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 $f$.

Proof
Let the leading coefficient of:
 * $\map f X$ be $a_n$
 * $\map g X$ be $b_n$.

Then:

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 $\struct {R, +, \circ}$:


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

Also see

 * Degree of Sum of Polynomials