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

Theorem
Let $\left({F, +, \circ}\right)$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $X$ be transcendental in $F$.

Let $F \left[{X}\right]$ be the ring of polynomial forms in $X$ over $F$.

Then $F \left[{X}\right]$ is an integral domain.

Proof
We already have from Ring of Polynomial Forms is Ring 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.