Field has no Proper Zero Divisors

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Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Then $\struct {F, +, \times}$ has no proper zero divisors.

That is:

$a \times b = 0_F \implies a = 0_F \lor b = 0_F$

Proof 1

By definition, $F$ is a division ring.

Again by definition, a division ring is a ring with unity with no proper zero divisors.


Proof 2

\(\ds a\) \(\ne\) \(\ds 0_F\)
\(\ds \leadsto \ \ \) \(\ds a^{-1} \times \paren {a \times b}\) \(=\) \(\ds a^{-1} \times 0_F\) $a^{-1}$ exists because $a \ne 0_F$
\(\ds \leadsto \ \ \) \(\ds \paren {a^{-1} \times a} \times b\) \(=\) \(\ds a^{-1} \times 0_F\) Field Axiom $\text M1$: Associativity of Product
\(\ds \leadsto \ \ \) \(\ds 1_F \times b\) \(=\) \(\ds a^{-1} \times 0_F\) Field Axiom $\text M4$: Inverses for Product
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds a^{-1} \times 0_F\) Field Axiom $\text M3$: Identity for Product
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds 0_F\) Field Product with Zero