# Degree of Product of Polynomials over Ring/Corollary 1

 It has been suggested that this page or section be merged into Degree of Product of Polynomials over Integral Domain. (Discuss)

## Corollary to Degree of Product of Polynomials over Ring

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

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

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

Let $R$ have no proper zero divisors.

Then:

$\forall f, g \in R \left[{X}\right]: \deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$

## Proof

$f \left({X}\right)$ be $a_n$
$g \left({X}\right)$ be $b_m$.
$\deg \left({f g}\right) \le \deg \left({f}\right) + \deg \left({g}\right)$

From the definition of polynomial multiplication:

$f \left({X}\right) g \left({X}\right) = a_n b_m X^{n+m} + \cdots + a_0 b_0$

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

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

$\blacksquare$