Ramification of primes without knowing the discriminant Let $\mathbb{K} = \mathbb{Q}[\sqrt[3]{5}] \ $, and let $\mathbb{L}$ be the normal closure of $\mathbb{K}$. 
Let $\mathbb{O}_{\mathbb{K}} \ $ be the integral closure of $\mathbb{Z}$ in $K$ and $\mathbb{O}_{\mathbb{L}} \ $ be the integral closure of $\mathbb{Z}$ in $\mathbb{L}$. I want to find the factorization of the primes $7, 11 \ $ and $13$ as ideals in $\mathbb{O}_{\mathbb{K}} \ $ and $\mathbb{O}_{\mathbb{L}} \ $. My question is : how can I do that without knowing an integral basis for $\mathbb{K}$ and $\mathbb{L}$ ?
 A: For a prime $p$ in $\mathbb Z$ unramified in an extension $K=\mathbb Q(\beta)$, if we are merely close to the true ring of algebraic integers, meaning that $p$ does not divide the discriminate of $\mathbb Z[\beta]$, then the localization at $p$ of the true ring of integers is the same as the localization of this good approximation. Thus, looking how the minimal polynomial for $\beta$ factors mod $p$ will show how $p$ splits.
A: I tend to work on such things on a fairly mindless and ad hoc basis. Let me show you my approach just for the case $p=7$:
The extension $\mathbb K\supset\mathbb Q$ is of degree $3$, and in the downstairs residue field $\mathbb F_7$, $5$ is a generator of the multiplicative group so that $\lambda=5^{1/3}$ has period $18$. You need to go all the way to $\mathbb F_{243}$ to get $18|(7^n-1)$, so that the residue-field extension degree is three, and the prime $7$ does not split at all in $\mathbb K$. On the other hand, we have $(\frac{-3}7)=(\frac47)=1$, so that in $\mathbb Q(\omega)$, where $\omega$ is a cube root of unity, $7$ should split, indeed $7=(2-\omega)(2-\omega^2)$. So there are at least two primes above $7$ in $\mathbb L$, but we know that each of these has residue-field extension degree $3$. So there you are, you can now try the same kind of argument at $11$ and $13$.
A: A usual method is to look at $f=x^3-5$ in $\mathbb F_p[x]$ ($p$ the prime you want to factorize). If $f$ has no multiple factors then the factorization of $f$ gives you the factorization of $p$; $p$ is unramified. Namely, $f=f_1f_2...f_k$ means $p=P_1P_2...P_k$, the norm of $P_i$ is $p$ to the degree of $f_i$, and $P_i=(p,f_i(\alpha))$ ($\alpha=\sqrt[3]{5}$).This is the case for all $p$'s you consider, as $f'=3x^2$.
As an example, for $p=7$, $x^3-5$ has no root in $\mathbb F_7[x]$ (if I didn't miss it), i.e. $(p)$ stays a prime ideal on $O_K$.
Should it fail (i.e. should $f$ have multiple factors in $\mathbb F_p[x]$), one can always look at the factorization of $f$ in $\mathbb Q_p[x]$ + some (not so small) details.
