Proof.
Assume (2) holds. Then I is contained in the Jacobson radical of A, since otherwise there would be a nonunit f \in A congruent to 1 modulo 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 10.41.6 and 10.36.22. 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/IB lifts to an idempotent of B 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 generating the unit ideal in A/I[T]. Set B = A[T]/(f). Let \overline{e} be the idempotent of B/IB corresponding to the decomposition
B/IB = A/I[T]/(g_0) \times A/I[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
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
\begin{align*} f & = \text{CharPol}_ A(T : B \to B) \\ & = \text{CharPol}_ A(T : eB \to eB) \text{CharPol}_ A(T : (1 - e)B \to (1 - e)B) \end{align*}
is a factorization into monic polynomials reducing to the given factorization modulo I. Here \text{CharPol}_ A denotes the characteristic polynomial of an endomorphism of a finite locally free module over A. If the module is free the \text{CharPol}_ A is defined as the characteristic polynomial of the corresponding matrix and in general one uses Algebra, Lemma 10.24.2 to glue. Details omitted. 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. Finally, I is contained in the Jacobson radical by the definition of a henselian pair. 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/I \times C \end{equation}
of A/I-algebras. Namely, the surjective ring map A'/IA' \to A/I is étale by Algebra, Lemma 10.143.8 and then we get the desired idempotent by Algebra, Lemma 10.143.9. 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 10.143.3).
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/I-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
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