Lemma 15.12.5 (0A04)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.12: Henselization of pairs
Lemma 15.12.5 (cite)

Lemma 15.12.5. Let $(A, I) = \mathop{\mathrm{colim}}\nolimits (A_ i, I_ i)$ be a filtered colimit of pairs. The functor of Lemma 15.12.1 gives $A^ h = \mathop{\mathrm{colim}}\nolimits A_ i^ h$ and $I^ h = \mathop{\mathrm{colim}}\nolimits I_ i^ h$.

Proof. By Categories, Lemma 4.24.5 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

\[ \mathop{\mathrm{Mor}}\nolimits ((A^ h, I^ h), (B, J)) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Mor}}\nolimits ((A_ i^ h, I_ i^ h), (B, J)) = \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{colim}}\nolimits (A_ i^ h, I_ i^ h), (B, J)) \]

Here the colimit is in the category of pairs. Since the colimit is filtered we obtain $\mathop{\mathrm{colim}}\nolimits (A_ i^ h, I_ i^ h) = (\mathop{\mathrm{colim}}\nolimits A_ i^ h, \mathop{\mathrm{colim}}\nolimits I_ i^ h)$ in the category of pairs; details omitted. Again using the colimit is filtered, this is a henselian pair (Lemma 15.11.13). Hence by the Yoneda lemma we find $(A^ h, I^ h) = (\mathop{\mathrm{colim}}\nolimits A_ i^ h, \mathop{\mathrm{colim}}\nolimits I_ i^ h)$. $\square$

Comments (9)

Comment #4896 by Laurent Moret-Bailly on January 27, 2020 at 12:47

Lemma 0038 says that $(A,I)^h$ is the colimit of the system $(A_i^h,I_i^h)$ in the category of henselian pairs. This colimit is (looking at the universal properties) the henselization of $(\operatorname{colim} A_i^h, \operatorname{colim} I_i^h)$ . The latter is henselian if the colimit is filtered, but in general this is not clear to me.

Comment #4897 by Laurent Moret-Bailly on January 27, 2020 at 15:04

Follow-up to Comment #4896: consider for instance, for an odd prime $p$ , the coproduct of $(\mathbb{Z}_p,(p))$ with itself: in the category of pairs, this is $(A,(p))$ with $A:=\mathbb{Z}_p \otimes \mathbb{Z}_p$ ; however $A/(p)=\mathbb{F}_p$ and $A$ is not connected, so $(A,I)$ is not even a Zariski pair. (To see that $A$ is not connected, let $t:=\sqrt{1+p}\in 1+p\mathbb{Z}_p$ . Then $\mathbb{Z}_p$ is faithfully flat over $A_0:=\mathbb{Z}[t]\left[\frac{1}{t+1}\right]$ , hence $A$ is faithfully flat over $A_0\otimes_\mathbb{Z} A_0$ which is not connected.)

Comment #4901 by Johan on January 28, 2020 at 10:07

Good catch! I will fix this later today or tomorrow. Thanks very much!

Comment #4902 by Johan on January 28, 2020 at 16:52

OK, this is now done. See changes here.

Comment #5019 by 羽山籍真 on April 09, 2020 at 12:33

In the comment 4897, if the prime $p=3$ , then $t=2$ . So $A_0=\mathbb{Z}[t]\left[\frac{1}{1+t} \right]=\mathbb{Z}[2]\left[\frac{1}{3}\right]=\mathbb{Z}\left[\frac{1}{3}\right]$ . So $A_0\rightarrow \mathbb{Z}_3$ is $\mathbb{Z}\left[\frac{1}{3}\right]\rightarrow \mathbb{Z}_3$ , which is problematic because the image $x$ of $\frac{1}{3}$ should satisfy $3\cdot x=1$ , but such $x$ does not exist in $\mathbb{Z}_3$ .

Comment #5020 by Laurent Moret-Bailly on April 09, 2020 at 15:15

@ 羽山籍真: No, $t$ is the square root of $1+p$ which is $\equiv1$ mod $p$ , so in this case it is $-2$ , not $2$ , and $A_0=\mathbb{Z}$ .

Comment #5021 by 羽山籍真 on April 09, 2020 at 15:46

@Laurent Moret-Bailly Thanks. But in this case, $A_0\rightarrow \mathbb{Z}_p$ is $\mathbb{Z}\rightarrow \mathbb{Z}_3$ , which is not faithfully flat.

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