---- > [!definition] Definition. ([[integral scheme]]) > A [[scheme]] $X$ is said to be **integral** if for every open $U \subset X$, $\mathcal{O}_{X}(U)$ is an [[integral domain]]. ^definition > [!equivalence] > A [[scheme]] $X$ is integral if and only if $X$ is [[reduced scheme|reduced]] and [[irreducible scheme|irreducible]]. ^equivalence > [!basicexample] > If $X=\text{Spec }A$ is an [[affine scheme]], then $X$ is integral if and only if $A$ is an [[integral domain]]. > > > [!proof]- Proof. > > Will use properties in e.g. [[structure sheaf on a ring spectrum]]. > > > > $\to.$ If $X=\text{Spec }A$ is integral then in particular $\mathcal{O}_{\text{Spec }A}(\text{Spec }A)=A$ is an [[integral domain]]. > > > > $\leftarrow.$ Suppose $A$ an [[integral domain]]. Note this implies $A_{\mathfrak{p}}$ is an [[integral domain]] for all $\mathfrak{p} \in \text{Spec }A$. Let $U \subset X$ be open. We want to show $\mathcal{O}_{X}(U)$ is an [[integral domain]]. The elements of $\mathcal{O}_{X}(U)$ are sections $s: U \to \coprod_{\mathfrak{p} \in U}A_{\mathfrak{p}}$ and they satisfy $s(\mathfrak{p}) \in A_{\mathfrak{p}}$ for all $\mathfrak{p}$. Since $A_{\mathfrak{p}}$ is an [[integral domain]], $(s_{1}s_{2})(\mathfrak{p})=s_{1}(\mathfrak{p})s_{2}(\mathfrak{p})=0 \implies s_{1}(\mathfrak{p})=0 \text{ or }s_{2}(\mathfrak{p})=0$ for all $\mathfrak{p} \in U$, and the result can be deduced from this observation. > > [!basicnonexample] > $X=\text{Spec }k[X,Y] / \langle XY \rangle$ is not an integral scheme, because $\Gamma(X, \mathcal{O}_{X})=k[X,Y] / \langle XY \rangle$ is not an integral domain ($XY=0$). $X$ *is* reduced (the [[ideal]] $\langle XY \rangle$ is [[radical of an ideal|radical]]), but it's not irreducible: as a [[topological space]], $X=V(X) \cup V(Y)$. ^nonexample > [!basicproperties] > >- The [[(pre)sheaf stalk|stalks]] $\mathcal{O}_{X, x}$ are all [[integral domain|integral domains]] when $X$ is an [[integral scheme]]. Indeed, each stalk looks like a [[localization]] $A_{\mathfrak{p}}$ of some [[integral domain]] $A=\mathcal{O}_{U}(U)$ whenever $U=\text{Spec }A$ is an [[affine scheme|open affine]] about $x$. [[localization|Localization]] of an [[integral domain]] is again an [[integral domain]]. ^properties > [!proof] Proof of Equivalence. > $\to.$ Suppose $X$ integral. Integral $\implies$ Reduced always, so just need to show $X$ is irreducible. We'll use the [[irreducible topological space|equivalence]]. Suppose $X$ is not irreducible: there exist open $U_{1},U_{2} \subset X$ such that $U_{1} \cap U_{2}=\emptyset$. Then the space $U_{1} \cup U_{2}$ is [[connected|disconnected]], hence $\mathcal{O}_{X}(U_{1} \cup U_{2})=\mathcal{O}_{X}(U_{1}) \times \mathcal{O}_{X}(U_{2})$ which is not an [[integral domain]]. Hence $X$ is irreducible. (See [[sheaf on disconnected space]].) > > $\leftarrow$. Suppose $X$ is reduced + irreducible. Let $U \subset X$ be any open subset, and suppose there are $s,s' \in \Gamma(U, \mathcal{O}_{X} |_{U})$ with $ss'=0$. Define > - $Y=\{ x \in U : \underbrace{ [U, s] }_{ s_{x} } \in \mathfrak{m}_{\mathcal{O}_{X, x}} \}$ > - $Y'=\{ x \in U : \underbrace{ [U, s'] }_{ s'_{x} } \in \mathfrak{m}_{\mathcal{O}_{X, x}} \}$ > > Then $Y$ and $Y'$ are [[closed set|closed subsets]] of $U$ (their complements are basic open sets $U\cap X_{f} \cong D(\overline{f})$, per [[untitled 538]]). Moreover, $Y \cup Y'=U$.[^1] But $X$ is irreducible, so $U$ is irreducible, so either $U=Y$ or $U=Y'$, say $U=Y$. In particular, $s_{x} \in \mathfrak{m}_{\mathcal{O}_{X}, x}$ for all $x \in U$. > > > But then the restriction of $s$ to any open affine subset of $U$ will be [[nilpotent element of a ring|nilpotent]]: if $V=\text{Spec }A$ is such, and $\mathfrak{p} \in V$ is arbitrary, then under the [[(pre)sheaf stalk|germ morphism]] $\begin{align} > \mathcal{O}_{X}(V) & \to \mathcal{O}_{X, \mathfrak{p}} \\ > \shortparallel \ \ \ \ & \ \ \ \ \ \ \ \ \ \shortparallel \\ > A \ \ \ \ \ & \to \ \ \ A_{\mathfrak{p}} > \end{align}$ > $s |_{V} \in A$ maps to an element of $\mathfrak{p}A_{\mathfrak{p}}$, because $s_{\mathfrak{p}}$ must live in the [[maximal ideal]] $\mathfrak{m}_{\mathcal{O}_{X}, \mathfrak{p}}=\mathfrak{p}A_{\mathfrak{p}}$. This implies $s |_{V} \in \mathfrak{p}$, and since $\mathfrak{p}$ was arbitrary and [[nilradical equals intersection of all prime ideals]], implies $s |_{V}$ is [[nilpotent element of a ring|nilpotent]]. Since $X$ is [[reduced scheme|reduced]], this means $s |_{V}=0$. Since affine schemes such as $V$ cover $X$, the [[sheaf|sheaf locality]] axiom implies $s=0$. > > ---- #### [^1]: Indeed, if $x \in U$ then $s_{x}s_{x}'=0 \in \mathcal{O}_{X ,x}$. This implies that either $s_{x}$ or $s_{x}'$ live in the (unique) [[maximal ideal]] (if not, then $s_{x}$, $s'_{x}$ are both [[unit|units]], but the product of units is never zero). Thus $x \in Y$ or $x \in Y'$. ---- #### 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 > ```