[Chapter XI, Henselian] and [Proposition 1, Gabber-henselian]

Lemma 15.11.6. Let $(A, I)$ be a pair. The following are equivalent

1. $(A, I)$ is a henselian pair,

2. given an étale ring map $A \to A'$ and an $A$-algebra map $\sigma : A' \to A/I$, there exists an $A$-algebra map $A' \to A$ lifting $\sigma$,

3. for any finite $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents,

4. for any integral $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents, and

5. (Gabber) $I$ is contained in the Jacobson radical of $A$ and every monic polynomial $f(T) \in A[T]$ 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 $n \ge 1$ has a root $\alpha \in 1 + I$.

Moreover, in part (5) the root is unique.

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. Set $B = A[T]/(f)$. Let $\overline{e}$ be the idempotent of $B/IB$ corresponding to the decomposition

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

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

\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). $\square$

Comment #2171 by JuanPablo on

There is a couple of points that could be clearer. I don't remember a definition of determinant for endomorphisms of locally finite modules of constant rank here, so maybe it could be said that the definition comes from lemma 10.22.1 (tag 00EJ).

"we see that $a$ satisfies a monic polynomial $f \in A[T]$ whose reduction modulo $I$ factors as $\overline{f} = g_0 T^n$ where $T, g_0$ generate the unit ideal in $A/I[T]$."

Here lemma 15.8.5 (tag 09XG) is used, but it seems in this case the lemma applies to an element corresponding to $(0,1)$ not $(1,0)$ so instead $\overline{f}=g_0(1-T)^n$ applying lemma 15.8.5 to $1-a$.

"By construction we have $A''_1/IA''_1 = A/I$ and $A''_2/IA''_2 = C'$"

I did not understand why this is so. I have an argument, but maybe I'm missing something simpler. This is because the idempotent corresponding to $A_2''$ in $A''$ is $b=h(a)s(a)$ where $hs+gr=1$ in $A[T]$, so reducing modulo $I$ you get $b=(1-a)^ns(a)$ modulo $I$. Now using that $a$ is an idempotent modulo $I$, you get $b=(1-a)(s(0)+(s(1)-s(0))a)=(1-a)s(0)$ modulo $I$. And using that, modulo $I$, $f(a)=(1-a)^ng_0(a)=(1-a)g_0(0)+(g(1)-g(0))a)=(1-a)g_0(0)=0$, obtain from $hs+gr=1$ evaluating in $0$ and multiplying by $1-a$ that $(1-a)s(0)=(1-a)$ (modulo $I$). So the idempotent corresponding to $A_2''/IA_2''$ is $1-a=(0,1)$ as required.

"It follows that $A \to A''_1$ is integral as well as \'etale, hence finite locally free. However, $A''_1/IA''_1 = A/I$ thus $A''_1$ has rank $1$ as an $A$-module along $V(I)$. Since $I$ is contained in the Jacobson radical of $A$ we conclude that $A''_1$ has rank $1$ everywhere and it follows that $A \to A''_1$ is an isomorphism."

I don't understand why integral and étale implies finite locally free, or why locally finite of rank 1 implies that $A \to A''_1$ is an isomorphism. I could see that $A \to A''_1$ is an isomorphism using Nakayama's lemma 10.19.1 (tag 00DV) to see that $A \to A''_1$ is onto, then lemma 10.141.9 (tag 00U8) to see that $A \to A''_1$ is projection into a factor, and then lemma 15.8.4 (tag 09XF) to see from $A/I=A''_1/IA''_1$ that the idempotent in $A$ associated to $A''_1$ is $1$ (so $A \to A''_1$ is an isomorphism).

Comment #2200 by on

OK, this proof is a bit of a mess. I've fixed the error of applying Lemma 15.9.9 wrong and I've fixed the other arguments in a way a bit more in line with the original intent of the proof. I wanted to use the helper lemma 15.9.9, but maybe there is a more straightforward proof?

In any case, many thanks! Edits can be found here but should be online soon.

Comment #3259 by Kestutis Cesnavicius on

In the last paragraph of the proof "maps $A/I$ isomorphically to $A'/IA'$" should be "maps $A/IA$ isomorphically to $A/IA$".

Comment #3635 by Brian Conrad on

In the statement of (5), replace "radical" with "Jacobson radical" (a property used in the proof that (5) implies (2), as it must be, via the fact that $A\to A'_1$ is integral etale and an isomorphism modulo $I$ forcing its locally constant fiber degree to be constant due to $I$ being in the Jacobson radical).

You can replace $A/IA$ with $A/I$ throughout.

In the first paragraph of the proof, replace "not contained in $I$" with "which is congruent to 1 modulo $I$", and near the end of it replace "idempotent of $B$ lifts to an idempotent of $B/IB$" with "idempotent of $B/IB$ lifts to an idempotent of $B$".

In the proof of $(3) \Rightarrow (1)$, replace "$\overline{e}$ be the nontrivial idempotent" with "$\mathbf{e}$ be the idempotent" (maybe $g_0=1$ or $h_0=1$), and near the end of that paragraph the displayed expression for factoring $f$ makes no sense as written: you mean to speak in terms of characteristic polynomials of linear endomorphisms of a finite locally free $A$-module of constant rank, not determinants as written (which are elements of $A$!).

In the proof that $(1) \Rightarrow (5)$, you should mention that $I$ is contained in the Jacobson radical of $A$ by the definition of "henselian pair".

Comment #4053 by Matthieu Romagny on

In Condition (5) of the statement of the Lemma: has a root

Comment #5013 by Laurent Moret-Bailly on

In the references, the link to "Gabber-henselian" doesn't work.

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