My try and where I am stuck :

(note:A zero-divisor is a nonzero element $a$ of a commutative ring R such that there is a nonzero element $b\in R$ with $ab=0_{R}$)


Assume $uv=1_{R}$ then clearly ($u\ne 0_{R}$) , ($v\ne 0_{R}$) and ($uv$ is non zero divisor because $uv=1_{R}$).

now ($v \cdot(uv)=v \cdot 1_{R}=v$) (Multiplying $uv=1_{R}$ by $v$ from left.)

$\rightarrow$ ($vuv=v)$

$\rightarrow$ ($vuv-v=0_{R}$) (Subtracting $v$ from both sides)

$\rightarrow$ $(vu-1_{R})v=0_{R}$ (By right distributive law)

$\rightarrow$ $(vu-1_{R}=0_{R}$ which is $vu=1_{R} )$ because $v\neq0_{R}$.Here is my problem : How to remove the probability that $v\neq 0_{R}$ may be zero divisor and thus $vu \neq 1_{R}$?


1 Answer 1


The strategy of your proof is correct, but it seems you confused the roles for $u$ and $v$ in the sense that in your proof you are unable to make use of the uniqueness of $v$.

Here is a variant of your proof: We know that $v$ is the unique element of $R$ with $uv = 1$. Put $v':= vu-1$ and compute $$ u\cdot (v+v') = uv + u(vu-1) = 1 + uvu-u = 1 +u-u = 1. $$ By the uniqueness property, we conclude $v+v' = v$, hence $v' = 0$. Therefore, $vu = 1$.

  • $\begingroup$ Many thanks to you for your helpful answer. All respect to you. $\endgroup$ Commented Jul 7, 2022 at 7:24

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