Inverse ideal of $(6,2+\sqrt{10})$ in $\mathbb{Q}(\sqrt{10})$

Question

I am trying to find the inverse ideal of $I=(6,2+\sqrt{10})$ in $\mathbb{Q}(\sqrt{10})$.

By definition, we have that $I^{-1}=\{\alpha\in\mathbb{Q}(\sqrt{10}):\alpha I\subseteq \mathbb{Z}[\sqrt{10}]\}$.

Perhaps I'm getting confused by the definition, but wouldn't $I^{-1}$ be all of $\mathbb{Z}[\sqrt{10}]$? Or do we want $\alpha I=\mathbb{Z}[\sqrt{10}]$?

Answer 1

The inverse ideal $I^{-1}$ consists of all the elements $z$ of the number field $K$ with the property $zI\subseteq \mathcal{O}_K$. When $I$ is contained in $\mathcal{O}_K$ we trivially have $\mathcal{O}_K\subseteq I^{-1}$, and the question is about finding all the non-integral elements of $K$ with the property $zI\subseteq \mathcal{O}_K$.

The case when $I$ is a principal ideal $a\mathcal{O}_K$ is easy. In this case $z\in I^{-1}$ if and only if $za\in\mathcal{O}_K$. It follows that $I^{-1}$ is the $\mathcal{O}_K$-module generated by $1/a$.

Because $(2+\sqrt{10})(2-\sqrt{10})=-6$ it follows that your ideal $I$ is principal, generated by $2+\sqrt{10}$. Consequently $I^{-1}$ is the $\Bbb{Z}[\sqrt{10}]$-module generated by $$\alpha=1/(2+\sqrt{10})=(\sqrt{10}-2)/6.$$