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