Milnor square

A cartesian square of unital rings $A=A_1{\times }_{A_0}{A}_2$ is a Milnor square when at least one map $A_i\to A_0$ is surjective. Finitely generated projective modules over $A$ can then be assembled from projective modules over the two upper rings together with an isomorphism of their reductions over $A_0$.

Evaluation of the cyclic generator at $1$ and at a primitive $p$th root gives the displayed pullback. The maps to the common field reduce integers modulo $p$ and set ${\zeta }_p=1$. If a representative polynomial for the second component has value $c$ at one, adding an integer multiple of ${\Phi }_p$ adjusts that value by a multiple of $p$ without changing the second component. This proves surjectivity onto compatible pairs.

The diagonal maps are extension of scalars; the maps to $A_0$ are differences. The connecting map sends $g$ to the difference of its patched free module and the corresponding free $A$-module. Elementary lifting and common free stabilization prove exactness at every internal term. The displayed fragment does not assert injectivity at its first term or surjectivity at its last.

A compatible triple $(P_1,P_2,\varphi )$ patches to $\{(x_1,x_2):\varphi ({\bar{x}}_1)={\bar{x}}_2\}$. It is finitely generated projective and recovers $P_i$ after extension of scalars. For free modules glued by $g$, the block identity $\mathrm{diag}(g,g^{-1})\in E_{\infty }(A_0)$ lifts through the surjective side and makes the direct sum with inverse gluing free. Complements reduce general projective triples to that free case. Conversely an idempotent over $A$ is exactly a compatible pair of idempotents over its two factors.