From Linear Algebra Done Right, Third Edition, Ch4 Ex11 by Sheldon Axler:
Suppose $p\in\mathcal P(\Bbb F)$ with $p\ne 0$. Let $U=\{pq : q\in\mathcal P(\Bbb F)\}$.
(a) Show that $\dim \mathcal P(\Bbb F)/U = \deg p$.
(b) Find a basis of $\mathcal P(\Bbb F)/U$.
I can clearly see that the dimension of $\mathcal P(\Bbb F)/U$ must be at least the degree of $p$ because any polynomial with degree less than $p$ will not be in $U$ and there are $\deg p$ of those. But then not every polynomial of degree at least $\deg p$ will have $p$ as a factor. So intuitively I'd think it was infinite, but that's obviously wrong.
Another idea I had was to figure out the dimension of $\mathcal P_m(\Bbb F)/U$ and then take a limit as $m\to \infty$. But I can't figure out what the dimension of $U$ is.
So I'm having trouble figuring out how to get started on this one.
If $q$ is a polynomial with bigger degree than $p$, I can divide $q$ by $p$ and get $q = \lambda p + r$, with $\lambda$ a polynomial and $r$ of polynomial of degree $ 0 \leq \deg(r) < \deg(p)$. So $q = r$ in $P(\mathbb F)/U$. This gives us that $P(\mathbb F)/U \cong \mathbb F\{1,x,x^2, \dots x^n\}$.