A Weil divisor that is not Cartier at the vertex of the quadric cone

The substitution $\phi (x)=u^2$, $\phi (y)=v^2$ and $\phi (z)=uv$ kills $xy-z^2$, so it induces a $k$-algebra homomorphism $\bar{\phi }:R\to k[u,v]$, given by $\bar{\phi }([p])=\phi (p)$. Since $z^2=xy$ in $R$, writing $c=2m+ϵ$ with $ϵ\in \{0,1\}$ shows that each monomial class $x^a{y}^b{z}^c$ equals $x^{a+m}{y}^{b+m}{z}^{ϵ}$; hence every class in $R$ is represented by $A(x,y)+zB(x,y)$ for some $A,B\in k[x,y]$. Its image is $A(u^2,v^2)+uvB(u^2,v^2)$. The first summand is supported on monomials $u^{2i}{v}^{2j}$, while the second is supported on monomials $u^{2i+1}{v}^{2j+1}$. These supports are disjoint, and each exponent pair in either support uniquely determines $(i,j)$; since monomials form a $k$-basis of $k[u,v]$, a zero image forces every coefficient of $A$ and $B$ to vanish. Thus $\bar{\phi }$ is injective. Its image is the subring generated by $u^2,uv,v^2$, namely $C:=k[u^2,uv,v^2]$, so $R\cong C$.

The ring $R\mathrm{/}(x)$ is isomorphic to $k[y,z]\mathrm{/}(z^2)$; there the ideal $(z)\mathrm{/}(z^2)$ is the nilradical, because $z^2=0$ makes it nilpotent and $f^n\in (z^2)\subseteq (z)$ with $(z)$ prime forces $f\in (z)$, so $(z)$ is the unique minimal prime of $k[y,z]\mathrm{/}(z^2)$ and its preimage $P=(x,z)$ is the unique minimal prime of $R$ over the principal ideal $(x)$.

Grade $R$ by $\mathrm{deg}x=\mathrm{deg}y=\mathrm{deg}z=1$, so that $xy-z^2$ is homogeneous of degree $2$ and $\mathfrak{m}=(x,y,z)$ is the set $R_{\ge 1}$ of elements of positive degree. Then $\mathfrak{m}P=(x^2,xy,xz,yz,z^2)=(x^2,xz,yz,z^2)$ using $xy=z^2$, so $\mathfrak{m}P\subseteq R_{\ge 2}$; every element of $P$ has the form $a_{0}x+b_{0}z+(a_{1}x+b_{1}z)$ with $a_0,b_0\in k$ and $a_1,b_1\in \mathfrak{m}$, so $P=kx+kz+\mathfrak{m}P$; since $x$ and $z$ have degree $1$ they do not lie in $\mathfrak{m}P$, so their images form a $k$-basis of $P\mathrm{/}\mathfrak{m}P$ over $R\mathrm{/}\mathfrak{m}=k$ and ${\dim }_{k}P\mathrm{/}\mathfrak{m}P=2$. Moreover, if $cx+dz\in \mathfrak{m}P$ with $c,d\in R$, decomposing $c=c_0+c_1$ and $d=d_0+d_1$ with $c_0,d_0\in k$ and $c_1,d_1\in \mathfrak{m}$ gives $c_{0}x+d_{0}z\in \mathfrak{m}P\cap R_1=0$, hence $c_0=d_0=0$ and $c,d\in \mathfrak{m}$.

Let $G=\{\pm 1\}$ act on $k[u,v]$ by $(u,v)\mapsto (-u,-v)$; since $\mathrm{char}k\mathrm{\ne }2$, a polynomial $\sum c_{pq}{u}^p{v}^q$ is fixed by $G$ exactly when $c_{pq}=(-1{)}^{p+q}{c}_{pq}$ for all $(p,q)$, that is, when $c_{pq}=0$ whenever $p+q$ is odd, and the monomials with $p+q$ even are precisely the products of $u^2$, $uv$ and $v^2$. Hence $C=k[u,v{]}^G$; in particular $C$ is a subring of the domain $k[u,v]$, so $C$ and $R\cong C$ are domains.

The images of $x$ and $z$ generate $P_{\mathfrak{m}}$ over $R_{\mathfrak{m}}$ by [F13], and they are linearly independent modulo $\mathfrak{m}{R}_{\mathfrak{m}}{P}_{\mathfrak{m}}$: if $(a\mathrm{/}s)x+(b\mathrm{/}t)z\in \mathfrak{m}{R}_{\mathfrak{m}}{P}_{\mathfrak{m}}$ with $a,b\in R$ and $s,t\mathrm{\notin }\mathfrak{m}$, then multiplying by $st\mathrm{\notin }\mathfrak{m}$ and clearing denominators inside $\mathfrak{m}{R}_{\mathfrak{m}}{P}_{\mathfrak{m}}=(\mathfrak{m}P)R_{\mathfrak{m}}$ produces $u\mathrm{\notin }\mathfrak{m}$ with $u(atx+bsz)\in \mathfrak{m}P$, so $uat,ubs\in \mathfrak{m}$ by step 1.3, whence $at,bs\in \mathfrak{m}$ and $a,b\in \mathfrak{m}$ because $s,t\mathrm{\notin }\mathfrak{m}$ and $\mathfrak{m}$ is prime; thus $a\mathrm{/}s,b\mathrm{/}t\in \mathfrak{m}{R}_{\mathfrak{m}}$. Hence the classes of $x$ and $z$ are a $k$-basis of $P_{\mathfrak{m}}\mathrm{/}\mathfrak{m}{R}_{\mathfrak{m}}{P}_{\mathfrak{m}}$ and this space is $2$-dimensional.

Let $\alpha \in \mathrm{Frac}(C)$ be integral over $C$. The same monic equation exhibits $\alpha$ as integral over $k[u,v]$, which is an integrally closed domain by [F10], so $\alpha \in k[u,v]$; and every element of $\mathrm{Frac}(C)$ is a quotient of $G$-invariant elements of $C$, hence is $G$-invariant, so $\mathrm{Frac}(C)\subseteq k(u,v{)}^G$. Therefore $\alpha \in k[u,v]\cap k(u,v{)}^G=k[u,v{]}^G=C$ by step 2.1, so $C$ is integrally closed, and by step 1.1 so is $R$.

If $P_{\mathfrak{m}}=(g)$ for some $g\in R_{\mathfrak{m}}$, then the quotient $P_{\mathfrak{m}}\mathrm{/}\mathfrak{m}{R}_{\mathfrak{m}}{P}_{\mathfrak{m}}$ is generated as a $k$-vector space by the image of $g$ alone and has dimension at most $1$, contradicting step 2.2. Hence $P_{\mathfrak{m}}$ is not a principal ideal of $R_{\mathfrak{m}}$.

The polynomial ring $k[x,y,z]$ is of finite type over the field $k$, hence Noetherian by [F11], and its quotient $R$ is Noetherian by [F11]. As $R$ is a domain by step 2.1, $X=\mathrm{Spec}R$ is an integral scheme by [F1] and [F2]. Its local rings at the points $\mathfrak{p}\in X$ are the prime localisations $R_{\mathfrak{p}}$ [F1], and these are integrally closed by step 3.1 and [F12]; so $X$ is normal by [F3].

The ring $R_{\mathfrak{m}}$ is a localisation of the Noetherian domain $R$ of step 4.1, hence a Noetherian ring by [F11] and a domain with fraction field $K=K(X)$: a product of two fractions $a\mathrm{/}s$, $b\mathrm{/}t$ with $s,t\mathrm{\notin }\mathfrak{m}$ is zero only if $ab=0$ in the domain $R$, and each fraction $a\mathrm{/}s$ with $s\mathrm{\notin }\mathfrak{m}$ and $a\mathrm{\ne }0$ is inverted by $s\mathrm{/}a\in K(X)$ [F13]; and $R_{\mathfrak{m}}$ is integrally closed by [F12] because $R$ is integrally closed by step 3.1 and $\mathfrak{m}$ is prime; so $R_{\mathfrak{m}}$ satisfies $(S_2)$ by [F8]. The intersection theorem of [F8] therefore gives $R_{\mathfrak{m}}={\bigcap }_{\mathrm{ht}\mathfrak{q}=1}{R}_{\mathfrak{q}}$ inside $K$, the intersection running over the height-one primes $\mathfrak{q}\subseteq \mathfrak{m}$ of $R$: primes of $R_{\mathfrak{m}}$ correspond to the primes $\mathfrak{q}\subseteq \mathfrak{m}$ of $R$, with $(R_{\mathfrak{m}}{)}_{\mathfrak{q}}=R_{\mathfrak{q}}$ and $\dim R_{\mathfrak{q}}=\mathrm{ht}\mathfrak{q}$ by [F12].

The ring $R$ is Noetherian by step 4.1 and $P$ is minimal over the principal ideal $(x)$ by step 1.2, so $\mathrm{ht}P\le 1$ by [F9]; and $P\mathrm{\ne }0$ because $0\mathrm{\ne }x\in P$ while $R$ is a domain by step 2.1, so $\mathrm{ht}P=1$.

The quotient $R\mathrm{/}P\cong k[y]$ is a domain, so $Z=V(P)$ is an integral closed subscheme of $X$ with generic point $P$; its codimension is $\dim {\mathcal{O}}_{X,P}=\dim R_P=\mathrm{ht}P=1$ by step 5.2, so $Z$ is a prime divisor of $X$ and $[Z]\in \mathrm{Div}(X)$.

Suppose for contradiction that $Z$ is Cartier at the vertex: there are an open neighbourhood $U$ of $\mathfrak{m}$ and a Cartier divisor $D$ on $U$ with ${\mathrm{cyc}}_U(D)=[Z\cap U]$, the point $\mathfrak{m}$ lying in the nonempty open set $U$, and the prime divisor $Z$ being the one of step 6.1. Then $U$, being an open subscheme of the integral scheme $X$ of step 4.1, is integral by [F2], normal by [F3], and Noetherian by [F14], so the Cartier-to-Weil construction of [F6] applies to it; let $\{(U_i,f_i)\}$ be a local-equation datum of $D$ [F7] and choose an index $i_0$ with $\mathfrak{m}\in U_{i_0}$. Since $U$ is integral, ${\mathcal{K}}_U$ is the constant sheaf $K(X)$ by [F5], so $f:=f_{i_0}$ is an element of $K(X{)}^{\times }$.

Every height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$ of $R$ lies in $U_{i_0}$: the closure of $\{\mathfrak{q}\}$ in $X$ is $V(\mathfrak{q})$, which contains $\mathfrak{m}$ because $\mathfrak{q}\subseteq \mathfrak{m}$, and if $\mathfrak{q}$ were not in $U_{i_0}$ then the closed set $X\setminus U_{i_0}$ would contain $\mathfrak{q}$, hence its closure and the point $\mathfrak{m}$, contradicting $\mathfrak{m}\in U_{i_0}$.

For every height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$ the point $\mathfrak{q}$ lies in $U_{i_0}$ by step 8.1, so the closure of $\{\mathfrak{q}\}$ in $U$ is a prime divisor $W_{\mathfrak{q}}$ of $U$ with generic point $\mathfrak{q}$, and the coefficient of $[W_{\mathfrak{q}}]$ in ${\mathrm{cyc}}_U(D)=[Z\cap U]$ is, by the coefficient formula of [F6] applied with the index $i_0$, the value $v_{\mathfrak{q}}(f_{\mathfrak{q}})$ of the normalized valuation of ${\mathcal{O}}_{U,\mathfrak{q}}={\mathcal{O}}_{X,\mathfrak{q}}=R_{\mathfrak{q}}$, which is $v_{\mathfrak{q}}(f)$ by [F4]. This coefficient is $1$ for $\mathfrak{q}=P$, because the closure of $\{P\}$ in $U$ is the restricted prime divisor $Z\cap U$ of step 6.1, and it is $0$ for every other height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$. Hence $v_P(f)=1$ and $v_{\mathfrak{q}}(f)=0$ for every other height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$.

For every height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$ we have $v_{\mathfrak{q}}(f)\ge 0$ by step 9.1, and $R_{\mathfrak{q}}$ is the valuation ring of $v_{\mathfrak{q}}$ by [F4], so $f\in R_{\mathfrak{q}}$; the intersection presentation of step 5.1 gives $f\in R_{\mathfrak{m}}$.

The case $g=0$ gives $g\in f{R}_{\mathfrak{m}}$ directly. Let $0\mathrm{\ne }g\in P_{\mathfrak{m}}=P{R}_{\mathfrak{m}}$. Then $g\in R_{\mathfrak{m}}\subseteq R_{\mathfrak{q}}$ for every height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$, so $v_{\mathfrak{q}}(g)\ge 0$; and $v_P(g)\ge 1$ because $g\in P{R}_P$, the maximal ideal of the discrete valuation ring $R_P$ by [F4]. With $v_P(f)=1$ and $v_{\mathfrak{q}}(f)=0$ for $\mathfrak{q}\mathrm{\ne }P$ from step 9.1 this gives $v_{\mathfrak{q}}(g\mathrm{/}f)\ge 0$ for every height-one prime $\mathfrak{q}\subseteq \mathfrak{m}$, so $g\mathrm{/}f\in R_{\mathfrak{m}}$ by the intersection presentation of step 5.1, that is, $g\in f{R}_{\mathfrak{m}}$. Hence $P_{\mathfrak{m}}\subseteq f{R}_{\mathfrak{m}}$.

Since $v_P(f)=1\ge 1$, the element $f$ lies in the maximal ideal $P{R}_P$ of the discrete valuation ring $R_P$ by [F4], and $f\in R_{\mathfrak{m}}$ by step 10.1, so $f\in P{R}_P\cap R_{\mathfrak{m}}=P{R}_{\mathfrak{m}}=P_{\mathfrak{m}}$, this contraction being computed inside $R_{\mathfrak{m}}$ along the localisation $R_{\mathfrak{m}}\to (R_{\mathfrak{m}}{)}_P=R_P$ of a prime ideal by [F12]. With step 10.2 this gives $P_{\mathfrak{m}}=f{R}_{\mathfrak{m}}$, a principal ideal of $R_{\mathfrak{m}}$.

The principal-ideal conclusion of step 11.1 contradicts the non-principality of step 3.2, so no open neighbourhood $U$ of $\mathfrak{m}$ carries a Cartier divisor $D$ with ${\mathrm{cyc}}_U(D)=[Z\cap U]$: the height-one prime divisor $Z=V(x,z)$ on the normal quadric cone is not Cartier at the vertex.