TITLE: Can someone explain tame and wild ramification in cubic integer rings?
QUESTION [14 upvotes]: From my slow study of quadratic integer rings, it seems that there is only one "level" of ramification in those, so no one bothers to say if the ramification is "tame" or "wild," but I have seen those terms in regards to cubic algebraic integers.
To try to understand this, I thought I'd look at $\mathbb{Z}[\root 3 \of {12}]$. Obviously $2$ and $3$ are ramified. But then I ran into the problem that I can't factor either of those numbers as readily as I can in a ring like $\mathbb{Z}[\sqrt{6}]$ or $\mathbb{Z}[\sqrt{30}]$ (I don't yet dare try to make sense of cubic integer rings not having unique factorization).
Although I do want to factor $2$ and $3$ in $\mathbb{Z}[\root 3 \of {12}]$, I can determine without factoring the "level" of their ramification, right? And if I make that determination, does it provide any clues for factoring ramified primes?

REPLY [8 votes]: It is a simple and well known fact that if $f$ is an Eisenstein polynomial for the prime number $p$, then $p$ is completely ramified in the extension generated by a root of $f$. Clearly $f(x) = x^3 - 12$ is Eisenstein for $3$,
and since ${\mathbb Q}(\sqrt[3]{12}) =  {\mathbb Q}(\sqrt[3]{18})$ and $g(x) = x^3-18$ is Eisenstein for $p = 2$, the prime $2$ is also completely ramified.
According to pari, the generators of the prime ideals are
$2 + \sqrt[3]{12} + \sqrt[3]{18}$ for the prime ideal above $2$ and
$3 + \sqrt[3]{12} + \sqrt[3]{18}$ for the prime ideal above $3$. This can be verified by hand in principle: cubing these elements must yield, up to a small power of the fundamental unit $\varepsilon = 1 + 3\sqrt[3]{12} - 3\sqrt[3]{18}$, the elements $2$ and $3$.
In fact we find 
$(2 + \sqrt[3]{12} + \sqrt[3]{18})^3 
   = 2(55 + 24\sqrt[3]{12} + 21\sqrt[3]{18}) = 2/\varepsilon$ and
$(3 + \sqrt[3]{12} + \sqrt[3]{18})^3 = 3/\varepsilon$.