Find the structure of $\mathbb{Z}[\sqrt[3]{2}]/(4+\sqrt[3]{4})$ Let $A=\mathbb{Z}[\sqrt[3]{2}]$ and $I=(4+\sqrt[3]{2^2})$. Elements in $A$ have the form $a\cdot 1+b\cdot 2^{\frac{1}{3}}+c\cdot 2^{\frac{2}{3}} \Rightarrow$ elements in $I$ have the form
$$
(a\cdot 1+b\cdot 2^{\frac{1}{3}}+c\cdot 2^{\frac{2}{3}})(4+2^{\frac{2}{3}}) = a(4+2^{\frac{2}{3}})+b(2+4\cdot 2^{\frac{1}{3}})+c(2\cdot 2^{\frac{1}{3}}+4\cdot 2^{\frac{2}{3}})
$$
Let $v_1=1, v_2= 2^{\frac{1}{3}}$, and $v_3=2^{\frac{2}{3}}$. Then elements in $I$ are linear combinations of $4v_1+v_3, 2v_1+4v_2$, and $2v_2+4v_3$. To find the number of elements in the group $A/I$, we can take the determinant of the matrix that represents these elements:
$$
\begin{vmatrix}
4&0&1\\ 2&4&0 \\0&2&4
\end{vmatrix}=68
$$
I'd like to conclude that $A/I$ is an abelian group with 68 elements. However, when I was looking at this problem from another perspective, I noticed that the surjection $A\to A/I$ contains 34 in the kernel which makes me think that $A/I$ is actually a cyclic group with 34 elements. I want to know if the fact that 34 is in the kernel says anything about the structure of the group itself, or if it's just a coincidence that $34\cdot 2=68$ turned out to be the determinant calculated above.
 A: No, it's not (entirely) a coincidence.
As an abelian group, $A$ is isomorphic to a direct sum/product of three copies of $\mathbb{Z}$, $A\cong \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$. The fact that $34\in I$ tells you that, again as abelian groups, the map $A\to A/I$ factors through
$$\frac{\mathbb{Z}}{34\mathbb{Z}}\times \mathbb{Z}\times\mathbb{Z}.$$
So of course the $34$ is intimately connected to the size of the group.
As an abelian group, $A$ is generated by the elements $4 + \sqrt[3]{4}$, $-8+\sqrt[3]{2}$, and $1$; the ideal $I$ is generated, again as an abelian group, by $4+\sqrt[3]{4}$, $-16+2\sqrt[3]{2}$, and $34$. So, as an abelian group (I'm not taking the multiplicative structure into account here):
$$\frac{A}{I}\cong \frac{\langle 4+\sqrt[3]{4}\rangle}{\langle 4+\sqrt[3]{4}\rangle} \times \frac{\langle -8+\sqrt[3]{2}\rangle}{2(-8+\sqrt[3]{2})} \times \frac{\langle 1\rangle}{\langle 34\rangle} \cong \frac{\mathbb{Z}}{\mathbb{Z}}\times\frac{\mathbb{Z}}{2\mathbb{Z}}\times \frac{\mathbb{Z}}{34\mathbb{Z}},$$
which tells you that as an abelian group $A/I$ is isomorphic to $C_2\oplus C_{34}$.
