Ring Element is Zero Divisor iff not Cancellable
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Theorem
Let $\struct {R, +, \circ}$ be a ring which is not null.
Let $z \in R^*$.
Then $z$ is a zero divisor if and only if $z$ is not cancellable for $\circ$.
Proof
Sufficient Condition
Let $z$ be a zero divisor.
Then either $z \circ x = 0_R$ or $x \circ z = 0_R$ for some $x \in R^*$.
Then:
- $z \circ 0_R = 0_R = 0_R \circ z$
and so $z$ is not cancellable.
$\Box$
Necessary Condition
Let $z$ not be cancellable in $R$.
Then there exists $x, y \in R$ such that $x \ne y$ and:
- $z \circ x = z \circ y$
Then:
\(\ds z \circ \paren {x + \paren {-y} }\) | \(=\) | \(\ds z \circ x + z \circ \paren {-y}\) | $\circ$ distributes over $+$ | |||||||||||
\(\ds \) | \(=\) | \(\ds z \circ x + \paren {-z \circ y}\) | Product with Ring Negative | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_R\) | as $z \circ x = z \circ y$ |
But $x \ne y$, so $x + \paren {-y} \ne 0$.
Thus $z$ is a zero divisor.
Similarly if $x \circ z = y \circ z$ where $x \ne y$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.2$
- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings: Lemma $1.4$