Degree of Product of Polynomials over Ring/Corollary 1

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Corollary to Degree of Product of Polynomials over Ring

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$.

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

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

Let $R$ have no proper zero divisors.

Then:

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

Proof

Let the leading coefficient of:

$\map f X$ be $a_n$

$\map g X$ be $b_m$.

From Degree of Product of Polynomials over Ring:

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

From the definition of polynomial multiplication:

$\map f X \map g X = a_n b_m X^{n + m} + \cdots + a_0 b_0$

As $\struct {R, +, \circ}$ has no proper zero divisors the $X^{n + m}$ term can not equal $0_R$ and so:

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

$\blacksquare$