## Tag `09XI`

Chapter 15: More on Algebra > Section 15.10: Henselian pairs

Lemma 15.10.8. Let $(A, I)$ be a pair. The following are equivalent

- $(A, I)$ is a henselian pair,
- 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 $\sigma$,
- for any finite $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents, and
- for any integral $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents.

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 15.10.5. On the other hand, since (2) holds, every idempotent of $B$ lifts to an idempotent of $B/IB$ by Lemma 15.9.9. 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 $A \to A'$ be an étale ring map and let $\sigma : A' \to A/I$ be an $A$-algebra map. This implies that $A'/IA' = A/I \times C$ for some ring $C$. Let $A'' \subset A'$ be the integral closure of $A$ in $A'$. By Lemma 15.10.7 we can write $A''/IA'' = A/I \times C'$ such that $A''/IA'' \to A'/IA'$ maps $A/I$ isomorphically to $A'/IA'$ and $C'$ to $C$ and such that there exists a $a \in A''$ mapping to $(1, 0)$ in $A/I \times C'$ such that $A''_a \cong A'_a$. By Lemma 15.10.6 we see that $1 - a$ satisfies a monic polynomial $f \in A[T]$ with $f \bmod I = g_0 T^n$ where $T, g_0$ generate the unit ideal in $A/I[T]$. As we assumed $(A, I)$ is a henselian pair we can factor $f$ as $f = g h$ where $g$ is a monic lift of $g_0$ and $h$ is a monic lift of $T^n$. Because $I$ is contained in the Jacobson radical of $A$, we find that $g$ and $h$ generate the unit ideal in $A[T]$ (details omitted; hint: use that $A[T]/(g, h)$ is finite over $A$). Thus $A[T]/(f) = A[T]/(h) \times A[T]/(g)$. It follows that $h(1 - a)$ and $g(1 - a)$ generate the unit ideal in $A''$ and we find a product decomposition $A'' = A''_1 \times A''_2$ such that $h(1 - a)$ acts as zero on $A''_1$ and $g(1 - a)$ acts as zero on $A''_2$. As $h \bmod I = T^n$, as $T, g_0$ generate the unit ideal, and as $a \bmod I = (1, 0)$ we conclude that the two disjoint union decompositions $$ \mathop{\mathrm{Spec}}(A''/IA'') = \mathop{\mathrm{Spec}}(A/I) \amalg \mathop{\mathrm{Spec}}(C') = \mathop{\mathrm{Spec}}(A''_1/IA''_1) \amalg \mathop{\mathrm{Spec}}(A''_2/IA''_2) $$ are the same: in both cases the first summand corresponds exactly to those primes of $A''$ containing $IA''$ such that $a$ maps to $1$ in the residue field. This implies that $A''_1/IA''_1 = A/I$ as factor rings of $A''/IA''$ (because product decompositions of rings correspond $1$-to-$1$ to disjoint union decompositions of the spectra, see Algebra, Lemmas 10.20.2 and 10.22.3). Moreover, it follows that $a$ maps to a unit in $A''_1$ (see Algebra, Lemma 10.18.1 and use that $IA''_1$ is contained in the radical of $A''_1$ as $A''_1$ is integral over $A$). Hence $A''_1 = (A''_1)_a$ is a factor of $A''_a = A'_a$ which is étale over $A$. Then $A \to A''_1$ is integral, of finite presentation, and flat (Algebra, Section 10.141) hence finite (Algebra, Lemma 10.35.5) hence finitely presented as an $A$-module (Algebra, Lemma 10.35.23) hence $A''_1$ is finite projective as an $A$-module (Algebra, Lemma 10.77.2). Since $A''_1/IA''_1 = A/I$ the module $A''_1$ has rank $1$ as an along $V(I)$ (see rank function described in Algebra, Lemma 10.77.2). Since $I$ is contained in the Jacobson radical of $A$ we conclude that $A''_1$ has rank $1$ everywhere. It follows that $A \to A''_1$ is an isomorphism (exercise: a ring map $R \to S$ such that $S$ is a locally free $R$-module of rank $1$ is an isomorphism). Thus $A' \to A'_a \cong A''_a \to (A''_1)_a = A''_1 = A$ is the desired lift of $\sigma$. In this way we see that (1) implies (2). $\square$

The code snippet corresponding to this tag is a part of the file `more-algebra.tex` and is located in lines 2377–2390 (see updates for more information).

```
\begin{lemma}
\label{lemma-characterize-henselian-pair}
Let $(A, I)$ be a pair. The following are equivalent
\begin{enumerate}
\item $(A, I)$ is a henselian pair,
\item given an \'etale 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 $\sigma$,
\item for any finite $A$-algebra $B$ the map $B \to B/IB$ induces
a bijection on idempotents, and
\item for any integral $A$-algebra $B$ the map $B \to B/IB$ induces
a bijection on idempotents.
\end{enumerate}
\end{lemma}
\begin{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 $\Spec(B) \to \Spec(A)$ is closed by
Algebra, Lemmas \ref{algebra-lemma-going-up-closed} and
\ref{algebra-lemma-integral-going-up}.
Thus the map from idempotents of $B$ to idempotents of $B/IB$
is injective by Lemma \ref{lemma-idempotents-determined-modulo-radical}.
On the other hand, since (2) holds, every idempotent
of $B$ lifts to an idempotent of $B/IB$
by Lemma \ref{lemma-lift-idempotent-upstairs}.
In this way we see that (2) implies (4).
\medskip\noindent
The implication (4) $\Rightarrow$ (3) is trivial.
\medskip\noindent
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 $\Spec(A)$. 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 $I$\footnote{Here we use determinants of endomorphisms
of finite locally free modules; if the module is free the determinant
is defined as the determinant of the corresponding matrix and in
general one uses Algebra, Lemma \ref{algebra-lemma-standard-covering} to
glue.}.
Thus (3) implies (1).
\medskip\noindent
Assume (1). Let $A \to A'$ be an \'etale ring map and let
$\sigma : A' \to A/I$ be an $A$-algebra map. This implies that
$A'/IA' = A/I \times C$ for some ring $C$. Let $A'' \subset A'$ be the integral
closure of $A$ in $A'$. By Lemma \ref{lemma-helper-finite-type}
we can write $A''/IA'' = A/I \times C'$ such that $A''/IA'' \to A'/IA'$
maps $A/I$ isomorphically to $A'/IA'$ and $C'$ to $C$ and such that
there exists a $a \in A''$ mapping to $(1, 0)$ in $A/I \times C'$
such that $A''_a \cong A'_a$.
By Lemma \ref{lemma-helper-integral}
we see that $1 - a$ satisfies a monic polynomial $f \in A[T]$
with $f \bmod I = g_0 T^n$ where
$T, g_0$ generate the unit ideal in $A/I[T]$. As we assumed
$(A, I)$ is a henselian pair we can factor $f$ as $f = g h$
where $g$ is a monic lift of $g_0$ and $h$ is a monic lift of $T^n$.
Because $I$ is contained in the Jacobson radical
of $A$, we find that $g$ and $h$ generate the unit ideal in $A[T]$
(details omitted; hint: use that $A[T]/(g, h)$ is finite over $A$).
Thus $A[T]/(f) = A[T]/(h) \times A[T]/(g)$. It follows that
$h(1 - a)$ and $g(1 - a)$ generate the unit ideal in $A''$ and
we find a product decomposition $A'' = A''_1 \times A''_2$
such that $h(1 - a)$ acts as zero on $A''_1$ and $g(1 - a)$ acts as zero on
$A''_2$. As $h \bmod I = T^n$, as $T, g_0$ generate the unit ideal, and
as $a \bmod I = (1, 0)$ we conclude that the two disjoint union decompositions
$$
\Spec(A''/IA'') = \Spec(A/I) \amalg \Spec(C') =
\Spec(A''_1/IA''_1) \amalg \Spec(A''_2/IA''_2)
$$
are the same: in both cases the first summand corresponds exactly
to those primes of $A''$ containing $IA''$ such that $a$ maps to $1$
in the residue field. This implies that $A''_1/IA''_1 = A/I$
as factor rings of $A''/IA''$ (because product decompositions of
rings correspond $1$-to-$1$ to disjoint union decompositions
of the spectra, see Algebra, Lemmas \ref{algebra-lemma-spec-product} and
\ref{algebra-lemma-disjoint-implies-product}). Moreover, it
follows that $a$ maps to a unit in $A''_1$
(see Algebra, Lemma \ref{algebra-lemma-contained-in-radical}
and use that $IA''_1$ is contained in the radical of $A''_1$
as $A''_1$ is integral over $A$).
Hence $A''_1 = (A''_1)_a$ is a factor of $A''_a = A'_a$ which is
\'etale over $A$. Then $A \to A''_1$ is integral, of finite presentation,
and flat (Algebra, Section \ref{algebra-section-etale}) hence finite
(Algebra, Lemma \ref{algebra-lemma-characterize-finite-in-terms-of-integral})
hence finitely presented as an $A$-module
(Algebra, Lemma \ref{algebra-lemma-finite-finitely-presented-extension})
hence $A''_1$ is finite projective as an $A$-module
(Algebra, Lemma \ref{algebra-lemma-finite-projective}).
Since $A''_1/IA''_1 = A/I$ the module $A''_1$ has rank $1$ as an
along $V(I)$ (see rank function described in
Algebra, Lemma \ref{algebra-lemma-finite-projective}).
Since $I$ is contained in the Jacobson radical
of $A$ we conclude that $A''_1$ has rank $1$ everywhere.
It follows that $A \to A''_1$ is an isomorphism
(exercise: a ring map $R \to S$ such that $S$ is a locally free $R$-module
of rank $1$ is an isomorphism). Thus
$A' \to A'_a \cong A''_a \to (A''_1)_a = A''_1 = A$ is the
desired lift of $\sigma$. In this way we see that (1) implies (2).
\end{proof}
```

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