zero times an element is zero in a ring

Lemma 1.

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

Proof.

$0\cdot a$$=$$(0+0)\cdot a,\text{ by definition of zero}$

$=$$0\cdot a+0\cdot a,\text{ by the distributive law}$

Thus $0\cdot a=0\cdot a+0\cdot a$. Let $b$ be the additive inverse of $0\cdot a\in R$. Hence:

$b+0\cdot a=b+(0\cdot a+0\cdot a)$

$(b+0\cdot a)=(b+0\cdot a)+0\cdot a$

$0=0+0\cdot a$

$0=0\cdot a$

as claimed. The proof of $a\cdot 0=0$ is done analogously.∎