Question about geometric interpretation of modules I would like to understand the accepted answer to this MO question about the geometric interpretation of modules. In particular, I would like clarification on the following excerpt.

Let $R$ be the coordinate ring of a variety and $I$ a radical. Then the
  $R$ module $R/I$ corresponds to attaching a one dimensional vector
  space on each point of $Z(I)$ and the zero vector space everywhere
  else. For Example $R=\mathbb k[x,y]$ and $I=(x,y)$ gives the
  skyscraper sheaf at the origin. $I=(x)$ gives the trivial one
  dimensional bundle on the $y$-axis etc. If your Ideal is not a radical,
  the situation is slightly more complicated. $R/I$ can be thought of
  as the trival bundle on an infinitesimal neighborhood of $Z(I)$.



*

*Why does the $R$-module $R/I$ correspond to attaching a one dimensional vector space on each point of $Z(I)$ and zero everywhere else? Does this mean some kind of line bundle?

*What's the meaning of "infinitesimal neighborhood" in the non-radical case?


Lastly, where can I find detailed geometric explanations of these ideas?
 A: Answer to Question 1. The "vector space at each point" means the following:
Given a ring $R$ and a module $M$, for each prime $\mathfrak p \subseteq R$ (which is a point of $\operatorname{Spec} R$), we get a $\kappa(\mathfrak p)$-vector space
$$M \otimes_R \kappa(\mathfrak p),$$
where $\kappa(\mathfrak p)$ denotes the residue field $R_\mathfrak p/\mathfrak p R_\mathfrak p$. If $M$ is finitely generated, then all these vector spaces are finite-dimensional. (Note: if $R$ is an algebra of finite type over an algebraically closed field $k$, then $\kappa(\mathfrak m)$ is just $k$ whenever $\mathfrak m$ is maximal; thus we really just get a $k$-vector space at each closed point.)
One can prove for example that on an integral scheme (think: spectrum of a domain), a finitely presented $R$-module $M$ is locally free of rank $n$ if and only if each vector space $M \otimes_R \kappa(\mathfrak p)$ has [the same] dimension $n$. This is what algebraic geometers mean by a vector bundle.
On the other hand, if $M = R/I$, then at some points the vector space will have dimension $1$, and at others dimension $0$. Thus, it can not be seen as a vector bundle: somehow the dimensions jump when we compare different points.
One place to consult about such matters is Eisenbud's Commutative Algebra with a View Towards Algebraic Geometry, which is just a fantastic book in general. See for example Exercises 4.11, 4.12, and 6.2 for some statements about locally free modules, and Corollary A3.3 for a sketch of the connection with vector bundles.
Answer to Question 2. The remark about the radical has to do with the fact that the assignment
\begin{align*}
\{I \subseteq R \text{ ideal}\} &\to \{V \subseteq \operatorname{Spec} R \text{ closed}\}\\
I & \mapsto V(I)
\end{align*}
is only injective if we restrict the left hand side to radical ideals. An equally acceptable statement would have been

$R/I$ can be thought of as the trivial bundle on the (possibly nonreduced) scheme $\operatorname{Spec} R/I$.

But this is somehow less geometric: in differential geometry there are no nonreduced spaces, so this doesn't help us understand how to picture this.
However, $\operatorname{Spec} R/I$ is just an infinitesimal thickening of the scheme $\operatorname{Spec} R/\sqrt I$, and the latter is often what people mean by the set $V(I)$ (again: the underlying sets $V(I)$ and $V(\sqrt I)$ are the same, but their scheme [i.e., ring] structures are different).
A good example to keep in mind is the ideal $(x^2) \subseteq k[x]$. The vanishing locus of $x^2$ is just the origin, but the ring $k[x]/(x^2)$ is nonreduced. The element $x$ is so small that its square is zero; hence the name infinitesimal. Strictly speaking, this is probably abuse of language, but it is commonplace in algebraic geometry.
