$K[X]$-Module $K[X]/(p)$ explanation for a beginner I am having trouble understanding $K[X]/(p)$ as a $K[X]$-Module, where $K[X]$ is polynomial ring, and $p$ is a normed polynomial in $K[X]$. 
It was used in the following context:
Let $\phi: K[T]\to K[T]/(p), \phi(h):=[h]$, that is the respective equivalence class of an element in $K[T]$ with respect to $(p)$.
It states that $[1]$ is a generating system of $K[T]/(p)$ as a $K[X]$-Module  
I have a good understanding of a vector space $V$ as a $K[X]$-Module:
Namely, a $K$-Vector Space $V$ and an endomorphism $f$, with scalar multiplication as well as, and most importantly:
for $P \in K[X]$, $v \in V$:
$P\cdot_{f}v=P(f)(v)$.
But what would f0r example a $K[X]$-Module $K[X]$ look like? In terms of multiplication, we'd get $P\cdot_{f}=P(f)(Q)$, but that does not make sense in my view.
A great explanation would be needed here.
 A: It is exactly the same as this general situation: You have a commutative ring $R$, an ideal $I\subset R$, the quotient $R/I$ and the canonical map:
\begin{align}
R&\longrightarrow R/I,\\
r&\longmapsto r+I.
\end{align}
$R/I$ is an $R$-module via$\;\lambda\cdot(r+I)\overset{\text{def}}{=}\lambda r+I$, and it is obvious that, as an $R$-module, $R/I$ is generated by the class of $1$, since for any  $\;r\in R$, we have
$$r+I=r\cdot(1+I)$$
A: More generally, any ring homomorphism $f:R\to S$ determines an $R$-module structure on $S$, by letting $r\cdot s:=f(r) s$. 
If $f$ is surjective, then $f(1)=1\in S$ will be a generator. 
Your confusion might have arose by wanting to attach some linear endomorphism to the action of $X$, which is irrelevant in this setting. 
Instead, here the action is given by multiplication of polynomials: $f\cdot[g] :=[f\cdot g] $. 
A: We can view both $K[X]$ and $K[X]/(p)$ as vector spaces with an endomorphism:
A basis of $K[X]$ is $\{1,X,X^2,\ldots\}$ (it's infinite-dimensional), and $X$ acts as the endomorphism which is of course multiplication by $X$, which maps $X^n$ to $X^{n+1}$.
A basis of $K[X]/(p)$ is $\{[1],[X],\ldots,[X^{\deg p-1}]\}$, and $X$ acts as the endomorphism whose matrix in this basis is the companion matrix of $p$.
In either case, $1$ is a generator of the module.
