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 $F \left[{X}\right]$ is a ring.

Suppose $f$ and $g$ are polynomials in $F \left[{X}\right]$ such that $f \ne 0_F, g \ne 0_F$.

If $\deg \left({f}\right) = \deg \left({g}\right) = 0$ 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:
 * $\deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$

and so:
 * $\deg \left({f g}\right) > 0$

which means $f g \ne 0_F$

Hence the result, by definition of integral domain.