Degree of Product of Polynomials
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 $R$ has no proper zero divisors 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 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 $(R, +, \circ)$:
- $\deg \left({f g}\right) \le \deg \left({f}\right) + \deg \left({g}\right)$
Finally, note that when $\left({R, +, \circ}\right)$ has no proper zero divisors the $X^{n+m}$ term can not equal $0_R$ and equality holds:
- $\deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$
$\blacksquare$
Sources
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.25$: Theorem $48$
