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

1. $(A, I)$ and $(A/I, J/I)$ are henselian pairs, and

2. $(A, J)$ 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). $\square$

Comment #2826 by Kestutis Cesnavicius on

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

Comment #2980 by Kestutis Cesnavicius on

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

Comment #3636 by Brian Conrad on

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 #5011 by Laurent Moret-Bailly on

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

Comment #5249 by on

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

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