# The Stacks Project

## 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 pair is a pair $(A, I)$ where $A$ is a ring and $I \subset A$ is an ideal. A morphism 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 pair is a pair $(A, I)$ satisfying

1. $I$ is contained in the Jacobson radical of $A$, and
2. 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

1. $(A, I)$ is a henselian pair,
2. 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 $\sigma$,
3. for any finite $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents, and
4. 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

1. $(A, I)$ and $(A/I, J/I)$ are henselian pairs, and
2. $(A, J)$ 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. $\square$

1. 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 10.22.1 to glue.

The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 2158–2838 (see updates for more information).

\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
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}
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

\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 $e = 0$
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
\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}

Comment #2997 by Peng Du on November 14, 2017 a 10:41 am UTC

In Lemma 15.10.11 (1), the pair (J/I, A/I) should be written as (A/I, J/I).

Comment #3120 by Johan (site) on February 1, 2018 a 1:04 am UTC

Already got fixed. Please leave comment on a lemma on the page of the lemma. Somebody else already left the comment there...

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