---- Assume $(X, \mathcal{O}_{X})$ is a [[locally Noetherian scheme|Noetherian]], [[integral scheme|integral]] [[scheme]] [[scheme over a field|over]] a [[field]] $k$ which is [[regular scheme|regular]] in [[codimension of a closed subspace|codimension]] $1$. (This is the default for this chapter of the course.) 'p.d.' stands for [[prime divisor in a scheme|prime divisor]]. The following result is useful for computing the [[divisor (of zeros and poles)|class group]] of $X$. > [!theorem] Theorem. ([[computing class group by deleting a subspace]]) > Take $Z \subsetneq X$ a proper [[closed set|closed subset]]. Put $U:=X-Z$. > > 1. There is a [[well-defined]] [[surjection|surjective]] [[group homomorphism|homomorphism]] of [[divisor (of zeros and poles)|class groups]] $\begin{align} > \text{Cl }X &\twoheadrightarrow \text{Cl }U \\ > \sum_{Y_{i} \subset X \text{ p.d.}} n_{i}Y_{i} & \mapsto \sum_{Y_{i} \subset X \text{ p.d.}} n_{i}(Y_{i} \cap U) > \end{align}$ > interpreting as zero if $Y_{i} \cap U=\emptyset$. > 2. If $\text{codim}(Z,X) \geq 2$, this is in fact a [[group isomorphism]]. That is, removing codimension-2 or higher subsets does not affect the class group: $\text{Cl }{X}$ 'can't see beyond codimension-1'. > 3. If $Z$ is in fact a [[prime divisor in a scheme|prime divisor]], then there is an [[exact sequence]] $\begin{align} > \mathbb{Z} & \to \text{Cl }X \to \text{Cl }U \to 0 \\ > 1 &\mapsto [Z]. > \end{align}$ > > > [!proof]- Proof. ([[computing class group by deleting a subspace]]) > **1.** At the level of [[prime divisor in a scheme|Weil divisors]] (i.e. before [[quotient group|quotienting]]) this map is defined, since the intersection $Y_{i} \cap U$ (if nonempty) is a [[prime divisor in a scheme|prime divisor]] of $U$ when $Y$ is of $X$.[^1] We want to show this map $\text{Div }X \to \text{Div }U$ to descend to a map $\text{Cl }X \to \text{Cl }U$, i.e., to show that its kernel contains all [[divisor (of zeros and poles)|principal divisors]], i.e., that $\sum_{Y} \nu_{Y}(f) \ (Y \cap U)$ is a principal divisor of $U$ for all $f \in K(U)^{*}=K(X)^{*}$. > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB12cYAPHYACJYaAAgC+ADRBjS6TLnyEUARnJVajFm07c+g4eICq02SAzY8BIgCY11es1aIOXXvwDCDIybkXFRMmV1By1nHTdgT3EpMXUYKABzeCJQADMAJwgAWyQyEBwIJABmagALGDooNhwAdwhyyoQZNMycxFV8wsRbEAaq51r6iqgm0wzs3OoCpGVmkHG2ks6kHoYsMCcQKDo4cqr7TU3OXiw4HDgRAEIRTjRSrAByEGoGOgAjGAYABXlLJRB0lgEqUcD55q1JstutQPmB+kU8iEjuw7lhpBQxEA > \begin{tikzcd} > \text{Div }X \arrow[d, two heads] \arrow[r] \arrow[rr, "\phi", bend left] & \text{Div }U \arrow[r, two heads] & \text{Cl }U \\ > \text{Cl }X \arrow[rru, "\exists ! \phi'"', dashed] & & > \end{tikzcd} > \end{document} > ``` > > This is true: it equals $(f |_{U})$, since $\mathcal{O}_{X, \eta}=\mathcal{O}_{U, \eta}$ for $\eta\in U$ [[generic point of an integral scheme|generic]], meaning. Done once doing surjectivity and the rest (on paper). ---- #### [^1]: Recall $\text{codim}(Y,X)=\text{codim}(Y \cap U, U)$, so $\text{codim}(Y \cap U, U)=1$. Certainly $Y \cap U$ is closed in $U$. On the other hand, it is open in $Y$, and certainly restricting a scheme to an open subscheme preserves being integral. So $Y \cap U$ indeed is a prime divisor of $U$. ----- #### 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 > ```