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(Quadratic) Gauss sums express square root of any integer as a sum of roots of unity (or of cosines of rational multiples of $2\pi$, if you will) with rational coefficients.

But Kronecker-Weber guarantees that any root of any integer can be expressed as a sum of that kind. What is the Is there corresponding formula for, say, $\sqrt[3]p$?

Upd. I'm sorry, but original question doesn't make much sense. The question I, perhaps, meant to ask is (as Matt E kindly points out) discussed in David Speyer's answer.

In particular, for $p=3k+1$ the cubic sum $\displaystyle\sum_{t\in\mathbb Z/p}\cos\left(\frac{2\pi t^3}p\right)$ is a root of the equation $x^3-3px-Ap=0$ where $4p=A^2+27B^2$ and $A\equiv 1\pmod 3$ (the discriminant of the cubic equation is $(27pB)^2$).

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up vote 6 down vote accepted

The Kronecker--Weber theorem does not guarantee what you claim. Indeed, $\mathbb Q(p^{1/3})$ is not Galois over $\mathbb Q$, and its Galois closure is equal to $\mathbb Q(\sqrt{-3},p^{1/3})$, which is an $S_3$ (and hence non-abelian) extension of $\mathbb Q$. It is an abelian extension of $\mathbb Q(\sqrt{-3})$, but not all extensions of this field are cyclotomic; rather, they are described by the theory of complex multiplication (Kronecker's Jugendtraum).

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(Actually, first I wanted to ask about "Gauss-sum-type" formulas for roots of a cubic polynomial with perfect square discriminant -- but then thought that just cube roots should be more tractable.) – Grigory M Jan 26 '12 at 20:19
@Grigory: Dear Grigory, If the discriminant is square, then you do get an abelian extension of $\mathbb Q$, which is thus cyclotomic. David Scheyer has a post explaining a very concrete approach to Kronecker--Weber in this case: Regards, – Matt E Jan 26 '12 at 20:22

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