<s>[INST] $\langle 3 \rangle$ in $\textbf{Z}[\omega]$ ramifies, not splits, right? Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.
As you know, $$\omega = \frac{-1 + \sqrt{-3}}{2}$$ is a complex cubit root of unity, that is, $\omega^3 = 1$. This slightly obscures the fact that $1 + 2 \omega = \sqrt{-3}$, and since $(\sqrt{-3})^2 = -3$, it follows that $\langle 3 \rangle = \langle \sqrt{-3} \rangle^2$. Meaning that $\langle 3 \rangle$ ramifies, it doesn't split. This even though $$\left( \frac{3 - \sqrt{-3}}{2} \right) \left( \frac{3 + \sqrt{-3}}{2} \right) = 3.$$
That's because $$\left( \frac{-1 + \sqrt{-3}}{2} \right) \sqrt{-3} = \frac{-\sqrt{-3} - 3}{2},$$ or $\omega(1 + 2 \omega) = \omega + 2 \omega^2$. Since $\omega$ is a unit, the ideals $\langle 1 + 2 \omega \rangle$ and $\langle \omega + 2 \omega^2 \rangle$ are in fact the same.
Are these calculations correct? Have I drawn the right conclusion?
 [/INST] Yes your conclusion is correct, and the factorisation
$$\left(\frac{3 - \sqrt{-3}}{2}\right) \left(\frac{3 + \sqrt{-3}}{2}\right) = 3$$
 is no problem since the factors on the left side are associate. Indeed
$$\frac{\frac{3 - \sqrt{-3}}{2}}{\frac{3 + \sqrt{-3}}{2}}=\frac{(3 - \sqrt{-3})^2}{(3 + \sqrt{-3})(3 - \sqrt{-3})}=\frac{6-6\sqrt{-3}}{12}=\frac12-\frac{\sqrt{-3}}2,$$
is a unit since it has norm $1$.
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