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

Theorem

Let $\struct {R, +, \circ}$ be an integral domain whose zero is $0_R$.

Let $R \sqbrk X$ be the ring of polynomial forms over $R$ in the indeterminate $X$.

For $f \in R \sqbrk X$ let $\map \deg f$ be the degree of $f$.

Then if neither $f$ nor $g$ are the null polynomial:

$\forall f, g \in R \sqbrk X: \map \deg {f g} \ge \map \deg f$

Proof

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

$\map \deg {f g} = \map \deg f + \map \deg g$

But $\map \deg g \ge 0$ by definition of degree, as $g$ is not null.

Hence the result.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 27$. Euclidean Rings: Theorem $51$