Coprime positive-degree plane forms form a regular sequence

Since $S$ is an integral domain by [L3] and $F\mathrm{\ne }0$, multiplication by $F$ is injective on $S=S\mathrm{/}(0)S$, and $S\mathrm{\ne }0$.

A nonconstant element $h\in S$ divides both $F$ and $G$ if and only if some irreducible element $p\in S$ divides both: given such $h$, factor $h$ into irreducibles using [L1]; every irreducible factor is a nonzero nonunit, hence nonconstant, since the constants of $S$ are $0$ and the units of $k$; conversely an irreducible common divisor is a common nonconstant factor by [L5].

Suppose no irreducible element divides both $F$ and $G$, and let $G,H\in S$ with $GH\in (F)$, that is, $F\mid GH$. Write $F=u{p}_1^{a_1}\cdots p_r^{a_r}$ with $u$ a unit and the $p_i$ irreducible, using [L1] and $F\mathrm{\ne }0$; no $p_i$ divides $G$, and $p_i\mid GH$, so $p_i\mid H$ for every $i$ because $p_i$ is prime by [L1]; hence $F\mid H$ and $H\in (F)$. Therefore multiplication by $G$ is injective on $S\mathrm{/}(F)$.

Both $F$ and $G$ are nonzero homogeneous of positive degree, so by [L4] every monomial of $F$ and of $G$ lies in $\mathfrak{m}=(x_0,x_1,x_2)$; hence $(F,G)\subseteq \mathfrak{m}$, which is a proper ideal, and $S\mathrm{/}(F,G)\mathrm{\ne }0$.

If $p\in S$ is irreducible, then $(p)$ has height one. By [L1] the element $p$ is prime, so $(p)$ is a prime ideal, nonzero and proper; and the only prime ideals of $S$ contained in $(p)$ are $0$ and $(p)$: indeed if $0\mathrm{\ne }\mathfrak{q}\subseteq (p)$ is prime, factor a nonzero element $q\in \mathfrak{q}$ into irreducibles by [L1], so that some irreducible $r$ divides $q$ and lies in $\mathfrak{q}\subseteq (p)$, which forces $p\mid r$ and hence $r$ associate to $p$, so $p\in \mathfrak{q}$ and $(p)\subseteq \mathfrak{q}$. Hence $\mathrm{Spec}{S}_{(p)}$ consists of the two primes corresponding to $0$ and $(p)$, a chain of length one, so $\dim S_{(p)}=1$ and $\mathrm{ht}(p)=1$ by [L6].

By 1.3, multiplication by $G$ is injective on the module $S\mathrm{/}(F)=(S\mathrm{/}(F))S$. Moreover $S\mathrm{/}(F)\mathrm{\ne }0$: otherwise $1\in (F)$, so $F$ would be a unit of the domain $S$ by [L3], contradicting that $F$ is a nonzero nonunit, being homogeneous of positive degree.

The regular condition is equivalent to the height-one condition. If a height-one prime $\mathfrak{p}$ contains both $F$ and $G$, then $\mathfrak{p}=(p)$ with $p$ irreducible by [L1], so $p$ divides both and by 1.2 the forms have a common nonconstant factor. Conversely, if a nonconstant $h$ divides both, then by 1.2 an irreducible $p$ divides both; by 1.5 the principal prime $(p)$ has height one, and it contains both $F$ and $G$. Hence no common nonconstant factor is equivalent to: no height-one prime of $S$ contains both forms.

The sequence $(F,G)$ is $S$-regular: $S\mathrm{\ne }0$ and multiplication by $F$ is injective on $S$ by 1.1; multiplication by $G$ is injective on $S\mathrm{/}(F)$, which is nonzero, by 2.1; and $S\mathrm{/}(F,G)\mathrm{\ne }0$ by 1.4. This meets the definition [L2] in both slots.

Steps 3.1 and 2.2 prove the two equivalent formulations of the statement: the pair $(F,G)$ is an $S$-regular sequence, and no prime ideal of height one contains both $F$ and $G$. No hypothesis beyond the stated ones was used, and the arguments are valid over an arbitrary field $k$.