The Stacks project

Proof. We saw above that our construction of $\widetilde S$, resp. $\widetilde M$ gives a sheaf of $S_0$-algebras, resp. $S_0$-modules. Hence we get a morphism by Schemes, Lemma 26.6.4. This morphism, when restricted to $D_{+}(f)$ comes from the canonical ring map $S_0 \to S_{(f)}$. The maps $S \to S_ f$, $S_{(f)} \to S_ f$ are $S_0$-algebra maps, see Lemma 27.8.1. Hence if the homogeneous prime $\mathfrak p \subset S$ corresponds to the $\mathbf{Z}$-graded prime $\mathfrak p' \subset S_ f$ and the (usual) prime $\mathfrak p'' \subset S_{(f)}$, then each of these has the same inverse image in $S_0$. $\square$