Field is Integral Domain

Theorem

Every field is an integral domain.

Proof 1

Let $\left({F, +, \circ}\right)$ 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.

$\blacksquare$

Note that this is equivalent to the statement $\forall x, y \in F: x \ne 0_F \land y \ne 0_F \implies x \circ y \ne 0_F$, so the set $F^*$ is closed under ring product.

Proof 2

This result follows directly from:

• Field has No Proper Zero Divisors
• By definition, that a field is a commutative ring.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 23$
• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 4.14$
• B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra (1970): $\S 1.3$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 55.1$