Henselization commutes with integral base change

Lemma 15.12.7. Let $(A, I) \to (B, J)$ be a map of pairs such that $V(J) = V(IB)$. Let $(A^ h , I^ h) \to (B^ h, J^ h)$ be the induced map on henselizations (Lemma 15.12.1). If $A \to B$ is integral, then the induced map $A^ h \otimes _ A B \to B^ h$ is an isomorphism.

Proof. By Lemma 15.12.6 we may assume $J = IB$. By Lemma 15.11.8 the pair $(A^ h \otimes _ A B, I^ h(A^ h \otimes _ A B))$ is henselian. By the universal property of $(B^ h, IB^ h)$ we obtain a map $B^ h \to A^ h \otimes _ A B$. We omit the proof that this map is the inverse of the map in the lemma. $\square$

Comment #3644 by Brian Conrad on

At the start of the proof, the wrong lemma is cited: should cite Lemma 09XK. In the second sentence, the invocation of the universal property doesn't make any sense unless $J$ lands in $I^h(A^h \otimes_A B) = I(A^h \otimes_A B)$, which would hold if $J=IB$. But there is the flexibility to replace $J$ with any bigger ideal of $B$ without affecting the henselian hypothesis on $(B, J)$, so you should really assume $V(J)=V(IB)$: this would ensure $(B,IB)$ is also henselian by Lemma 09XJ and would ensure that when $A$ is local with maximal ideal $I$ we can take $J$ to be the Jacobson radical of $B$ (rather than demanding $J=IB$, which would be an unpleasant requirement for such cases).

Comment #3645 by Brian Conrad on

I should have also mentioned that currently this Lemma is only mentioned in two proofs, neither of which is impacted by imposing the requirement $V(J)=V(IB)$ (though the second proof which claims to invoke this Lemma doesn't really clearly indicate where it is used -- that proof says this Lemma is used, but never explicitly invokes it with a cross-reference at any step).

Comment #3832 by slogan_bot on

Suggested slogan: "Henselization commutes with base change along integral maps"

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