Field is Integral Domain

Theorem
Every field is an integral domain.

Proof
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.

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

We can also get this from the fact that $$F$$ is a division ring, which is a ring with unity with no divisors of zero, and also commutative.