# Gauss-type sums for cube roots

(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, $\sqrtp$?

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 $(27pB)^2$).

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).
• @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: math.stackexchange.com/a/31600/221 Regards, – Matt E Jan 26 '12 at 20:22