Let $\mathbb{C}[X]$ be a module over itself. Given $z \in \mathbb{C}$ define $M_z = \{ f(X) \in \mathbb{C}[X] \ | \ f(z) = 0\}$. $M_z$ is a submodule of $\mathbb{C}[X]$. Give a $\mathbb{C}$-basis for $M_z$.
Now the way I understand it, to provide a $\mathbb{C}$-basis for $M_z$ means that we consider $M_z$ as a $\mathbb{C}$-module and then provide a basis for it as a $\mathbb{C}$ module. However I'm not sure if it's possible to do that.
I know that if we consider $M_z$ as a $\mathbb{C}[X]$ module then $X-z$ is a basis for $M_z$, the reason for that is that any $f(X) \in M_z$ can be expressed in the form $f(X) = (X-z)g(X)$ for some $g(X) \in \mathbb{C}[X]$. However when considering $M_z$ as a $\mathbb{C}$-module I don't have any ideas as to what I basis should even look like.
Is it possible to obtain a $\mathbb{C}$-basis for $M_z$?
An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$