The unity of a ring $(R,\,+,\,\cdot)$ is the multiplicative identity of the ring, if it has such. The unity is often denoted by $e$ $u$ or 1. Thus, the unity satisfies $$e\cdot a = a\cdot e = a\quad\forall a\in R.$$

If $R$ consists of the mere 0, then $0$ is its unity, since in every ring, $0\cdot a = a\cdot 0 = 0$ Conversely, if 0 is the unity in some ring $R$ then $R = \{0\}$ , (because $a = 0\cdot a = 0\,\,\forall a\in R$ .

Note. When considering a ring $R$ it is often mentioned that ``...having $1 \neq 0$ ' or that ``...with non-zero unity'', sometimes only ``...with unity'' or ``...with identity element''; all these exclude the case $R = \{0\}$

Proof. Let $u$ be an idempotent and regular element. For any element $x$ of $R$ we have $$ux = u^2x = u(ux),$$ and because $u$ is no left zero divisor, it may be cancelled from the equation; thus we get $x = ux$ Similarly, $x = xu$ So $u$ is the unity of the ring. The other half of the theorem is apparent.