Consider the ring $\mathcal O$ of all algebraic integers and a subring $\mathcal A\subset \mathcal O$. If $u\in\mathcal A$ is invertible in $\mathcal O$, then is the inverse of $u$ necessarily in $\mathcal A$?
In other words, if $u\in\mathcal A$ is invertible in $\mathcal O$, then is $u$ invertible in $\mathcal A$?
This seems like a very fundamental question. Are there any contexts I could refer to so I could better understand the related topics?
The answer is yes. The reason is that $u\in\mathcal{O}$ is a root of its minimal polynomial $$ X^n+a_{n-1}X^{n-1}+\dots+a_0\in\mathbb{Z}[X]. $$ where $n=[\mathbb{Q}(u):\mathbb{Q}]$. Now for $u\in\mathcal{A}$ invertible in $\mathcal{O}$, we get $$ a_0(u^{-1})^n+a_1(u^{-1})^{n-1}+\dots + a_{n-1}u^{-1}+1=0 $$ and so $a_0X^n+a_1X^{n-1}+\dots+a_{n-1}+1$ is up to constant multiple the minimal (monic integer-coefficient) polynomial of $u^{-1}\in\mathcal{O}$ (since $\mathbb{Q}(u^{-1})=\mathbb{Q}(u)$). So we have $a_0=\pm 1$ and $u^{-1}=\pm (u^{n-1}+a_{n-1}u^{n-2}+\dots + a_1)\in\mathbb{Z}[u]\subseteq\mathcal{A}$.