Let $\mathbb{F}$ be a cubic field, i.e, $\mathbb{F} = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of a cubic irreducible polynomial over $\mathbb{Q}$, satisfying $disc(\mathbb{F}/\mathbb{Q})$ is a prime or a power of a prime, say $q$. By standard number theory, we know that $\mathbb{F}$ is ramified only at $q$. My question is how does $q$ factors in $\mathbb{F}$, more precisely, what is the factorization of $q\mathcal{O}\_{\mathbb{F}}$ given just the mentioned above assumptions. In this case, there are only 2 possibilities of factorizations, namely: 


$q\mathcal{O}\_{\mathbb{F}} = \mathfrak{q}^{3}$,


And $q\mathcal{O}\_{\mathbb{F}} = \mathfrak{q}^{2} \mathfrak{p}$ 


I check with GP/Pari with all such fields whose discriminant is less than 5000 and is a prime (not a power of prime) (there are 168 such fields) and it turns out that all of them have the last kind of factorization. I wonder if this is true in general and if there is a theorem telling us that it is so.


I guess the motivation is to study the behaviors (factorizations) of primes in number fields with very limited ramifications, I wonder what tools are often used in addressing such questions.



Splitting of primes in cubic fields with limited ramifications.
Yes, certainly. As long as p is tamely ramified (which is the case for you once p > 3) then the power of p in the discriminant is e-1, where e is the ramification degree. By requiring that the discriminant is exactly p, you are forcing e-1 = 1, i.e. you are restricting yourself to the second of your two options.


