Help with proof regarding degrees of polynomials

Question

How do you prove that if $f(x)\mid g(x)$ in $F[x]$, then either $g(x) = 0$ or $\deg(g(x)) \geq \deg(f(x))$

I'm not really sure how to prove these types of statements

Answer 1

First, prove that $\operatorname{deg}(hg)=\operatorname{deg} h + \operatorname{deg} g$. For any $h,g \in F[x]$. Combine this with the definition of divisibility to get your result.

Answer 2

Let $f(x) | g(x)$. Then, $\exists h(x)$ such that $f(x) \cdot h(x) = g(x)$. $h(x)$ is a polynomial of degree $n$ or $h(x) = 0$. $f(x)$ is a polynomial of degree $m$. Which means that $g(x)$ is a polynomial of degree either $m+n$ or $g(x) = 0$.

Proof that $deg(pq) = deg(p) + deg(q)$ :

Let $p$ be a polynomial of degree $n$ and $q$ be a polynomial of degree $m$. Then, $p(x) = p_0 + p_1 x + p_2 x^2 + ..... p_n x^n$ where $p_n \not= 0$. And $q(x) = q_0 + q_1 x + q_2 x^2 + ..... q_n x^n$, $q_n \not= 0$. So, the highest order term of $p$ and $q$ will be $p_n q_n x^n x^m = p_n q_n x^{n+m}$ [note that $p_n$ and $q_n$ are nonzero]. Thus, $deg(pq)= deg( p ) + deg(q)$.