In order to understand a proof of a book I try to get I need to deal with Quadratic Integer Rings. As far as I got till now if I look at $\mathbb{Q}(\sqrt(d))$, $O_{\sqrt(d)}$ and $p \equiv \eta_p \mod{4}$ an odd prime. Then in my opinion $I:=p\mathbb{Z}+p\mathbb{Z}\sqrt(d)$ is an Ideal in $O_{\sqrt(d)}$ therefore $\frac{O_{\sqrt(d)}}{I}$ a factorring.
If I look know at $\beta=a+b\sqrt(d)$, then $\beta^p \equiv \beta \mod{p}$ when $(\frac{d}{p})=1$(Legendre Symbol) and $\beta^p \equiv \overline{\beta} \mod{p}$ when $(\frac{d}{p})=-1$. My proof now tells me, that in the case $(\frac{d}{p})=1 \Longrightarrow \beta^{p-1}=1$. (There might also be some connection between $p \mod{4}$ and $(\frac{d}{p})$, but it is hard to get in that book since I found already an error and am not sure if there aren't any more) That would mean that $\beta$ has a inverse therefore that $\frac{O_{\sqrt(d)}}{I}$ in this case is a field... Is that true or did I get something wrong? And if yes is there any nice way to show that?