Lemma 1   Let $R$ be a ring with zero element $0$ (i.e. $0$ is the additive identity of $R$ . Then for any element $a\in R$ we have $0\cdot a = a\cdot 0 = 0$

Proof. \begin{eqnarray*} 0\cdot a &=& (0+0)\cdot a , \quad \text{ by definition of zero}\\ &=& 0\cdot a + 0\cdot a, \quad \text{ by the distributive law}\\ \end{eqnarray*}Thus $0\cdot a=0\cdot a + 0\cdot a$ Let $b$ be the additive inverse of $0\cdot a \in R$ Hence: \begin{eqnarray*} b+0\cdot a=b+(0\cdot a + 0\cdot a)\\ (b+0\cdot a)=(b+0\cdot a) + 0\cdot a\\ 0=0 + 0\cdot a\\ 0=0\cdot a \end{eqnarray*}as claimed. The proof of $a\cdot 0=0$ is done analogously.