# Let $R$ be a ring with unity and $u\in R$.If there is a unique $v\in R$ such that $uv=1_{R}$,show that $vu=1_{R}$,and therefore u is a unit.

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 $$ab=0_{R}$$)

Proof:

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 $$vu \neq 1_{R}$$?

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, $$vu = 1$$.