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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.

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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.

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I see it now. Thank you very much. –  T.B. Feb 17 '11 at 3:37
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I find out that there is a possibly a more elementary to see this. By an exercise in Marcus (ex 21c, chapter 3), $disc(\mathbb{F}/\mathbb{Q})$ is divisible by $q^{k}$ where $k = \sum(e_{i}-1)f_{i}$ with $e_{i}, f_{i}$ are the ramification indices and inertia degrees of the primes lying over $q$. Then it must follow that only the second option for factorization satisfies this if we force the discriminant to be prime. –  T.B. Feb 17 '11 at 4:47
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The property of the tamely ramified extension $\mathbb{F}/\mathbb{Q}$ that JSE used in his answer certainly implies this, however, it's probably a deeper fact that the difference $diff(\mathcal{O}_{\mathbb{F}})$ is divisible by $q^{e-1}$ and the difference happens to be equal to the discriminant in this case. –  T.B. Feb 17 '11 at 4:47
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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.

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Sorry, I did not say accurately what I have done, I just looked at the fields where the discriminant is prime (not a power of a prime like in your example), and for all those cases, only the last option occurs (for as much as the calculations I've done). So, I wonder if it's true that only the last possibility happens if the discriminant is a prime. In any case, thank you very much for the nice example and insight. –  T.B. Feb 17 '11 at 1:52
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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, $$disc(K/\mathbb{Q})=disc^{3}(\mathbb{Q}(\sqrt{d})/\mathbb{Q})N_{\mathbb{Q}(\sqrt{d})/\mathbb{Q})}(disc(K/\mathbb{Q}(\sqrt{d})))=d^3.$$

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Great idea. Thank you. I actually happened to compute the discriminant for $\mathbb{K}$ and it turns out to be $p^{3}$ in all the cases I've done the calculations (assuming discriminant of $\mathbb{F}$ is $p$). Do you think this is a coincidence or is it true? I know that discriminant of $\mathbb{K}$ is divisible by $disc(\mathbb{F})^{[\mathbb{K}:\mathbb{F}]} = p^{2}$. But I wonder why it happened that all of these disc of $K$ are $p^{3}$? –  T.B. Feb 17 '11 at 6:37
    
To clarify this a bit, in my calculations for the $\mathbb{F}$ constructed from irreducible cubic polynomials with the assumption of the original question, all the $\mathbb{K}$ happened to be $S_{3}$-extensions. –  T.B. Feb 17 '11 at 6:44
    
I'm not so sure what was you original assumption, but if it was that the discriminant is a prime power then there are examples of F such that K/Q is not an $S_3$ extension- e.g. Felipe Voloch's example above. –  Guillermo Mantilla Feb 17 '11 at 7:50
    
The assumption is that $disc(\mathbb{F}/\mathbb{Q})$ is prime. In Felipe Voloch's example, it is a power of a prime. –  T.B. Feb 17 '11 at 8:05
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OK, as long as the discriminant is not a perfect square you'll always get a $S_3$ extension. –  Guillermo Mantilla Feb 17 '11 at 8:14
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