Find two elements $u$ and $v$ on a ring $R$ satisfying that are the left-inverse for some $a \in R$ yet $(u-v)^n = (u-v)^m$ for some $n$ and $m$

Question

I was trying to prove Kaplansky's theorem on infinitely many left inverses in a ring: Let R be a ring and $r \in R$ an element with two left-inverses, i.e, there exists $u$ and $v$ satisfying that $ur=1=vr$ but $u \neq v$. Then, it has infinitely many left-inverses.

I know of this approach. Yet, while trying to prove me myself, I though of considering elements of the forms $$u_n=u+(u-v)^n.$$ If I am not mistaken $$u_nr = ur+(u-v)^{(n-1)}(u-v)r=1,$$ but I cannot prove that $(u-v)^n \neq (u-v)^m$ whenever $n \neq m$. So, that made me think that maybe it is not true. Also, it would make sense it's false since the common proof considers a more complicated $u_n$. But I cannot think of a counterexample right now, any help?

Answer 1

To expand on Martin's comment: If you let $R$ be the ring of the endomorphisms of $\mathbb Z^\mathbb N$ acting on the right with $$ ((t_n)_n)r_0 = ((t_{n+1})_n, \quad ((t_n)_n)a = \left\{\begin{array}{cc}0, & \text{ if } n=1, \\ t_{n-1}, & \text{if } n>1,\end{array}\right. $$ then $ar_0=1$. If $\delta_k\colon \mathbb Z^{\mathbb N} \to \mathbb Z$ is the map $((t_n)_n)\delta_k = t_k$ and $e_n \in \mathbb Z^{\mathbb N}$ is the function $e_n(m) = \delta_{m,n}$, and we set $\eta_k\in R$ to be $\delta_k.e_1$, then clearly for any $k\geq 1$ the element $a+\eta_k\in R$ is also a left-inverse of $r_0$, but $\eta_k\circ \eta_l =0$ if $k,l\geq 2$. In particular, if $u=a$ and $v=a+\eta_2$ then $u-v=\eta_2$ and $\{\eta_2^n: n \geq 1\} = \{\eta_2,0\}$, so in this example one does not obtain any additional left inverses by considering $\{(u-v)^n: n \in \mathbb N\}$ (though of course the theorem is true, as the elements $a+\eta_k$, $(k \geq 1)$ demonstrate).