# Zero Product with Proper Zero Divisor is with Zero Divisor

## Theorem

Let $\struct {R, +, \circ}$ be a ring.

Let $x \in R$ be a proper zero divisor of $R$.

Then:

$\paren {x \divides 0_R} \land \paren {x \circ y = 0_R} \land \paren {y \ne 0_R} \implies y \divides 0_R$

That is, if $x$ is a proper zero divisor, then whatever non-zero element you form the product with it by to get zero must itself be a zero divisor.

## Proof

Follows directly from the definition of proper zero divisor.

If $y \ne 0_R$ and $x \circ y = 0_R$ and $x \in R^*$ (which is has to be if it's a proper zero divisor), then all the criteria of being a zero divisor are fulfilled by $y$.

$\blacksquare$