ScalingStacks

Proof. Let $L$ be a non-empty finite subset of $I$ such that $\operatorname{supp}\nolimits(\zeta)\cap\operatorname{supp}\nolimits(\zeta^{\prime})$ is finite for any two distinct elements $\zeta$ and $\zeta^{\prime}$ in $L$. Let $r=\sum_{\zeta\in L}a_{\zeta}\llbracket\zeta\rrbracket$ where $a_{\zeta}\in{\mathbf{Z}}-\{0\}$ for $\zeta\in L$. Let $\zeta_{0}\in L$. There is $x\in\operatorname{supp}\nolimits(\zeta_{0})\cap U$ with $x{\not\in}\{\zeta_{0}(0),\zeta_{0}(1)\}$ and $x{\not\in}\bigcup_{\zeta\in L-\{\zeta_{0}\}}\operatorname{supp}\nolimits(\zeta)$. Let $c\in C(x)$ and $\iota(c)$ be the other element of $C(x)$. We have $m_{c}(\zeta_{0})=-m_{\iota(c)}(\zeta_{0})=\pm 1$, while $m_{c}(\zeta^{\prime})=m_{\iota(c)}(\zeta^{\prime})=0$ for $\zeta^{\prime}\in L-\{\zeta_{0}\}$. It follows that $m_{c}(r)=-m_{\iota(c)}(r)=\pm a_{\gamma}$. Consequently, $\bigl(l_{x}\circ(m_{c},m_{\iota(c)})\bigr)(r)=\pm l_{x}(a_{\gamma},-a_{\gamma})\neq 0$. Since every non-zero element of $R(X)$ is of the form $r$ as above, the lemma follows. ∎