# Degree of Product of Polynomials over Ring

## Theorem

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

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

Let $f,g\in R[X]$.

Then:

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

where:

$\deg \left({f}\right)$ denotes the degree of $f$.

### Corollary 1

Let $R$ have no proper zero divisors.

Then:

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

### Corollary 2

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

• $f \left({X}\right)$ be $a_n$
• $g \left({X}\right)$ be $b_n$.

Then:

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

Consider the leading coefficient of the product $f \left({X}\right) g \left({X}\right)$: call it $c$.

From the definition of polynomial addition and polynomial multiplication:

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

Clearly the highest term of $f \left({X}\right) g \left({X}\right)$ can have an index no higher than $n+m$.

Hence the result:

$\deg \left({f g}\right) \not > \deg \left({f}\right) + \deg \left({g}\right)$

Next, note that the general ring with unity $(R, +, \circ)$ may have proper zero divisors.

Therefore it is possible that $X^{n+m}$ may equal $0_R$.

If that is the case, then the highest term will have an index definitely less than $n+m$.

That is, in that particular case:

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

Thus, for a general ring with unity $(R, +, \circ)$:

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

$\blacksquare$