The Stacks Project


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

  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, and
  4. 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). $\square$

  1. 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 10.22.1 to glue.

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}

Comments (3)

Comment #2171 by JuanPablo on August 21, 2016 a 8:21 pm UTC

There is a couple of points that could be clearer. $$ f = \det\nolimits_A(T : B \to B) = \det\nolimits_A(T : eB \to eB) \det\nolimits_A(T : (1 - e)B \to (1 - e)B) $$ 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 ├ętale 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 Johan (site) on August 29, 2016 a 4:50 pm UTC

OK, this proof is a bit of a mess. I've fixed the error of applying Lemma 09XG 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 09XG, but maybe there is a more straightforward proof?

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

Comment #3275 by Kestutis Cesnavicius on April 11, 2018 a 8:58 pm UTC

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

There are also 2 comments on Section 15.10: More on Algebra.

Add a comment on tag 09XI

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

This captcha seems more appropriate than the usual illegible gibberish, right?