Lemma 15.9.2. Let $A$ be a ring, let $I \subset A$ be an ideal, let $\overline{e} \in A/I$ be an idempotent. There exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an idempotent $e' \in A'$ lifting $\overline{e}$.
Proof. Choose any lift $x \in A$ of $\overline{e}$. Set
\[ A' = A[t]/(t^2 - t)\left[\frac{1}{t - 1 + x}\right]. \]
The ring map $A \to A'$ is étale because $(2t - 1)\text{d}t = 0$ and $(2t - 1)(2t - 1) = 1$ which is invertible. We have $A'/IA' = A/I[t]/(t^2 - t)[\frac{1}{t - 1 + \overline{e}}] \cong A/I$ the last map sending $t$ to $\overline{e}$ which works as $\overline{e}$ is a root of $t^2 - t$. This also shows that setting $e'$ equal to the class of $t$ in $A'$ works. $\square$
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