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Let $d$ be a square free integer. What is the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{d})$?

Do we have to do a case analysis for $d \equiv 1,2,3 \pmod{4}$? Let $S$ denote the integral closure of $\mathbb{Q}(\sqrt{d})$. Then for $d \equiv 1 \pmod{4}$, I see that $\frac{1 + \sqrt{d}}{2} \in S$, since $x^{2} - x - \frac{d-1}{4} = f(x)$ satisfies $f(\frac{1 + \sqrt{d}}{2}) = 0$, but how to prove that $\mathbb{Z}[\frac{1 + \sqrt{d}}{2}] = S$? I do not know how to proceed for the other cases. Clearly, $\mathbb{Z}[\sqrt{d}] \subset S$ regardless of the residue of $d$ modulo 4.

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  • $\begingroup$ There are many proofs on the web. $\endgroup$ – rogerl Mar 26 at 15:41
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The only prime that can be singular in $\Bbb{Z}[\sqrt{d}]$ is $2$. This happens only when $d\equiv1\pmod{4}$. Now verify that $2$ is not singular in $\Bbb{Z}[\tfrac{1+\sqrt{d}}{2}]$.

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