How to factor in cubic extensions? Working within the field $K=\mathbb{Q}(\sqrt[3]{n})$, for any cube root of $n$, how does one factor the unramified rational prime ideals $(p)$? 
For starters, I'm relatively new to this and not too sure I completely understand factoring in these extensions, and so I'll instead talk about factoring $x^3-n$ over $\mathbb{F}_{p}$.
Assuming for simplicity that $\mathcal{O}^{K}=\mathbb{Z}(\sqrt[3]{n})$, it's obvious that if $p \equiv 2 \pmod{3}$, then $x^3-n=0$ has exactly one solution in $\mathbb{F}_{p}$. For $p \equiv 1 \pmod{3}$, I have no clue what is supposed to be done. 
I know that there must be three solutions or no solutions, but getting there is the problem. All I've been able to think of that might be key to figuring this out is assigning the Frobenius a conjugacy clas s of $S_{3}$ due to the fact that $\mid\rho(p)\mid=\chi_{3}(p)$, where $\rho(p)$ is some representation of an element within the conjugacy class and $\chi_{3}(p)$ is the non-trivial Dirichlet Character of modulus 3. 
This seems nice since it  alludes to working within $\mathbb{Z}(\frac{-1+\sqrt{-3}}{2})$ for the answer, and for one case, $\mathbb{Q}(\sqrt[3]{2})$, the solution does indeed involve $\mathbb{Z}(\frac{-1+\sqrt{-3}}{2})$ (understanding how this connection is made is difficult).
 A: The whole process is given in Şaban Alaca and Kenneth Williams introductory book Introductory Algebraic Number Theory $([A1])$ that contains many examples, particularly in cubic rings. In chapter $10$ this process is explained. But there are many other good books that explain this process.
Let $\varphi\in\mathcal{O}_{K}$ be an algebraic integer that generates the number field $K=\mathbb Q(\varphi)$. The discriminant of this algebraic integer $\varphi$ (definition $6.4.2, [A1]$) divided by the discriminant of the number field $K$ (definition $7.1.2, [A1]$) is the square of the index of this algebraic integer $\varphi$ (definition $7.1.4, [A1]$). If the prime p does not divide this index Dedekind has given a procedure of factorizing this prime to prime ideals (theorem $10.3.1$ and $10.5.1, [A1]$). You simply factorize the minimal monic polynomial $f(\varphi)\in\mathbb Z[\varphi]$ of the rational integer $\varphi$ over the field $\mathbb F_p$
$$f(\varphi)\equiv g_1(\varphi)^{e_1}\cdot g_2(\varphi)^{e_2}\cdots g_k(\varphi)^{e_k}\;\;\;\;(mod\;p)$$
to irreducible polynomials $g_k(\varphi)$ over $\mathbb F_p$. Then the prime $p$ factorizes to the prime ideals
$$\langle p\rangle=\bigl\langle p,g_1(\varphi)\bigr\rangle ^{e_1}\,\bigl\langle p,g_2(\varphi)\bigl\rangle^{e_2}\cdots\bigl\langle p,g_k(\varphi)\bigl\rangle^{e_k}.$$
If the prime p divides the index of the algebraic integer $\varphi$ you can try to find another algebraic integer $\vartheta$ generating the number field $K$ with an index not divisible by $p$. This does not always help because there are algebraic rings of integers in which the index of every algebraic integer is divisible by this prime (see example $7.4.3, [A1]$). Those primes will normally be small so that you can determine the possible ideals by simply running through the ideals
$$\langle p, \vartheta\rangle$$
(in a brute force attack) with an integral basis $(\omega_k)$ of $\mathcal{O}_{K}$, the algebraic integer $\vartheta=\sum_k a_k\omega_k$ and the coeffcients $0\le a_k\lt p$. An ideal with norm $p$ will certainly be prime ($10.1.6, [A1]$).
If the polynomial $f(\varphi)$ factorizes in a cubic field over the field $\mathbb F_p$ it must have a linear factor $\varphi-z_0$. The linear factor is quickly found because $z_0$ is a zero of the polynomial $f(\varphi)$ over the field $\mathbb F_p$. The factorization of the remaining quadratic polynomial is just as easy. If the polynomial is irreducible over the field $\mathbb F_p$, the ideal $\langle p\rangle$ is prime. You cannot rely on a factorization to only one prime ideal $\langle p\rangle$ or to three prime ideals in a cubic ring. In example $10.4.1, [A1],$ the cubic ring of integers of the field $\mathbb Q(\sqrt[3]{2})$ factorizes the prime $5$ to only $2$ prime ideals.
