Lemma 115.5.6. Let $R$ be a ring. Let $F(X, Y) \in R[X, Y]$ be homogeneous of degree $d$. Let $S = R[X, Y]/(F)$ as a graded ring. Let $\mathfrak p \subset R$ be a prime such that some coefficient of $F$ is not in $\mathfrak p$. There exists an $f \in R$ $f \not\in \mathfrak p$, an integer $e$, and a $G \in R[X, Y]_ e$ such that multiplication by $G$ induces isomorphisms $(S_ n)_ f \to (S_{n + e})_ f$ for all $n \geq d$.

**Proof.**
During the course of the proof we may replace $R$ by $R_ f$ for $f\in R$, $f\not\in \mathfrak p$ (finitely often). As a first step we do such a replacement such that some coefficient of $F$ is invertible in $R$. In particular the modules $S_ n$ are now locally free of rank $d$ for $n \geq d$ by Lemma 115.5.4. Pick any $G \in R[X, Y]_ e$ such that the image of $G$ in $\kappa (\mathfrak p)[X, Y]$ is relatively prime to the image of $F(X, Y)$ (this is possible for some $e$). Apply Algebra, Lemma 10.79.4 to the map induced by multiplication by $G$ from $S_ d \to S_{d + e}$. By our choice of $G$ and Lemma 115.5.5 we see $S_ d \otimes \kappa (\mathfrak p) \to S_{d + e} \otimes \kappa (\mathfrak p)$ is bijective. Thus, after replacing $R$ by $R_ f$ for a suitable $f$ we may assume that $G : S_ d \to S_{d + e}$ is bijective. This in turn implies that the image of $G$ in $\kappa (\mathfrak p')[X, Y]$ is relatively prime to the image of $F$ for all primes $\mathfrak p'$ of $R$. And then by Algebra, Lemma 10.79.4 again we see that all the maps $G : S_ d \to S_{d + e}$, $n \geq d$ are isomorphisms.
$\square$

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