Degree of Sum 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 polynomials over $R$ in the indeterminate $X$.

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

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

Proof
First we associate to $f = a_0 + a_1 X + \cdots + a_n X^n \in R \left[{X}\right]$ a formal vector $x_f = \left({ a_0, a_1, \ldots, a_n, 0_R, \ldots}\right) \in R^\infty$.

Let $x_f^i \in R$ denote the element at the $i$th position.

Then:
 * $\displaystyle \deg \left({f}\right) = \sup \left\{{i \in \N : x_f^i \ne 0_R}\right\}$

The sum $+$ in the polynomial ring $R \left[{X}\right]$ gives rise to the following identity in $R^\infty$:


 * $x_{f+g}^i = x_f^i + x_g^i$

Next, let $f, g \in R \left[{X}\right]$, and let $d = \max \left\{{\deg(f), \deg(g)}\right\}$.

Then $x_f^i = 0_R = x_g^i$ for all $i > d$, so we have:


 * $\displaystyle \deg \left({f+g}\right) = \sup \left\{{i \in \N : x_{f+g}^i \ne 0_R}\right\} = \sup \left\{{i \in \N : x_f^i + x_g^i \ne 0_R}\right\} \le d$