# Equivalence of two cubic congruences

I am working on problem 16 in chapter 9 of Ireland and Rosen, and have gotten stumped on the one part of the problem that is not immediately obvious. I am trying to show that $x^3\equiv11 (2-3\omega)$ is solvable in $\mathbb{Z}[\omega]$ iff $x^3\equiv11(7)$ is solvable in $\mathbb{Z}$ (the book writes $x^3=11(7)$ but I believe they intend equivalence).

Things I've considered: letting $D=Z[\omega]$, the ring of Eisenstein integers, and $\pi=2-3\omega$, $D/\pi D$ is a finite field with $N\pi = 19$ elements, with $\{0,1,...,18\}$ a complete set of coset representatives, where $N\pi=\pi\bar\pi$ (so for $n\in\mathbb{Z}$, $Nn=n^2$). $7\equiv1(3)$, so $D/7D$ is not a finite field since $7$ is not prime in $D$, but since $7=(3+\omega)(3+\omega^2)$ in $D$, letting $\rho=3+\omega$ or $3+\omega^2=2-\omega$, $D/\rho D$ is a finite field with $7$ elements, with $\{0,...,6\}$ a complete set of coset representatives. We know the value of the cubic character evaluated at $11$ is given by $\chi_\pi(11)=(\frac{11}{\pi})_3\equiv11^{(N\pi-1)/3}\equiv11^6(\pi)$, using the notation in I&R. Likewise, $\chi_7(11)=(\frac{11}{7})_3\equiv11^{(N7-1)/3}\equiv11^{16}(7)$. The connection between these two congruences, however, still escapes me. I'm not even sure how a congruence in $D$ is reduced in this case to a congruence in $\mathbb{Z}$; $11\equiv2(3)$ so it is a rational, primary prime in $D$; that probably comes into play. And to get to this congruence, we have already used cubic reciprocity, so I don't think that flipping the congruence will prove fruitful.

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Is $x^3\equiv2-3\omega$ solvable? Since $D/11D$ has 121 elements this is hard to resolve by straightforward checking. Fill in the details of the following proof that it is not solvable. $\chi_\pi(2-3\omega)=\chi_{2-3\omega}(11)$ and so we shall have a solution iff $x^3\equiv11(2-3\omega)$ is solvable. This congruence is solvable iff $x^3=11(7)$ is solvable in $\mathbb{Z}$. However, $x^3\equiv a(7)$ is solvable in $\mathbb{Z}$ iff $a\equiv1$ or $6(7)$.
Well, $x^3\equiv2-3\omega(11)$ is solvable: try $1+8\omega$ or $7+10\omega$, both of which are equal to $2+8\omega$ when cubed. $2+8\omega-(2-3\omega)\equiv11\omega\equiv0(11)$. But $x^3\equiv2+3\omega(11)$ is not solvable, since $N(2+3\omega)=7$, and hence for $\pi=2+3\omega$, $D/\pi D \simeq D/7D$ and so the rest follows.