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Fix erroneous lemma on colimits henselian pairs

Overlooked that the colimit is in the category of henselian pairs

Thanks to Moret-Bailly who found the error, see
https://stacks.math.columbia.edu/tag/0A04#comment-4896
https://stacks.math.columbia.edu/tag/0A04#comment-4897
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aisejohan committed Jan 28, 2020
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  1. +68 −3 more-algebra.tex
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@@ -2846,6 +2846,54 @@ \section{Henselian pairs}
to the effect that $\alpha'$ defines a root of $f(T)$ in $1 + I$, as desired.
\end{proof}

\begin{lemma}
\label{lemma-filtered-colimits-henselian}
The property of being Henselian is preserved under filtered colimits of pairs.
More precisely, let $J$ be a directed set and let $(A_j, I_j)$
be a system of henselian pairs over $J$.
Then $A = \colim A_j$ equipped with the ideal $I = \colim I_j$
is a henselian pair $(A, I)$.
\end{lemma}

\begin{proof}
If $u \in 1 + I$ then for some $j \in J$ we see that $u$ is the
image of some $u_j \in 1 + I_j$. Then $u_j$ is invertible in $A_j$ by
Algebra, Lemma \ref{algebra-lemma-contained-in-radical}
and the assumption that $I_j$ is contained in the Jacobson radical of $A_j$.
Hence $u$ is invertible in $A$. Thus $I$
is contained in the Jacobson radical of $A$ (by the lemma).

\medskip\noindent
Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0 h_0$
be a factorization with $g_0, h_0 \in A/I[T]$ monic generating the unit
ideal in $A/I[T]$. Write $1 = g_0 g'_0 + h_0 h'_0$ for some
$g'_0, h'_0 \in A/I[T]$. Since $A = \colim A_j$ and $A/I = \colim A_j/I_j$
are filterd colimits we can find a $j \in J$ and $f_j \in A_j$ and
a factorization $\overline{f}_j = g_{j, 0} h_{j, 0}$
with $g_{j, 0}, h_{j, 0} \in A_j/I_j[T]$ monic
and $1 = g_{j, 0} g'_{j, 0} + h_{j, 0} h'_{j, 0}$
for some $g'_{j, 0}, h'_{j, 0} \in A_j/I_j[T]$ with
$f_j, g_{j, 0}, h_{j, 0}, g'_{j, 0}, h'_{j, 0}$
mapping to $f, g_0, h_0, g'_0, h'_0$.
Since $(A_j, I_j)$ is a henselian pair, we can lift
$\overline{f}_j = g_{j, 0} h_{j, 0}$ to a factorization
over $A_j$ and taking the image in $A$ we obtain a
corresponding factorization in $A$. Hence $(A, I)$ is henselian.
\end{proof}

\begin{example}[Moret-Bailly]
\label{example-moret-bailly}
Lemma \ref{lemma-filtered-colimits-henselian} is wrong if the colimit
isn't filtered. For example, if we take the coproduct of the
henselian pairs $(\mathbf{Z}_p, (p))$ and $(\mathbf{Z}_p, (p))$, then we
obtain $(A, pA)$ with $A = \mathbf{Z}_p \otimes_\mathbf{Z} \mathbf{Z}_p$.
This isn't a henslian pair: $A/pA = \mathbf{F}_p$ hence if $(A, pA)$
where henselian, then $A$ would have to be local. However, $\Spec(A)$
is disconnected; for example for odd primes $p$ we have the nontrivial
idempotent $(1 + p)^{-1}u \otimes u$ where $u \in \mathbf{Z}_p$ is a square
root of $1 + p$. Some details omitted.
\end{example}

\begin{lemma}
\label{lemma-irreducible-henselian-pair-connected}
Let $(A, I)$ be a henselian pair. Let $\mathfrak p \subset A$
@@ -3081,14 +3129,31 @@ \section{Henselization of pairs}

\begin{lemma}
\label{lemma-henselization-colimit}
Let $(A, I) = \colim (A_i, I_i)$ be a colimit of pairs. The functor of
Let $(A, I) = \colim (A_i, I_i)$ be a filtered colimit of pairs. The functor of
Lemma \ref{lemma-henselization} gives
$A^h = \colim A_i^h$ and $I^h = \colim I_i^h$.
\end{lemma}

\noindent
This lemma is false for non-filtered colimits, see
Example \ref{example-moret-bailly}.

\begin{proof}
This is true for any left adjoint, see
Categories, Lemma \ref{categories-lemma-adjoint-exact}.
By Categories, Lemma \ref{categories-lemma-adjoint-exact}
we see that $(A^h, I^h)$ is the colimit of the system $(A_i^h, I_i^h)$
in the category of henselian pairs. Thus for a henselian pair $(B, J)$
we have
$$
\Mor((A^h, I^h), (B, J)) =
\lim \Mor((A_i^h, I_i^h), (B, J)) =
\Mor(\colim (A_i^h, I_i^h), (B, J))
$$
Here the colimit is in the category of pairs. Since the colimit is
filtered we obtain $\colim (A_i^h, I_i^h) = (\colim A_i^h, \colim I_i^h)$
in the category of pairs; details omitted. Again using the colimit is filtered,
this is a henselian pair (Lemma \ref{lemma-filtered-colimits-henselian}).
Hence by the Yoneda lemma we
find $(A^h, I^h) = (\colim A_i^h, \colim I_i^h)$.
\end{proof}

\begin{lemma}
@@ -19855,3 +19855,5 @@
0FWQ,exercises-exercise-dimension
0FWR,exercises-exercise-tor-computation
0FWS,exercises-exercise-depth-goes-up
0FWT,more-algebra-lemma-filtered-colimits-henselian
0FWU,more-algebra-example-moret-bailly

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