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Let $\mathcal{O}=\mathbb{Z}[\alpha]$ be a number ring, where $\alpha \in \mathbb{C}, \alpha^3=\alpha+1$. I have proved that $$23\mathcal{O}=(23, \alpha -10)^2(23, \alpha-3)$$ and that $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$. Now I am asked to prove that $(23, \alpha -10, \alpha -3)= \mathcal{O}$. A basis for $\mathcal{O}$ would be $\{1, \alpha, \alpha^2\}$. It would be enough to show that $1\in (23, \alpha -10, \alpha -3)$. However I've made several attempts and I cannot produce $1$ from a combination of $(23, \alpha -10, \alpha -3)$. Can someone help me with this?

Bill Dubuque
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kubo
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  • By the dupe: $(a!-!3,f(a),23) = (a!-!3,\color{#c00}{f(3),23}) = 1,$ if $,(\color{#c00}{f(3),23})=1,,$ as in OP where $f(a) = a-10,$ so $,\color{#c00}{f(3)} = -7\ \ $ – Bill Dubuque Oct 28 '24 at 06:50

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Observe that not only 23 but also $7=\alpha-3-(\alpha-10)$ is an element of $\mathcal{O}$. Note also that $23$ and $7$ are coprime. Thus there are integers $u,v$ such that $1=23u+7v$ implying that $1\in \mathcal{O}$.

Jens Schwaiger
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