The Stacks project

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 10.143.3) 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

which is étale over $A$ by Algebra, Lemmas 10.143.3 and 10.143.8. 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 10.143.3. Since $A^ h/I^ h = A/I$, 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

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

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 10.151.4. 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

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. $\square$

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 10.143.3) we conclude that $A \to A^ h$ is flat by Algebra, Lemma 10.39.3. 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 10.101.3). Taking the colimit we find that $A/I^ n = A^ h/I^ nA^ h$ as desired. $\square$

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 10.155.1 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 10.154.7. $\square$

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 10.97.2 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 10.39.6. 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 10.39.15. By Algebra, Lemma 10.97.6 the ring $A^\wedge $ is Noetherian. Hence we conclude that $A^ h$ is Noetherian by Algebra, Lemma 10.164.1. $\square$

Proof. By Categories, Lemma 4.24.5 we see that $(A^ h, I^ h)$ is the colimit of the system $(A_ i^ h, I_ i^ h)$ in the category of henselian pairs. Thus for a henselian pair $(B, J)$ we have

Here the colimit is in the category of pairs. Since the colimit is filtered we obtain $\mathop{\mathrm{colim}}\nolimits (A_ i^ h, I_ i^ h) = (\mathop{\mathrm{colim}}\nolimits A_ i^ h, \mathop{\mathrm{colim}}\nolimits I_ i^ h)$ in the category of pairs; details omitted. Again using the colimit is filtered, this is a henselian pair (Lemma 15.11.13). Hence by the Yoneda lemma we find $(A^ h, I^ h) = (\mathop{\mathrm{colim}}\nolimits A_ i^ h, \mathop{\mathrm{colim}}\nolimits I_ i^ h)$. $\square$

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. $\square$

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. $\square$

Proof. The map comes from the functoriality of the construction. If $n = 1$ the result is immediate. If $n > 1$, observe that $(I_1 \cap \ldots \cap I_{n - 1}) + I_ n = A$ by our condition on the ideals. Thus if we prove the lemma for $n = 2$, then the lemma follows for all $n$ by induction.

The case $n = 2$. Choose $f_ i \in I_ i$ such that $f_1 + f_2 = 1$. Recall that $A^ h$ is constructed as the colimit of $A$-algebras $B$ such that $A \to B$ is étale and $A/I_1 \cap I_2 \to B/(I_1 \cap I_2)B$ is an isomorphism. See proof of Lemma 15.12.1. Denote $Hens$ the category of such $A$-algebras $B$. Similarly, denote $Hens_ i$, $i = 1, 2$ the category of étale $A$-algebras $B$ such that $A/I_ i \to B/I_ iB$ is an isomorphism.

Consider the subcategory $Hens_1(f_2) \subset Hens_1$ consisting of those $B$ such that $f_2$ is invertible. This is a cofinal subcategory since if $B$ is in $Hens_1$, then $B_{f_2}$ is in $Hens_1$ as well (hint: $B_{f_2}/I_1B_{f_2} = (B/I_1B)_{f_2} = B/I_1B$ because $f_2$ is a unit modulo $I_1$). Similarly, the subcategory $Hens_2(f_1) \subset Hens_2$ is cofinal. Hence to compute $A_1^ h$ and $A_2^ h$ we may work with this smaller categories, see Categories, Lemma 4.17.2.

Consider the functor

This functor makes sense as

The second equality holds because $(I_1 \cap I_2)B_1 = I_1B_1 \cap I_2B_1 = I_1B_1$ as $f_2 \in I_2$ is invertible in $B_1$ (we have also used Algebra, Lemma 10.39.2). The last equality holds by Algebra, Lemma 10.15.4. To finish the proof, it suffices to prove that the displayed functor is cofinal (see lemma quoted above). The first condition of Categories, Definition 4.17.1 holds because an object $B$ of $Hens$ maps to $B_{f_2} \times B_{f_1}$ which is in the image of the functor. The second one always holds because if we have $B \to B_1 \times B_2$ and $B \to B'_1 \times B'_2$ with $B_1, B'_1 \in Hens_1(f_2)$ and $B_2, B'_2 \in Hens_2(f_1)$, then we can set $B''_1 = B_1 \otimes _ B B'_1$ which is in $Hens_1(f_2)$ and similarly $B''_2 = B_2 \otimes _ B B''_2$ in $Hens_2(f_1)$ to find a map $B \to B''_1 \times B''_2$ fitting into the commutative diagram

This finishes the proof. $\square$