The Stacks project

Lemma 27.11.8. Let $S$ be a graded ring. Let $d \geq 1$. Set $S' = S^{(d)}$ with notation as in Algebra, Section 10.56. Set $X = \text{Proj}(S)$ and $X' = \text{Proj}(S')$. There is a canonical isomorphism $i : X \to X'$ of schemes such that

  1. for any graded $S$-module $M$ setting $M' = M^{(d)}$, we have a canonical isomorphism $\widetilde{M} \to i^*\widetilde{M'}$,

  2. we have canonical isomorphisms $\mathcal{O}_{X}(nd) \to i^*\mathcal{O}_{X'}(n)$

and these isomorphisms are compatible with the multiplication maps of Lemma 27.9.1 and hence with the maps (, (, (, (, (, and ( (see proof for precise statements.

Proof. The injective ring map $S' \to S$ (which is not a homomorphism of graded rings due to our conventions), induces a map $j : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S')$. Given a graded prime ideal $\mathfrak p \subset S$ we see that $\mathfrak p' = j(\mathfrak p) = S' \cap \mathfrak p$ is a graded prime ideal of $S'$. Moreover, if $f \in S_+$ is homogeneous and $f \not\in \mathfrak p$, then $f^ d \in S'_+$ and $f^ d \not\in \mathfrak p'$. Conversely, if $\mathfrak p' \subset S'$ is a graded prime ideal not containing some homogeneous element $f \in S'_+$, then $\mathfrak p = \{ g \in S \mid g^ d \in \mathfrak p'\} $ is a graded prime ideal of $S$ not containing $f$ whose image under $j$ is $\mathfrak p'$. To see that $\mathfrak p$ is an ideal, note that if $g, h \in \mathfrak p$, then $(g + h)^{2d} \in \mathfrak p'$ by the binomial formula and hence $g + h \in \mathfrak p'$ as $\mathfrak p'$ is a prime. In this way we see that $j$ induces a homeomorphism $i : X \to X'$. Moreover, given $f \in S_+$ homogeneous, then we have $S_{(f)} \cong S'_{(f^ d)}$. Since these isomorphisms are compatible with the restrictions mappings of Lemma 27.8.1, we see that there exists an isomorphism $i^\sharp : i^{-1}\mathcal{O}_{X'} \to \mathcal{O}_ X$ of structure sheaves on $X$ and $X'$, hence $i$ is an isomorphism of schemes.

Let $M$ be a graded $S$-module. Given $f \in S_+$ homogeneous, we have $M_{(f)} \cong M'_{(f^ d)}$, hence in exactly the same manner as above we obtain the isomorphism in (1). The isomorphisms in (2) are a special case of (1) for $M = S(nd)$ which gives $M' = S'(n)$. Let $M$ and $N$ be graded $S$-modules. Then we have

\[ M' \otimes _{S'} N' = (M \otimes _ S N)^{(d)} = (M \otimes _ S N)' \]

as can be verified directly from the definitions. Having said this the compatibility with the multiplication maps of Lemma 27.9.1 is the commutativity of the diagram

\[ \xymatrix{ \widetilde M \otimes _{\mathcal{O}_ X} \widetilde N \ar[d]_{(1) \otimes (1)} \ar[r] & \widetilde{M \otimes _ S N} \ar[d]^{(1)} \\ i^*(\widetilde{M'} \otimes _{\mathcal{O}_{X'}} \widetilde{N'}) \ar[r] & i^*(\widetilde{M' \otimes _{S'} N'}) } \]

This can be seen by looking at the construction of the maps over the open $D_+(f) = D_+(f^ d)$ where the top horizontal arrow is given by the map $M_{(f)} \times N_{(f)} \to (M \otimes _ S N)_{(f)}$ and the lower horizontal arrow by the map $M'_{(f^ d)} \times N'_{(f^ d)} \to (M' \otimes _{S'} N')_{(f^ d)}$. Since these maps agree via the identifications $M_{(f)} = M'_{(f^ d)}$, etc, we get the desired compatibility. We omit the proof of the other compatibilities. $\square$

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