
Lemma 15.9.10. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $A \to B$ be an integral ring map. Let $\overline{e} \in B/IB$ be an idempotent. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an idempotent $e' \in B \otimes _ A A'$ lifting $\overline{e}$.

Proof. Choose an element $y \in B$ lifting $\overline{e}$. Choose $f \in A[x]$ as in Lemma 15.9.9 for $y$. By Lemma 15.9.6 we can find an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and such that $f = gh$ in $A[x]$ with $g(x) = x^ d \bmod IA'$ and $h(x) = (x - 1)^ d \bmod IA'$. After replacing $A$ by $A'$ we may assume that the factorization is defined over $A$. In that case we see that $b_1 = g(y) \in B$ is a lift of $\overline{e}^ d = \overline{e}$ and $b_2 = h(y) \in B$ is a lift of $(\overline{e} - 1)^ d = (-1)^ d (1 - \overline{e})^ d = (-1)^ d(1 - \overline{e})$ and moreover $b_1b_2 = 0$. Thus $(b_1, b_2)B/IB = B/IB$ and $V(b_1, b_2) \subset \mathop{\mathrm{Spec}}(B)$ is disjoint from $V(IB)$. Since $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is closed (see Algebra, Lemmas 10.35.22 and 10.40.6) we can find an $a \in A$ which maps to an invertible element of $A/I$ whose image in $B$ lies in $(b_1, b_2)$, see Lemma 15.9.8. After replacing $A$ by the localization $A_ a$ we get that $(b_1, b_2) = B$. Then $\mathop{\mathrm{Spec}}(B) = D(b_1) \amalg D(b_2)$; disjoint union because $(b_1, b_2) = B$ and covers $\mathop{\mathrm{Spec}}(B)$ because $b_1b_2 = 0$. Let $e \in B$ be the idempotent corresponding to the open and closed subset $D(b_1)$, see Algebra, Lemma 10.20.3. Since $b_1$ is a lift of $\overline{e}$ and $b_2$ is a lift of $\pm (1 - \overline{e})$ we conclude that $e$ is a lift of $\overline{e}$ by the uniqueness statement in Algebra, Lemma 10.20.3. $\square$

Comment #3633 by Brian Conrad on

"disjoint union because $b_1 b_2 = 0$" should say "disjoint union because $(b_1,b_2)=B$, and covers ${\rm{Spec}}(B)$ because $b_1b_2=0$".

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