TITLE: $x$ is a left zero-divisor $\iff$ $x$ is a right zero-divisor.
QUESTION [5 upvotes]: Let $R$ be a ring with unity. Show that $x$ is a left zero-divisor if and only if $x$ is a right zero-divisor.
Suppose, $x$ is a left zero divisor. Then, $\exists y \in R$ such that $xy = 0 \Rightarrow (xy)\cdot 1_R = 0 \Rightarrow (xy)(x^{-1}x)= 0 \Rightarrow (xyx^{-1})x = 0$. Hence, $x$ is a right zero divisor. 
Now, suppose, $x$ is a right zero divisor. Then,$\exists y \in R$ such that $yx = 0 \Rightarrow 1_R\cdot(yx) = 0 \Rightarrow (xx^{-1})(yx)= 0 \Rightarrow x(x^{-1}yx) = 0$. Thus, $x$ is a left zero divisor. 
But, this isn't correct. So, please tell me what's wrong with my proof.

REPLY [2 votes]: Although the claim is incorrect, the following is nonetheless true:
Proposition. Let $R$ denote a ring (with unity). Then if every non-zero element of $R$ has a left multiplicative inverse, then every non-zero element of $R$ has a two-sided multiplicative inverse.
Proof. Suppose $x_0$ is non-zero. Then $(*) \;\; x_1x_0=1$ for some non-zero $x_1$. Hence $(**)\;\; x_2x_1 = 1$ for some non-zero $x_2$. Now consider the product $x_2 x_1 x_0$. We have:
$$x_2 x_1 x_0 = (x_2 x_1) x_0 = 1x_0 = x_0$$
$$x_2 x_1 x_0 = x_2 (x_1 x_0) = x_2$$
So $x_0=x_2$. Hence from $(**)$ we have $x_0x_1=1$. But we already know that $x_1 x_0 = 1$ from $(*)$.