I am working on quotient rings of $\mathbb{Z}[x]$ and I encounter some troubles when there are questions about integrally closed (w.r.t. its field of fraction).
For example, when I need to prove (or disprove) $\mathbb{Z}[x]/(9x^2+1)$ is integrally closed, I start off by considering a polynomial $p(x)\in \mathbb{Z}\left[\frac{i}{3}\right][x]$ and assume it is satisfied by some elements $q\in \mathbb{Q}\left[\frac{i}{3}\right]$. However, when I try to show that all roots of $p(x)$ cannot have factors other than $3$ in their denominators, I failed.
I tried the following work:
Let $p(x)=\sum_{i=0}^na_ix^i$, then $a_i\in \mathbb{Z}\left[\frac{i}{3}\right]$. Now homogeneous algebraic expression of roots (like $r_1r_2+r_2r_3+r_3r_1$) has denominator with only $3$ as their factor. (...) So we can conclude that every $r_i$ can only have $3$ as factor of their denominators.
I suspect whether the (...) works but currently I cannot find any counter example. So I'd like to seek for some guidance on how to work on the proof (e.g. some ideas) or even sketch out the proof (avoiding using very difficult tools [if possible] e.g. tensor products, projective modules etc). Thank you very much.
$\Bbb Z[\frac{i}{3}]$ is the localization of $\Bbb Z[i]$ at the multiplicatively closed subset given by powers of $3$. Thus, as a localization of a PID, it is a PID, in particular integrally closed.
In general, localizations of integrally closed integral domains are again integrally closed. (Though we don't need this in full generality here.)