Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors

Theorem
Let $(R, +, \circ)$ be an integral domain whose zero is $0_R$.

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

For $f \in R \left[{X}\right]$ let $\deg \left({f}\right)$ be the degree of $f$.

Then if neither $f$ nor $g$ are the null polynomial:
 * $\forall f, g \in R \left[{X}\right]: \deg \left({f g}\right) \ge \deg \left({f}\right)$

Proof
From Degree of Product of Polynomials over Integral Domain, we have:


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

But $\deg \left({g}\right) \ge 0$ by definition of degree, as $g$ is not null.

Hence the result.