$CL(O_S) \cong \mathbb{Z}/3\mathbb{Z}$.

Question

Let $F = \mathbb{Q}(T)$ and let $X$ be the set of all places of $F$, and let $S = \{w\} \subset X$ where $w$ is the place of $F$ corresponding to the maximal ideal $(T^3 - 2)$ of $\mathbb{Q}[T]$. Let $$O_S = \{f \in F: \text{ord}_v(f) \ge 0 \text{ for all }v \in X \setminus S\}$$$$= \{f(T)/(T^3 - 2)^n : n \ge 0,\text{ }f(T) \in \mathbb{Q}[T],\text{ deg}(f(T)) \le 3n\}.$$ How do I show that $\text{Cl}(O_s) \cong \mathbb{Z}/3\mathbb{Z}$?

Answer 1

Define the degree of a valuation $v\in X$ to be the degree of the residue field of $v$ over $\mathbb{Q}$. (It is equal to the degree of the corresponding irreducible polynomial for finite valuations) Then it can be easily seen that a divisor $D=\sum n_i v_i$ is principal (i.e. of the form $\mathrm{div}(f)$ for some $f\in \mathbb{Q}(T)^*$ ) iff $\mathrm{deg}(D) := \sum n_i \mathrm{deg}(v_i) = 0.$

Now $Cl(O_S)$ is equal to the space of divisors in $X\backslash S$ modulo $\{\mathrm{div}_{X\backslash S}(f) = \sum_{v\in X\backslash S}v(f).v: f\in \mathbb{Q}(T)^*\}$. But it is clear that a divisor $D$ in $X\backslash S$ is of the form $\mathrm{div}_{X\backslash S}(f)$ iff it is a restriction of a principal divisor in $X$ which in turn is equivalent to the existence of an integer $n$ s.t. $0=\mathrm{deg} (D+n w) =\mathrm{deg} D+ 3n $. So degree mod 3 gives an isomorphism between $Cl(O_S)$ and $\mathbb Z/3\mathbb Z$.