# Quadratic Integer Ring mod p - Field?

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?

• It's hard to understand point of your proof. You could write \beta^{p^2}=\beta for all \beta, it's true and what you will derive from this? \beta need to be invertible... Let p>2, then O/I ~ Z+\sqrt{d}Z/p ~ Z[x]/(x^2-d,p) ~ F_p[x]/(x^2-d) and the last one is field <=> Legendre Symbol equal to -1 (the very first isomorphisms comes from description of O and invertibility of 2 mod p) – Bad English Jun 2 at 16:04
• What is the book? – Will Jagy Jun 2 at 18:14
• Solving the pell equation - Jacobson, p. 357, it id the first question I am asking here and I wasn't sure how much I can cite here – Sunshine_007 Jun 2 at 18:55
• Actually this isomorphism " O/I ~ Z+\sqrt{d}Z/p ~ Z[x]/(x^2-d,p) ~ F_p[x]/(x^2-d) and the last one is field <=> Legendre Symbol equal to -1" would be everything I would need. Could you give me any more detail why this works/ some ressource where I could read about it? I have problems to understand why the first and second isomorphisms work...? – Sunshine_007 Jun 2 at 19:11