Lemma 15.12.1. The inclusion functor

has a left adjoint $(A, I) \mapsto (A^ h, I^ h)$.

We continue the discussion started in Section 15.11.

Lemma 15.12.1. The inclusion functor

\[ \text{category of henselian pairs} \longrightarrow \text{category of pairs} \]

has a left adjoint $(A, I) \mapsto (A^ h, I^ h)$.

**Proof.**
Let $(A, I)$ be a pair. Consider the category $\mathcal{C}$ consisting of étale ring maps $A \to B$ such that $A/I \to B/IB$ is an isomorphism. We will show that the category $\mathcal{C}$ is directed and that $A^ h = \mathop{\mathrm{colim}}\nolimits _{B \in \mathcal{C}} B$ with ideal $I^ h = IA^ h$ gives the desired adjoint.

We first prove that $\mathcal{C}$ is directed (Categories, Definition 4.19.1). It is nonempty because $\text{id} : A \to A$ is an object. If $B$ and $B'$ are two objects of $\mathcal{C}$, then $B'' = B \otimes _ A B'$ is an object of $\mathcal{C}$ (use Algebra, Lemma 10.141.3) and there are morphisms $B \to B''$ and $B' \to B''$. Suppose that $f, g : B \to B'$ are two maps between objects of $\mathcal{C}$. Then a coequalizer is

\[ (B' \otimes _{f, B, g} B') \otimes _{(B' \otimes _ A B')} B' \]

which is étale over $A$ by Algebra, Lemmas 10.141.3 and 10.141.8. Thus the category $\mathcal{C}$ is directed.

Since $B/IB = A/I$ for all objects $B$ of $\mathcal{C}$ we see that $A^ h/I^ h = A^ h/IA^ h = \mathop{\mathrm{colim}}\nolimits B/IB = \mathop{\mathrm{colim}}\nolimits A/I = A/I$.

Next, we show that $A^ h = \mathop{\mathrm{colim}}\nolimits _{B \in \mathcal{C}} B$ with $I^ h = IA^ h$ is a henselian pair. To do this we will verify condition (2) of Lemma 15.11.6. Namely, suppose given an étale ring map $A^ h \to A'$ and $A^ h$-algebra map $\sigma : A' \to A^ h/I^ h$. Then there exists a $B \in \mathcal{C}$ and an étale ring map $B \to B'$ such that $A' = B' \otimes _ B A^ h$. See Algebra, Lemma 10.141.3. Since $A^ h/I^ h = A/IB$, the map $\sigma $ induces an $A$-algebra map $s : B' \to A/I$. Then $B'/IB' = A/I \times C$ as $A/I$-algebra, where $C$ is the kernel of the map $B'/IB' \to A/I$ induced by $s$. Let $g \in B'$ map to $(1, 0) \in A/I \times C$. Then $B \to B'_ g$ is étale and $A/I \to B'_ g/IB'_ g$ is an isomorphism, i.e., $B'_ g$ is an object of $\mathcal{C}$. Thus we obtain a canonical map $B'_ g \to A^ h$ such that

\[ \vcenter { \xymatrix{ B'_ g \ar[r] & A^ h \\ B \ar[u] \ar[ur] } } \quad \text{and}\quad \vcenter { \xymatrix{ B' \ar[r] \ar[rrd]_ s & B'_ g \ar[r] & A^ h \ar[d] \\ & & A/I } } \]

commute. This induces a map $A' = B' \otimes _ B A^ h \to A^ h$ compatible with $\sigma $ as desired.

Let $(A, I) \to (A', I')$ be a morphism of pairs with $(A', I')$ henselian. We will show there is a unique factorization $A \to A^ h \to A'$ which will finish the proof. Namely, for each $A \to B$ in $\mathcal{C}$ the ring map $A' \to B' = A' \otimes _ A B$ is étale and induces an isomorphism $A'/I' \to B'/I'B'$. Hence there is a section $\sigma _ B : B' \to A'$ by Lemma 15.11.6. Given a morphism $B_1 \to B_2$ in $\mathcal{C}$ we claim the diagram

\[ \xymatrix{ B'_1 \ar[rr] \ar[rd]_{\sigma _{B_1}} & & B'_2 \ar[ld]^{\sigma _{B_2}} \\ & A' } \]

commutes. This follows once we prove that for every $B$ in $\mathcal{C}$ the section $\sigma _ B$ is the unique $A'$-algebra map $B' \to A'$. We have $B' \otimes _{A'} B' = B' \times R$ for some ring $R$, see Algebra, Lemma 10.147.4. In our case $R/I'R = 0$ as $B'/I'B' = A'/I'$. Thus given two $A'$-algebra maps $\sigma _ B, \sigma _ B' : B' \to A'$ then $e = (\sigma _ B \otimes \sigma _ B')(0, 1) \in A'$ is an idempotent contained in $I'$. We conclude that $e = 0$ by Lemma 15.10.2. Hence $\sigma _ B = \sigma _ B'$ as desired. Using the commutativity we obtain

\[ A^ h = \mathop{\mathrm{colim}}\nolimits _{B \in \mathcal{C}} B \to \mathop{\mathrm{colim}}\nolimits _{B \in \mathcal{C}} A' \otimes _ A B \xrightarrow {\mathop{\mathrm{colim}}\nolimits \sigma _ B} A' \]

as desired. The uniqueness of the maps $\sigma _ B$ also guarantees that this map is unique. Hence $(A, I) \mapsto (A^ h, I^ h)$ is the desired adjoint. $\square$

Lemma 15.12.2. Let $(A, I)$ be a pair. Let $(A^ h, I^ h)$ be as in Lemma 15.12.1. Then $A \to A^ h$ is flat, $I^ h = IA^ h$ and $A/I^ n \to A^ h/I^ nA^ h$ is an isomorphism for all $n$.

**Proof.**
In the proof of Lemma 15.12.1 we have seen that $A^ h$ is a filtered colimit of étale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism and we have seen that $I^ h = IA^ h$. As an étale ring map is flat (Algebra, Lemma 10.141.3) we conclude that $A \to A^ h$ is flat by Algebra, Lemma 10.38.3. Since each $A \to B$ is flat we find that the maps $A/I^ n \to B/I^ nB$ are isomorphisms as well (for example by Algebra, Lemma 10.100.3). Taking the colimit we find that $A/I^ n = A^ h/I^ nA^ h$ as desired.
$\square$

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 10.150.1 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 10.149.6. $\square$

Lemma 15.12.4. Let $(A, I)$ be a pair with $A$ Noetherian. Let $(A^ h, I^ h)$ be as in Lemma 15.12.1. Then the map of $I$-adic completions

\[ A^\wedge \to (A^ h)^\wedge \]

is an isomorphism. Moreover, $A^ h$ is Noetherian, the maps $A \to A^ h \to A^\wedge $ are flat, and $A^ h \to A^\wedge $ is faithfully flat.

**Proof.**
The first statement is an immediate consequence of Lemma 15.12.2 and in fact holds without assuming $A$ is Noetherian. In the proof of Lemma 15.12.1 we have seen that $A^ h$ is a filtered colimit of étale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism. For each such $A \to B$ the induced map $A^\wedge \to B^\wedge $ is an isomorphism (see proof of Lemma 15.12.2). By Algebra, Lemma 10.96.2 the ring map $B \to A^\wedge = B^\wedge = (A^ h)^\wedge $ is flat for each $B$. Thus $A^ h \to A^\wedge = (A^ h)^\wedge $ is flat by Algebra, Lemma 10.38.6. Since $I^ h = IA^ h$ is contained in the Jacobson radical of $A^ h$ and since $A^ h \to A^\wedge $ induces an isomorphism $A^ h/I^ h \to A/I$ we see that $A^ h \to A^\wedge $ is faithfully flat by Algebra, Lemma 10.38.15. By Algebra, Lemma 10.96.6 the ring $A^\wedge $ is Noetherian. Hence we conclude that $A^ h$ is Noetherian by Algebra, Lemma 10.158.1.
$\square$

Lemma 15.12.5. Let $(A, I) = \mathop{\mathrm{colim}}\nolimits (A_ i, I_ i)$ be a colimit of pairs. The functor of Lemma 15.12.1 gives $A^ h = \mathop{\mathrm{colim}}\nolimits A_ i^ h$ and $I^ h = \mathop{\mathrm{colim}}\nolimits I_ i^ h$.

**Proof.**
This is true for any left adjoint, see Categories, Lemma 4.24.5.
$\square$

Lemma 15.12.6. Let $A$ be a ring with ideals $I$ and $J$. If $V(I) = V(J)$ then the functor of Lemma 15.12.1 produces the same ring for the pair $(A, I)$ as for the pair $(A, J)$.

**Proof.**
Let $(A', IA')$ be the pair produced by Lemma 15.12.1 starting with the pair $(A, I)$, see Lemma 15.12.2. Let $(A'', JA'')$ be the pair produced by Lemma 15.12.1 starting with the pair $(A, J)$. By Lemma 15.11.7 we see that $(A', JA')$ is a henselian pair and $(A'', IA'')$ is a henselian pair. By the universal property of the construction we obtain unique $A$-algebra maps $A'' \to A'$ and $A' \to A''$. The uniqueness shows that these are mutually inverse.
$\square$

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.
$\square$

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