Let $V$ be a vector space of all real-valued polynomials in a fixed variable $t$. Let $\textbf{p}$ be a fixed polynomial of degree $n$, and let $W$ be all polynomials divisible by $\textbf{p}$.
What is a basis for the quotient space $V/W$? My guess would be that a basis for $V/W$ is $\{[1], [t], [t^2], \dotsc, [t^{n-1}]\}$.
I understand the elements of $V/W$ are equivalence classes of the form $$[\textbf{q}] = \{\textbf{x} \in V | \textbf{x} - \textbf{q} \in W\}.$$
This means that $\textbf{x} - \textbf{q} = \textbf{p}\textbf{y}$ for some polynomial $\textbf{y}$. I do notice that $\textbf{q}$ is a remainder term, and $\operatorname{deg}(\textbf{q}) < \operatorname{deg}(\textbf{p})$. But I still don't see how I can determine a basis for $V/W$. If anyone can help me that would be great!
Spanning Let $f\in V$, then $f(t)=q(t)p(t)+\sum_{i=0}^{n-1} a_i t^i$. So $[f(t)]=\sum_{i=0}^{n-1} a_i [t^i]$.
Linear Independence Suppose some $\sum_{i=0}^{n-1} b_i [t^i]=[0]$. Then $[\sum_{i=0}^{n-1} b_i t^i]=[0]$. That means $p(t)$, a polynomial of degree $n$, divides a polynomial $\sum_{i=0}^{n-1} b_i t^i$ of degree at most $n-1$. That can only happen if we have that $\sum_{i=0}^{n-1} b_i t^i$ is the zero polynomial, and so all $b_i=0$.