**Proof.**
Assume (2) holds. Then $I$ is contained in the Jacobson radical of $A$, since otherwise there would be a nonunit $f \in A$ not contained in $I$ and the map $A \to A_ f$ would contradict (2). Hence $IB \subset B$ is contained in the Jacobson radical of $B$ for $B$ integral over $A$ because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is closed by Algebra, Lemmas 10.40.6 and 10.35.22. Thus the map from idempotents of $B$ to idempotents of $B/IB$ is injective by Lemma 15.10.2. On the other hand, since (2) holds, every idempotent of $B$ lifts to an idempotent of $B/IB$ by Lemma 15.9.10. In this way we see that (2) implies (4).

The implication (4) $\Rightarrow $ (3) is trivial.

Assume (3). Let $\mathfrak m$ be a maximal ideal and consider the finite map $A \to B = A/(I \cap \mathfrak m)$. The condition that $B \to B/IB$ induces a bijection on idempotents implies that $I \subset \mathfrak m$ (if not, then $B = A/I \times A/\mathfrak m$ and $B/IB = A/I$). Thus we see that $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be monic and suppose given a factorization $\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic. Set $B = A[T]/(f)$. Let $\overline{e}$ be the nontrivial idempotent of $B/IB$ corresponding to the decomposition

\[ B/IB = A/I[T]/(g_0) \times A[T]/(h_0) \]

of $A$-algebras. Let $e \in B$ be an idempotent lifting $\overline{e}$ which exists as we assumed (3). This gives a product decomposition

\[ B = eB \times (1 - e)B \]

Note that $B$ is free of rank $\deg (f)$ as an $A$-module. Hence $eB$ and $(1 - e)B$ are finite locally free $A$-modules. However, since $eB$ and $(1 - e)B$ have constant rank $\deg (g_0)$ and $\deg (h_0)$ over $A/I$ we find that the same is true over $\mathop{\mathrm{Spec}}(A)$. We conclude that

\[ f = \det \nolimits _ A(T : B \to B) = \det \nolimits _ A(T : eB \to eB) \det \nolimits _ A(T : (1 - e)B \to (1 - e)B) \]

is a factorization into monic polynomials reducing to the given factorization modulo $I$^{1}. Thus (3) implies (1).

Assume (1). Let $f$ be as in (5). The factorization of $f \bmod I$ as $T^ n$ times $T - 1$ lifts to a factorization $f = gh$ with $g$ and $h$ monic by Definition 15.11.1. Then $h$ has to have degree $1$ and we see that $f$ has a root reducing to $1$ modulo $1$. Thus (1) implies (5).

Before we give the proof of the last step, let us show that the root $\alpha $ in (5), if it exists, is unique. Namely, Due to the explicit shape of $f(T)$, we have $f'(\alpha ) \in 1 + I$ where $f'$ is the derivative of $f$ with respect to $T$. An elementary argument shows that

\[ f(T) = f(\alpha + T - \alpha ) = f(\alpha ) + f'(\alpha ) \cdot (T - \alpha ) \bmod (T - \alpha )^2 A[T] \]

This shows that any other root $\alpha ' \in 1 + I$ of $f(T)$ satisfies $0 = f(\alpha ') - f(\alpha ) = (\alpha ' - \alpha )(1 + i)$ for some $i \in I$, so that, since $1 + i$ is a unit in $A$, we have $\alpha = \alpha '$.

Assume (5). We will show that (2) holds, in other words, that for every étale map $A \to A'$, every section $\sigma : A' \to A/I$ modulo $I$ lifts to a section $A' \to A$. Since $A \to A'$ is étale, the section $\sigma $ determines a decomposition

15.11.6.1
\begin{equation} \label{more-algebra-equation-GCHP} A'/IA' \cong A/IA \times C \end{equation}

of $A/IA$-algebras. Namely, the surjective ring map $A'/IA' \to A/I$ is étale by Algebra, Lemma 10.141.8 and then we get the desired idempotent by Algebra, Lemma 10.141.9. We will show that this decomposition lifts to a decomposition

15.11.6.2
\begin{equation} \label{more-algebra-equation-GCHP-want} A' \cong A'_1 \times A'_2 \end{equation}

of $A$-algebras with $A'_1$ integral over $A$. Then $A \to A'_1$ is integral and étale and $A/I \to A'_1/IA'_1$ is an isomorphism, thus $A \to A'_1$ is an isomorphism by Lemma 15.10.3 (here we also use that an étale ring map is flat and of finite presentation, see Algebra, Lemma 10.141.3).

Let $B'$ be the integral closure of $A$ in $A'$. By Lemma 15.11.5 we may decompose

15.11.6.3
\begin{equation} \label{more-algebra-equation-dec-mod-I} B'/IB' \cong A/I \times C' \end{equation}

as $A/IA$-algebras compatibly with (15.11.6.1) and we may find $b \in B'$ that lifts $(1, 0)$ such that $B'_ b \to A'_ b$ is an isomorphism. If the decomposition (15.11.6.3) lifts to a decomposition

15.11.6.4
\begin{equation} \label{more-algebra-equation-want-2} B' \cong B'_1 \times B'_2 \end{equation}

of $A$-algebras, then the induced decomposition $A' = A'_1 \times A'_2$ will give the desired (15.11.6.2): indeed, since $b$ is a unit in $B'_1$ (details omitted), we will have $B'_1 \cong A'_1$, so that $A'_1$ will be integral over $A$.

Choose a finite $A$-subalgebra $B'' \subset B'$ containing $b$ (observe that any finitely generated $A$-subalgebra of $B'$ is finite over $A$). After enlarging $B''$ we may assume $b$ maps to an idempotent in $B''/IB''$ producing

15.11.6.5
\begin{equation} \label{more-algebra-equation-again-dec-mod-I} B''/IB'' \cong C''_1 \times C''_2 \end{equation}

Since $B'_ b \cong A'_ b$ we see that $B'_ b$ is of finite type over $A$. Say $B'_ b$ is generated by $b_1/b^ n, \ldots , b_ t/b^ n$ over $A$ and enlarge $B''$ so that $b_1, \ldots , b_ t \in B''$. Then $B''_ b \to B'_ b$ is surjective as well as injective, hence an isomorphism. In particular, we see that $C''_1 = A/I$! Therefore $A/I \to C''_1$ is an isomorphism, in particular surjective. By Lemma 15.10.4 we can find an $f(T) \in A[T]$ of the form

\[ f(T) = T^ n(T - 1) + a_ n T^ n + \ldots + a_1 T + a_0 \]

with $a_ n, \ldots , a_0 \in I$ and $n \ge 1$ such that $f(b) = 0$. In particular, we find that $B'$ is a $A[T]/(f)$-algebra. By (5) we deduce there is a root $a \in 1 + I$ of $f$. This produces a product decomposition $A[T]/(f) = A[T]/(T - a) \times D$ compatible with the splitting (15.11.6.3) of $B'/IB'$. The induced splitting of $B'$ is then a desired (15.11.6.4).
$\square$

## Comments (4)

Comment #2171 by JuanPablo on

Comment #2200 by Johan on

Comment #3259 by Kestutis Cesnavicius on

Comment #3355 by Johan on

There are also: