# stacks / stacks-project

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
 @@ -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}