Lemma 15.9.11. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $\overline{P}$ be a finite projective $A/I$-module. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite projective $A'$-module $P'$ lifting $\overline{P}$.

**Proof.**
We can choose an integer $n$ and a direct sum decomposition $(A/I)^{\oplus n} = \overline{P} \oplus \overline{K}$ for some $R/I$-module $\overline{K}$. Choose a lift $\varphi : A^{\oplus n} \to A^{\oplus n}$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$. Let $f \in A[x]$ be the characteristic polynomial of $\varphi $. Set $B = A[x]/(f)$. By Cayley-Hamilton (Algebra, Lemma 10.16.1) there is a map $B \to \text{End}_ A(A^{\oplus n})$ mapping $x$ to $\varphi $. For every prime $\mathfrak p \supset I$ the image of $f$ in $\kappa (\mathfrak p)$ is $(x - 1)^ rx^{n - r}$ where $r$ is the dimension of $\overline{P} \otimes _{A/I} \kappa (\mathfrak p)$. Hence $(x - 1)^ nx^ n$ maps to zero in $B \otimes _ A \kappa (\mathfrak p)$ for all $\mathfrak p \supset I$. Thus $x(1 - x)$ is contained in every prime ideal of $B/IB$. Hence $x^ N(1 - x)^ N$ is contained in $IB$ for some $N \geq 1$. It follows that $x^ N + (1 - x)^ N$ is a unit in $B/IB$ and that

is an idempotent as both assertions hold in $\mathbf{Z}[x]/(x^ N(x - 1)^ N)$. The image of $\overline{e}$ in $\text{End}_{A/I}((A/I)^{\oplus n})$ is

as $\overline{p}$ is an idempotent. After replacing $A$ by an étale extension $A'$ as in the lemma, we may assume there exists an idempotent $e \in B$ which maps to $\overline{e}$ in $B/IB$, see Lemma 15.9.10. Then the image of $e$ under the map

is an idempotent element $p$ which lifts $\overline{p}$. Setting $P = \mathop{\mathrm{Im}}(p)$ we win. $\square$

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