Degree of Product of Polynomials over Ring

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

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

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

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

If $R$ is an integral domain then equality holds:
 * $\forall f, g \in R \left[{X}\right]: \deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$

Proof
Let the leading coefficient of:
 * $f \left({X}\right)$ 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 multiplication of polynomials:


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

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 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 $(R, +, \circ)$:


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

Finally, note that when $(R, +, \circ)$ is an integral domain by definition there are no zero divisors.

In that case $X^{n+m}$ can not equal $0_R$ and the equality holds:


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