Degree of sum of polynomials

Dependencies:

  1. Polynomial

Let $p, q \in R[x]$. Then $\deg(p+q) \le \max(\deg(p), \deg(q))$.

Proof

Let $k = \max(\deg(p), \deg(q))$.

Case 1: $p = q = 0$

\[ p = q = 0 \implies \deg(p) = \deg(q) = -\infty \implies k = -\infty \]

$\deg(p+q) = \deg(0) = -\infty = k$.

Therefore, $p+q \in R[x]$ and $\deg(p+q) \le k$.

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

\[ p \neq 0 \vee q \neq 0 \implies \deg(p) \ge 0 \vee \deg(q) \ge 0 \implies k \ge 0 \]

$\forall i > k, (p+q)_i = p_i + q_i = 0 + 0 = 0$.

Therefore, $p+q \in R[x]$ and $\deg(p+q) \le k$.

Dependency for:

  1. F[x]/p(x) is isomorphic to F[x]/p(x)F[x]
  2. Polynomial division theorem
  3. F[x]/p(x): A ring
  4. p(x)F[x] = F[x] iff p is a non-zero constant
  5. Polynomials of a ring form a ring

Info:

Transitive dependencies:

  1. Group
  2. Ring
  3. Polynomial