The Stacks project

Proof. Observe that the local rings of $X = \mathbf{P}^ n_ R$ are UFDs because $X$ is covered by affine pieces isomorphic to $\mathbf{A}^ n_ R$ and $R[x_1, \ldots , x_ n]$ is a UFD (Algebra, Lemma 10.120.10). Thus $Z$ is an effective Cartier divisor by Lemma 31.15.9. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the quasi-coherent sheaf of ideals corresponding to $Z$. Choose an isomorphism $\mathcal{O}(m) \to \mathcal{I}$ for some $m \in \mathbf{Z}$, see Lemma 31.28.5. Then the composition

is nonzero. We conclude that $m \leq 0$ and that the corresponding section of $\mathcal{O}_ X(m)^{\otimes -1} = \mathcal{O}_ X(-m)$ is given by some $F \in R[T_0, \ldots , T_ n]$ of degree $-m$, see Cohomology of Schemes, Lemma 30.8.1. Thus on the $i$th standard open $U_ i = D_+(T_ i)$ the closed subscheme $Z \cap U_ i$ is cut out by the ideal

Thus the homogeneous elements of the graded ideal $I(Z) = \mathop{\mathrm{Ker}}(R[T_0, \ldots , T_ n] \to \bigoplus \Gamma (\mathcal{O}_ Z(m)))$ is the set of homogeneous polynomials $G$ such that

for $i = 0, \ldots , n$. Clearing denominators, we see there exist $e_ i \geq 0$ such that

for $i = 0, \ldots , n$. As $R$ is a UFD, so is $R[T_0, \ldots , T_ n]$. Then $F | T_0^{e_0}G$ and $F | T_1^{e_1}G$ implies $F | G$ as $T_0^{e_0}$ and $T_1^{e_1}$ have no factor in common. Thus $I(Z) = (F)$. $\square$