Degree of product of polynomials

Dependencies:

  1. Polynomial

Let $p, q \in R[x]$. Then $\deg(pq) \le \deg(p) + \deg(q)$. If $R$ has no zero-divisors, $\deg(pq) = \deg(p) + \deg(q)$.

Proof

Case 1: $p = 0$ or $q = 0$

$\deg(pq) = \deg(0) = -\infty$

Since one of $\deg(p)$ and $\deg(q)$ is $-\infty$ and the other is either finite or $-\infty$, their sum is $-\infty$.

Therefore, $\deg(pq) = \deg(p) + \deg(q)$.

Case 2: $p \neq 0$ and $q \neq 0$

Therefore, $\deg(p)$ and $\deg(q)$ are finite.

$(pq)_i = \sum_{j=0}^i p_jq_{i-j} \in R$.

Let $k = \deg(p) + \deg(q)$. Let $i > k$

\begin{align} & j \le \deg(p) \\ &\Rightarrow i-j \ge k+1-j \\ &= \deg(p) + \deg(q) + 1 - j \\ &= \deg(q) + 1 + (\deg(p) - j) \\ &> \deg(q) \\ &\Rightarrow q_{i-j} = 0 \end{align}

So, if $j \le \deg(p)$, $q_{i-j} = 0$ and if $j > \deg(p)$, $p_j = 0$. Therefore, $p_jq_{i-j} = 0$ for all $0 \le j \le i$.

$(pq)_i = \sum_{j=0}^i p_jq_{i-j} = 0$ for all $i > k$. Therefore, $pq \in R[x]$ and $\deg(pq) \le \deg(p) + \deg(q)$.

\begin{align} & j < \deg(p) \\ &\Rightarrow k-j \\ &= \deg(q) + \deg(p) - j \\ &> \deg(q) \\ &\Rightarrow q_{k-j} = 0 \end{align}

\begin{align} & (pq)_k = \sum_{j=0}^{\deg(p)-1} p_jq_{k-j} + p_{\deg(p)}q_{k-\deg(p)} + \sum_{j=\deg(p)+1}^k p_jq_{k-j} \\ &= \sum_{j=0}^{\deg(p)-1} p_j0 + p_{\deg(p)}q_{\deg(q)} + \sum_{j=\deg(p)+1}^k 0q_{k-j} \\ &= p_{\deg(p)}q_{\deg(q)} \end{align}

By definition, $p_{\deg(p)} \neq 0$ and $q_{\deg(q)} \neq 0$. If $R$ has no zero-divisors, their product is non-zero. Therefore, $\deg(pq) = k = \deg(p) + \deg(q)$.

Dependency for:

  1. Degree of factor is less than degree of polynomial
  2. Eisenstein's criterion
  3. Zero divisors of a polynomial
  4. Polynomial GCD theorem
  5. The ideal generated by an irreducible polynomial is maximal
  6. Polynomial division theorem
  7. F[x]/p(x): A ring
  8. The ring F[x]/p(x) is a field iff p is irreducible
  9. Polynomials of a ring form a ring
  10. Gauss' Lemma

Info:

Transitive dependencies:

  1. Group
  2. Ring
  3. Polynomial