Find prime ideals of the ring $\Bbb Z [ \sqrt[3]2]$ which contain $5$ 
Find all prime ideals $p$ of the ring $ R= \Bbb Z [ \sqrt[3]2] $ such that $ 5 \in p$ and find $R/p$ for all of them.

I know that of course $R$ is prime and $R/R = \{0 \}$.  
Unfortunately I have no idea what else can be done there.
 A: Primes in the ring $\mathbb Z[\sqrt[3]{2}]$ over $5$ correspond to primes in $\mathbb Z[T]$ over $(5,T^3-2)$, which correspond to primes in $\mathbb F_5[T]$ over $(T^3-2)$, which correspond to the irreducible factors of $T^3-2$. So all you have to do is decomposing $T^3-2$ into irreducible factors over $\mathbb F_5$ and once you got this, the primes are given by $(5,f(\sqrt[3]{2}))$, where $f$ runs through the irreducible factors.
Edit: Some more details. It is well known that prime ideals, which contain an ideal $I$, correspond to prime ideals of $R/I$. If you are not aware of this, you should read about this. So we have to look at the ring $\mathbb Z[\sqrt[3]{2}]/(5)$ to find primes, that contain $5$. Then we look at the chain of isomorphisms
$$\mathbb Z[\sqrt[3]{2}]/(5) \cong \mathbb Z[T]/(5,T^3-2) \cong \mathbb Z/5\mathbb Z[T]/(T^3-2)$$
and see that we have to find prime ideals in $\mathbb Z/5\mathbb Z[T]$, that contain $(T^3-2)$. But then it is also well known that in a principal ideal domain, primes over an ideal correspond to the prime factors of the generator of that ideal.
A: First of all $R$ is not a prime ideal as the zero ring is not a domain. Recall that you define what a domain is, and then what a prime ideal is, by saying that the quotient of $R$ by it is a domain, or equivalently that its complement in $R$ is stable by finite products.
Note that $R \simeq \mathbf{Z}[T] / (T^3 - 2)$ by $\sqrt[3]{2}\mapsto \overline{T}$ so that prime ideals of $R$ containing $5$ are given by prime ideals of $\mathbf{Z}[T]$ containing $(T^3 - 2)$ such that their image in $\mathbf{Z}[T] / (T^3 - 2)$ contains (the image of) $5$, and these simply correspond to prime ideals of $\mathbf{Z}[T]$ containing $(T^3 - 2)$ and $5$. What could you now say about elements $T^3 - 2$ and $5$ in $\mathbf{Z}[T]$ ?
A: Start with $$\Bbb Z[\sqrt[3]2]\simeq\mathbb Z[X]/(X^3-2).$$ Let $P$ be a prime ideal of $\mathbb Z[X]$ containing $5$ and $X^3-2$, so $(5,X^3-2)\subseteq P$. Now look at $\mathbb Z[X]/(5,X^3-2)$ and notice that it is isomorphic to $(\mathbb Z/5\mathbb Z)[X]/(X^3-2)$. In $(\mathbb Z/5\mathbb Z)[X]$ we have $X^3-2=(X-3)(X^2+3X+4)$, and then, by CRT, we get $$(\mathbb Z/5\mathbb Z)[X]/(X^3-2)\simeq(\mathbb Z/5\mathbb Z)[X]/(X-3)\times(\mathbb Z/5\mathbb Z)[X]/(X^2+3X+4).$$ Since $(\mathbb Z/5\mathbb Z)[X]\simeq \mathbb Z/5\mathbb Z$, and $X^2+3X+4$ is irreducible, we get that $\mathbb Z[X]/(X^3-2)$ is a product of two finite fields, so it has exactly two prime (maximal) ideals, and their factor rings are the fields with $5$, respectively $25$ elements.
The above correspondence can tell you which one are the two prime ideals in $ \Bbb Z[\sqrt[3]2]$: $P_1=(5,\sqrt[3]2-2)$, respectively $P_2=(5,\sqrt[3]4+3\sqrt[3]2+4)$.
