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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?

user108580
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  • One approach: if $u$ is in $\mathcal{A}$, then $u$ is in some number field $K/\mathbb{Q}$. Consider the Galois closure $\overline{K}$ with Galois group $G$. In $\overline{K}$, we can write $u^{-1} = \pm \prod_{\sigma \neq 1} (\sigma u)$, where $\sigma$ ranges over $G$. Each $\sigma u$ is also an algebraic integer, as the Galois action preserves the algebraic polynomial. There's probably a simpler way, but this jumped out at me. – davidlowryduda Oct 18 '23 at 03:12
  • @davidlowryduda I don't quite follow your argument. Certainly, the Galois conjugates of $u$ belong to $\mathcal{O}$, but why should they belong to $\mathcal{A}$? And if they don't, why does their product? – Alex Wertheim Oct 18 '23 at 03:15
  • See here in the dupe for this and generalizations (including how to characterize integral extensions by a closely related "survival" property). – Bill Dubuque Oct 18 '23 at 03:43
  • I strongly vote for removing the [duplicate] tag. First thing is that OP may have never heard about localizations and may therefore not be able to follow the comments and answers in the linked post (which is not a fault on the OP's side). Second, if a different question may have answers containing some ideas which are useful for OP's question doesn't make it "duplicate". @BillDubuque, that's not a criticism of your comment. Just mentioning in the comments (or even in an answer) that there is another post with useful ideas is totally fine of course. – russoo Oct 31 '23 at 21:43
  • ... and forgot the third point: the questions are simply not the same. – russoo Oct 31 '23 at 22:01
  • @russo The OP does not need to know anything about localizations to understand this answer in the dupe (the answer they accepted below is essentially the same). Duplicate closure doesn't require that the questions are the same (cf. the SE banner: "this question already has answers here"). See also abstract duplicates. – Bill Dubuque Oct 31 '23 at 22:13

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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 $u^{-1}=\pm (u^{n-1}+a_{n-1}u^{n-2}+\dots + a_1)\in\mathbb{Z}[u]\subseteq\mathcal{A}$.

user10354138
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