The Stacks project

27.11 Functoriality of Proj

A graded ring map $\psi : A \to B$ does not always give rise to a morphism of associated projective homogeneous spectra. The reason is that the inverse image $\psi ^{-1}(\mathfrak q)$ of a homogeneous prime $\mathfrak q \subset B$ may contain the irrelevant prime $A_{+}$ even if $\mathfrak q$ does not contain $B_{+}$. The correct result is stated as follows.

Lemma 27.11.1. Let $A$, $B$ be two graded rings. Set $X = \text{Proj}(A)$ and $Y = \text{Proj}(B)$. Let $\psi : A \to B$ be a graded ring map. Set

\[ U(\psi ) = \bigcup \nolimits _{f \in A_{+}\ \text{homogeneous}} D_{+}(\psi (f)) \subset Y. \]

Then there is a canonical morphism of schemes

\[ r_\psi : U(\psi ) \longrightarrow X \]

and a map of $\mathbf{Z}$-graded $\mathcal{O}_{U(\psi )}$-algebras

\[ \theta = \theta _\psi : r_\psi ^*\left( \bigoplus \nolimits _{d \in \mathbf{Z}} \mathcal{O}_ X(d) \right) \longrightarrow \bigoplus \nolimits _{d \in \mathbf{Z}} \mathcal{O}_{U(\psi )}(d). \]

The triple $(U(\psi ), r_\psi , \theta )$ is characterized by the following properties:

  1. For every $d \geq 0$ the diagram

    \[ \xymatrix{ A_ d \ar[d] \ar[rr]_{\psi } & & B_ d \ar[d] \\ \Gamma (X, \mathcal{O}_ X(d)) \ar[r]^-\theta & \Gamma (U(\psi ), \mathcal{O}_ Y(d)) & \Gamma (Y, \mathcal{O}_ Y(d)) \ar[l] } \]

    is commutative.

  2. For any $f \in A_{+}$ homogeneous we have $r_\psi ^{-1}(D_{+}(f)) = D_{+}(\psi (f))$ and the restriction of $r_\psi $ to $D_{+}(\psi (f))$ corresponds to the ring map $A_{(f)} \to B_{(\psi (f))}$ induced by $\psi $.

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

\[ A_ f \otimes _{A_{(f)}} B_{(\psi (f))} \longrightarrow B_{\psi (f)} \]

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

\[ a/f^ m \otimes b/\psi (f)^ e \longmapsto \psi (a)b/\psi (f)^{m + e}. \]

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. $\square$

Lemma 27.11.2. Let $A$, $B$, and $C$ be graded rings. Set $X = \text{Proj}(A)$, $Y = \text{Proj}(B)$ and $Z = \text{Proj}(C)$. Let $\varphi : A \to B$, $\psi : B \to C$ be graded ring maps. Then we have

\[ U(\psi \circ \varphi ) = r_\varphi ^{-1}(U(\psi )) \quad \text{and} \quad r_{\psi \circ \varphi } = r_\varphi \circ r_\psi |_{U(\psi \circ \varphi )}. \]

In addition we have

\[ \theta _\psi \circ r_\psi ^*\theta _\varphi = \theta _{\psi \circ \varphi } \]

with obvious notation.

Proof. Omitted. $\square$

Lemma 27.11.3. With hypotheses and notation as in Lemma 27.11.1 above. Assume $A_ d \to B_ d$ is surjective for all $d \gg 0$. Then

  1. $U(\psi ) = Y$,

  2. $r_\psi : Y \to X$ is a closed immersion, and

  3. the maps $\theta : r_\psi ^*\mathcal{O}_ X(n) \to \mathcal{O}_ Y(n)$ are surjective but not isomorphisms in general (even if $A \to B$ is surjective).

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

\[ A_ f \to B_{\psi (f)} \]

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.) $\square$

Lemma 27.11.4. With hypotheses and notation as in Lemma 27.11.1 above. Assume $A_ d \to B_ d$ is an isomorphism for all $d \gg 0$. Then

  1. $U(\psi ) = Y$,

  2. $r_\psi : Y \to X$ is an isomorphism, and

  3. the maps $\theta : r_\psi ^*\mathcal{O}_ X(n) \to \mathcal{O}_ Y(n)$ are isomorphisms.

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. $\square$

Lemma 27.11.5. With hypotheses and notation as in Lemma 27.11.1 above. Assume $A_ d \to B_ d$ is surjective for $d \gg 0$ and that $A$ is generated by $A_1$ over $A_0$. Then

  1. $U(\psi ) = Y$,

  2. $r_\psi : Y \to X$ is a closed immersion, and

  3. the maps $\theta : r_\psi ^*\mathcal{O}_ X(n) \to \mathcal{O}_ Y(n)$ are isomorphisms.

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. $\square$

Lemma 27.11.6. With hypotheses and notation as in Lemma 27.11.1 above. Assume there exists a ring map $R \to A_0$ and a ring map $R \to R'$ such that $B = R' \otimes _ R A$. Then

  1. $U(\psi ) = Y$,

  2. the diagram

    \[ \xymatrix{ Y = \text{Proj}(B) \ar[r]_{r_\psi } \ar[d] & \text{Proj}(A) = X \ar[d] \\ \mathop{\mathrm{Spec}}(R') \ar[r] & \mathop{\mathrm{Spec}}(R) } \]

    is a fibre product square, and

  3. the maps $\theta : r_\psi ^*\mathcal{O}_ X(n) \to \mathcal{O}_ Y(n)$ are isomorphisms.

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

Lemma 27.11.7. With hypotheses and notation as in Lemma 27.11.1 above. Assume there exists a $g \in A_0$ such that $\psi $ induces an isomorphism $A_ g \to B$. Then $U(\psi ) = Y$, $r_\psi : Y \to X$ is an open immersion which induces an isomorphism of $Y$ with the inverse image of $D(g) \subset \mathop{\mathrm{Spec}}(A_0)$. Moreover the map $\theta $ is an isomorphism.

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

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. $\square$

Comments (2)

Comment #1561 by Armando on

In the proof of the lemma 26.11.8: why is the set an ideal?

How I can conclude that ?

Thak you for your attention, Armando

Comment #1577 by on

This is true because and then you get it. I have added a sentence in the text to explain this, see here. Thanks!

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