S(3) field extensions I'm reading a paper on the statistics of number fields.  How does one build an extension of Q with Galois group, S(3)?  How is it possible to find all isomorphism classes of S(3) field extensions of Q with given discriminant?
 A: There are basically three different approaches.


*

*An approach based on cubic forms, which was used by 
Karim Belabas to quickly
list cubic fields with discriminant up to a certain bound. 

*The approach via class field theory mentioned above: look at all quadratic number fields
with small discriminant, find all of its cubic cyclic extensions via class field theory,
and check which of these are not abelian over the rationals (this set includes all whose 
conductor is not invariant under the Galois group of the quadratic field).

*The approach via Kummer theory: start as in 2., but then adjoin a cube root of unity;
cyclic cubic extensions will simply be Kummer extensions over this extension. Standard
methods (this should be in Gras' book on class field theory) allow you to step down once
the construction is done.
Approaches 2 and 3 go back to Hasse [Arithmetische Theorie der kubischen Zahlkörper auf klassenkörpertheoretischer Grundlage; Math. Z. 31 (1930), 565-582] and Reichardt
[Arithmetische Theorie der kubischen Körper als Radikalkörper; Monatsh. Math. Phys. 40 (1933), 323-350].
Edit: I also should have given the origin of approach 1, since it is most often (incorrectly) credited to Delone and Faddeev. In the preface to their book Irrationalities of the third degree, however, they do credit F.W. Levi with this observation: see
Kubische Zahlkörper und binäre kubische Formenklassen 
(Cubic number fields and classes of binary cubic forms), Leipz. Ber. 66 (1914), 26-37
A: If $f$ is a cubic, irreducible over the rationals, then its splitting field has Galois group cyclic of order three if the discriminant is a square, symmetric on three letters otherwise. I don't know enough to answer the 2nd question. 
A: One way (I don't know whether this is done in practice) is
to go via the intermediate quadratic extension.
Let $L$ be the $S_3$ extension. Then it has a quadratic subfield $K$.
There is a relation between the discrimiants of $K$ and $L$ and
the relative discriminant of $L/K$. This is easy to remember in terms
of the differents of the extensions:
$$\mathcal{D}_{L/\mathbb{Q}}=\mathcal{D}_{L/K}\mathcal{D}_{K/\mathbb{Q}}.$$
Taking norms gives an equation involving discriminants.
One upshot to this is that for a given discriminant $d_{L/\mathbb{Q}}$
there will be only finitely many possible $K$. Given $K$ then there
are at most finitely many cubic extensions $L/K$ having the right discriminant,
by class field theory. Not all of these will have $L/\mathbb{Q}$ $S_3$-Galois
of course, but for each $K$ one can be sure one has found all admissible $L$.
Added Another correspondent has mentioned Henri Cohen's
Advanced Topics in Computations Number Theory. While I couldn't
find a treatment of this exact question in Cohen's book, he does
devote a whole chapter to the construction of cubic fields. To
solve the problem at hand, it suffices to construct all cubic
fields whose different divides that of the $S_3$-extension
(this gives a bound on the discriminant of the cubic) and see which of
these fit inside $S_3$-extensions of the sought discriminant.
I should add that Cohen's books should be the first resort for questions
of this kind. They exhibit a wealth of technique and also a plethora
of useful references.
A: I'm not sure about this, so please correct me (or just downvote?).
Finding fields of a given discriminant has two main algorithms, depending on the context. If you want to find all fields up to a certain discriminant in order to build a table, this is done using theorems for bounds on the coefficients of a minimal element. This direction, particular to cubic fields, is one of the chapters in the Advanced book mentioned below. If you want to find all fields of a given (large) discriminant, there's the following.
Given a non-square discriminant $D$, find it's square part: $D = ab^2$ ($b$ largest possible). So $F = \mathbb{Q}(\sqrt{a})$ is the quadratic subfield of the $S_3$ closure, and the closure is a cyclic extension of this. Hence, by class field theory, it corresponds to a ray-class of order 3 in some ray class group of $F$.
To settle the correct ramification, the ray class group must have modulus $b$. So, compute the  ray class group of modulus $b$, and for each class of order 3 there's a cubic extension of the quadratic field. This will be the closure of your desired cubic over the rationals, so just look for the cubic subfield.
This algorithm has subexponential complexity in $D$. To learn how to compute class groups of quadratic fields:

"Computational Algebraic Number Theory", H. Cohen

To learn how to compute ray class groups and how to find fields corresponding to classes:

"Advanced Topics in Computational Number Theory", H. Cohen

