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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. $\square$

Comments (4)

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 lands in , which would hold if . But there is the flexibility to replace with any bigger ideal of without affecting the henselian hypothesis on , so you should really assume : this would ensure is also henselian by Lemma 09XJ and would ensure that when is local with maximal ideal we can take to be the Jacobson radical of (rather than demanding , 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 (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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