical linear algebra to compute (approximate) linear relations among $\alpha_1, 1, \sqrt[3]{2}, \sqrt[3]{2}^2$ and among $\alpha_2, 1, \sqrt[3]{2}, \sqrt[3]{2}^2$. Using the following code in Magma

QQ := Rationals();
QQx<x> := PolynomialRing(QQ);
f := x^3 - 2;
K<nu> := NumberField(f);
OK := Integers(K);
places := InfinitePlaces(K);
P1 := ideal< OK | 11, nu - 7>;
P2 := ideal< OK | 11, nu^2 + 7*nu + 5>;
L1 := Lattice(P1);
L2 := Lattice(P2);
v1 := ShortestVectors(L1)[1];
v2 := ShortestVectors(L2)[1];
nu_RR := Evaluate(nu, places[1]);
LinearRelation([ComplexField() | v1[1], 1, nu_RR, nu_RR^2]);
LinearRelation([ComplexField() | v2[1], 1, nu_RR, nu_RR^2]);

we find relations $$ [ 1, 1, -1, -1 ], [ 1, 1, -3, -2 ] $$ which yield exactly the generators we eyeballed before. You can try the code out on the online Magma calculator here: http://magma.maths.usyd.edu.au/calc/ .

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.