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 4.24.5 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

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"


Post a comment

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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

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 input field. As a reminder, this is tag 0A04. Beware of the difference between the letter 'O' and the digit '0'.