State/Additivity of polynomial degrees

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Proposition
Proof

Proposition: (Additivity of polynomial degrees) Suppose that $R$ is an integral domain, such as $\ZZ$ or $\RR$, for example. Suppose that $P(X)$ and $Q(X)$ are nonzero polynomials with coefficients in $R$.

Then, $deg(P(X) \cdot Q(X)) = deg(P(X)) + deg(Q(X))$.

Logical Connections

This statement logically relies on the following definitions and statements: Def/Integral domain, Def/Polynomial

The following statements and definitions rely on the material of this page: State/Root counting over a field

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State/Additivity of polynomial degrees

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	Let $R$ be an integral domain. Suppose that $P(X)$ and $Q(X)$ are nonzero polynomials with coefficients in $R$.
	Let $p_0, \ldots, p_d$ and $q_0, \ldots, q_e$ be the coefficients of $P$ and $Q$, respectively. In other words, $$P(X) = p_0 + p_1 X + \cdots + p_d X^d, \mbox{ and } Q(X) = q_0 + q_1 X + \cdots + q_e X^e,$$ and $d = deg(P)$ and $e = deg(Q)$. For $i \in \NN$, let $$S_i = \{ (j,k) \in \NN^2 \mbox{ such that } 0 \leq j \leq d, 0 \leq k \leq e, j + k = i \}.$$ Multiplying $P$ and $Q$ yields: $$P(X) \cdot Q(X) = \sum_{i=0}^{d+e} \left( \sum_{(j,k) \in S_i} p_j q_k \right) X^i.$$ The coefficient of $X^{d+e}$ is precisely $p_d q_e$, since $S_{d+e} = \{(d,e) \}$. Since $deg(P) = d$, and $deg(q) = e$, it follows that $p_d \neq 0$, and $q_e \neq 0$. Since $R$ is an integral domain, this implies that $p_d q_e \neq 0$.
	Hence $deg(PQ) = d + e$.