---- > [!justification] Motivation. > The generalization from [[affine variety|affine varieties]] to [[affine scheme|affine schemes]] proceeds by > - Introducing the set [[prime ideal|Spec]] $A$, where $A$ is any ([[commutative ring|commutative]]) [[ring]], >- Then upgrading to a [[topological space]] with the [[Zariski topology on a ring spectrum|Zariski topology]], >- Then upgrading to a [[locally ringed space]] via the [[structure sheaf on a ring spectrum|the Spec structure sheaf]] construction, and hence to an [[affine scheme]]. > >The ensuing **proj construction** similarly generalizes [[projective variety|projective varieties]] to the [[scheme|schematic]] setting. ^justification > [!definition] Definition. ([[proj construction|proj]] *set*) > Let $S$ be a [[graded ring]] with [[irrelevant ideal]] $S_{+}$. As a set, one defines $\text{Proj }S:=\{ \mathfrak{p} \in \text{Spec } S : \mathfrak{p} \text{ is homogeneous and }\mathfrak{p} \not \supset S_{+} \} \subset \text{Spec }S.$ ^definition > [!definition] Definition. ([[proj construction|proj]] *space*) > The **Zariski topology** on $\text{Proj }S$ is specified by taking as [[closed set|closed sets]] $V(I):=\{ \mathfrak{p} \in \text{Proj }S : \mathfrak{p} \supset I\}$ for $I$ a [[homogeneous ideal]] of $S$. > The open sets $(f \in S \text{ homogeneous})$ $D_{+}(f):=\{ \mathfrak{p} \in \text{Proj }S : \mathfrak{p} \not \ni f \}=\text{Proj }S-V(\langle f \rangle )$ form a [[basis for a topology|basis]] [[topology generated by a basis|for]] the Zariski topology on $\text{Proj }S$. > >[!justification]- Verifying this topology/basis. >> > >We verify that the Zariski topology on $\text{Proj }S$, defined by taking the [[closed set|closed sets]] to be $V(I)=\{ \mathfrak{p} \in \text{Proj } S: \mathfrak{p} \supset I \}$ for [[ideal|ideals]] $I$ of $S$, is indeed a [[topological space|topology]] on $\text{Proj }S$. Clearly $\text{Proj }S=V\big( (0) \big)$ and $\emptyset=V(\text{Proj }S)$ are closed. > > > >A finite union $V(I_{1}) \cup \dots \cup V(I_{r})$ of closed sets is closed because $\begin{align} > >V(I_{1}) \cup \dots \cup V(I_{r}) &= \{ \mathfrak{p} \in \text{Proj }S: \mathfrak{p} \supset I_{i} \text{ for some }i \in [r] \} \\ > &= \{ \mathfrak{p} \in \text{Proj } S : \mathfrak{p} \supset I_{1} \cap \dots \cap I_{r}\} \\ > &= V(I_{1} \cap \dots \cap I_{r}). > \end{align}$ > >where we have used the easy [[prime ideal|prime containment]] result. >> >> An arbitrary intersection $\bigcap_{\alpha}^{}V(I_{\alpha})$ of closed sets is closed because $\begin{align} > \bigcap_{\alpha}^{}V(I_{\alpha})&=\{ \mathfrak{p} \in \text{Proj } S: \mathfrak{p} \supset I_{\alpha} \text{ for all } \alpha \} \\ > &= \left\{ \mathfrak{p} \in \text{Proj }S : \mathfrak{p} \supset \sum_{\alpha} I_{\alpha} \right\} \\ > &= V\left( \sum_{\alpha}I_{\alpha} \right). > \end{align}$ > To verify basis, we check the [[condition for obtaining a basis from a topology]]. Each $D_{+}(f)$ is open because $\text{Proj }S-D_{+}(f)=V(\langle f \rangle)$. Let $U \subset \text{Proj }S$ be an open set, by definition of the form $U=\text{Proj }S-V(I)=\{ \mathfrak{p} \in \text{Proj }S : \mathfrak{p} \not \supset I \}$ for some [[homogeneous ideal]] $I$, and let $\mathfrak{q} \in U$ be arbitrary: $\mathfrak{q} \not \supset I$. Since $I \not \subset \mathfrak{q}$, there exists homogeneous $i \in I$ such that $i \notin \mathfrak{q}$. Now, $\mathfrak{q} \in D_{+}(i) \subset U:$ certainly $\mathfrak{q} \in D_{+}(i)$, and if $\mathfrak{h} \in D_{+}(i)$ then $\mathfrak{h} \not \ni i$ which means that $\mathfrak{h}$ cannot contain $I$ and hence $\mathfrak{h} \in U$. > [!definition] Definition. ([[proj construction|proj]] *scheme*) > $\text{Proj }S$ is made into a [[locally ringed space|locally]] [[ringed space]] as follows. > - For $\mathfrak{p} \in \text{Proj }S$, consider the [[multiplicative subset of a ring|multiplicative subset]] $T:=\{ f \in S - \mathfrak{p}: f \text{ is homogeneous} \}.$ Define $S_{(\mathfrak{p})}$ to be the [[subring]] [[localization|of]] $T ^{-1} S$ of $\textcolor{thistle}{\text{degree-zero elements}}$: $S_{(\mathfrak{p})}:=\left\{ \frac{s}{t} : \begin{aligned} &s \in S, \; t \in T \text{ homogeneous, \textcolor{thistle}{with} }\textcolor{thistle}{s \text{ also}} \\ & \textcolor{thistle}{\text{homogeneous and of the same degree as } t}\end{aligned}\right\} \subset T ^{-1} S.$ We call $S_{(\mathfrak{p})}$ **$S$ localized at $\mathfrak{p}$ in degree zero**. > >- Define, for $f \in S$ [[graded ring|homogeneous]], $S_{(f)}$ to be the [[subring]] [[localization|of]] $S_{f}$ of $\textcolor{thistle}{\text{degree-zero elements}}$: $S_{(f)}:=\left\{ \frac{s}{f^{n}} : n \geq 0, \begin{aligned} \ & s \in S, \; { \textcolor{thistle}{\text{with }} }\textcolor{thistle}{s \text{ homogeneous}} \\ & \textcolor{thistle}{\text{and of the same degree as } f^{n}}\end{aligned}\right\} \subset S_{f}.$ We call $S_{(f)}$[^2] **$S$ localized at $f$ in degree zero**. > >The [[ringed space|structure sheaf]] $\mathcal{O}_{\text{Proj }S}$ is then constructed as follows. Given $U \subset \text{Proj }S$ open, define $\mathcal{O}_{\text{Proj }S}(U):=\left\{ s:U \to \coprod_{\mathfrak{p} \in U} S_{(\mathfrak{p})} : \begin{align} &\textcolor{Thistle}{(1) \ s(\mathfrak{p}) \in S_{(\mathfrak{p})} \text{ for all }\mathfrak{p} \in U} \\ & \textcolor{Skyblue}{(2) \ \forall \mathfrak{p} \in U, \exists \text{nbhd } \mathfrak{p} \in V \subset U } \\ & \quad \quad \textcolor{Skyblue}{\text{ and } a, f \in S \text{ s.t. } \forall \mathfrak{q} \in V, } \\ & \textcolor{Skyblue}{\quad \quad f \notin \mathfrak{q} \text{ and } s(\mathfrak{q})=\frac{a}{f} \in S_{(\mathfrak{q})}} \end{align} \right\}.$ The space $(\text{Proj }S, \mathcal{O}_{\text{Proj } S})$ is a [[locally ringed space]], since $\mathcal{O}_{\text{Proj }S, \mathfrak{p}}=S_{(\mathfrak{p})}$. It is in fact a [[scheme]], [[cover|covered by]] basic open [[affine scheme|affines]] $D_{+}(f) \cong \text{Spec }S_{(f)}$, $f \in S$. > [!justification]- Verifying that $S_{(\mathfrak{p})}$ is a [[local ring]]. > *Claim: the [[ideal]] $\mathfrak{m}:=\left\{ \frac{p}{t}: p \in \mathfrak{p}, t \notin \mathfrak{p} , p, t \text{ homogeneous of equal degree} \right\}$ is uniquely [[maximal ideal|maximal]].* Indeed, any $\frac{u}{v} \in S_{(\mathfrak{p})}-\mathfrak{m}$ is [[unit|invertible]], and therefore not is contained in any proper ideal at all. Contrapositively, if an element of $S_{(\mathfrak{p})}$ is contained in a proper ideal, then it is contained in $\mathfrak{m}$. ^justification > [!basicproperties] > [[(pre)sheaf stalk|Stalks]]: For any $\mathfrak{q} \in \text{Proj }S$, $\mathcal{O}_{\text{Proj }S,\mathfrak{q}}=S_{(\mathfrak{q})}$. > [[Basis for a topology|Basic]] [[affine scheme|open affines]]: $D_{+}(f) \cong \text{Spec }S_{(f)}$. ^properties > [!basicexample] ($\text{Proj }k[T_{0},\dots,T_{n}]$ as the scheme-y version of [[projective space]] $k\mathbb{P}^{n}$) > $k=\overline{k}$ is an [[algebraically closed]] [[field]]. > > > > [[prime ideal|We know]] $\text{Spec }k[X,Y]=\underbrace{ \{ \langle X-a , Y-b\rangle : a,b \in k\} }_{ \text{maximal ideals} } \cup \{ \langle f \rangle : f \text{ irred.} \} \cup \{ (0) \}.$ > Which of these are in $\text{Proj }k[X,Y] \subset \text{Spec }k[X,Y]$? > - None of the [[maximal ideal|maximal ideals]] are, because none of them except $\langle X,Y \rangle=S_{+}$ is homogeneous and $\langle X,Y \rangle$ is the [[irrelevant ideal]]. > - $\langle f \rangle$, $f$ irreducible, will be homogeneous if $f$ is a homogeneous irreducible polynomial. In $k[X,Y]$, any homogeneous polynomial factors *linearly* into irreducibles[^2], so $f$ must be a linear polynomial: $f=aX + bY$ for $a,b$ not both zero > - Certainly $(0)$ is homogeneous and $(0) \not \supset S_{+}$ > > Thus, $\text{Proj }k[X,Y]=\{ \langle bX+aY \rangle : a,b \in k \text{ not both }0 \} \cup \{ (0) \}.$ > So the points of $\text{Proj }k[X,Y]-\{ 0 \}$ are in [[bijection]] with elements of $\frac{k^{2} - \{ 0 \}}{k^{*}},$ > i.e., pairs of points in $k$ up to rescaling. That's the set $k\mathbb{P}^{1}$! > > More generally, can show using upcoming results that the points of $\text{Proj }k[T_{0},\dots,T_{n}]$ which correspond to primes not containing other primes of $\text{Proj }k[T_{0},\dots,T_{n}]$[^1] are of the form $\langle a_{1}T_{0} - a_{0}T_{1}, a_{2}T_{0} - a_{0}T_{2}, \dots, a_{n}T_{0}-a_{0}T_{n} \rangle .$ > This corresponds to a point $(a_{0},\dots,a_{n}) \in k^{n+1}$ up to scaling, i.e., a point $[a_{0}:\dots: a_{n}] \in k\mathbb{P}^{n}$. ^basic-example > [!proof]- Proof of Stalks. > The elements of $\mathcal{O}_{\text{Proj } S, \mathfrak{q}}$ are [[equivalence relation|classes]] $[U,s]$ such that $\mathfrak{q} \in U$ and $s \in \mathcal{\mathcal{O}}_{\text{Proj }S}(U)$. Shrinking $U$ if necessary, we may assume that $s(\mathfrak{p})=\frac{a}{f}$, $f \notin \mathfrak{p}$, for all $\mathfrak{p} \in U$, where $a,f \in S$ are [[graded ring|homogeneous]] of equal degree. This gives rise to a map $\begin{align} > \mathcal{O}_{\text{Proj }S, \mathfrak{q}} & \to S_{\mathfrak{(q)}} \\ > [U, s] & \mapsto s(\mathfrak{q}) > \end{align}$ > which we want to show is an [[isomorphism]]. > > **Surjectivity.** Let $\frac{a}{f} \in S_{(\mathfrak{q})}$. Can define a (constant) map > > $\begin{align} > s: D_{+}(f) & \to \coprod_{\mathfrak{p} \in U} S_{(\mathfrak{\mathfrak{p})}} \\ > \mathfrak{p} & \mapsto \frac{a}{f} \in S_{(\mathfrak{p})} > \end{align}$ > Satisfies $\textcolor{Thistle}{(1)}$ because $\frac{a}{f}$ is an element of $S_{(\mathfrak{p})}$ for all $\mathfrak{p} \in D_+(f)$, since by definition $\mathfrak{p} \in D_{+}(f)$ implies $f \in S-\mathfrak{p}$. Satisfies $\textcolor{Skyblue}{(2)}$ by construction. Thus, $s \in \mathcal{O}_{\text{Proj }S}\big( D_{+}(f) \big)$, and under our proposed map, the germ $[D_{+}(f), s] \in \mathcal{O}_{\text{Proj } S}, \mathfrak{q}$ maps to $s(\mathfrak{q})=\frac{a}{f}$. [[surjection|Surjectivity]] follows. > > **Injectivity.** Suppose $[U,s] \in \mathcal{O}_{\text{Proj }S,\mathfrak{q}}$ is such that $s(\mathfrak{q})=0$. We want to show that this implies $[U,s]=0$. We know that, shrinking $U \ni \mathfrak{q}$ if necessary, we may assume that $s(\mathfrak{p})=\frac{a}{f}$, $f \notin \mathfrak{p}$, for all $\mathfrak{p} \in U$, for some $a,f \in S$ [[graded ring|homogeneous]] of equal degree. Since $s(\mathfrak{q})=0$, we have $\frac{a}{f}=\frac{0}{1}$ *in $S_{(\mathfrak{q})}$*, meaning there exists homogeneous $h \in S-\mathfrak{q}$ satisfying $ha=0$. > > Just because something is zero in one [[localization]] does not mean it is zero in others, however. Our goal is now to show that $\frac{a}{f}=0$ in $S_{(\mathfrak{\mathfrak{p}})}$ for all $\mathfrak{p} \in U$. > > Use $h$ to shrink $U$ even more... Let $V:=D_{+}(f) \cap D_{+}(h) \ni \mathfrak{q}$ be the (open) set of elements in $\text{Proj }S$ containing neither $f$ nor $h$. Then for all $\mathfrak{p} \in V$, $\frac{a}{f} \in S_{(\mathfrak{p})}$ equals $\frac{0}{1}$, because $h$ is 'available to annihilate with': $ha=0$. This is implying that $s |_{V}=0$, and hence the germ $[U,s]=0$. > > [^1]: Can't say 'maximal ideals' because, as we've just seen, $\text{Proj }k[T_{0},\dots,T_{n}]$ might not contain any maximal ideals of $\text{}k[T_{0},\dots,T_{n}]$ so that'd be confusing. ---- #### [^2]: Take care regarding the notation. This is *not* $S$ localized at the [[principal ideal]] [[ideal generated by a subset|generated by]] $f$. - [ ] obviously need to show the $D_{+}(f)$ are affine schemes and make proj S a scheme, that is on next homework (update: now in ipad) ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```