The Stacks project

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 4.24.6 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

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

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 15.11.12). 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

Lemma 0038 says that is the colimit of the system in the category of henselian pairs. This colimit is (looking at the universal properties) the henselization of . 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

Follow-up to Comment #4896: consider for instance, for an odd prime , the coproduct of with itself: in the category of pairs, this is with ; however and is not connected, so is not even a Zariski pair. (To see that is not connected, let . Then is faithfully flat over , hence is faithfully flat over which is not connected.)

Comment #4901 by on

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

Comment #5019 by 羽山籍真 on

In the comment 4897, if the prime , then . So . So is , which is problematic because the image of should satisfy , but such does not exist in .

Comment #5020 by Laurent Moret-Bailly on

@ 羽山籍真: No, is the square root of which is mod , so in this case it is , not , and .

Comment #5021 by 羽山籍真 on

@Laurent Moret-Bailly Thanks. But in this case, is , which is not faithfully flat.

Comment #5035 by slogan_bot on

Suggested slogan: "Henselization of pairs commutes with filtered colimits"


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