Is there any number $n$ such that $nm=0$, $n\neq 0$, and $m\neq 0$?

Question

I answered a question about whether zero is prime or composite on Khan Academy a while ago. Since then, two people have commented on my answer, asking another question that I don't know the answer to.

Are there zero divisors (numbers such that, when multiplied by some nonzero number, the product is zero) other than zero? If so, what are they? Wouldn't that go against the Zero Product Property, which states that if you take two numbers, $n$ and $m$, and multiply them together to get zero, then either $n$ or $m$ is equal to zero? I.e. if $nm=0$, then either $n=0$ and $m=0$.

Answer 1

In fields (e.g., $\mathbb{Q}$,$\mathbb{R}$,$\mathbb{C}$) and integral domains (e.g. $\mathbb{Z},\mathbb{Z}[x]$), no such pair of "numbers" $n,m$ can exist.

For fields, this follows from the invertibility of multiplication. If $nm = 0$ but $n \neq 0$, then $m = n^{-1}0 = 0$, and vice versa.

For integral domains, this is an axiom - and it's precisely what differentiates integral domains from rings.

In arbitrary rings, such numbers can exists. E.g., in the ring $\mathbb{Z_4} = \{0,1,2,3\}$ you have $2\cdot 2 = 4 = 0$.

Answer 2

That depends on what you mean by a "number". Are we talking about real numbers? Complex numbers? Split-complex? Quaternions? Octonions? Arbitrary rings?

Certainly, if we state that $n$ and $m$ are real numbers, then this statement must hold.