Polynomial Forms over Field form Integral Domain/Formulation 1/Proof 1

Proof
We already have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \sqbrk X$ is a ring.

Suppose $f$ and $g$ are polynomials in $F \sqbrk X$ such that $f \ne 0_F, g \ne 0_F$.

If $\map \deg f = \map \deg g = 0$ then $f$ and $g$ are elements of $F$.

As $F$ is a field and a field is an integral domain, $f g \ne 0_f$.

Otherwise from Degree of Product of Polynomials over Integral Domain:
 * $\map \deg {f g} = \map \deg f + \map \deg g$

and so:
 * $\map \deg {f g} > 0$

which means:
 * $f g \ne 0_F$

Hence the result, by definition of integral domain.