---- > [!definition] Definition. ([[structure sheaf on a ring spectrum]]) > Let $A$ be a [[commutative ring|commutative]] [[ring]]. Endow $\text{Spec }A$ with the [[Zariski topology on a ring spectrum|Zariski topology]]. For a [[prime ideal]] $\mathfrak{p} \in \text{Spec }A$, denote by $A_{\mathfrak{p}}$ the [[localization]] of $A$ at $\mathfrak{p}$, $A_{\mathfrak{p}}:=S^{-1} A, \text{ where } S=A-\mathfrak{p}.$ > The **structure sheaf** of $\text{Spec }A$, denoted by $\mathcal{O}_{\text{Spec }A}$ or sometimes just $\mathcal{O}$, is the [[sheaf]] of [[ring|rings]] $\mathcal{O}_{\text{Spec }A}(U)=\left\{ s:U \to \coprod_{\mathfrak{p} \in U} A_{\mathfrak{p}} : \begin{align} > &\textcolor{Thistle}{(1) \ s(\mathfrak{p}) \in A_{\mathfrak{p}} \text{ for all }\mathfrak{p}} \\ > & \textcolor{Skyblue}{(2) \ \forall \mathfrak{p} \in U, \exists \text{nbhd } \mathfrak{p} \in V \subset U } \\ > & \quad \quad \textcolor{Skyblue}{\text{ and } a, f \in A \text{ s.t. } \forall \mathfrak{q} \in V, } \\ > & \textcolor{Skyblue}{\quad \quad f \notin \mathfrak{q} \text{ and } s(\mathfrak{q})=\frac{a}{f} \in A_{\mathfrak{q}}} > \end{align} \right\}.$ > [!basicproperties] > - [[(pre)sheaf stalk|Stalks]]: For any $\mathfrak{p} \in \text{Spec }A$, $\mathcal{O}_{\mathfrak{p}}=A_{\mathfrak{p}}$. > - For any $f \in A$, $\mathcal{O}\big(D(f)\big)=A_{f}$. As a special case, the space of global sections $\mathcal{O}(\text{Spec }A)=\mathcal{O}\big(D(1)\big)$ equals $A_{1}=A$. > - (Easy + used explicitly often; made explicit once here.) As one'd hope, the [[(pre)sheaf stalk|germ morphism]] $\mathcal{O}(\text{Spec }A) \to \mathcal{\mathcal{O}_{\mathfrak{p}}}$ is [[isomorphism|isomorphic to]] the [[localization|localization map]] $A \to A_{\mathfrak{p}}$.[^1] ^properties > [!basicexample] Example. ($\mathcal{O}_{\text{Spec } A}$ for $A=\{ * \}$ a singleton) > Suppose $A$ is such that $\text{Spec }A=\{ * \}$ is a singleton. (For example, $A$ may be a [[field]] $k$, or the [[quotient ring|quotient]] $k[t] / \langle t^{2} \rangle$.) What does $\mathcal{O}_{\text{Spec }A}$ look like? > To begin, singletons carry a unique [[topological space|topology]], namely, $\{ \{ * \}, \emptyset \}$. As with any [[sheaf]], $\mathcal{O}_{\text{Spec }A}(\emptyset)=\{ 0 \}$. So the data is all in $\mathcal{O}_{\text{Spec } A}(\{ * \})$. From the properties above: $\mathcal{O}_{\text{Spec } A}(\{ * \})=\mathcal{O}_{\text{Spec } A}(\text{Spec } A)=A.$ So $\mathcal{O}_{\text{Spec }A}$ is a copy of $A$ attached to a point. It is often implicitly regarded as $A$ itself. There is one [[(pre)sheaf stalk|stalk]], that being $\mathcal{O}_{\text{Spec } A, *}=A_{*} = A$, where we have [[local ring#^properties|used that]] localizing a local ring at its maximal ideal does nothing. ^basic-example > [!proof] Proofs of Properties. > **$\mathcal{O}_{\mathfrak{p}}\cong A_{\mathfrak{p}}$.** We have a map $\begin{align} > \mathcal{O}_{\mathfrak{p}}& \to A_{\mathfrak{p}} \\ > [U, s] & \mapsto s(\mathfrak{p}). > \end{align}$ > The claim is that it is [[bijection|bijective]]. > > **Surjective.** Consider an arbitrary element $\frac{a}{f} \in A_{\mathfrak{p}}$, where $a \in A$ and $f \in A-\mathfrak{p}$, i.e., $f \notin \mathfrak{p}$. Define the constant map $s:D(f) \to \coprod_{\mathfrak{q } \in D(f)}A_{\mathfrak{q}}$ by $s(\mathfrak{q}):= \frac{a}{f} \in A_{\mathfrak{p}}$.[^1] Satisfies $\textcolor{Thistle}{(1)}$ because $\frac{a}{f} \in A_{\mathfrak{q}}$ for all $\mathfrak{q} \in D(f)$, by the very definition of $D(f)$.[^2] Satisfies $\textcolor{Skyblue}{(2)}$ by construction. So, indeed, $s \in \mathcal{O}\big( D(f) \big)$. The germ $[D(f), s] \in \mathcal{O}_{\mathfrak{p}}$ maps to $s(\mathfrak{p})=\frac{a}{f}$, witnessing surjectivity. > > **Injective.** Let $\mathfrak{p} \in U \subset \text{Spec }A$ and $s \in \mathcal{O}(U)$, defining a germ $[U, s] \in \mathcal{O}(\mathfrak{p})$. Suppose $s(\mathfrak{p})=0=\frac{0}{1}$. WTS $[U,s]=0$. Shrinking $U$ if necessary, we may assume $s(\mathfrak{q})=\frac{a}{f} \in A_{\mathfrak{q}}$ for all $\mathfrak{q} \in U$ for some $a,f \in A$ with $f$ not in any $\mathfrak{q}$. In particular, $s(\mathfrak{p})=\frac{a}{f}$ and so we have $\frac{a}{f}=\frac{0}{1}$ *in $A_{\mathfrak{p}}$* meaning that there exists $h \in A-\mathfrak{p}$ such that $ha=0$. Just because $\frac{a}{f}=0$ in $A_{\mathfrak{p}}$, though, doesn't mean it is zero in other localizations. > > Now use $h$ to shrink $U$ even more.... Let $V:=D(f) \cap D(h)$ be the set of [[prime ideal|prime ideals]] containing neither $f$ nor $h$. Then for $\mathfrak{q} \in V$, $\frac{a}{f}=\frac{0}{1}$ in $A_{\mathfrak{q}}$, because now $h \in A-\mathfrak{q}=S$ 'is available to annihilate with' and $ha=0$. This is saying that $s |_{V}=0$. So $[U,s]=0$ in $\mathcal{O}_{\mathfrak{p}}$. > > **$\mathcal{O}\big( D(f) \big) \cong A_{f}$.** We claim there is an isomorphism $\begin{align} > \Psi: A_{f} &\xrightarrow{\sim} \mathcal{O}_{\text{Spec } A}\big( D(f) \big) \\ > \frac{a}{f^{n}} & \mapsto \left( \underbrace{ \mathfrak{p} }_{ \not \ni f } \mapsto \frac{a}{f^{n}} \in A_{\mathfrak{p}} \right) > \end{align}$ > > **Injective.** Suppose $\Psi\left( \frac{a}{f^{n}} \right)=0$. Then for all $\mathfrak{p} \in D(f)$, $\frac{a}{f^{n}}=0$ as an element of $A_{\mathfrak{p}}$. We have to show that this implies $\frac{a}{f^{n}}=0$ as an element of $A_{f}$. We know (by definition of [[localization]]) that for all $\mathfrak{p} \in D(f)$ there exists $h_{\mathfrak{p}} \in A-\mathfrak{p}$ such that $h_{\mathfrak{p}} a=0$. In other words, $h_{\mathfrak{p}} \in \text{Ann}(a)=\{ g \in A : ga=0 \}$. Since $h_{\mathfrak{p}} \not \in \mathfrak{p}$, we know $\text{Ann }a \not \subset \mathfrak{p}$ for every $\mathfrak{p} \in D(f)$, so $V(\text{Ann }a) \cap D(f)=\emptyset$. So any [[prime ideal]] containing $\text{Ann }a$ must contain $f$, meaning that > > $f \in \bigcap_{\mathfrak{q} \in V(\text{Ann }a)}^{}= \sqrt{ \text{Ann }a }.$ > Thus, $f^{m} \in \text{Ann }a$ for some $m>0$. By definition of $\text{Ann }a$, $f^{m}a=0$. So $\frac{a}{f^{n}}=\frac{af^{m}}{f^{n}f^{m}}=0$ in $A_{f}$. > [^1]: Sets of the form $D(f)$ are [[basis for a topology|basic open sets in]] the [[Zariski topology on a ring spectrum|Zariski topology]] on $\text{Spec }A$; see [[Zariski topology on a ring spectrum|Zariski topology]]. [^2]: $\frac{a}{f}$ is defined initially to be an element of $A_{\mathfrak{p}}$. What do we mean when we say it also belongs to $A_{\mathfrak{q}}$? The point is that, because $\mathfrak{q} \in D(f)$, $f \notin \mathfrak{q}$. And so it is available to divide by when forming $A_{\mathfrak{q}}=(A-\mathfrak{q})^{-1}A$. > [!note] Remark. > This construction is not unlike that of [[sheafification]]. ^note > [!justification] > As the condition is local (as it is in [[sheafification]]), it is clear $\mathcal{O}$ is a [[sheaf]]. [[covariant functor|Functorality]] and locality hold merely because they hold for general set-functions functions and function restriction. As for gluing, given a [[cover|open cover]] $\{ U_{i} \}=U$ and $s_{i} \in \mathcal{O}(U_{i})$ with agreement on pairwise overlaps, it is clear that the $s_{i}$ glue together into a well-defined section $s \in \mathcal{O}(U)$: given $\mathfrak{p} \in U$, can pick any $U_{i} \ni \mathfrak{p}$ and put $s(\mathfrak{p}):=s_{i}(\mathfrak{p}).$ It just needs to be verified that $\textcolor{Skyblue}{(2)}$ is satisfied by $s$. Since the condition is local, this is immediate: any $\mathfrak{p} \in U$ in particular belongs to some $U_{i}$, $s(\mathfrak{p})=s_{i}(\mathfrak{p})$, and then use that the condition holds for $s_{i}$. ^justification ---- #### ---- #### 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 > ```