Field is Integral Domain
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Theorem
Every field is an integral domain.
Proof 1
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Suppose $\exists x, y \in F: x \circ y = 0_F$.
Suppose $x \ne 0_F$.
Then, by the definition of a field, $x^{-1}$ exists in $F$ and:
- $y = 1_F \circ y = x^{-1} \circ x \circ y = x^{-1} \circ 0_F = 0_F$.
Otherwise $x = 0_F$.
So if $x \circ y = 0_F$, either $x = 0_F$ or $y = 0_F$ as we were to show.
$\blacksquare$
Proof 2
This result follows directly from:
- Field has no Proper Zero Divisors
- By definition, that a field is a commutative ring.
$\blacksquare$
Sources
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): integral domain