The Stacks project

Proof. Clearly condition (2) uniquely determines the morphism of schemes and the open subset $U(\psi )$. Pick $f \in A_ d$ with $d \geq 1$. Note that $\mathcal{O}_ X(n)|_{D_{+}(f)}$ corresponds to the $A_{(f)}$-module $(A_ f)_ n$ and that $\mathcal{O}_ Y(n)|_{D_{+}(\psi (f))}$ corresponds to the $B_{(\psi (f))}$-module $(B_{\psi (f)})_ n$. In other words $\theta$ when restricted to $D_{+}(\psi (f))$ corresponds to a map of $\mathbf{Z}$-graded $B_{(\psi (f))}$-algebras

Condition (1) determines the images of all elements of $A$. Since $f$ is an invertible element which is mapped to $\psi (f)$ we see that $1/f^ m$ is mapped to $1/\psi (f)^ m$. It easily follows from this that $\theta$ is uniquely determined, namely it is given by the rule

To show existence we remark that the proof of uniqueness above gave a well defined prescription for the morphism $r$ and the map $\theta$ when restricted to every standard open of the form $D_{+}(\psi (f)) \subset U(\psi )$ into $D_{+}(f)$. Call these $r_ f$ and $\theta _ f$. Hence we only need to verify that if $D_{+}(f) \subset D_{+}(g)$ for some $f, g \in A_{+}$ homogeneous, then the restriction of $r_ g$ to $D_{+}(\psi (f))$ matches $r_ f$. This is clear from the formulas given for $r$ and $\theta$ above. $\square$

Proof. Omitted. $\square$

Proof. Part (1) follows from the definition of $U(\psi )$ and the fact that $D_{+}(f) = D_{+}(f^ n)$ for any $n > 0$. For $f \in A_{+}$ homogeneous we see that $A_{(f)} \to B_{(\psi (f))}$ is surjective because any element of $B_{(\psi (f))}$ can be represented by a fraction $b/\psi (f)^ n$ with $n$ arbitrarily large (which forces the degree of $b \in B$ to be large). This proves (2). The same argument shows the map

is surjective which proves the surjectivity of $\theta$. For an example where this map is not an isomorphism consider the graded ring $A = k[x, y]$ where $k$ is a field and $\deg (x) = 1$, $\deg (y) = 2$. Set $I = (x)$, so that $B = k[y]$. Note that $\mathcal{O}_ Y(1) = 0$ in this case. But it is easy to see that $r_\psi ^*\mathcal{O}_ X(1)$ is not zero. (There are less silly examples.) $\square$

Proof. We have (1) by Lemma 27.11.3. Let $f \in A_{+}$ be homogeneous. The assumption on $\psi$ implies that $A_ f \to B_ f$ is an isomorphism (details omitted). Thus it is clear that $r_\psi$ and $\theta$ restrict to isomorphisms over $D_{+}(f)$. The lemma follows. $\square$

Proof. By Lemmas 27.11.4 and 27.11.2 we may replace $B$ by the image of $A \to B$ without changing $X$ or the sheaves $\mathcal{O}_ X(n)$. Thus we may assume that $A \to B$ is surjective. By Lemma 27.11.3 we get (1) and (2) and surjectivity in (3). By Lemma 27.10.3 we see that both $\mathcal{O}_ X(n)$ and $\mathcal{O}_ Y(n)$ are invertible. Hence $\theta$ is an isomorphism. $\square$

Proof. This follows immediately by looking at what happens over the standard opens $D_{+}(f)$ for $f \in A_{+}$. $\square$

Proof. This is a special case of Lemma 27.11.6 above. $\square$

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

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

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. $\square$