Field is Integral Domain/Proof 1

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