Zero Product with Proper Zero Divisor is with Zero Divisor
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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 $y$.
$\blacksquare$