Degree of Sum of Polynomials

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Theorem

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

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

For $f \in R \sqbrk X$ let $\map \deg f$ be the degree of $f$.


Then:

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


Proof

First we associate to $f = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$ a formal vector $x_f = \tuple {a_0, a_1, \ldots, a_n, 0_R, \ldots} \in R^\infty$.

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

Then:

$\displaystyle \map \deg f = \sup \set {i \in \N : x_f^i \ne 0_R}$

The sum $+$ in the polynomial ring $R \sqbrk X$ gives rise to the following identity in $R^\infty$:

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


Next, let $f, g \in R \sqbrk X$, and let $d = \max \set {\map \deg f, \map \deg g}$.

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

$\displaystyle \map \deg {f + g} = \sup \set {i \in \N : x_{f + g}^i \ne 0_R} = \sup \set {i \in \N : x_f^i + x_g^i \ne 0_R} \le d$

$\blacksquare$


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