Compatibility henselization of pairs and of local rings.

Lemma 15.12.3. The functor of Lemma 15.12.1 associates to a local ring $(A, \mathfrak m)$ its henselization.

Proof. Let $(A^ h, \mathfrak m^ h)$ be the henselization of the pair $(A, \mathfrak m)$ constructed in Lemma 15.12.1. Then $\mathfrak m^ h = \mathfrak m A^ h$ is a maximal ideal by Lemma 15.12.2 and since it is contained in the Jacobson radical, we conclude $A^ h$ is local with maximal ideal $\mathfrak m^ h$. Having said this there are two ways to finish the proof.

First proof: observe that the construction in the proof of Algebra, Lemma 10.154.1 as a colimit is the same as the colimit used to construct $A^ h$ in Lemma 15.12.1. Second proof: Both the henselization $A \to S$ and $A \to A^ h$ of Lemma 15.12.1 are local ring homomorphisms, both $S$ and $A^ h$ are filtered colimits of étale $A$-algebras, both $S$ and $A^ h$ are henselian local rings, and both $S$ and $A^ h$ have residue fields equal to $\kappa (\mathfrak m)$ (by Lemma 15.12.2 for the second case). Hence they are canonically isomorphic by Algebra, Lemma 10.153.6. $\square$

Comment #988 by on

Suggested slogan: Henselization is left adjoint to the inclusion of Henselian rings in rings.

Comment #3643 by Brian Conrad on

In the second proof, which at the end invokes a lemma that has locality hypotheses, you need to first indicate why $A^h$ is local. One way based on general principles beyond the local setting is to recall that for any pair $(B,J)$ the ideal $J^h$ of $B^h$ is contained in the Jacobson radical (by the henselian property for the pair $(B^h, J^h)$) and by construction $J^h = JB^h$ yet always $B^h/JA^h = B/J$ (so when $B$ is local with maximal ideal $J$, the ideal $J^h$ is maximal and so is the unique maximal ideal of $J^h$, moreover with the same residue field as $B$).

Note that in this argument, we are using two facts currently recorded in Lemma 0AGU, which currently appears just after the present Lemma in section 0EM7 (no circularity, since Lemma 0AGU doesn't use the present lemma, so their order should be swapped).

Also, this shows that before swapping the order of the two lemmas, the "first proof" isn't entirely satisfactory since it isn't giving that $\mathfrak{m}^h$ is also the maximal ideal (which would however be "known" if Lemma 0AGU were put before the present lemma, as recommended above).

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