# Sixth Power Character on Q(\sqrt{-3})

We define the sixth power character on $Z[w]$, where $w=\frac{-1+\sqrt{-3}}{2}$, the cubic root of unity, like follows:

Let $u,v \in Z[w]$, such that $u$ is a prime in $Z[w]$, and $(v,u)=1$, then exists a only one integer m such that, $v^{\frac{Nu-1}{6}}\equiv \zeta_{6}^{m} (mod\ u).$

We define $( \frac{v}{u})_{6} = \zeta _{6}^{m}$ like above.

Now, $3=-w^{2}(1-w)^{2}$, and let be $u$ a primary prime in $Z[w]$, $u\neq 2$.

Does exist a criteria to compute $(\frac{1-w}{u})_{6}$, which depents only on the residue class of $u$ modulo some rational integer $n$?

What is that $n$? What is the criteria?

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By Artin's reciprocity law, the symbol only depends on the residue class of u modulo the conductor of the extension generated by a sixth root of the numerator, which is easy to compute. Alternatively, write the sixth power residue symbol in terms of the quadratic and the cubic symbol and used cubic and quadratic reciprocity in this ring. Except for applications to root numbers etc. this has nothing to do with elliptic curves. –  Franz Lemmermeyer Jan 25 at 10:42