Let $F$ be a field and $f(x), g(x) \in F[x]$. Show that $f(x)$ divides $g(x)$ if and only if $g(x) \in \langle f(x) \rangle$.
This seems... almost trivial to me (which is usually a sign that I'm missing something):
We first assume that $f(x) \neq 0$ because zero divides everything, and $\langle 0 \rangle = \{0\}$.
If $g(x) \in \langle f(x) \rangle$, then $f(x)~|~g(x)$ by the definition of a principal ideal (that is, if $g(x)$ is in the ideal generated by $f(x)$, it's just a multiple of $f(x)$).
If $f(x)~|~g(x)$, then $g(x) = q(x)f(x)$ for $q(x) \in F[x]$. But $q(x)f(x) \in \langle f(x) \rangle$, so $g(x) \in \langle f(x) \rangle$.
As often (often) happens, a problem that seems so simple usually means I'm missing something—especially when I made no use of the fact that $F$ is a field. Am I going wrong with a definition, or...
As has been pointed out, this exercise is as easy as you thought. In particular, your argument is complete. The only mistake is when you said "zero divides everything"; I believe you meant everything (other than the zero) divides zero.