Lemma 15.11.9 (0DYD)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.11: Henselian pairs
Lemma 15.11.9 (cite)

Lemma 15.11.9. Let $I \subset J \subset A$ be ideals of a ring $A$. The following are equivalent

$(A, I)$ and $(A/I, J/I)$ are henselian pairs, and
$(A, J)$ is an henselian pair.

Proof. Assume (1). Let $B$ be an integral $A$-algebra. Consider the ring maps

\[ B \to B/IB \to B/JB \]

By Lemma 15.11.6 we find that both arrows induce bijections on idempotents. Hence so does the composition. Whence $(A, J)$ is a henselian pair by Lemma 15.11.6.

Conversely, assume (2) holds. Then $(A/I, J/I)$ is a henselian pair by Lemma 15.11.8. Let $B$ be an integral $A$-algebra. Consider the ring maps

\[ B \to B/IB \to B/JB \]

By Lemma 15.11.6 we find that the composition and the second arrow induce bijections on idempotents. Hence so does the first arrow. It follows that $(A, I)$ is a henselian pair (by the lemma again). $\square$

Comments (8)

Comment #2826 by Kestutis Cesnavicius on September 26, 2017 at 18:53

The assertion may be strengthened to an "if and only if."

Comment #2927 by Johan on October 10, 2017 at 01:41

Thanks for this. Added the change here.

Comment #2980 by Kestutis Cesnavicius on October 30, 2017 at 10:54

Thanks for the change. In the statement of (1), one should have $(A/I, J/I)$ instead of $(J/I, A/I)$ .

Comment #3104 by Johan on February 01, 2018 at 00:13

Thanks, fixed here.

Comment #3636 by Brian Conrad on October 19, 2018 at 23:59

The proof that (1) implies (2) seems to be incomplete as written since to do the second factorization lifting you'd need to know that the lifted monic factorization in $(A/I)[T]$ satisfies the "generate the unit ideal" condition, which is not addressed. But anyway it seems easier to just argue by the same method as in the proof of the converse: using notation as set up there, in the composition $B \to B/IB \to B/JB$ the first and second maps induce bijections on sets of idempotents bu the assumption (1), so the composite is also such a bijection and hence (2) holds (by one of the equivalent characterizations of "henselian pair").

Comment #3734 by Johan on October 23, 2018 at 19:37

Yes, thanks very much. Fixed here.

Comment #5011 by Laurent Moret-Bailly on April 04, 2020 at 09:11

Suggested corollary: If $(A,I)$ and $(A,I')$ are henselian pairs, so is $(A,I+I')$ .

Comment #5249 by Johan on June 15, 2020 at 11:11

Yes, this is a good suggestion! Thanks and I've added it here. Should be online pretty soon.

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