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

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

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