uniqueness of multiplicative zero of a ring

Question

Let $R$ be a ring. Fix $a\in R$. If $ab=0$ for all $b\in R$, must $a=0$? Here, $0$ denotes the unique element such that $0+b=b$ for all $b$.

Answer 1

By definition, a ring has a multiplicative identity $1$ that satisfies $a\cdot 1=a$ for all $a$. If $a\cdot b=0$ for all $b$, then in particular $a\cdot 1=0$, which implies $a=0$.

One the other hand, a "rng" is an object that satisfies all of the ring axioms except for the existence of $1$. It is easy to construct examples of rngs for which $a\ne 0$.

Take any abelian group $(R,+)$. Let $0$ denote the additive identity. Define a multiplication rule $a\cdot b=0$ for all $a,b\in R$. This satisfies associativity and distributivity. Therefore $(R,+,\cdot)$ is a rng.