Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors
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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 $51$