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 (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. $\square$

Comment #1561 by Armando on

In the proof of the lemma 26.11.8: why is the set $\mathfrak{p}=\{g\in S\mid g^d\in\mathfrak{p}^{\prime}\}$ an ideal?

How I can conclude that $\forall g,h\in\mathfrak{p},\,(g+h)^d=\sum_{k=0}^d\binom{d}{k}g^kh^{d-k}\in\mathfrak{p}^{\prime}$?

Thak you for your attention, Armando

Comment #1577 by on

This is true because $(g + h)^{2d} \in \mathfrak p'$ and then you get it. I have added a sentence in the text to explain this, see here. Thanks!

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