Show that $\langle\,23 \rangle\ =\langle\, 23 , \alpha-10\rangle^2 \langle\, 23, \alpha-3 \rangle$

Question

Let $\alpha$ be a root of $x^3-x-1$ and $K=\mathbb{Q}(\alpha)$.
(i) Show that $\langle\,23 \rangle\ =\langle\, 23 , \alpha-10\rangle^2 \langle\, 23, \alpha-3 \rangle$
(ii) Show that $\langle\,1 \rangle\ =\langle\, 23, \alpha-10, \alpha-3 \rangle\ $

I have shown $\mathcal{O}_K=\mathbb{Z}[\alpha]$. It's quite messy to simplify $\langle\, 23 , \alpha-10\rangle^2 \langle\, 23, \alpha-3 \rangle$ using $\langle a,b \rangle \langle c, d \rangle=\langle ac, bd, ad, bc \rangle$ unlike some quadratic extensions.
How do I show (i), (ii) using some theorems?

Answer 1

To find the ideal prime decomposition of $23\mathcal O_K$ all you need is to factorize $x^3-x-1$ in $\mathbb{F}_{23}$. We have the following:

$$x^3-x-1 = (x-10)^2(x-3) \text{ in } \mathbb{F}_{23}$$

Hence we get $23\mathcal O_K = (23,\alpha-10)^2(23,\alpha-3)$

This result is known as the Dedekind's Theorem and you can find more about it here.

For the second part, as noted in the comments we have that:

$$-7 = (\alpha - 10) - (\alpha - 3) \in (23, \alpha-10, \alpha-3)$$ $$1 = 4\cdot23 + 13 \cdot (-7) \in (23, \alpha-10, \alpha-3)$$

Hence we get that $(1) \subseteq (23,\alpha-10)^2(23,\alpha-3)$. The other inclusion is trivial.