In Weibel's book on K-theory, he introduces Milnor squares and Milnor patching as follows:
I was wondering if someone might be able and willing to help me a little by constructing a nice friendly example of a projective module over some specific ring constructed by means of Milnor patching. Something a bit more sophisticated than constructing it over the trivial ring, but not something too sophisticated if that would be possible.
Ideally, I'd like to see an example where one constructs the projective module over the ring $\mathbb{Z}/n\mathbb{Z}$ for some $n$, which is what I've been trying to do myself, but have so far been unsuccessful at doing.
Thank you in advance!
Let $k$ be your favorite field and consider the square
$$\matrix{k[t^2,t^3]&\rightarrow&k\cr \downarrow&&\downarrow\cr k[t]&\rightarrow&k[\epsilon]/(\epsilon^2)}$$
The top map takes all powers of $t$ to zero. The bottom map takes $t$ to $\epsilon$. The down arrows are inclusions.
Then to construct a projective module on $k[t^2,t^3]$, you start with projective modules over $k$ and $k[t]$ (which are necessarily free, because these rings are PID's), and patch them together over $k[\epsilon]/(\epsilon^2)$.
In particular, if your module is rank 1 (so you are starting with rank 1 free modules in the upper left and lower right) then a patching is given by a unit in $k[\epsilon]/(\epsilon^2)$. Such units are of the form $a+b\epsilon$ where $\neq 0$. Up to multiplication by units that come from $k$ (or $k[t]$), these are all of the form $u=1+b\epsilon$, and for any two such units $u,v$ you can check that $uv^{-1}$ does not split as a product of units coming from $k$ and $k[t]$. Therefore each $b\in k$ gives a unit $1+b\epsilon$ and each of these units gives a distinct rank one projective module over $k[t^2,t^3]$.
In short, the moral of this Milnor square is that rank one projective modules over $k[t^2,t^3]$ are in natural one-one correspondence with elements of $k$.
In fact, more can be said --- rank one projective modules (over any ring) form a group under tensor product (this is called the Picard group), and in this case the Picard group of $k[t^2,t^3]$ is isomorphic as a group (not just as a set) to the additive group of $k$. This too follows from a somewhat more careful analysis of how patching works.