Analyzing singularity of $V(xz-y^2)$ Suppose we are given the variety $V(xz-y^2)$. Checking partial derivatives, we can see that the only place where all derivatives vanish is the origin.
We form the quotient $\mathfrak{m}/\mathfrak{m}^2$, where $\mathfrak{m}$ is the image of $(x,y,z)$ in the ring $\mathcal{O}_p:=(k[x,y,z]/(xz-y^2))_\mathfrak{m}$. Since $xz-y^2 \subseteq \mathfrak{m}^2$, we can see that it gets killed in the quotient. Hence, we can deduce that $\mathfrak{m}/\mathfrak{m}^2 \cong k^3$ and that $\mathrm{dim}_{A/\mathfrak{m}} (\mathfrak{m}/\mathfrak{m}^2)=3 > 2$, so this point cannot be regular.


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*Is this reasoning correct, and is there more that can be said, as in a finer classification for the type of singularity?

*How can one find the krull dimension of $\mathcal{O}_p$ formally?
As for part $2$, it is heuristically true that the krull dimension i s greaater than $1$ since we have $(0) \subset (x) \subset (x,z)$ is a chain. I'm just not sure how to bound the dimension above, does it suffice to find the number of generators for $\mathfrak{m}$?
 A: (1) Yes, the point $(0,0,0)$ is a singularity because $\dim_{k} (\mathfrak m / \mathfrak m^2) = 3 > 2$.
Notice that if you resolve this singularity by blowing up the ambient $\mathbb A^3$ at $(0,0,0)$ and taking the strict transform of the surface $V(xz - y^2)$, then the preimage of the singular point will be a single $\mathbb P^1$. This singularity is called an $A_1$ singularity. There is a classification of du Val surface singularities, whose resolutions consist of a network of $\mathbb P^1$'s intersecting in a pattern resembling a Dynkin diagram of ADE type.
(2) To determine the Krull dimension of $\mathcal O_{\mathfrak p}$, observe that there is a correspondence between the prime ideals in $A = k[x,y,z]/(xz - y^2)$ contained inside $(x,y,z)$ and the prime ideals in $\mathcal O_{\mathfrak p}$. [The correspondence sends a given prime ideal $\mathfrak q \subset \mathfrak p$ to the prime ideal $\mathfrak q A_{\mathfrak p}\subset A_{\mathfrak p}$.] Geometrically, this is saying that there is a correspondence between the irreducible closed subvarieties of the surface $V(xz - y^2)$ that contain $(0,0,0)$ and the prime ideals in $\mathcal O_{\mathfrak p}$.
Anyway, the conclusion is that the Krull dimension of $\mathcal O_{\mathfrak p}$ is the height of the maximal ideal $(x,y,z)$. Since the Krull dimension of $A$ is the supremum of the heights of all its maximal ideals, it follows that the Krull dimension of $\mathcal O_{\mathfrak p}$ is less than or equal to the Krull dimension of $A$, i.e. it is less than or equal to $2$. Since you have a chain $(0) \subset (x) \subset (x,y,z)$, this shows that it is precisely $2$.
Alternatively, use this fact:
$$ {\rm ht \ } \mathfrak p + {\rm dim \ }A/\mathfrak p = {\rm dim \ } A,$$
which holds when $A$ is an integral domain that is also a finitely generated $k$-algebra and $\mathfrak p \subset A$ is a prime ideal. "ht" denotes "height" and "dim" denotes Krull dimension.
