A plane intersection with no common component is nonempty and zero-dimensional

The ring $R$ has dimension $3$ by [L1], and $0⊊(x_0)⊊(x_0,x_1)⊊\mathfrak{m}$ is a strict chain of primes of $R$: each displayed ideal is prime, since the successive quotients are $k[x_1,x_2]$, $k[x_1]$, $k$ and polynomial rings over a domain are domains by [L2]. Hence $\mathrm{ht}(\mathfrak{m})\ge 3$, while $\mathrm{ht}(\mathfrak{m})\le \dim R=3$ because chains of primes below $\mathfrak{m}$ are chains of primes of $R$; so $\mathrm{ht}(\mathfrak{m})=3$, and every prime $P⊊\mathfrak{m}$ satisfies $\mathrm{ht}(P)+1\le \mathrm{ht}(\mathfrak{m})$, that is, $\mathrm{ht}(P)\le 2$. Finally, every proper homogeneous prime $P$ of $R$ lies in $\mathfrak{m}$ by [L3].

Let $\mathfrak{q}\subseteq R$ be a prime with $(F,G)\subseteq \mathfrak{q}$. Then $\mathfrak{q}\mathrm{\ne }0$ because $F\in \mathfrak{q}$ and $F\mathrm{\ne }0$, so $\mathrm{ht}(\mathfrak{q})\ge 1$ by the chain $0⊊\mathfrak{q}$ of primes of the domain $R$; if $\mathrm{ht}(\mathfrak{q})=1$, then $\mathfrak{q}$ is a height-one prime containing both $F$ and $G$, contradicting [L4]. Hence every prime of $R$ containing $(F,G)$ has height at least two.

By [L3] the primes of $S=R\mathrm{/}(F,G)$ correspond inclusion-preservingly to the primes of $R$ containing $(F,G)$, and homogeneous primes to homogeneous primes; moreover the irrelevant ideal is $S_{+}=\mathfrak{m}\mathrm{/}(F,G)$. Hence for a prime $\mathfrak{q}\subseteq S$ with preimage $Q\subseteq R$ one has $S_{+}⊈\mathfrak{q}$ if and only if $\mathfrak{m}⊈Q$; since a proper homogeneous prime $Q$ of $R$ satisfies $Q\subseteq \mathfrak{m}$ by 1.1, this is equivalent to $Q⊊\mathfrak{m}$.

By [L9], $R$ is Noetherian. The ideal $(F,G)$ is proper because positive-degree homogeneous forms lie in $\mathfrak{m}=(x_0,x_1,x_2)\mathrm{\ne }R$. Let $\mathfrak{P}$ be a prime minimal over $(F,G)$, which exists by [L5]. Then $\mathfrak{P}$ is homogeneous. Indeed, let ${\mathfrak{P}}^{\mathrm{\prime }}$ be the ideal generated by all homogeneous elements of $\mathfrak{P}$; it is a homogeneous ideal with $(F,G)\subseteq {\mathfrak{P}}^{\mathrm{\prime }}\subseteq \mathfrak{P}$, since $F,G$ are homogeneous elements of $\mathfrak{P}$. It is prime: for homogeneous $a,b$ with $ab\in {\mathfrak{P}}^{\mathrm{\prime }}$ we have $ab\in \mathfrak{P}$, so $a\in \mathfrak{P}$ or $b\in \mathfrak{P}$, hence $a\in {\mathfrak{P}}^{\mathrm{\prime }}$ or $b\in {\mathfrak{P}}^{\mathrm{\prime }}$; an ideal generated by homogeneous elements with this property is prime, because for arbitrary $x,y$ with $xy\in {\mathfrak{P}}^{\mathrm{\prime }}$ one inducts on ${\mathrm{deg}}_{\max }(x)+{\mathrm{deg}}_{\max }(y)$, where ${\mathrm{deg}}_{\max }$ is the largest degree of a homogeneous component: the components of top degree $d$ of $x$ and $e$ of $y$ multiply to the top-degree component $x_d{y}_e\in {\mathfrak{P}}^{\mathrm{\prime }}$, so $x_d\in {\mathfrak{P}}^{\mathrm{\prime }}$ or $y_e\in {\mathfrak{P}}^{\mathrm{\prime }}$, and in the first case $(x-x_d)y\in {\mathfrak{P}}^{\mathrm{\prime }}$ with ${\mathrm{deg}}_{\max }(x-x_d)<d$ gives $x-x_d\in {\mathfrak{P}}^{\mathrm{\prime }}$ or $y\in {\mathfrak{P}}^{\mathrm{\prime }}$ by induction, whence $x\in {\mathfrak{P}}^{\mathrm{\prime }}$ or $y\in {\mathfrak{P}}^{\mathrm{\prime }}$, the other case being symmetric. Thus ${\mathfrak{P}}^{\mathrm{\prime }}$ is a prime containing $(F,G)$ and contained in $\mathfrak{P}$, so minimality forces ${\mathfrak{P}}^{\mathrm{\prime }}=\mathfrak{P}$ and $\mathfrak{P}$ is homogeneous. By 1.2, $\mathrm{ht}(\mathfrak{P})\ge 2$, and by the Noetherian Krull height theorem in [L5] applied to the two generators $F,G$ we have $\mathrm{ht}(\mathfrak{P})\le 2$, so $\mathrm{ht}(\mathfrak{P})=2$; since $\mathrm{ht}(\mathfrak{m})=3$ by 1.1, the inclusion $\mathfrak{P}\subseteq \mathfrak{m}$ of 1.1 is strict: $\mathfrak{P}⊊\mathfrak{m}$.

Every standard chart ring $A_i$ is either zero or of Krull dimension $0$. By [L6] the primes of $A_i$ correspond inclusion-preservingly to the homogeneous primes of $S$ avoiding $x_i$, so it suffices to rule out a strict chain ${\mathfrak{p}}_0⊊{\mathfrak{p}}_1$ of homogeneous primes of $S$ with $x_i\mathrm{\notin }{\mathfrak{p}}_1$. By 2.1 such a chain lifts to homogeneous primes $P_0⊊P_1$ of $R$ containing $(F,G)$ and, since $x_i$ lies in neither ${\mathfrak{p}}_j$, with $x_i\mathrm{\notin }{P}_1$. Then $\mathrm{ht}(P_0)\ge 2$ by 1.2; on the other hand $P_1$ is a proper homogeneous prime with $P_1\mathrm{\ne }\mathfrak{m}$, so $P_1⊊\mathfrak{m}$ by 1.1 and $\mathrm{ht}(P_1)\le 2$; and $\mathrm{ht}(P_0)+1\le \mathrm{ht}(P_1)$ by [L1]. Hence $2\le \mathrm{ht}(P_0)<\mathrm{ht}(P_1)\le 2$, a contradiction. Therefore $A_i$ admits no strict chain of two primes.

By 2.2 the prime $\mathfrak{P}$ is homogeneous with $\mathfrak{P}⊊\mathfrak{m}$ and $(F,G)\subseteq \mathfrak{P}$, so by 2.1 its image $\mathfrak{p}=\mathfrak{P}\mathrm{/}(F,G)$ is a homogeneous prime of $S$ with $S_{+}⊈\mathfrak{p}$; hence $\mathfrak{p}$ is a point of $X$ by [L6] and $X\mathrm{\ne }\mathrm{\varnothing }$.

$\dim S=1$. For the lower bound, $\mathfrak{P}⊊\mathfrak{m}$ from 2.2 gives via 2.1 a strict chain of primes of $S$, so $\dim S\ge 1$ and $S\mathrm{\ne }0$. For the upper bound, suppose ${\mathfrak{q}}_0⊊{\mathfrak{q}}_1⊊{\mathfrak{q}}_2$ is a strict chain of primes of $S$; by 2.1 it lifts to primes $Q_0⊊Q_1⊊Q_2$ of $R$ all containing $(F,G)$. By 2.2 we have $\mathrm{ht}(Q_0)\ge 2$, so there is a strict chain of length two of primes below $Q_0$; adjoining $Q_0⊊Q_1⊊Q_2$ gives a strict chain of length $4$ in $R$, contradicting $\dim R=3$ from 1.1. Hence $\dim S=1$, and in particular $S$ is not zero-dimensional: it is also not Artinian, because in an Artinian ring every prime is maximal (Every prime ideal of an Artinian ring is maximal), which would force $\dim S=0$.

By [L10], $X$ is Noetherian, so its chain dimension in [L7] is defined. $X$ has no strict chain $Z_0⊊Z_1$ of nonempty irreducible closed subsets. Suppose such a chain is given. Some standard chart $D=D_{+}(x_i)$ meets $Z_0$, since the charts cover $X$ by [L6]; then the subsets $W_j=Z_j\cap D$ of the affine chart $D=\mathrm{Spec}(A_i)$ are nonempty, closed in $D$, and satisfy $W_0\subseteq W_1$. A nonempty open subset $U$ of an irreducible space $Z$ is irreducible and dense: if $U=F_1\cup F_2$ with $F_1,F_2$ closed in $U$, then $Z=\overline{F_1}\cup \overline{F_2}\cup (Z\setminus U)$ is a union of closed subsets, so irreducibility of $Z$ forces one of the three to equal $Z$, and since $Z\setminus U\mathrm{\ne }Z$ while $\overline{F_k}\cap U=F_k$, this gives $F_1=U$ or $F_2=U$; and if the closure of $U$ in $Z$ were a proper closed subset, then $Z=\bar{U}\cup (Z\setminus U)$ would be a union of two proper closed subsets. Applying this to the open subset $W_j$ of the irreducible space $Z_j$ shows that $W_j$ is irreducible and dense in $Z_j$, hence that its closure in $X$ is $Z_j$; thus $W_0\mathrm{\ne }{W}_1$, since $W_0=W_1$ would give $Z_0=Z_1$. The $W_j$ are nonempty irreducible closed subsets of $\mathrm{Spec}(A_i)$, so by [L7] their radical defining ideals are distinct primes and form a strict chain of two primes of $A_i$, contradicting 3.1. Hence no such chain $Z_0⊊Z_1$ exists, and the chain dimension of $X$ is $0$.

Claim 1 is 3.2, claim 2 is 3.1 together with 4.1, and claim 3 is 3.3. The hypotheses actually used are: $F,G$ nonzero homogeneous of positive degrees with no common nonconstant factor, over an arbitrary field $k$; the Axiom of Choice enters only through the minimal-prime and height-theorem suppliers of [L5] the irreducible-closed-subset characterisation of [L7], and the Noetherian-spectrum supplier of [L10], and it is the standing assumption [L8]. Noetherianity of the chart rings is used in [L10] to justify the topological dimension convention in 4.1. Finiteness of $X$ is not assumed, and the minimal prime chosen in 2.2 is proved homogeneous rather than chosen inside the homogeneous locus.