A normal singular surface: the quadric cone

By [F1] the element $f$ is prime, so $R=k[x,y,z]\mathrm{/}(f)$ is a domain and $X=V(f)$ is an irreducible closed subset, hence a classical affine variety with coordinate ring $R$ [F2]; $R$ is Noetherian by [F6]. Since $(f)$ is a nonzero principal prime, $\mathrm{ht}(f)=1$ [F6]; the height formula [F6] then gives $\dim R=3-1=2$ and, for every point $x\in X$, ${\dim }_{x}X=\dim X=2$ [F6]. The Jacobian matrix of $f$ is the row $(y,x,-2z)$, so by [F3] a point $x\in X$ is regular exactly when that row has rank $3-2=1$, and singular exactly when $y=x=z=0$; as $\mathrm{char}k\mathrm{\ne }2$ this is the origin and nothing else. Hence every point $x\mathrm{\ne }(0,0,0)$ of $X$ is regular, and its local ring $R_{{\mathfrak{m}}_x}$ is a regular local ring.

The sequence $x,y$ is a regular sequence in $R$: $x$ is a nonzerodivisor because $R$ is a domain, and $R\mathrm{/}xR\cong k[y,z]\mathrm{/}(z^2)$, in which $y$ is again a nonzerodivisor. Hence $\mathrm{depth}{R}_{{\mathfrak{m}}_0}\ge 2=\dim R_{{\mathfrak{m}}_0}$, where ${\mathfrak{m}}_0=(x,y,z)$ is the maximal ideal of the origin [F5, F6].

$R$ satisfies condition $(R_1)$. Let $\mathfrak{p}$ be a prime with $\mathrm{ht}\mathfrak{p}\le 1$. If $\mathfrak{p}=0$ then $R_{\mathfrak{p}}$ is a field, hence regular. If $\mathrm{ht}\mathfrak{p}=1$, the quotient $R\mathrm{/}\mathfrak{p}$ has dimension $2-1=1$ by [F6], so the closed subvariety $V(\mathfrak{p})\subseteq X$, whose coordinate ring is the domain $R\mathrm{/}\mathfrak{p}$ [F2], has dimension $1$; a one-dimensional variety is not a single point, so $V(\mathfrak{p})$ contains a point $x\mathrm{\ne }(0,0,0)$. The corresponding maximal ideal ${\mathfrak{m}}_x$ contains $\mathfrak{p}$, and $R_{{\mathfrak{m}}_x}$ is regular by step 1.1, so $R_{\mathfrak{p}}=(R_{{\mathfrak{m}}_x}{)}_{\mathfrak{p}{R}_{{\mathfrak{m}}_x}}$ is regular by [F5]. Thus $(R_1)$ holds.

The origin is singular. At the origin the Jacobian row $(y,x,-2z)$ is the zero row, of rank $0$, while $3-{\dim }_{{\mathfrak{m}}_0}{R}_{{\mathfrak{m}}_0}=3-2=1$; by the criterion [F3] the local ring $R_{{\mathfrak{m}}_0}$ is not regular, so the origin is a singular point of $X$.

$R$ satisfies condition $(S_2)$. Let $\mathfrak{p}$ be a prime. If $\mathfrak{p}={\mathfrak{m}}_0$, step 1.2 gives $\mathrm{depth}{R}_{\mathfrak{p}}\ge 2=\dim R_{\mathfrak{p}}$. If $\mathfrak{p}\mathrm{\ne }{\mathfrak{m}}_0$ and $\mathrm{ht}\mathfrak{p}\le 1$, then $R_{\mathfrak{p}}$ is regular by step 2.1 and therefore Cohen--Macaulay, so $\mathrm{depth}{R}_{\mathfrak{p}}=\dim R_{\mathfrak{p}}=\min \{2,\dim R_{\mathfrak{p}}\}$. If $\mathrm{ht}\mathfrak{p}=2$ then $\mathfrak{p}$ is maximal, hence $\mathfrak{p}={\mathfrak{m}}_x$ for a point $x\mathrm{\ne }(0,0,0)$, and $R_{\mathfrak{p}}$ is regular by step 1.1, so again $\mathrm{depth}{R}_{\mathfrak{p}}=2=\min \{2,\dim R_{\mathfrak{p}}\}$. Thus every prime satisfies $\mathrm{depth}{R}_{\mathfrak{p}}\ge \min \{2,\dim R_{\mathfrak{p}}\}$, that is, $(S_2)$ holds.

By steps 2.1 and 3.1 the ring $R$ satisfies $(R_1)$ and $(S_2)$, so $R$ is normal by Serre's criterion [F4]; consequently each prime localisation $R_{\mathfrak{p}}$, in particular each local ring $R_{{\mathfrak{m}}_x}$ at a point of $X$, is an integrally closed domain, and $X$ is a normal variety [F7]. By step 2.2 the origin is a singular point. Therefore $X=V(xy-z^2)$ is a normal classical variety that is singular at the origin: normal does not imply nonsingular in dimension two, and the characteristic hypothesis $\mathrm{char}k\mathrm{\ne }2$ was used only to identify the singular locus with the origin.