Non-regular points of $\mathbb{Z}\left[\sqrt 5\right]$ I want to calculate the non-regular points in $\mathbb{Z}\left[\sqrt 5\right]$. I know the definition of regular point but I don't know what the process is for finding these points in general.
Remember: If $X$ is a variety of dimension $n$, then a closed point $x \in X$ is regular if and only if there exist $n$ hypersurfaces that are cut with multiplicity one.
Example: $(0,0,0) \in \mathbb{R}^3$ is regular because the hypersurfaces $xy = 0$, $xz = 0$, and $yz = 0$  are cut in $(0,0,0).$
Would someone please give me a hint?
Thanks to all for your help.
 A: The key to your question is the following observation:

Fact: Let $R$ be a one-dimensional local ring. Then, $R$ is a regular local ring if and only if $R$ is a DVR.

For a proof see [1, Tag00PD].
From this, and the fact that $\mathbb{Z}[\sqrt{5}]$ is a one-dimensional domain (being finite over $\mathbb{Z}$) to answer your questions it suffices to find the maximal ideals $\mathfrak{m}$ of $\mathbb{Z}[\sqrt{5}]$ such that $\mathbb{Z}[\sqrt{5}]_\mathfrak{m}$ is not a DVR.
Now, on the surface this seems difficult. But, there is a wonderfully simple way to do this which is explained in [2, §III.6].
Namely, we have the following

Fact: Let $R$ be a Dedekind domain with field of fractions $F$, and let $f(x)\in R[x]$ be a monic irreducible polynomial. Let $\alpha$ be a root
in $\overline{F}$ of $f(x)$ and set $S:=R[\alpha]$. Then, we have an
equality
$$\mathrm{Max}(S)=\{(\mathfrak{p},g(\alpha)):\mathfrak{p}\in\mathrm{Max}(R)\text{
 and }g(x)\text{ is an irreducible factor of }f(x)\text{ in
 }(R/\mathfrak{p})[x]\}$$
Moreover, if $\mathfrak{m}$ in $\mathrm{Max}(S)$ is such that
$f'(\alpha)\notin \mathfrak{m}$, then $S_\mathfrak{m}$ is a DVR.

Note the similarity to the variety case you mentioned in your post.
In particular, for your case we can take $R=\mathbb{Z}$ and $f(x)=x^2-5$ so then $S=\mathbb{Z}[\sqrt{5}]$. Note then that $\alpha=\sqrt{5}$ and $f'(\alpha)=2\sqrt{5}$. So, we see that if $\mathfrak{m}$ does not contain $2\sqrt{5}$ then automatically $S_\mathfrak{m}$ is a DVR. So, the only primes for you to explicitly check are those maximal ideals containing $2\sqrt{5}$. The above theorem gives you a pretty explicit description of those maximal ideals. So, I leave the rest to you.
[1] Various authors, 2020. Stacks project. https://stacks.math.columbia.edu/
[2] Lorenzini, D., 1996. An invitation to arithmetic geometry (Vol. 9). American Mathematical Soc..
