Degree of factor is less than degree of polynomial

Dependencies:

1. Degree of product of polynomials

Let $R$ be a ring with no zero-divisors. Let $p(x) = a(x)b(x)$ where $a(x), b(x), p(x) \in R[x]$ and $p \neq 0$. Then $0 \le \deg(a) \le \deg(p)$ and $0 \le \deg(b) \le \deg(p)$.

Proof

\[ p \neq 0 \implies (a \neq 0 \wedge b \neq 0) \implies (\deg(a) \ge 0 \wedge \deg(b) \ge 0) \]

Since $R$ has no zero-divisors, $\deg(p) = \deg(a) + \deg(b)$.

\[ 0 \le \deg(a) \le \deg(a) + \deg(b) = \deg(p) \] \[ 0 \le \deg(b) \le \deg(a) + \deg(b) = \deg(p) \]

Dependency for:

1. A polynomial of degree n has at most n zeros
2. Gauss' Lemma

Info:

• Depth: 4
• Number of transitive dependencies: 4

Transitive dependencies:

1. Group
2. Ring
3. Polynomial
4. Degree of product of polynomials