# Degree of Product of Polynomials over Ring

## Theorem

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

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

Let $f, g \in R \sqbrk X$.

Then:

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

where:

$\map \deg f$ denotes the degree of $f$.

### Corollary 1

Let $R$ have no proper zero divisors.

Then:

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

### 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

$\map f X$ be $a_n$
$\map g X$ be $b_n$.

Then:

 $\ds \map f X$ $=$ $\ds a_n X^n + \cdots + a_0$ $\ds \map g X$ $=$ $\ds b_n X^n + \cdots + b_0$

Consider the leading coefficient of the product $\map f X \map g X$: call it $c$.

From the definition of polynomial addition and polynomial multiplication:

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

Clearly the highest term of $\map f X \map g X$ can have an index no higher than $n + m$.

Hence the result:

$\map \deg {f g} \not > \map \deg f + \map \deg g$

Next, note that the general ring with unity $\struct {R, +, \circ}$ 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:

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

Thus, for a general ring with unity $\struct {R, +, \circ}$:

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

$\blacksquare$