Let $\alpha=\frac{1+{\sqrt{-23}}}{2}$. Let $I$ be an ideal in $\mathbb{Z}[\alpha]$ generated by $2$ and $\alpha$. Show that $I^3$ is a principal ideal.
I know that the ring is a Dedekind domain, and I think may be I can use structure theorems of module to solve this. But I don't know how to continue.
The ring $\mathbb Z[\alpha]$ has a norm given by $$\mathbb N(x+y\alpha) = (x+y\alpha)(x+y\overline\alpha) = x^2+xy+6y^2.$$
Using the fact that $I=(2,\alpha)$ is a prime ideal lying above $2$, we find that the ideal norm $\mathbb N(I) = 2$, so $\mathbb N(I^3) = 8$. Hence, if $I^3$ is to be principal, it should be generated by an element of norm $8$. An obvious candidate for this is $(2-\alpha)$.
With the number theoretic motivation over with, we can now work purely manually. We have $$(2,\alpha)^3 = (4,2\alpha,\alpha^2)(2,\alpha) = (8,4\alpha,2\alpha^2,\alpha^3).$$
Using the fact that $\alpha^2-\alpha + 6 = 0$, observe that $$2-\alpha = \alpha^3+4\alpha + 8,$$ so $(2-\alpha)\subset(2,\alpha)^3$.
We can deduce equality either by an argument using norms, or by observing that $$8 = -(\alpha^2-\alpha -2) = (\alpha+1)(2-\alpha)\\4\alpha = 8 + 4(\alpha-2) = (\alpha-3)(2-\alpha)\\ 2\alpha^2=2\alpha-12=2(\alpha-2)-8=(-\alpha-3)(2-\alpha)\\ \alpha^3 = \alpha^2-6\alpha=\alpha(\alpha-2)-4\alpha=(3-2\alpha)(2-\alpha) $$ so $(2,\alpha)^3\subset (2-\alpha)$.