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What's a good example where computing the generators of a real cubic field extension of $\mathbb{Q}$ is nontrivial?

I usually see these fields specified in terms of generators, is there a good example where computing the generator is nontrivial, but possible?

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Let $K$ be the splitting field over the rationals of some quartic with real roots and Galois group $S_4$, the symmetric group on 4 letters ($x^4-5x^3+6x^2-1$ will probably do). Now $S_4$ has a subgroup $G$ of order 8 (isomorphic to the group of symmetries of a square), and the fixed field $E$ of that subgroup will be a real cubic extension of the rationals. It is possible to compute a generator of $E$; whether it is trivial to do so is a somewhat subjective question. Personally, I'd say it's nontrivial, but I'm open to counterarguments.

EDIT: Let $a,b,c,d$ be the roots of the quartic. Then there are three cubic subfields, each generated by one of the numbers $ab+cd$, $ac+bd$, and $ad+bc$. These three numbers are conjugate over the rationals, and are the roots of the resolvent cubic, as Hurkyl asked, and as I forgot. There are formulas for the coefficients of the resolvent cubic in terms of the coefficients of the original quartic. The old theory-of-equations textbooks will discuss the resolvent cubic, as will the more computationally inclined of the abstract algebra texts, and of course it's bound to be discussed here and there on the web.

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Is this given by a root of the resolvent cubic? – Hurkyl May 20 '12 at 9:40
@Hurkyl, I don't think so. If I have it right, the resolvent cubic is unique, but there is more than one 8-element subgroup of $S_4$, and different subgroups have different fixed fields. – Gerry Myerson May 20 '12 at 12:57
@Hurkyl, sorry, I think you're exactly right. There are three fields, one for each root of the resolvent cubic. I'll edit. – Gerry Myerson May 22 '12 at 1:55

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