Question.
Let $X=\mathrm{Spec}\mathbb{Z}[x]/(x^2-p)$.
Show that $X$ is a regular scheme (i.e., all the local rings $\mathcal{O}_{X,x}$ are regular local rings) if and only if $p=2$, or $p\equiv 3\pmod{4}.$
Thoughts.
It seems to have something to do with the quadratic reciprocity law.
Since $X$ is locally noetherian, it suffices to check that $\mathcal{O}_{X,x}$ is regular for all closed points $x$ of $X$, which corresponds to maximal ideals of $\mathbb{Z}[x]/(x^2-p)$.
So we should determine them.
If $\mathfrak{p}$ is prime, then its intersection with $\mathbb{Z}$ is either $p, 0,$ or a prime $\ell\neq p$.
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if $\mathfrak{p}\cap\mathbb{Z}=(0)$, then $\mathfrak{p}$ is of the form $(f(x))$ where $f$ is divisible by $x^2-p$. But $x^2-p$ is irreducible, so $f(x)=(x^2-p)$, and $\mathfrak{p}$ is not maximal — for example, $(x^2-p)\subset (x,p)$.
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if $\mathfrak{p}\cap \mathbb{Z}=(p)$, then $\mathfrak{p}=(x,p)=(x)$.
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if $\mathfrak{p} \cap \mathbb{Z} = (\ell)$ for $\ell\neq p$, then $\mathfrak{p}$ can be viewed as a prime ideal in $\mathbb{F}_\ell[x]/(x^2-p)$. If $p$ is not a quadratic residue mod $\ell$, then $x^2-p$ is irreducible over $\mathbb{F}_\ell$, and $\mathfrak{p}=(\ell, x^2-p)$. If $p$ is a quadratic residue mod $\ell$, then $\mathfrak{p}=(\ell,x+g(x))$ or $\mathfrak{p}=(\ell,x-g(x))$, where $g(x)\in \mathbb{Z}[x]/(x^2-p)$ is chosen such that its reduction mod $\ell$ (inside $\mathbb{F}_\ell[x]/(x^2-p)$) is a square root of $p$.
Having classified all closed points of $X$, I am not sure how to show regularity at all these points under the assumption that $p=2$ or $p\equiv 3\pmod{4}$. Could someone give me a proof or hint? Thanks in advance!
Since $X$ is clearly irreducible of dimension $1$, we know $\dim \mathcal{O}_{X,x}=1$. So, we just need to show that $\dim_{k(x)}\mathfrak{m}_x/\mathfrak{m}_x^2=1$ for each closed point $x$ as classified above.
For example, when $\mathfrak{p}=(x)$, we have $k(\mathfrak{p})=\mathbb{F}_p$ and $\mathfrak{m}_\mathfrak{p}/\mathfrak{m}_\mathfrak{p}^2$ is one-dimensional over the residue field with basis $x$. So, the point $(x)$ is always regular.
Best Answer
Hint: show that the ring $\mathbb{Z}[\sqrt{p}]$ is integrally closed in its fraction field unless $p$ is congruent to $1$ mod $4$, in which case it never is.
To do this, you can be quite explicit: figure out which elements of $\mathbb{Q}(\sqrt{p})$ are solutions to quadratic polynomials of the form $x^2 + ax + b$ by noting that you can recover $a$ and $b$ from some quadratic number $\alpha$ by taking its (negative) trace $(\alpha + \overline{\alpha})$ and norm $\alpha\overline{\alpha}$ respectively.