Splitting of primes in cubic fields with limited ramifications.

Question

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.

Answer 1

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.

Answer 2

Since you are assuming that $q$ ramifies, you will not see the first and third cases in your list. You will definitely see the second case if $F/Q$ is Galois. Now you know where to look: take $p$ prime, $p \equiv 1 \mod 3$ and look at the cubic subfield of the field generated by the $p$-th roots of unity. You must have done something wrong in your calculation, as the first example $p=7$ leads to $x^3 + x^2 - 2x - 1$ of discriminant $49$ and $q=7$ factors as in your second option (i.e. first option in the revised version of the question).

Edit: Question was changed while I was typing the answer. Second option means $(q)$ is the cube of a prime, while fourth option means $(q)$ is the square of a prime times another prime.

Answer 3

Let $K$ be the Galois closure of $F$. Since $F$ has prime discriminant, say $p$, the extension $K/\mathbb{Q}(\sqrt{p})$ is an unramified extension. In particular the ramification degree of $p$ in $K$ is 2, so it is impossible to have a sub-extension with ramification at $p$ not equal to $2$ or $1$.

Added to adress T.B comments below: What I said above can be generalized to the case when $F$ has fundamental discriminant $d$, i.e. if $p$ is a prime dividing $d$ it must happen that $pO_{F}=\mathcal{P}^2\mathcal{Q}$ for some $\mathcal{P} \neq \mathcal{Q}$. The proof is exactly the same as the above argument.

"I actually happened to compute the discriminant for $K$ and it turns out to be $p^3$ in all the cases I've done the calculations (assuming discriminant of $F$ is $p$). Do you think this is a coincidence or is it true?" Yes it is true, even in the general case that $disc(F)=d$ is fundamental. The point is that $K/\mathbb{Q}(\sqrt{d})$ is unramified, in other words its relative discriminant is trivial. Hence, $$disc(K/\mathbb{Q})=disc^{3}(\mathbb{Q}(\sqrt{d})/\mathbb{Q})N_{\mathbb{Q}(\sqrt{d})/\mathbb{Q})}(disc(K/\mathbb{Q}(\sqrt{d})))=d^3.$$