# Field has no Proper Zero Divisors

## Theorem

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.

$\blacksquare$

## 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

$\blacksquare$