## Tag `09XD`

## 15.10. Henselian pairs

Some of the results of Section 15.9 may be viewed as results about henselian pairs. In this section a

pairis a pair $(A, I)$ where $A$ is a ring and $I \subset A$ is an ideal. Amorphism of pairs$(A, I) \to (B, J)$ is a ring map $\varphi : A \to B$ with $\varphi(I) \subset J$. As in Section 15.9 given an object $\xi$ over $A$ we denote $\overline{\xi}$ the ''base change'' of $\xi$ to an object over $A/I$ (provided this makes sense).Definition 15.10.1. A

henselian pairis a pair $(A, I)$ satisfying

- $I$ is contained in the Jacobson radical of $A$, and
- for any monic polynomial $f \in A[T]$ and factorization $\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$, there exists a factorization $f = gh$ in $A[T]$ with $g, h$ monic and $g_0 = \overline{g}$ and $h_0 = \overline{h}$.

Observe that if $A$ is a local ring and $I = \mathfrak m$ is the maximal ideal, then $(A, I)$ is a henselian pair if and only if $A$ is a henselian local ring, see Algebra, Lemma 10.148.3. In Lemma 15.10.8 we give a number of equivalent characterizations of henselian pairs (and we will add more as time goes on).

Lemma 15.10.2. Let $(A, I)$ be a pair with $I$ locally nilpotent. Then the functor $B \mapsto B/IB$ induces an equivalence between the category of étale algebras over $A$ and the category of étale algebras over $A/I$. Moreover, the pair is henselian.

Proof.Essential surjectivity holds by Algebra, Lemma 10.141.10. If $B$, $B'$ are étale over $A$ and $B/IB \to B'/IB'$ is a morphism of $A/I$-algebras, then we can lift this by Algebra, Lemma 10.136.17. Finally, suppose that $f, g : B \to B'$ are two $A$-algebra maps with $f \mod I = g \mod I$. Choose an idempotent $e \in B \otimes_A B$ generating the kernel of the multiplication map $B \otimes_A B \to B$, see Algebra, Lemmas 10.147.4 and 10.147.3 (to see that étale is unramified). Then $(f \otimes g)(e) \in IB$. Since $IB$ is locally nilpotent (Algebra, Lemma 10.31.3) this implies $(f \otimes g)(e) = 0$ by Algebra, Lemma 10.31.6. Thus $f = g$.It is clear that $I$ is contained in the radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0h_0$ be a factorization of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. By Lemma 15.9.5 there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ such that the factorization lifts to a factorization into monic polynomials over $A'$. By the above we have $A = A'$ and the factorization is over $A$. $\square$

Lemma 15.10.3. Let $A = \mathop{\mathrm{lim}}\nolimits A_n$ where $(A_n)$ is an inverse system of rings whose transition maps are surjective and have locally nilpotent kernels. Then $(A, I_n)$ is a henselian pair, where $I_n = \mathop{\mathrm{Ker}}(A \to A_n)$.

Proof.Fix $n$. Let $a \in A$ be an element which maps to $1$ in $A_n$. By Algebra, Lemma 10.31.4 we see that $a$ maps to a unit in $A_m$ for all $m \geq n$. Hence $a$ is a unit in $A$. Thus by Algebra, Lemma 10.18.1 the ideal $I_n$ is contained in the radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_nh_n$ be a factorization of $\overline{f} = f \bmod I_n$ with $g_n, h_n \in A_n[T]$ monic generating the unit ideal in $A_n[T]$. By Lemma 15.10.2 we can successively lift this factorization to $f \bmod I_m = g_m h_m$ with $g_m, h_m$ monic in $A_m[T]$ for all $m \geq n$. As $A = \mathop{\mathrm{lim}}\nolimits A_m$ this finishes the proof. $\square$Lemma 15.10.4. Let $(A, I)$ be a pair. If $A$ is $I$-adically complete, then the pair is henselian.

Proof.By Algebra, Lemma 10.95.6 the ideal $I$ is contained in the radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0h_0$ be a factorization of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. By Lemma 15.10.2 we can successively lift this factorization to $f \bmod I^n = g_n h_n$ with $g_n, h_n$ monic in $A/I^n[T]$ for all $n \geq 1$. As $A = \mathop{\mathrm{lim}}\nolimits A/I^n$ this finishes the proof. $\square$Lemma 15.10.5. Let $(A, I)$ be a pair. If $I$ is contained in the Jacobson radical of $A$, then the map from idempotents of $A$ to idempotents of $A/I$ is injective.

Proof.An idempotent of a local ring is either $0$ or $1$. Thus an idempotent is determined by the set of maximal ideals where it vanishes, by Algebra, Lemma 10.23.1. $\square$Lemma 15.10.6. Let $(A, I)$ be a pair. Let $A \to B$ be an integral ring map such that $B/IB = C_1 \times C_2$ as $A/I$-algebra with $A/I \to C_1$ injective. Any element $b \in B$ mapping to $(0, 1)$ in $B/IB$ is the zero of a monic polynomial $f \in A[T]$ with $f \bmod I = g T^n$ and $g(0)$ a unit in $A/I$.

Proof.Let $b \in B$ map to $(0, 1)$ in $C_1 \times C_2$. Let $J \subset A[T]$ be the kernel of the map $A[T] \to B$, $T \mapsto b$. Since $B$ is integral over $A$, it is integral over $A[T]$. Hence the image of $\mathop{\mathrm{Spec}}(B)$ in $\mathop{\mathrm{Spec}}(A[T])$ is closed by Algebra, Lemmas 10.40.6 and 10.35.22. Hence this image is equal to $V(J) = \mathop{\mathrm{Spec}}(A[T]/J)$ by Algebra, Lemma 10.29.5. Intersecting with the inverse image of $V(I)$ our choice of $b$ shows we have $V(J + IA[T]) \subset V(T^2 - T)$. Hence there exists an $n \geq 1$ and $g \in J$ with $g \bmod IA[T] = (T^2 - T)^n$. On the other hand, as $A \to B$ is integral there exists a monic polynomial $h \in J$. Note that $h(0) \bmod I$ maps to zero under the composition $A[T] \to B \to B/IB \to C_1$. Since $A/I \to C_1$ is injective we conclude $h \bmod IA[T] = h_0 T$ for some $h_0 \in A/I[T]$. Set $$ f = g + h^m $$ for $m > n$. If $m$ is large enough, this is a monic polynomial and $$ f \bmod IA[T] = (T^2 - T)^n + h_0^m T^m = T^n((T - 1)^n + h_0^m T^{m - n}) $$ and hence the desired conclusion. $\square$Lemma 15.10.7. Let $(A, I)$ be a pair. Let $A \to B$ be a finite type ring map such that $B/IB = C_1 \times C_2$ with $A/I \to C_1$ finite. Let $B'$ be the integral closure of $A$ in $B$. Then we can write $B'/IB' = C_1 \times C'_2$ such that the map $B'/IB' \to B/IB$ preserves product decompositions and there exists a $g \in B'$ mapping to $(1, 0)$ in $C_1 \times C'_2$ with $B'_g \to B_g$ an isomorphism.

Proof.Observe that $A \to B$ is quasi-finite at every prime of the closed subset $T = \mathop{\mathrm{Spec}}(C_1) \subset \mathop{\mathrm{Spec}}(B)$ (this follows by looking at fibre rings, see Algebra, Definition 10.121.3). Consider the diagram of topological spaces $$ \xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[rr]_\phi \ar[rd]_\psi & & \mathop{\mathrm{Spec}}(B') \ar[ld]^{\psi'} \\ & \mathop{\mathrm{Spec}}(A) } $$ By Algebra, Theorem 10.122.13 for every $\mathfrak p \in T$ there is a $h_\mathfrak p \in B'$, $h_\mathfrak p \not \in \mathfrak p$ such that $B'_h \to B_h$ is an isomorphism. The union $U = \bigcup D(h_\mathfrak p)$ gives an open $U \subset \mathop{\mathrm{Spec}}(B')$ such that $\phi^{-1}(U) \to U$ is a homeomorphism and $T \subset \phi^{-1}(U)$. Since $T$ is open in $\psi^{-1}(V(I))$ we conclude that $\phi(T)$ is open in $U \cap (\psi')^{-1}(V(I))$. Thus $\phi(T)$ is open in $(\psi')^{-1}(V(I))$. On the other hand, since $C_1$ is finite over $A/I$ it is finite over $B'$. Hence $\phi(T)$ is a closed subset of $\mathop{\mathrm{Spec}}(B')$ by Algebra, Lemmas 10.40.6 and 10.35.22. We conclude that $\mathop{\mathrm{Spec}}(B'/IB') \supset \phi(T)$ is open and closed. By Algebra, Lemma 10.22.3 we get a corresponding product decomposition $B'/IB' = C'_1 \times C'_2$. The map $B'/IB' \to B/IB$ maps $C'_1$ into $C_1$ and $C'_2$ into $C_2$ as one sees by looking at what happens on spectra (hint: the inverse image of $\phi(T)$ is exactly $T$; some details omitted). Pick a $g \in B'$ mapping to $(1, 0)$ in $C'_1 \times C'_2$ such that $D(g) \subset U$; this is possible because $\mathop{\mathrm{Spec}}(C'_1)$ and $\mathop{\mathrm{Spec}}(C'_2)$ are disjoint and closed in $\mathop{\mathrm{Spec}}(B')$ and $\mathop{\mathrm{Spec}}(C'_1)$ is contained in $U$. Then $B'_g \to B_g$ defines a homeomorphism on spectra and an isomorphism on local rings (by our choice of $U$ above). Hence it is an isomorphism, as follows for example from Algebra, Lemma 10.23.1. Finally, it follows that $C'_1 = C_1$ and the proof is complete. $\square$Lemma 15.10.8. Let $(A, I)$ be a pair. The following are equivalent

- $(A, I)$ is a henselian pair,
- given an étale ring map $A \to A'$ and an $A$-algebra map $\sigma : A' \to A/I$, there exists an $A$-algebra map $A' \to A$ lifting $\sigma$,
- for any finite $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents, and
- for any integral $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents.

Proof.Assume (2) holds. Then $I$ is contained in the Jacobson radical of $A$, since otherwise there would be a nonunit $f \in A$ not contained in $I$ and the map $A \to A_f$ would contradict (2). Hence $IB \subset B$ is contained in the Jacobson radical of $B$ for $B$ integral over $A$ because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is closed by Algebra, Lemmas 10.40.6 and 10.35.22. Thus the map from idempotents of $B$ to idempotents of $B/IB$ is injective by Lemma 15.10.5. On the other hand, since (2) holds, every idempotent of $B$ lifts to an idempotent of $B/IB$ by Lemma 15.9.9. In this way we see that (2) implies (4).The implication (4) $\Rightarrow$ (3) is trivial.

Assume (3). Let $\mathfrak m$ be a maximal ideal and consider the finite map $A \to B = A/(I \cap \mathfrak m)$. The condition that $B \to B/IB$ induces a bijection on idempotents implies that $I \subset \mathfrak m$ (if not, then $B = A/I \times A/\mathfrak m$ and $B/IB = A/I$). Thus we see that $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be monic and suppose given a factorization $\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic. Set $B = A[T]/(f)$. Let $\overline{e}$ be the nontrivial idempotent of $B/IB$ corresponding to the decomposition $$ B/IB = A/I[T]/(g_0) \times A[T]/(h_0) $$ of $A$-algebras. Let $e \in B$ be an idempotent lifting $\overline{e}$ which exists as we assumed (3). This gives a product decomposition $$ B = eB \times (1 - e)B $$ Note that $B$ is free of rank $\deg(f)$ as an $A$-module. Hence $eB$ and $(1 - e)B$ are finite locally free $A$-modules. However, since $eB$ and $(1 - e)B$ have constant rank $\deg(g_0)$ and $\deg(h_0)$ over $A/I$ we find that the same is true over $\mathop{\mathrm{Spec}}(A)$. We conclude that $$ f = \det\nolimits_A(T : B \to B) = \det\nolimits_A(T : eB \to eB) \det\nolimits_A(T : (1 - e)B \to (1 - e)B) $$ is a factorization into monic polynomials reducing to the given factorization modulo $I$

^{1}. Thus (3) implies (1).Assume (1). Let $A \to A'$ be an étale ring map and let $\sigma : A' \to A/I$ be an $A$-algebra map. This implies that $A'/IA' = A/I \times C$ for some ring $C$. Let $A'' \subset A'$ be the integral closure of $A$ in $A'$. By Lemma 15.10.7 we can write $A''/IA'' = A/I \times C'$ such that $A''/IA'' \to A'/IA'$ maps $A/I$ isomorphically to $A'/IA'$ and $C'$ to $C$ and such that there exists a $a \in A''$ mapping to $(1, 0)$ in $A/I \times C'$ such that $A''_a \cong A'_a$. By Lemma 15.10.6 we see that $1 - a$ satisfies a monic polynomial $f \in A[T]$ with $f \bmod I = g_0 T^n$ where $T, g_0$ generate the unit ideal in $A/I[T]$. As we assumed $(A, I)$ is a henselian pair we can factor $f$ as $f = g h$ where $g$ is a monic lift of $g_0$ and $h$ is a monic lift of $T^n$. Because $I$ is contained in the Jacobson radical of $A$, we find that $g$ and $h$ generate the unit ideal in $A[T]$ (details omitted; hint: use that $A[T]/(g, h)$ is finite over $A$). Thus $A[T]/(f) = A[T]/(h) \times A[T]/(g)$. It follows that $h(1 - a)$ and $g(1 - a)$ generate the unit ideal in $A''$ and we find a product decomposition $A'' = A''_1 \times A''_2$ such that $h(1 - a)$ acts as zero on $A''_1$ and $g(1 - a)$ acts as zero on $A''_2$. As $h \bmod I = T^n$, as $T, g_0$ generate the unit ideal, and as $a \bmod I = (1, 0)$ we conclude that the two disjoint union decompositions $$ \mathop{\mathrm{Spec}}(A''/IA'') = \mathop{\mathrm{Spec}}(A/I) \amalg \mathop{\mathrm{Spec}}(C') = \mathop{\mathrm{Spec}}(A''_1/IA''_1) \amalg \mathop{\mathrm{Spec}}(A''_2/IA''_2) $$ are the same: in both cases the first summand corresponds exactly to those primes of $A''$ containing $IA''$ such that $a$ maps to $1$ in the residue field. This implies that $A''_1/IA''_1 = A/I$ as factor rings of $A''/IA''$ (because product decompositions of rings correspond $1$-to-$1$ to disjoint union decompositions of the spectra, see Algebra, Lemmas 10.20.2 and 10.22.3). Moreover, it follows that $a$ maps to a unit in $A''_1$ (see Algebra, Lemma 10.18.1 and use that $IA''_1$ is contained in the radical of $A''_1$ as $A''_1$ is integral over $A$). Hence $A''_1 = (A''_1)_a$ is a factor of $A''_a = A'_a$ which is étale over $A$. Then $A \to A''_1$ is integral, of finite presentation, and flat (Algebra, Section 10.141) hence finite (Algebra, Lemma 10.35.5) hence finitely presented as an $A$-module (Algebra, Lemma 10.35.23) hence $A''_1$ is finite projective as an $A$-module (Algebra, Lemma 10.77.2). Since $A''_1/IA''_1 = A/I$ the module $A''_1$ has rank $1$ as an along $V(I)$ (see rank function described in Algebra, Lemma 10.77.2). Since $I$ is contained in the Jacobson radical of $A$ we conclude that $A''_1$ has rank $1$ everywhere. It follows that $A \to A''_1$ is an isomorphism (exercise: a ring map $R \to S$ such that $S$ is a locally free $R$-module of rank $1$ is an isomorphism). Thus $A' \to A'_a \cong A''_a \to (A''_1)_a = A''_1 = A$ is the desired lift of $\sigma$. In this way we see that (1) implies (2). $\square$

Lemma 15.10.9. Let $A$ be a ring. Let $I, J \subset A$ be ideals with $V(I) = V(J)$. Then $(A, I)$ is henselian if and only if $(A, J)$ is henselian.

Proof.For any integral ring map $A \to B$ we see that $V(IB) = V(JB)$. Hence idempotents of $B/IB$ and $B/JB$ are in bijective correspondence (Algebra, Lemma 10.20.3). It follows that $B \to B/IB$ induces a bijection on sets of idempotents if and only if $B \to B/JB$ induces a bijection on sets of idempotents. Thus we conclude by Lemma 15.10.8. $\square$Lemma 15.10.10. Let $(A, I)$ be a henselian pair and let $A \to B$ be an integral ring map. Then $(B, IB)$ is a henselian pair.

Proof.Immediate from the fourth characterization of henselian pairs in Lemma 15.10.8 and the fact that the composition of integral ring maps is integral. $\square$Lemma 15.10.11. Let $I \subset J \subset A$ be ideals of a ring $A$. The following are equivalent

- $(A, I)$ and $(A/I, J/I)$ are henselian pairs, and
- $(A, J)$ is an henselian pair.

Proof.Assume (1). Looking at Definition 15.10.1 we see that $V(I)$ contains all closed points of $\mathop{\mathrm{Spec}}(A)$ and that $V(J) = V(J/I)$ contains all closed points of $V(I) = \mathop{\mathrm{Spec}}(A/I)$. Hence $V(J)$ contains all closed points of $\mathop{\mathrm{Spec}}(A)$, i.e., $J$ is contained in the Jacobson radical of $A$. Next, let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0 h_0$ be a factorization in $A/J[T]$ with $g_0, h_0$ monic generating the unit ideal in $A/J[T]$. Then we can first lift this factorization to a factorization in $A/I[T]$ and then to a factorization in $A[T]$. Thus $(A, J)$ is a henselian pair.Conversely, assume (2) holds. Then $(A/I, J/I)$ is a henselian pair by Lemma 15.10.10. Let $B$ be an integral $A$-algebra. Consider the ring maps $$ B \to B/IB \to B/JB $$ By Lemma 15.10.8 we find that the composition and the second arrow induce bijections on idempotents. Hence so does the first arrow. It follows that $(A, I)$ is a henselian pair (by the lemma again). $\square$

Lemma 15.10.12. Let $J$ be a set and let $\{ (A_j, I_j)\}_{j \in J}$ be a collection of pairs. Then $(\prod_{j \in J} A_j, \prod_{j\in J} I_j)$ is Henselian if and only if so is each $(A_j, I_j)$.

Proof.For every $j \in J$, the projection $\prod_{j \in J} A_j \rightarrow A_j$ is an integral ring map, so Lemma 15.10.10 proves that each $(A_j, I_j)$ is Henselian if $(\prod_{j \in J} A_j, \prod_{j\in J} I_j)$ is Henselian.Conversely, suppose that each $(A_j, I_j)$ is a Henselian pair. Then every $1 + x$ with $x \in \prod_{j \in J} I_j$ is a unit in $\prod_{j \in J} A_j$ because it is so componentwise by Algebra, Lemma 10.18.1 and Definition 15.10.1. Thus, by Algebra, Lemma 10.18.1 again, $\prod_{j \in J} I_j$ is contained in the Jacobson radical of $\prod_{j \in J} A_j$. Continuing to work componentwise, it likewise follows that for every monic $f \in (\prod_{j \in J} A_j)[T]$ and every factorization $\overline{f} = g_0h_0$ with monic $g_0, h_0 \in (\prod_{j \in J} A_j / \prod_{j \in J} I_j)[T] = (\prod_{j \in J} A_j/I_j)[T]$ that generate the unit ideal in $(\prod_{j \in J} A_j / \prod_{j \in J} I_j)[T]$, there exists a factorization $f = gh$ in $(\prod_{j \in J} A_j)[T]$ with $g$, $h$ monic and reducing to $g_0$, $h_0$. In conclusion, according to Definition 15.10.1 $(\prod_{j \in J} A_j, \prod_{j\in J} I_j)$ is a Henselian pair. $\square$

Lemma 15.10.13. Let $(A, I)$ be a henselian pair. Let $\mathfrak p \subset A$ be a prime ideal. Then $V(\mathfrak p + I)$ is connected.

Proof.By Lemma 15.10.10 we see that $(A/\mathfrak p, I + \mathfrak p/\mathfrak p)$ is a henselian pair. Thus it suffices to prove: If $(A, I)$ is a henselian pair and $A$ is a domain, then $\mathop{\mathrm{Spec}}(A/I) = V(I)$ is connected. If not, then $A/I$ has a nontrivial idempotent by Algebra, Lemma 10.20.4. By Lemma 15.10.8 this would imply $A$ has a nontrivial idempotent. This is a contradiction. $\square$Lemma 15.10.14. 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 15.10.8. 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 15.10.8. 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 15.10.5. 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 15.10.15. The functor of Lemma 15.10.14 associates to a local ring $(A, \mathfrak m)$ its henselization.

Proof.First proof: in the proof of Algebra, Lemma 10.150.1 it is shown that the henselization of $A$ is given by the colimit used to construct $A^h$ in Lemma 15.10.14. Second proof: Both the henselization $S$ and the ring $A^h$ of Lemma 15.10.14 are filtered colimits of étale $A$-algebras, henselian, and have residue fields equal to $\kappa(\mathfrak m)$. Hence they are canonically isomorphic by Algebra, Lemma 10.149.6. $\square$Lemma 15.10.16. Let $(A, I)$ be a pair. Let $(A^h, I^h)$ be as in Lemma 15.10.14. 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 15.10.14 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 15.10.17. Let $(A, I)$ be a pair with $A$ Noetherian. Let $(A^h, I^h)$ be as in Lemma 15.10.14. 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 15.10.16 and in fact holds without assuming $A$ is Noetherian. In the proof of Lemma 15.10.14 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 15.10.16). 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 radical ideal 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 15.10.18. Let $(A, I) = \mathop{\mathrm{colim}}\nolimits (A_i, I_i)$ be a colimit of pairs. The functor of Lemma 15.10.14 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 15.10.19. Let $(A, I) \to (B, J)$ be a map of pairs. Let $(A^h , I^h) \to (B^h, J^h)$ be the induced map on henselizations (Lemma 15.10.14). If $A \to B$ is integral, then the induced map $A^h \otimes_A B \to B^h$ is an isomorphism.

Proof.By Lemma 15.10.8 the pair $(A^h \otimes_A B, I^h(A^h \otimes_A B))$ is henselian. By the universal property 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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```
\section{Henselian pairs}
\label{section-henselian-pairs}
\noindent
Some of the results of Section \ref{section-lifting} may be viewed as results
about henselian pairs. In this section a {\it pair} is a pair $(A, I)$
where $A$ is a ring and $I \subset A$ is an ideal. A {\it morphism of pairs}
$(A, I) \to (B, J)$ is a ring map $\varphi : A \to B$ with
$\varphi(I) \subset J$. As in
Section \ref{section-lifting} given an object $\xi$ over $A$ we denote
$\overline{\xi}$ the ``base change'' of $\xi$ to an object over $A/I$
(provided this makes sense).
\begin{definition}
\label{definition-henselian-pair}
A {\it henselian pair} is a pair $(A, I)$ satisfying
\begin{enumerate}
\item $I$ is contained in the Jacobson radical of $A$, and
\item for any monic polynomial $f \in A[T]$ and factorization
$\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic
generating the unit ideal in $A/I[T]$, there
exists a factorization $f = gh$ in $A[T]$ with $g, h$ monic
and $g_0 = \overline{g}$ and $h_0 = \overline{h}$.
\end{enumerate}
\end{definition}
\noindent
Observe that if $A$ is a local ring and $I = \mathfrak m$ is the maximal
ideal, then $(A, I)$ is a henselian pair if and only if $A$ is a henselian
local ring, see
Algebra, Lemma \ref{algebra-lemma-characterize-henselian}.
In Lemma \ref{lemma-characterize-henselian-pair} we give a number of
equivalent characterizations of
henselian pairs (and we will add more as time goes on).
\begin{lemma}
\label{lemma-locally-nilpotent-henselian}
Let $(A, I)$ be a pair with $I$ locally nilpotent. Then the functor
$B \mapsto B/IB$ induces an equivalence between the category of
\'etale algebras over $A$ and the category of \'etale algebras over $A/I$.
Moreover, the pair is henselian.
\end{lemma}
\begin{proof}
Essential surjectivity holds by Algebra, Lemma \ref{algebra-lemma-lift-etale}.
If $B$, $B'$ are \'etale over $A$ and $B/IB \to B'/IB'$ is a morphism
of $A/I$-algebras, then we can lift this by
Algebra, Lemma \ref{algebra-lemma-smooth-strong-lift}.
Finally, suppose that $f, g : B \to B'$ are two $A$-algebra
maps with $f \mod I = g \mod I$. Choose an idempotent $e \in B \otimes_A B$
generating the kernel of the multiplication map $B \otimes_A B \to B$,
see Algebra, Lemmas \ref{algebra-lemma-diagonal-unramified}
and \ref{algebra-lemma-unramified} (to see that \'etale is unramified).
Then $(f \otimes g)(e) \in IB$. Since $IB$ is locally nilpotent
(Algebra, Lemma \ref{algebra-lemma-locally-nilpotent}) this implies
$(f \otimes g)(e) = 0$ by Algebra, Lemma \ref{algebra-lemma-lift-idempotents}.
Thus $f = g$.
\medskip\noindent
It is clear that $I$ is contained in the radical of $A$.
Let $f \in A[T]$ be a monic polynomial and let
$\overline{f} = g_0h_0$ be a factorization
of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic
generating the unit ideal in $A/I[T]$. By
Lemma \ref{lemma-lift-factorization-monic}
there exists an \'etale ring map $A \to A'$ which
induces an isomorphism $A/I \to A'/IA'$ such that
the factorization lifts to a factorization into monic polynomials
over $A'$. By the above we have $A = A'$ and the factorization
is over $A$.
\end{proof}
\begin{lemma}
\label{lemma-limit-henselian}
Let $A = \lim A_n$ where $(A_n)$ is an inverse system of rings
whose transition maps are surjective and have locally nilpotent kernels.
Then $(A, I_n)$ is a henselian pair, where $I_n = \Ker(A \to A_n)$.
\end{lemma}
\begin{proof}
Fix $n$. Let $a \in A$ be an element which maps to $1$ in $A_n$.
By Algebra, Lemma \ref{algebra-lemma-locally-nilpotent-unit}
we see that $a$ maps to a unit in $A_m$ for all $m \geq n$.
Hence $a$ is a unit in $A$. Thus by
Algebra, Lemma \ref{algebra-lemma-contained-in-radical}
the ideal $I_n$ is contained in the radical of $A$.
Let $f \in A[T]$ be a monic polynomial and let
$\overline{f} = g_nh_n$ be a factorization
of $\overline{f} = f \bmod I_n$ with $g_n, h_n \in A_n[T]$ monic
generating the unit ideal in $A_n[T]$. By
Lemma \ref{lemma-locally-nilpotent-henselian}
we can successively lift this factorization to
$f \bmod I_m = g_m h_m$ with $g_m, h_m$ monic
in $A_m[T]$ for all $m \geq n$.
As $A = \lim A_m$ this finishes the proof.
\end{proof}
\begin{lemma}
\label{lemma-complete-henselian}
Let $(A, I)$ be a pair. If $A$ is $I$-adically complete, then
the pair is henselian.
\end{lemma}
\begin{proof}
By Algebra, Lemma \ref{algebra-lemma-radical-completion}
the ideal $I$ is contained in the radical of $A$.
Let $f \in A[T]$ be a monic polynomial and let
$\overline{f} = g_0h_0$ be a factorization
of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic
generating the unit ideal in $A/I[T]$. By
Lemma \ref{lemma-locally-nilpotent-henselian}
we can successively lift this factorization to
$f \bmod I^n = g_n h_n$ with $g_n, h_n$ monic
in $A/I^n[T]$ for all $n \geq 1$.
As $A = \lim A/I^n$ this finishes the proof.
\end{proof}
\begin{lemma}
\label{lemma-idempotents-determined-modulo-radical}
Let $(A, I)$ be a pair. If $I$ is contained in the Jacobson radical
of $A$, then the map from idempotents of $A$ to idempotents of
$A/I$ is injective.
\end{lemma}
\begin{proof}
An idempotent of a local ring is either $0$ or $1$.
Thus an idempotent is determined by the set of maximal ideals
where it vanishes, by
Algebra, Lemma \ref{algebra-lemma-characterize-zero-local}.
\end{proof}
\begin{lemma}
\label{lemma-helper-integral}
Let $(A, I)$ be a pair. Let $A \to B$ be an integral ring map
such that $B/IB = C_1 \times C_2$ as $A/I$-algebra with $A/I \to C_1$
injective. Any element $b \in B$ mapping to $(0, 1)$ in $B/IB$
is the zero of a monic polynomial $f \in A[T]$
with $f \bmod I = g T^n$ and $g(0)$ a unit in $A/I$.
\end{lemma}
\begin{proof}
Let $b \in B$ map to $(0, 1)$ in $C_1 \times C_2$.
Let $J \subset A[T]$ be the kernel of the map $A[T] \to B$, $T \mapsto b$.
Since $B$ is integral over $A$, it is integral over $A[T]$. Hence
the image of $\Spec(B)$ in $\Spec(A[T])$ is closed by
Algebra, Lemmas \ref{algebra-lemma-going-up-closed} and
\ref{algebra-lemma-integral-going-up}. Hence this image is
equal to $V(J) = \Spec(A[T]/J)$ by
Algebra, Lemma \ref{algebra-lemma-injective-minimal-primes-in-image}.
Intersecting with the inverse image of $V(I)$ our choice of $b$ shows
we have $V(J + IA[T]) \subset V(T^2 - T)$. Hence there exists an $n \geq 1$
and $g \in J$ with $g \bmod IA[T] = (T^2 - T)^n$.
On the other hand, as $A \to B$ is integral there exists a monic
polynomial $h \in J$. Note that $h(0) \bmod I$ maps to zero
under the composition $A[T] \to B \to B/IB \to C_1$. Since $A/I \to C_1$
is injective we conclude $h \bmod IA[T] = h_0 T$ for some $h_0 \in A/I[T]$.
Set
$$
f = g + h^m
$$
for $m > n$. If $m$ is large enough, this is a monic polynomial and
$$
f \bmod IA[T] = (T^2 - T)^n + h_0^m T^m =
T^n((T - 1)^n + h_0^m T^{m - n})
$$
and hence the desired conclusion.
\end{proof}
\begin{lemma}
\label{lemma-helper-finite-type}
Let $(A, I)$ be a pair. Let $A \to B$ be a finite type ring map
such that $B/IB = C_1 \times C_2$ with $A/I \to C_1$ finite.
Let $B'$ be the integral closure of $A$ in $B$.
Then we can write $B'/IB' = C_1 \times C'_2$ such that
the map $B'/IB' \to B/IB$ preserves product decompositions
and there exists a $g \in B'$ mapping to $(1, 0)$ in
$C_1 \times C'_2$ with $B'_g \to B_g$ an isomorphism.
\end{lemma}
\begin{proof}
Observe that $A \to B$ is quasi-finite at every prime of the
closed subset $T = \Spec(C_1) \subset \Spec(B)$ (this follows
by looking at fibre rings, see
Algebra, Definition \ref{algebra-definition-quasi-finite}).
Consider the diagram of topological spaces
$$
\xymatrix{
\Spec(B) \ar[rr]_\phi \ar[rd]_\psi & & \Spec(B') \ar[ld]^{\psi'} \\
& \Spec(A)
}
$$
By Algebra, Theorem \ref{algebra-theorem-main-theorem}
for every $\mathfrak p \in T$ there is a $h_\mathfrak p \in B'$,
$h_\mathfrak p \not \in \mathfrak p$ such that $B'_h \to B_h$ is
an isomorphism. The union $U = \bigcup D(h_\mathfrak p)$ gives an open
$U \subset \Spec(B')$ such that $\phi^{-1}(U) \to U$ is a homeomorphism
and $T \subset \phi^{-1}(U)$. Since $T$ is open in $\psi^{-1}(V(I))$
we conclude that $\phi(T)$ is open in $U \cap (\psi')^{-1}(V(I))$.
Thus $\phi(T)$ is open in $(\psi')^{-1}(V(I))$.
On the other hand, since $C_1$ is finite over $A/I$ it is
finite over $B'$. Hence $\phi(T)$ is a closed subset of $\Spec(B')$
by Algebra, Lemmas \ref{algebra-lemma-going-up-closed} and
\ref{algebra-lemma-integral-going-up}. We conclude that
$\Spec(B'/IB') \supset \phi(T)$ is open and closed. By
Algebra, Lemma \ref{algebra-lemma-disjoint-implies-product}
we get a corresponding product decomposition $B'/IB' = C'_1 \times C'_2$.
The map $B'/IB' \to B/IB$ maps $C'_1$ into $C_1$ and $C'_2$ into $C_2$
as one sees by looking at what happens on spectra (hint: the inverse
image of $\phi(T)$ is exactly $T$; some details omitted).
Pick a $g \in B'$ mapping to $(1, 0)$ in $C'_1 \times C'_2$
such that $D(g) \subset U$; this is possible because $\Spec(C'_1)$
and $\Spec(C'_2)$ are disjoint and closed in $\Spec(B')$ and
$\Spec(C'_1)$ is contained in $U$. Then $B'_g \to B_g$ defines a homeomorphism
on spectra and an isomorphism on local rings (by our choice of $U$ above).
Hence it is an isomorphism, as follows for example from
Algebra, Lemma \ref{algebra-lemma-characterize-zero-local}.
Finally, it follows that $C'_1 = C_1$ and the proof is complete.
\end{proof}
\begin{lemma}
\label{lemma-characterize-henselian-pair}
Let $(A, I)$ be a pair. The following are equivalent
\begin{enumerate}
\item $(A, I)$ is a henselian pair,
\item given an \'etale ring map $A \to A'$ and an $A$-algebra map
$\sigma : A' \to A/I$, there exists an $A$-algebra map $A' \to A$
lifting $\sigma$,
\item for any finite $A$-algebra $B$ the map $B \to B/IB$ induces
a bijection on idempotents, and
\item for any integral $A$-algebra $B$ the map $B \to B/IB$ induces
a bijection on idempotents.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (2) holds. Then $I$ is contained in the Jacobson radical of $A$, since
otherwise there would be a nonunit $f \in A$ not contained in $I$
and the map $A \to A_f$ would contradict (2). Hence $IB \subset B$
is contained in the Jacobson radical of $B$ for $B$ integral over $A$
because $\Spec(B) \to \Spec(A)$ is closed by
Algebra, Lemmas \ref{algebra-lemma-going-up-closed} and
\ref{algebra-lemma-integral-going-up}.
Thus the map from idempotents of $B$ to idempotents of $B/IB$
is injective by Lemma \ref{lemma-idempotents-determined-modulo-radical}.
On the other hand, since (2) holds, every idempotent
of $B$ lifts to an idempotent of $B/IB$
by Lemma \ref{lemma-lift-idempotent-upstairs}.
In this way we see that (2) implies (4).
\medskip\noindent
The implication (4) $\Rightarrow$ (3) is trivial.
\medskip\noindent
Assume (3). Let $\mathfrak m$ be a maximal ideal and consider the
finite map $A \to B = A/(I \cap \mathfrak m)$. The condition that
$B \to B/IB$ induces a bijection on idempotents implies that
$I \subset \mathfrak m$ (if not, then $B = A/I \times A/\mathfrak m$
and $B/IB = A/I$). Thus we see that $I$ is contained in the Jacobson
radical of $A$. Let $f \in A[T]$ be monic and suppose given a
factorization $\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic.
Set $B = A[T]/(f)$. Let $\overline{e}$ be the nontrivial idempotent
of $B/IB$ corresponding to the decomposition
$$
B/IB = A/I[T]/(g_0) \times A[T]/(h_0)
$$
of $A$-algebras. Let $e \in B$ be an idempotent lifting $\overline{e}$
which exists as we assumed (3). This gives a product decomposition
$$
B = eB \times (1 - e)B
$$
Note that $B$ is free of rank $\deg(f)$ as an $A$-module.
Hence $eB$ and $(1 - e)B$ are finite locally free $A$-modules.
However, since $eB$ and $(1 - e)B$ have constant rank
$\deg(g_0)$ and $\deg(h_0)$ over $A/I$ we find that the same
is true over $\Spec(A)$. We conclude that
$$
f = \det\nolimits_A(T : B \to B) =
\det\nolimits_A(T : eB \to eB) \det\nolimits_A(T : (1 - e)B \to (1 - e)B)
$$
is a factorization into monic polynomials reducing to the given
factorization modulo $I$\footnote{Here we use determinants of endomorphisms
of finite locally free modules; if the module is free the determinant
is defined as the determinant of the corresponding matrix and in
general one uses Algebra, Lemma \ref{algebra-lemma-standard-covering} to
glue.}.
Thus (3) implies (1).
\medskip\noindent
Assume (1). Let $A \to A'$ be an \'etale ring map and let
$\sigma : A' \to A/I$ be an $A$-algebra map. This implies that
$A'/IA' = A/I \times C$ for some ring $C$. Let $A'' \subset A'$ be the integral
closure of $A$ in $A'$. By Lemma \ref{lemma-helper-finite-type}
we can write $A''/IA'' = A/I \times C'$ such that $A''/IA'' \to A'/IA'$
maps $A/I$ isomorphically to $A'/IA'$ and $C'$ to $C$ and such that
there exists a $a \in A''$ mapping to $(1, 0)$ in $A/I \times C'$
such that $A''_a \cong A'_a$.
By Lemma \ref{lemma-helper-integral}
we see that $1 - a$ satisfies a monic polynomial $f \in A[T]$
with $f \bmod I = g_0 T^n$ where
$T, g_0$ generate the unit ideal in $A/I[T]$. As we assumed
$(A, I)$ is a henselian pair we can factor $f$ as $f = g h$
where $g$ is a monic lift of $g_0$ and $h$ is a monic lift of $T^n$.
Because $I$ is contained in the Jacobson radical
of $A$, we find that $g$ and $h$ generate the unit ideal in $A[T]$
(details omitted; hint: use that $A[T]/(g, h)$ is finite over $A$).
Thus $A[T]/(f) = A[T]/(h) \times A[T]/(g)$. It follows that
$h(1 - a)$ and $g(1 - a)$ generate the unit ideal in $A''$ and
we find a product decomposition $A'' = A''_1 \times A''_2$
such that $h(1 - a)$ acts as zero on $A''_1$ and $g(1 - a)$ acts as zero on
$A''_2$. As $h \bmod I = T^n$, as $T, g_0$ generate the unit ideal, and
as $a \bmod I = (1, 0)$ we conclude that the two disjoint union decompositions
$$
\Spec(A''/IA'') = \Spec(A/I) \amalg \Spec(C') =
\Spec(A''_1/IA''_1) \amalg \Spec(A''_2/IA''_2)
$$
are the same: in both cases the first summand corresponds exactly
to those primes of $A''$ containing $IA''$ such that $a$ maps to $1$
in the residue field. This implies that $A''_1/IA''_1 = A/I$
as factor rings of $A''/IA''$ (because product decompositions of
rings correspond $1$-to-$1$ to disjoint union decompositions
of the spectra, see Algebra, Lemmas \ref{algebra-lemma-spec-product} and
\ref{algebra-lemma-disjoint-implies-product}). Moreover, it
follows that $a$ maps to a unit in $A''_1$
(see Algebra, Lemma \ref{algebra-lemma-contained-in-radical}
and use that $IA''_1$ is contained in the radical of $A''_1$
as $A''_1$ is integral over $A$).
Hence $A''_1 = (A''_1)_a$ is a factor of $A''_a = A'_a$ which is
\'etale over $A$. Then $A \to A''_1$ is integral, of finite presentation,
and flat (Algebra, Section \ref{algebra-section-etale}) hence finite
(Algebra, Lemma \ref{algebra-lemma-characterize-finite-in-terms-of-integral})
hence finitely presented as an $A$-module
(Algebra, Lemma \ref{algebra-lemma-finite-finitely-presented-extension})
hence $A''_1$ is finite projective as an $A$-module
(Algebra, Lemma \ref{algebra-lemma-finite-projective}).
Since $A''_1/IA''_1 = A/I$ the module $A''_1$ has rank $1$ as an
along $V(I)$ (see rank function described in
Algebra, Lemma \ref{algebra-lemma-finite-projective}).
Since $I$ is contained in the Jacobson radical
of $A$ we conclude that $A''_1$ has rank $1$ everywhere.
It follows that $A \to A''_1$ is an isomorphism
(exercise: a ring map $R \to S$ such that $S$ is a locally free $R$-module
of rank $1$ is an isomorphism). Thus
$A' \to A'_a \cong A''_a \to (A''_1)_a = A''_1 = A$ is the
desired lift of $\sigma$. In this way we see that (1) implies (2).
\end{proof}
\begin{lemma}
\label{lemma-change-ideal-henselian-pair}
Let $A$ be a ring. Let $I, J \subset A$ be ideals with $V(I) = V(J)$.
Then $(A, I)$ is henselian if and only if $(A, J)$ is henselian.
\end{lemma}
\begin{proof}
For any integral ring map $A \to B$ we see that $V(IB) = V(JB)$.
Hence idempotents of $B/IB$ and $B/JB$ are in bijective correspondence
(Algebra, Lemma \ref{algebra-lemma-disjoint-decomposition}).
It follows that $B \to B/IB$ induces a bijection on sets of
idempotents if and only if $B \to B/JB$ induces a bijection on sets
of idempotents. Thus we conclude by
Lemma \ref{lemma-characterize-henselian-pair}.
\end{proof}
\begin{lemma}
\label{lemma-integral-over-henselian-pair}
Let $(A, I)$ be a henselian pair and let $A \to B$ be an integral ring
map. Then $(B, IB)$ is a henselian pair.
\end{lemma}
\begin{proof}
Immediate from the fourth characterization of henselian pairs in
Lemma \ref{lemma-characterize-henselian-pair} and the fact that the
composition of integral ring maps is integral.
\end{proof}
\begin{lemma}
\label{lemma-henselian-henselian-pair}
Let $I \subset J \subset A$ be ideals of a ring $A$.
The following are equivalent
\begin{enumerate}
\item $(A, I)$ and $(A/I, J/I)$ are henselian pairs, and
\item $(A, J)$ is an henselian pair.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1). Looking at Definition \ref{definition-henselian-pair}
we see that $V(I)$ contains all closed points of $\Spec(A)$
and that $V(J) = V(J/I)$ contains all closed points of $V(I) = \Spec(A/I)$.
Hence $V(J)$ contains all closed points of $\Spec(A)$, i.e.,
$J$ is contained in the Jacobson radical of $A$.
Next, let $f \in A[T]$ be a monic polynomial and
let $\overline{f} = g_0 h_0$ be a factorization in
$A/J[T]$ with $g_0, h_0$ monic generating the unit
ideal in $A/J[T]$. Then we can first lift this factorization
to a factorization in $A/I[T]$ and then to a factorization
in $A[T]$. Thus $(A, J)$ is a henselian pair.
\medskip\noindent
Conversely, assume (2) holds. Then $(A/I, J/I)$ is a henselian pair
by Lemma \ref{lemma-integral-over-henselian-pair}. Let $B$ be an
integral $A$-algebra. Consider the ring maps
$$
B \to B/IB \to B/JB
$$
By Lemma \ref{lemma-characterize-henselian-pair} we find that the composition
and the second arrow induce bijections on idempotents.
Hence so does the first arrow. It follows that $(A, I)$ is a henselian
pair (by the lemma again).
\end{proof}
\begin{lemma}
\label{lemma-product-henselian-pairs}
Let $J$ be a set and let $\{ (A_j, I_j)\}_{j \in J}$ be a collection
of pairs. Then $(\prod_{j \in J} A_j, \prod_{j\in J} I_j)$ is Henselian
if and only if so is each $(A_j, I_j)$.
\end{lemma}
\begin{proof}
For every $j \in J$, the projection $\prod_{j \in J} A_j \rightarrow A_j$
is an integral ring map, so Lemma \ref{lemma-integral-over-henselian-pair}
proves that each $(A_j, I_j)$ is Henselian if
$(\prod_{j \in J} A_j, \prod_{j\in J} I_j)$ is Henselian.
\medskip\noindent
Conversely, suppose that each $(A_j, I_j)$ is a Henselian pair.
Then every $1 + x$ with $x \in \prod_{j \in J} I_j$ is a unit
in $\prod_{j \in J} A_j$ because it is so componentwise by
Algebra, Lemma \ref{algebra-lemma-contained-in-radical} and
Definition \ref{definition-henselian-pair}.
Thus, by Algebra, Lemma \ref{algebra-lemma-contained-in-radical}
again, $\prod_{j \in J} I_j$ is contained in the Jacobson radical
of $\prod_{j \in J} A_j$. Continuing to work componentwise, it
likewise follows that for every monic $f \in (\prod_{j \in J} A_j)[T]$
and every factorization $\overline{f} = g_0h_0$ with monic
$g_0, h_0 \in (\prod_{j \in J} A_j / \prod_{j \in J} I_j)[T] =
(\prod_{j \in J} A_j/I_j)[T]$ that generate the unit ideal in
$(\prod_{j \in J} A_j / \prod_{j \in J} I_j)[T]$, there exists a
factorization $f = gh$ in $(\prod_{j \in J} A_j)[T]$ with $g$, $h$ monic
and reducing to $g_0$, $h_0$. In conclusion, according to
Definition \ref{definition-henselian-pair}
$(\prod_{j \in J} A_j, \prod_{j\in J} I_j)$ is a Henselian pair.
\end{proof}
\begin{lemma}
\label{lemma-irreducible-henselian-pair-connected}
Let $(A, I)$ be a henselian pair. Let $\mathfrak p \subset A$
be a prime ideal. Then $V(\mathfrak p + I)$ is connected.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-integral-over-henselian-pair} we see that
$(A/\mathfrak p, I + \mathfrak p/\mathfrak p)$ is a henselian pair.
Thus it suffices to prove: If $(A, I)$ is a henselian pair and
$A$ is a domain, then $\Spec(A/I) = V(I)$ is connected. If not,
then $A/I$ has a nontrivial idempotent by
Algebra, Lemma \ref{algebra-lemma-characterize-spec-connected}.
By Lemma \ref{lemma-characterize-henselian-pair}
this would imply $A$ has a nontrivial idempotent. This is a contradiction.
\end{proof}
\begin{lemma}
\label{lemma-henselization}
The inclusion functor
$$
\text{category of henselian pairs}
\longrightarrow
\text{category of pairs}
$$
has a left adjoint $(A, I) \mapsto (A^h, I^h)$.
\end{lemma}
\begin{proof}
Let $(A, I)$ be a pair. Consider the category $\mathcal{C}$ consisting
of \'etale 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
$A^h = \colim_{B \in \mathcal{C}} B$ with ideal $I^h = IA^h$ gives
the desired adjoint.
\medskip\noindent
We first prove that $\mathcal{C}$ is directed
(Categories, Definition \ref{categories-definition-directed}).
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 \ref{algebra-lemma-etale})
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 \'etale over $A$ by
Algebra, Lemmas \ref{algebra-lemma-etale} and
\ref{algebra-lemma-map-between-etale}.
Thus the category $\mathcal{C}$ is directed.
\medskip\noindent
Since $B/IB = A/I$ for all objects $B$ of $\mathcal{C}$ we
see that $A^h/I^h = A^h/IA^h = \colim B/IB = \colim A/I = A/I$.
\medskip\noindent
Next, we show that $A^h = \colim_{B \in \mathcal{C}} B$ with
$I^h = IA^h$ is a henselian pair. To do this we will verify
condition (2) of Lemma \ref{lemma-characterize-henselian-pair}.
Namely, suppose given an \'etale 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 \'etale ring map $B \to B'$ such that
$A' = B' \otimes_B A^h$. See Algebra, Lemma \ref{algebra-lemma-etale}.
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 \'etale 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.
\medskip\noindent
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 \'etale and induces
an isomorphism $A'/I' \to B'/I'B'$. Hence there is a section
$\sigma_B : B' \to A'$ by Lemma \ref{lemma-characterize-henselian-pair}.
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 \ref{algebra-lemma-diagonal-unramified}. 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 \ref{lemma-idempotents-determined-modulo-radical}.
Hence $\sigma_B = \sigma_B'$ as desired.
Using the commutativity we obtain
$$
A^h = \colim_{B \in \mathcal{C}} B \to
\colim_{B \in \mathcal{C}} A' \otimes_A B \xrightarrow{\colim \sigma_B} A'
$$
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.
\end{proof}
\begin{lemma}
\label{lemma-henselization-local-ring}
\begin{slogan}
Compatibility henselization of pairs and of local rings.
\end{slogan}
The functor of Lemma \ref{lemma-henselization} associates to a local ring
$(A, \mathfrak m)$ its henselization.
\end{lemma}
\begin{proof}
First proof: in the proof of
Algebra, Lemma \ref{algebra-lemma-henselization}
it is shown that the henselization of $A$ is given by
the colimit used to construct $A^h$ in
Lemma \ref{lemma-henselization}.
Second proof: Both the henselization $S$ and the ring $A^h$ of
Lemma \ref{lemma-henselization} are filtered colimits of \'etale
$A$-algebras, henselian, and have residue fields equal to
$\kappa(\mathfrak m)$. Hence they are canonically isomorphic by
Algebra, Lemma \ref{algebra-lemma-uniqueness-henselian}.
\end{proof}
\begin{lemma}
\label{lemma-henselization-flat}
Let $(A, I)$ be a pair. Let $(A^h, I^h)$ be as in
Lemma \ref{lemma-henselization}. 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 $n$.
\end{lemma}
\begin{proof}
In the proof of Lemma \ref{lemma-henselization} we have seen that
$A^h$ is a filtered colimit of \'etale $A$-algebras $B$ such that
$A/I \to B/IB$ is an isomorphism and we have seen that
$I^h = IA^h$. As an \'etale ring map is flat
(Algebra, Lemma \ref{algebra-lemma-etale}) we conclude that
$A \to A^h$ is flat by Algebra, Lemma \ref{algebra-lemma-colimit-flat}.
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 \ref{algebra-lemma-lift-basis}).
Taking the colimit we find that $A/I^n = A^h/I^nA^h$
as desired.
\end{proof}
\begin{lemma}
\label{lemma-henselization-Noetherian-pair}
Let $(A, I)$ be a pair with $A$ Noetherian. Let $(A^h, I^h)$ be as in
Lemma \ref{lemma-henselization}. 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.
\end{lemma}
\begin{proof}
The first statement is an immediate consequence of
Lemma \ref{lemma-henselization-flat}
and in fact holds without assuming $A$ is Noetherian.
In the proof of Lemma \ref{lemma-henselization} we have seen that
$A^h$ is a filtered colimit of \'etale $A$-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 \ref{lemma-henselization-flat}).
By Algebra, Lemma \ref{algebra-lemma-completion-flat} 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 \ref{algebra-lemma-colimit-rings-flat}.
Since $I^h = IA^h$ is contained in the radical ideal 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 \ref{algebra-lemma-ff}.
By Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian}
the ring $A^\wedge$ is Noetherian.
Hence we conclude that $A^h$ is Noetherian by
Algebra, Lemma \ref{algebra-lemma-descent-Noetherian}.
\end{proof}
\begin{lemma}
\label{lemma-henselization-colimit}
Let $(A, I) = \colim (A_i, I_i)$ be a colimit of pairs. The functor of
Lemma \ref{lemma-henselization} gives
$A^h = \colim A_i^h$ and $I^h = \colim I_i^h$.
\end{lemma}
\begin{proof}
This is true for any left adjoint, see
Categories, Lemma \ref{categories-lemma-adjoint-exact}.
\end{proof}
\begin{lemma}
\label{lemma-henselization-integral}
Let $(A, I) \to (B, J)$ be a map of pairs.
Let $(A^h , I^h) \to (B^h, J^h)$ be the induced map
on henselizations (Lemma \ref{lemma-henselization}).
If $A \to B$ is integral, then the induced map
$A^h \otimes_A B \to B^h$ is an isomorphism.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-characterize-henselian-pair}
the pair $(A^h \otimes_A B, I^h(A^h \otimes_A B))$ is henselian.
By the universal property 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.
\end{proof}
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