----- Let $k$ be a [[field]] and [[proj construction|denote]] by $\mathbb{P}^{n}_{k}=\text{Proj }k[X_{0},\dots,X_{n}]$ the [[scheme]]-theoretic [[projective space]] over $k$. > [!proposition] Proposition. ([[class group of projective space]]) > Let $H \subset \mathbb{P}^{n}_{k}$ be a hyperplane, say, $H=V(\langle X_{0} \rangle)$. $H$ is naturally an [[integral scheme|integral]] [[subscheme|closed subscheme]] of [[codimension of a closed subspace|codimension]] $1$, i.e., a [[prime divisor in a scheme|prime divisor]] on $\mathbb{P}^{n}_{k}$. > [[divisor (of zeros and poles)|We have]] $\text{Cl }\mathbb{P}^{n}_{k}=\mathbb{Z} \langle H \rangle$, [[submodule generated by a subset|generated by]] the [[equivalence class|class]] of $H$. ^proposition > [!proof]- Proof. ([[class group of projective space]]) > First note $U:=\mathbb{P}^{n}_{k} - H=\mathbb{P}^{n}_{k}-V(\langle X_{0} \rangle)=D_{+}(X_{0})$. Then recall from properties in the [[proj construction]] that $D_{+}(X_{0})$ is the degree-zero localization $D_{+}(X_{0})=\text{Spec }k[X_{0},\dots,X_{n}]_{(X_{0})}=\text{Spec }k\left[ \frac{X_{1}}{X_{0}},\dots, \frac{X_{n}}{X_{0}} \right]=\mathbb{A}^{n}_{k}$, [[affine scheme|scheme-theoretic affine space]] over $k$. So $U=\mathbb{A}^{n}_{k}$, and since [[polynomial 4|polynomial ring]] in $n$ indeterminates is a [[UFD]] ([[Gauss's lemma]]), [[a radical ideal in a Noetherian ring equals an intersection of finitely many prime ideals|we have]] $\text{Cl }U=0$. > > Per [[computing class group by deleting a subspace]], there is an [[exact sequence]] of [[abelian group|abelian groups]] $\begin{align} > \mathbb{Z} &\to \text{Cl }X \to \cancel{ \text{Cl }U }^{=0} \to 0 \\ > 1 & \mapsto [H] > \end{align}$ > implying the map $1 \mapsto [H]$ is [[surjection|surjective]]. Have to now show it is [[injection|injective]]. Suppose there is $k \in \mathbb{Z}$ such that $k[H]=0$, i.e., such that $kH=(f)$ for some [[divisor (of zeros and poles)|principal divisor]] $(f)$, $f \in K(\mathbb{P}^{n}_{k})^{*}$. We claim that $k=0$. > > What is $K(\mathbb{P}^{n}_{k})^{*}$? It is of course the same as $K(D_{_{+}}(X_{0}))^{*}$[^1], [[generic point of an integral scheme|which equals]] $\text{Frac }k\left[ \frac{X_{1}}{X_{0}}, \dots, \frac{X_{n}}{X_{0}} \right]$. So $f$ is a [[rational function|rational function]] in the variables $\frac{X_{1}}{X_{0}},\dots, \frac{X_{n}}{X_{0}}$, that is, a ratio $f=\frac{g}{h}$ of homogeneous polynomials of equal degree. [[factorization into irreducibles|Factor]]: $f=\frac{g}{h}=\frac{\prod_{i}^{}g_{i}^{a_{i}}}{\prod_{i}^{}h_{i}^{b_{i}}}, g_{i}, h_{i} \text{ \ irreducible}$ > - [ ] last sentence (no time to typeset) ----- #### [^1]: For the two [[integral scheme|integral schemes]] share the same [[generic point of an integral scheme|generic point]] $\eta$ and [[(pre)sheaf stalk|stalk]] $\mathcal{O}_{X, \eta}=K(X)$, since one is an [[subscheme|open subscheme]] of the other. [^2]: ~~Here using that [[discrete valuation]] $\nu_{Y}$ is a [[group homomorphism]], e.g. $\nu_{Y}\left( \prod_{i}^{}g_{i}^{a_{i}} \right)=\sum_{i}\nu_{Y}(g_{i}^{a_{i}})=\sum_{i} a_{i} \nu_{Y}(g_{i})$ and $\nu_{Y}\left( \frac{a}{b} \right)=\nu_{Y}(a)-\nu_{Y}(b)$. ~~ ---- #### 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 > ```