# Prove that for an order $R$, the ranks of $\sqrt{0_R}$ and $1+\sqrt{0_R}$ are equal.

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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I'll spare you the proof that $exp$ and $log$ are mutually inverse isomorphisms. All I need is the fact that they're injective, which I take for granted.
The ring $S:=R\otimes\mathbb{Q}$ is finitely generated, hence its nilradical $\sqrt{0_S}$ is also finitely generated and so it is nilpotent, i.e. there exists $m\in\mathbb{Z}_{>0}$ such that $x^m=0$ for all $x\in\sqrt{0_S}$. Of course $\sqrt{0_R}\subseteq\sqrt{0_S}$ is naturally a subgroup, hence so is $1+\sqrt{0_R}\subseteq1+\sqrt{0_S}$. Consider the chain of subgroups $$m!\sqrt{0_R}\subseteq\sqrt{0_R}\subseteq\frac{1}{m!}\sqrt{0_R}\subseteq\sqrt{0_S}.$$ The restriction of the map $exp$ to $m!\sqrt{0_R}$ is an injection mapping into $1+\sqrt{0_R}$; the factor $m!$ clears all denominators in the sum $\sum_{n=0}^{\infty}\frac{x^n}{n!}$. Similarly the restriction of the map $log$ to $1+\sqrt{0_R}$ is an injection mapping into $\frac{1}{m!}\sqrt{0_R}$, as the denominator in the sum $\sum_{n=1}^{\infty}\frac{x^n}{n}$ divides $m!$. This yields inequalities $$\operatorname{rank}(m!\sqrt{0_R})\leq\operatorname{rank}(1+\sqrt{0_R})\leq\operatorname{rank}(\frac{1}{m!}\sqrt{0_R}).$$ and of course we have isomorphisms $\frac{1}{m!}\sqrt{0_R}\cong\sqrt{0_R}\cong m!\sqrt{0_R}$ by multiplication by $m!$, yielding equalities $\operatorname{rank}(m!\sqrt{0_R})=\operatorname{rank}(\sqrt{0_R})=\operatorname{rank}(\frac{1}{m!}\sqrt{0_R})$, from which it is immediate that $$\operatorname{rank}(\sqrt{0_R})=\operatorname{rank}(1+\sqrt{0_R}).$$