Zero Divisor Product is Zero Divisor

Theorem

The ring product of a zero divisor with any ring element is a zero divisor.

Proof

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

Let $x \divides 0_R$ in $R$.

Then:

\(\ds \exists y \in R, y \ne 0_R: \, \)

\(\ds x \circ y\)

\(=\)

\(\ds 0_R\)

Definition of Zero Divisor of Ring

\(\ds \leadsto \ \ \)

\(\ds \forall z \in R: \, \)

\(\ds z \circ \paren {x \circ y}\)

\(=\)

\(\ds z \circ 0_R\)

\(\ds \)

\(=\)

\(\ds 0_R\)

Ring Product with Zero

\(\ds \leadsto \ \ \)

\(\ds \forall z \in R: \, \)

\(\ds \paren {z \circ x} \circ y\)

\(=\)

\(\ds 0_R\)

Associativity of $\circ$

So $z \circ x \divides 0_R$ in $R$.

The same thing happens if we form the product $\paren {x \circ y} \circ z$.

$\blacksquare$