Field has no Proper Zero Divisors
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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 |
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \gamma$