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. $\square$

Comment #2174 by JuanPablo on

In the second paragraph it says that the coequalizer of $f,g:B \to B'$ is in $\mathcal{C}$. I do not think this is obvious, but the proof is similar to unicity later in this proof.

That is, given $K$ the kernel of the multiplication map $B\otimes_A B\to B$, then $K$ is generated by an idempotent $e$ (by lemma 10.147.4 (tag 02FL)), and given the map $f\otimes g:B\otimes_A B\to B'$, then the coequalizer of $f$ an $g$ equals $B''=B'/f\otimes g(K)B'=B'_{1-f\otimes g(e)}$ is étale over $A$, and from $B/IB=B'/IB'=A/I$ one gets $f\otimes g(K)\subset IB'$ and $B''/IB''=A/I$.

Comment #2203 by on

@#2174: The formula for the coequalizer was wrong. Fixed here. Observe that if $A \to B$ is \'etale, then $B \otimes_A B \to B$ is \'etale as it is a map of \'etale $A$-algebras.

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