Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 113 / 3

A prime Weil divisor is an integral closed subscheme of codimension one. If $Y$ has generic point ${\eta }_Y$, regularity in codimension one makes the local ring ${\mathcal{O}}_{X,{\eta }_Y}$ a discrete valuation ring with fraction field $K(X)$. Its normalized discrete valuationis the order of vanishing along $Y$: writing $f=u{\pi }^m$ for a unit $u$ and uniformizer $\pi$ gives ${\nu }_Y(f)=m$. Additivity of exponents makes this a group homomorphism.

Solved by gpt-5.6-sol high.

Restriction sends a Weil divisor on $X$ to the sum of its components meeting $U$. Every prime divisor on $U$ has a codimension-one closure in the Noetherian scheme $X$, so the induced map $\mathrm{Cl}(X)\to \mathrm{Cl}(U)$ is surjective.

If the class of $D$ restricts to zero, then $D{|}_U={\mathrm{div}}_U(f)$ for some $f\in K(X{)}^{\ast }$. The divisor $D-{\mathrm{div}}_X(f)$ is supported on $Z$, hence is an integral combination of $Z_1,\dots ,Z_n$. Conversely, every such combination restricts to zero. Thereforewhere the first map sends the $i$th basis vector to $[Z_i]$, is exact.

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At the generic point of $L_0$, the functions $x_1,x_2$ are units and the equation gives $x_0=x_3^3/(x_1{x}_2)$. Taking $x_3$ as a uniformizer yieldsAt $L_1$, the functions $x_0,x_2$ are units and $x_1=x_3^3/(x_0{x}_2)$, so ${\nu }_{L_1}(x_0/x_3)=-1$. Symmetrically, ${\nu }_{L_2}(x_0/x_3)=-1$.

The complement of the three lines is $U=D_{+}(x_3)$, withThis unique factorization domain has trivial divisor class group, so part b says that $[L_0],[L_1],[L_2]$ generate $\mathrm{Cl}(X)$. The units on $U$ are scalar multiples of $u_0^a{u}_1^b$, and their boundary valuation vectors are generated byThe resulting integer matrix has Smith normal form $\mathrm{diag}(1,3,0)$. Consequently

Solved by gpt-5.6-sol high.