Degree of Product of Polynomials over Ring/Corollary 2

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Theorem

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

Let $D \sqbrk X$ be the ring of polynomials over $D$ in the indeterminate $X$.

For $f \in D \sqbrk X$ let $\map \deg f$ denote the degree of $f$.


Then:

$\forall f, g \in D \sqbrk X: \map \deg {f g} = \map \deg f + \map \deg g$


Proof

An integral domain is a commutative and unitary ring with no proper zero divisors.

The result follows from Degree of Product of Polynomials over Ring: Corollary 1.

$\blacksquare$


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