Let $\zeta = \frac12 +\frac{\sqrt{3}}{2}i$. I've proven that $\Bbb Z[\zeta]$ is a Euclidean domain with the norm given by multiplication by the complex conjugate. I'd now like to solve the system of congruences: $$\begin{align} x\equiv 1 \mod 2+5\zeta \\ x\equiv \zeta \mod 1+2\zeta\end{align}$$
In fact, I'm a bit rusty on simple integer congruences, so I'm aware that this may be trivial using the same techniques. If it helps, I've found in a previous question that $$\mathrm{gcd}(2+5\zeta,1+2\zeta) = \zeta$$ if I'm not mistaken. Since $\zeta$ is a unit this means that the ideals $(1+2\zeta),(2+5\zeta)$ are coprime, so I believe I can use the Chinese Remainder Theorem. I'm not sure how this would go in this case. Can someone help?
"The" gcd is $1$. Now we use a Chinese Remainder Theorem algorithm. Suppose that $(1+2\zeta)u\equiv 1\pmod{2+5\zeta}$ and $(2+5\zeta)v\equiv 1 \pmod{1+2\zeta}$. Then the solution of our system of congruences is $$x\equiv (1)(1+2\zeta)(u)+(\zeta)(2+5\zeta)(v)\pmod{(2+5\zeta)(1+2\zeta)}.$$
So two structurally identical problems remain, finding $u$ and finding $v$. They can each be solved using the Euclidean Algorithm.