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Let $R$ be a ring, and $u$ an element of $R$ such that :

  • $u$ is right invertible : there exists $v \in R$ such that $uv = 1$,
  • $u$ is a left zero divisor : there exists nonzero $z \in R$ such that $uz = 0$.

Show that there are infinitely many elements $w \in R$ such that $uw = 0$.

Edit: Here's my work so far.

Let $E = \{ w \in R \ \vert \ uw=0\}$. We want to show it has infinitely many elements.

I've found that $(vu-1) \in E$ since $u(vu-1) = (uv)u - u = 1u-u = 0$.

Hence $(vu-1)\alpha$ would also be an element of $E$ for any $\alpha \in R$. But I'm not sure that proves the set $E$ has infinitely many elements.

mathnoob44
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1 Answers1

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I claim that the elements $z_n = (vu-1)u^n$ are all distinct and all satisfy $uz_n = 0$. The second property is easy to check, as $$ u(vu-1)u^n = (u - u)u^n = 0. $$ To conclude the proof, we suppose that $z_n = z_m$ for some $n > m \geq 0$, so in other words, $$ (vu-1)u^n = (vu-1)u^m \iff vu^{n+1} - u^n = vu^{m+1} - u^m. $$ Since $n \geq m+1$, we can multiply this on the right by $v^n$ to find that $$ vu - 1 = 0. $$ However, now we have $uv = vu = 1$ so both $u$ and $v$ are units. This contradicts the fact that $u$ is also a left zero divisor, completing the proof.

Mike Daas
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