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Let $R$ be an order, that is, a ring of which the additive group is finitely generated and torsion-free. As an exercise I am trying to prove that $\operatorname{rank}(\sqrt{0_R})=\operatorname{rank}(1+\sqrt{0_R})$, where I have been given $\operatorname{rank}(A):=\dim_{\mathbb{Q}}(A\otimes\mathbb{Q})$ as the definition of the rank of an abelian group $A$. But so far I have only been able to displace the problem: I was excited to find isomorphisms $\exp$ and $\log$ given by \begin{eqnarray*} \exp:&\ &\sqrt{0_R}\otimes\mathbb{Q}\ \longrightarrow\ 1+\sqrt{0_R}\otimes\mathbb{Q}:\ x\ \longmapsto\ \sum_{n=0}^{\infty}\frac{x^n}{n!},\\ \log:&\ &1+\sqrt{0_R}\otimes\mathbb{Q}\ \longrightarrow\ \sqrt{0_R}\otimes\mathbb{Q}:\ 1-x\ \longmapsto\ -\sum_{n=1}^{\infty}\frac{x^n}{n}, \end{eqnarray*} but it turns out that proving that $1+\sqrt{0_R}\otimes\mathbb{Q}\cong(1+\sqrt{0_R})\otimes\mathbb{Q}$ is the essence of the problem. Any ideas on how to prove this? And am I even on the right path with these maps $\exp$ and $\log$?

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