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.

Let $\mathbb{F}$ be a cubic field, i.e, $\mathbb{F} = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of an 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. There In this case, there are only 4 2 possibilities of factorizations, namely:

$q\mathcal{O}_{\mathbb{F}} = \mathfrak{q}$ where the inertia degree of $\mathfrak{q}$ over $q$ is 3

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

$q\mathcal{O}_{\mathbb{F}} = \mathfrak{q}\mathfrak{p}$ with exactly one of these primes has inertia degree 2.

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

I check with GP/Pari with all such fields whose discriminant is less than 5000 (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 am not sure wonder what tools are often used in addressing such questions.

Let $\mathbb{F}$ be a cubic field, i.e, $\mathbb{F} = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of an 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. There are only 4 possibilities of factorizations, namely:

$q\mathcal{O}_{\mathbb{F}} = \mathfrak{q}$ where the inertia degree of $\mathfrak{q}$ over $q$ is 3

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

$q\mathcal{O}_{\mathbb{F}} = \mathfrak{q}\mathfrak{p}$ with exactly one of these primes has inertia degree 2.

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

I check with GP/Pari with all such fields whose discriminant is less than 5000 (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 am not sure what tools are often used in addressing such questions.