Milnor Squares and Milnor Patching: Examples?

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.