## 27.12 Morphisms into Proj

Let $S$ be a graded ring. Let $X = \text{Proj}(S)$ be the homogeneous spectrum of $S$. Let $d \geq 1$ be an integer. Consider the open subscheme

27.12.0.1
$$\label{constructions-equation-Ud} U_ d = \bigcup \nolimits _{f \in S_ d} D_{+}(f) \quad \subset \quad X = \text{Proj}(S)$$

Note that $d | d' \Rightarrow U_ d \subset U_{d'}$ and $X = \bigcup _ d U_ d$. Neither $X$ nor $U_ d$ need be quasi-compact, see Algebra, Lemma 10.57.3. Let us write $\mathcal{O}_{U_ d}(n) = \mathcal{O}_ X(n)|_{U_ d}$. By Lemma 27.10.2 we know that $\mathcal{O}_{U_ d}(nd)$, $n \in \mathbf{Z}$ is an invertible $\mathcal{O}_{U_ d}$-module and that all the multiplication maps $\mathcal{O}_{U_ d}(nd) \otimes _{\mathcal{O}_{U_ d}} \mathcal{O}_ X(m) \to \mathcal{O}_{U_ d}(nd + m)$ of (27.10.1.1) are isomorphisms. In particular we have $\mathcal{O}_{U_ d}(nd) \cong \mathcal{O}_{U_ d}(d)^{\otimes n}$. The graded ring map (27.10.1.3) on global sections combined with restriction to $U_ d$ give a homomorphism of graded rings

27.12.0.2
$$\label{constructions-equation-psi-d} \psi ^ d : S^{(d)} \longrightarrow \Gamma _*(U_ d, \mathcal{O}_{U_ d}(d)).$$

For the notation $S^{(d)}$, see Algebra, Section 10.56. For the notation $\Gamma _*$ see Modules, Definition 17.24.7. Moreover, since $U_ d$ is covered by the opens $D_{+}(f)$, $f \in S_ d$ we see that $\mathcal{O}_{U_ d}(d)$ is globally generated by the sections in the image of $\psi ^ d_1 : S^{(d)}_1 = S_ d \to \Gamma (U_ d, \mathcal{O}_{U_ d}(d))$, see Modules, Definition 17.4.1.

Let $Y$ be a scheme, and let $\varphi : Y \to X$ be a morphism of schemes. Assume the image $\varphi (Y)$ is contained in the open subscheme $U_ d$ of $X$. By the discussion following Modules, Definition 17.24.7 we obtain a homomorphism of graded rings

$\Gamma _*(U_ d, \mathcal{O}_{U_ d}(d)) \longrightarrow \Gamma _*(Y, \varphi ^*\mathcal{O}_ X(d)).$

The composition of this and $\psi ^ d$ gives a graded ring homomorphism

27.12.0.3
$$\label{constructions-equation-psi-phi-d} \psi _\varphi ^ d : S^{(d)} \longrightarrow \Gamma _*(Y, \varphi ^*\mathcal{O}_ X(d))$$

which has the property that the invertible sheaf $\varphi ^*\mathcal{O}_ X(d)$ is globally generated by the sections in the image of $(S^{(d)})_1 = S_ d \to \Gamma (Y, \varphi ^*\mathcal{O}_ X(d))$.

Lemma 27.12.1. Let $S$ be a graded ring, and $X = \text{Proj}(S)$. Let $d \geq 1$ and $U_ d \subset X$ as above. Let $Y$ be a scheme. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Let $\psi : S^{(d)} \to \Gamma _*(Y, \mathcal{L})$ be a graded ring homomorphism such that $\mathcal{L}$ is generated by the sections in the image of $\psi |_{S_ d} : S_ d \to \Gamma (Y, \mathcal{L})$. Then there exist a morphism $\varphi : Y \to X$ such that $\varphi (Y) \subset U_ d$ and an isomorphism $\alpha : \varphi ^*\mathcal{O}_{U_ d}(d) \to \mathcal{L}$ such that $\psi _\varphi ^ d$ agrees with $\psi$ via $\alpha$:

$\xymatrix{ \Gamma _*(Y, \mathcal{L}) & \Gamma _*(Y, \varphi ^*\mathcal{O}_{U_ d}(d)) \ar[l]^-\alpha & \Gamma _*(U_ d, \mathcal{O}_{U_ d}(d)) \ar[l]^-{\varphi ^*} \\ S^{(d)} \ar[u]^\psi & & S^{(d)} \ar[u]^{\psi ^ d} \ar[ul]^{\psi ^ d_\varphi } \ar[ll]_{\text{id}} }$

commutes. Moreover, the pair $(\varphi , \alpha )$ is unique.

Proof. Pick $f \in S_ d$. Denote $s = \psi (f) \in \Gamma (Y, \mathcal{L})$. On the open set $Y_ s$ where $s$ does not vanish multiplication by $s$ induces an isomorphism $\mathcal{O}_{Y_ s} \to \mathcal{L}|_{Y_ s}$, see Modules, Lemma 17.24.10. We will denote the inverse of this map $x \mapsto x/s$, and similarly for powers of $\mathcal{L}$. Using this we define a ring map $\psi _{(f)} : S_{(f)} \to \Gamma (Y_ s, \mathcal{O})$ by mapping the fraction $a/f^ n$ to $\psi (a)/s^ n$. By Schemes, Lemma 26.6.4 this corresponds to a morphism $\varphi _ f : Y_ s \to \mathop{\mathrm{Spec}}(S_{(f)}) = D_{+}(f)$. We also introduce the isomorphism $\alpha _ f : \varphi _ f^*\mathcal{O}_{D_{+}(f)}(d) \to \mathcal{L}|_{Y_ s}$ which maps the pullback of the trivializing section $f$ over $D_{+}(f)$ to the trivializing section $s$ over $Y_ s$. With this choice the commutativity of the diagram in the lemma holds with $Y$ replaced by $Y_ s$, $\varphi$ replaced by $\varphi _ f$, and $\alpha$ replaced by $\alpha _ f$; verification omitted.

Suppose that $f' \in S_ d$ is a second element, and denote $s' = \psi (f') \in \Gamma (Y, \mathcal{L})$. Then $Y_ s \cap Y_{s'} = Y_{ss'}$ and similarly $D_{+}(f) \cap D_{+}(f') = D_{+}(ff')$. In Lemma 27.10.6 we saw that $D_{+}(f') \cap D_{+}(f)$ is the same as the set of points of $D_{+}(f)$ where the section of $\mathcal{O}_ X(d)$ defined by $f'$ does not vanish. Hence $\varphi _ f^{-1}(D_{+}(f') \cap D_{+}(f)) = Y_ s \cap Y_{s'} = \varphi _{f'}^{-1}(D_{+}(f') \cap D_{+}(f))$. On $D_{+}(f) \cap D_{+}(f')$ the fraction $f/f'$ is an invertible section of the structure sheaf with inverse $f'/f$. Note that $\psi _{(f')}(f/f') = \psi (f)/s' = s/s'$ and $\psi _{(f)}(f'/f) = \psi (f')/s = s'/s$. We claim there is a unique ring map $S_{(ff')} \to \Gamma (Y_{ss'}, \mathcal{O})$ making the following diagram commute

$\xymatrix{ \Gamma (Y_ s, \mathcal{O}) \ar[r] & \Gamma (Y_{ss'}, \mathcal{O}) & \Gamma (Y_{s, '} \mathcal{O}) \ar[l]\\ S_{(f)} \ar[r] \ar[u]^{\psi _{(f)}} & S_{(ff')} \ar[u] & S_{(f')} \ar[l] \ar[u]^{\psi _{(f')}} }$

It exists because we may use the rule $x/(ff')^ n \mapsto \psi (x)/(ss')^ n$, which “works” by the formulas above. Uniqueness follows as $\text{Proj}(S)$ is separated, see Lemma 27.8.8 and its proof. This shows that the morphisms $\varphi _ f$ and $\varphi _{f'}$ agree over $Y_ s \cap Y_{s'}$. The restrictions of $\alpha _ f$ and $\alpha _{f'}$ agree over $Y_ s \cap Y_{s'}$ because the regular functions $s/s'$ and $\psi _{(f')}(f)$ agree. This proves that the morphisms $\psi _ f$ glue to a global morphism from $Y$ into $U_ d \subset X$, and that the maps $\alpha _ f$ glue to an isomorphism satisfying the conditions of the lemma.

We still have to show the pair $(\varphi , \alpha )$ is unique. Suppose $(\varphi ', \alpha ')$ is a second such pair. Let $f \in S_ d$. By the commutativity of the diagrams in the lemma we have that the inverse images of $D_{+}(f)$ under both $\varphi$ and $\varphi '$ are equal to $Y_{\psi (f)}$. Since the opens $D_{+}(f)$ are a basis for the topology on $X$, and since $X$ is a sober topological space (see Schemes, Lemma 26.11.1) this means the maps $\varphi$ and $\varphi '$ are the same on underlying topological spaces. Let us use $s = \psi (f)$ to trivialize the invertible sheaf $\mathcal{L}$ over $Y_{\psi (f)}$. By the commutativity of the diagrams we have that $\alpha ^{\otimes n}(\psi ^ d_{\varphi }(x)) = \psi (x) = (\alpha ')^{\otimes n}(\psi ^ d_{\varphi '}(x))$ for all $x \in S_{nd}$. By construction of $\psi ^ d_{\varphi }$ and $\psi ^ d_{\varphi '}$ we have $\psi ^ d_{\varphi }(x) = \varphi ^\sharp (x/f^ n) \psi ^ d_{\varphi }(f^ n)$ over $Y_{\psi (f)}$, and similarly for $\psi ^ d_{\varphi '}$. By the commutativity of the diagrams of the lemma we deduce that $\varphi ^\sharp (x/f^ n) = (\varphi ')^\sharp (x/f^ n)$. This proves that $\varphi$ and $\varphi '$ induce the same morphism from $Y_{\psi (f)}$ into the affine scheme $D_{+}(f) = \mathop{\mathrm{Spec}}(S_{(f)})$. Hence $\varphi$ and $\varphi '$ are the same as morphisms. Finally, it remains to show that the commutativity of the diagram of the lemma singles out, given $\varphi$, a unique $\alpha$. We omit the verification. $\square$

We continue the discussion from above the lemma. Let $S$ be a graded ring. Let $Y$ be a scheme. We will consider triples $(d, \mathcal{L}, \psi )$ where

1. $d \geq 1$ is an integer,

2. $\mathcal{L}$ is an invertible $\mathcal{O}_ Y$-module, and

3. $\psi : S^{(d)} \to \Gamma _*(Y, \mathcal{L})$ is a graded ring homomorphism such that $\mathcal{L}$ is generated by the global sections $\psi (f)$, with $f \in S_ d$.

Given a morphism $h : Y' \to Y$ and a triple $(d, \mathcal{L}, \psi )$ over $Y$ we can pull it back to the triple $(d, h^*\mathcal{L}, h^* \circ \psi )$. Given two triples $(d, \mathcal{L}, \psi )$ and $(d, \mathcal{L}', \psi ')$ with the same integer $d$ we say they are strictly equivalent if there exists an isomorphism $\beta : \mathcal{L} \to \mathcal{L}'$ such that $\beta \circ \psi = \psi '$ as graded ring maps $S^{(d)} \to \Gamma _*(Y, \mathcal{L}')$.

For each integer $d \geq 1$ we define

\begin{eqnarray*} F_ d : \mathit{Sch}^{opp} & \longrightarrow & \textit{Sets}, \\ Y & \longmapsto & \{ \text{strict equivalence classes of triples } (d, \mathcal{L}, \psi ) \text{ as above}\} \end{eqnarray*}

with pullbacks as defined above.

Lemma 27.12.2. Let $S$ be a graded ring. Let $X = \text{Proj}(S)$. The open subscheme $U_ d \subset X$ (27.12.0.1) represents the functor $F_ d$ and the triple $(d, \mathcal{O}_{U_ d}(d), \psi ^ d)$ defined above is the universal family (see Schemes, Section 26.15).

Proof. This is a reformulation of Lemma 27.12.1 $\square$

Lemma 27.12.3. Let $S$ be a graded ring generated as an $S_0$-algebra by the elements of $S_1$. In this case the scheme $X = \text{Proj}(S)$ represents the functor which associates to a scheme $Y$ the set of pairs $(\mathcal{L}, \psi )$, where

1. $\mathcal{L}$ is an invertible $\mathcal{O}_ Y$-module, and

2. $\psi : S \to \Gamma _*(Y, \mathcal{L})$ is a graded ring homomorphism such that $\mathcal{L}$ is generated by the global sections $\psi (f)$, with $f \in S_1$

up to strict equivalence as above.

Proof. Under the assumptions of the lemma we have $X = U_1$ and the lemma is a reformulation of Lemma 27.12.2 above. $\square$

We end this section with a discussion of a functor corresponding to $\text{Proj}(S)$ for a general graded ring $S$. We advise the reader to skip the rest of this section.

Fix an arbitrary graded ring $S$. Let $T$ be a scheme. We will say two triples $(d, \mathcal{L}, \psi )$ and $(d', \mathcal{L}', \psi ')$ over $T$ with possibly different integers $d$, $d'$ are equivalent if there exists an isomorphism $\beta : \mathcal{L}^{\otimes d'} \to (\mathcal{L}')^{\otimes d}$ of invertible sheaves over $T$ such that $\beta \circ \psi |_{S^{(dd')}}$ and $\psi '|_{S^{(dd')}}$ agree as graded ring maps $S^{(dd')} \to \Gamma _*(Y, (\mathcal{L}')^{\otimes dd'})$.

Lemma 27.12.4. Let $S$ be a graded ring. Set $X = \text{Proj}(S)$. Let $T$ be a scheme. Let $(d, \mathcal{L}, \psi )$ and $(d', \mathcal{L}', \psi ')$ be two triples over $T$. The following are equivalent:

1. Let $n = \text{lcm}(d, d')$. Write $n = ad = a'd'$. There exists an isomorphism $\beta : \mathcal{L}^{\otimes a} \to (\mathcal{L}')^{\otimes a'}$ with the property that $\beta \circ \psi |_{S^{(n)}}$ and $\psi '|_{S^{(n)}}$ agree as graded ring maps $S^{(n)} \to \Gamma _*(Y, (\mathcal{L}')^{\otimes n})$.

2. The triples $(d, \mathcal{L}, \psi )$ and $(d', \mathcal{L}', \psi ')$ are equivalent.

3. For some positive integer $n = ad = a'd'$ there exists an isomorphism $\beta : \mathcal{L}^{\otimes a} \to (\mathcal{L}')^{\otimes a'}$ with the property that $\beta \circ \psi |_{S^{(n)}}$ and $\psi '|_{S^{(n)}}$ agree as graded ring maps $S^{(n)} \to \Gamma _*(Y, (\mathcal{L}')^{\otimes n})$.

4. The morphisms $\varphi : T \to X$ and $\varphi ' : T \to X$ associated to $(d, \mathcal{L}, \psi )$ and $(d', \mathcal{L}', \psi ')$ are equal.

Proof. Clearly (1) implies (2) and (2) implies (3) by restricting to more divisible degrees and powers of invertible sheaves. Also (3) implies (4) by the uniqueness statement in Lemma 27.12.1. Thus we have to prove that (4) implies (1). Assume (4), in other words $\varphi = \varphi '$. Note that this implies that we may write $\mathcal{L} = \varphi ^*\mathcal{O}_ X(d)$ and $\mathcal{L}' = \varphi ^*\mathcal{O}_ X(d')$. Moreover, via these identifications we have that the graded ring maps $\psi$ and $\psi '$ correspond to the restriction of the canonical graded ring map

$S \longrightarrow \bigoplus \nolimits _{n \geq 0} \Gamma (X, \mathcal{O}_ X(n))$

to $S^{(d)}$ and $S^{(d')}$ composed with pullback by $\varphi$ (by Lemma 27.12.1 again). Hence taking $\beta$ to be the isomorphism

$(\varphi ^*\mathcal{O}_ X(d))^{\otimes a} = \varphi ^*\mathcal{O}_ X(n) = (\varphi ^*\mathcal{O}_ X(d'))^{\otimes a'}$

works. $\square$

Let $S$ be a graded ring. Let $X = \text{Proj}(S)$. Over the open subscheme scheme $U_ d \subset X = \text{Proj}(S)$ (27.12.0.1) we have the triple $(d, \mathcal{O}_{U_ d}(d), \psi ^ d)$. Clearly, if $d | d'$ the triples $(d, \mathcal{O}_{U_ d}(d), \psi ^ d)$ and $(d', \mathcal{O}_{U_{d'}}(d'), \psi ^{d'})$ are equivalent when restricted to the open $U_ d$ (which is a subset of $U_{d'}$). This, combined with Lemma 27.12.1 shows that morphisms $Y \to X$ correspond roughly to equivalence classes of triples over $Y$. This is not quite true since if $Y$ is not quasi-compact, then there may not be a single triple which works. Thus we have to be slightly careful in defining the corresponding functor.

Here is one possible way to do this. Suppose $d' = ad$. Consider the transformation of functors $F_ d \to F_{d'}$ which assigns to the triple $(d, \mathcal{L}, \psi )$ over $T$ the triple $(d', \mathcal{L}^{\otimes a}, \psi |_{S^{(d')}})$. One of the implications of Lemma 27.12.4 is that the transformation $F_ d \to F_{d'}$ is injective! For a quasi-compact scheme $T$ we define

$F(T) = \bigcup \nolimits _{d \in \mathbf{N}} F_ d(T)$

with transition maps as explained above. This clearly defines a contravariant functor on the category of quasi-compact schemes with values in sets. For a general scheme $T$ we define

$F(T) = \mathop{\mathrm{lim}}\nolimits _{V \subset T\text{ quasi-compact open}} F(V).$

In other words, an element $\xi$ of $F(T)$ corresponds to a compatible system of choices of elements $\xi _ V \in F(V)$ where $V$ ranges over the quasi-compact opens of $T$. We omit the definition of the pullback map $F(T) \to F(T')$ for a morphism $T' \to T$ of schemes. Thus we have defined our functor

\begin{eqnarray*} F : \mathit{Sch}^{opp} & \longrightarrow & \textit{Sets} \end{eqnarray*}

Lemma 27.12.5. Let $S$ be a graded ring. Let $X = \text{Proj}(S)$. The functor $F$ defined above is representable by the scheme $X$.

Proof. We have seen above that the functor $F_ d$ corresponds to the open subscheme $U_ d \subset X$. Moreover the transformation of functors $F_ d \to F_{d'}$ (if $d | d'$) defined above corresponds to the inclusion morphism $U_ d \to U_{d'}$ (see discussion above). Hence to show that $F$ is represented by $X$ it suffices to show that $T \to X$ for a quasi-compact scheme $T$ ends up in some $U_ d$, and that for a general scheme $T$ we have

$\mathop{\mathrm{Mor}}\nolimits (T, X) = \mathop{\mathrm{lim}}\nolimits _{V \subset T\text{ quasi-compact open}} \mathop{\mathrm{Mor}}\nolimits (V, X).$

These verifications are omitted. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).