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 (27.10.1.1), (27.10.1.2), (27.10.1.3), (27.10.1.4), (27.10.1.5), and (27.10.1.6) (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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