I want to consider a commutative ring $R$ (we can restrict to, say, $k[x,y,z]$ if needed) with an ideal $I$ and a maximal ideal $m$. Now, I think $I/mI$ is actually a $R/m$-vector space. I'm wondering if there's a slick way to prove this? I can define a scalar action
$$R/m \times I/mI \longrightarrow I/mI$$ $$(\bar{h},\bar{f}) \mapsto \overline{hf}$$
where the overbar denotes the respective projection to the quotient. You can then prove that this is well-defined. So you can define this scalar multiplication and show that it obeys the vector space axioms, but I'm hoping there's maybe a better way.
Also, if $R$ is Noetherian, we can write a finite list of generators for the ideal $I=(f_{1}, \ldots, f_{k})$. Is it then necessarily true that $(\bar{f}_{1}, \ldots, \bar{f}_{k})$ is a basis of the vector space $I/mI$?
I suppose I should also check, are all my hypothesis correct here? I know something like this is true, but maybe I need to be assuming more structure or something. Thanks!
First of all, the $R$-module structure of $I$ induces an $R/m$-module structure on $I/mI$. This is essentially your first argument. Secondly, it suffices to prove that $R/m$ is a field, for by definition a vector space is a module over a ring, which is actually a field. But $R/m$ is a field because $m$ is a maximal ideal. Finally, if $f_1,\dots,f_k$ is a set of generators of $I$, this by no means implies that $\bar{f}_1,\dots,\bar{f}_k$ is an $R/m$-basis of $I/mI$ (consider the counterexample $f_1=f_2$). It is though a spanning set. On the other hand, if $(R,m)$ is a local ring and $f_1,\dots,f_k$ a minimal set of generators, then $\bar{f}_1,\dots,\bar{f}_k$ is a basis.