The Stacks project

Proof. Essential surjectivity holds by Algebra, Lemma 10.143.10. 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 10.138.17. Finally, suppose that $f, g : B \to B'$ are two $A$-algebra maps with $f \bmod I = g \bmod I$. 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 10.151.4 and 10.151.3 (to see that étale is unramified). Then $(f \otimes g)(e) \in IB'$. Since $IB'$ is locally nilpotent (Algebra, Lemma 10.32.3) this implies $(f \otimes g)(e) = 0$ by Algebra, Lemma 10.32.6. Thus $f = g$.

It is clear that $I$ is contained in the Jacobson 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 $A$. $\square$

Proof. Fix $n$. Let $a \in A$ be an element which maps to $1$ in $A_ n$. By Algebra, Lemma 10.32.4 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 10.19.1 the ideal $I_ n$ is contained in the Jacobson 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 15.11.2 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 $m \geq n$. At each step we have to verify that our lifts $g_ m, h_ m$ generate the unit ideal in $A_ n[T]$; this follows from the corresponding fact for $g_ n, h_ n$ and the fact that $\mathop{\mathrm{Spec}}(A_ n[T]) = \mathop{\mathrm{Spec}}(A_ m[T])$ because the kernel of $A_ m \to A_ n$ is locally nilpotent. As $A = \mathop{\mathrm{lim}}\nolimits A_ m$ this finishes the proof. $\square$

Proof. By Algebra, Lemma 10.96.6 the ideal $I$ is contained in the Jacobson 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 15.11.2 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. $\square$

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 10.122.3). Consider the diagram of topological spaces

By Algebra, Theorem 10.123.12 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 10.41.6 and 10.36.22. We conclude that $\mathop{\mathrm{Spec}}(B'/IB') \supset \phi (T)$ is open and closed. By Algebra, Lemma 10.24.3 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. $\square$

Proof. Assume (2) holds. Then $I$ is contained in the Jacobson radical of $A$, since otherwise there would be a nonunit $f \in A$ congruent to $1$ modulo $I$ 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 10.41.6 and 10.36.22. Thus the map from idempotents of $B$ to idempotents of $B/IB$ is injective by Lemma 15.10.2. On the other hand, since (2) holds, every idempotent of $B/IB$ lifts to an idempotent of $B$ by Lemma 15.9.10. 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 generating the unit ideal in $A/I[T]$. Set $B = A[T]/(f)$. Let $\overline{e}$ be the idempotent of $B/IB$ corresponding to the decomposition

of $A$-algebras. Let $e \in B$ be an idempotent lifting $\overline{e}$ which exists as we assumed (3). This gives a product decomposition

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

is a factorization into monic polynomials reducing to the given factorization modulo $I$. Here $\text{CharPol}_ A$ denotes the characteristic polynomial of an endomorphism of a finite locally free module over $A$. If the module is free the $\text{CharPol}_ A$ is defined as the characteristic polynomial of the corresponding matrix and in general one uses Algebra, Lemma 10.24.2 to glue. Details omitted. Thus (3) implies (1).

Assume (1). Let $f$ be as in (5). The factorization of $f \bmod I$ as $T^ n$ times $T - 1$ lifts to a factorization $f = gh$ with $g$ and $h$ monic by Definition 15.11.1. Then $h$ has to have degree $1$ and we see that $f$ has a root reducing to $1$ modulo $1$. Finally, $I$ is contained in the Jacobson radical by the definition of a henselian pair. Thus (1) implies (5).

Before we give the proof of the last step, let us show that the root $\alpha $ in (5), if it exists, is unique. Namely, Due to the explicit shape of $f(T)$, we have $f'(\alpha ) \in 1 + I$ where $f'$ is the derivative of $f$ with respect to $T$. An elementary argument shows that

This shows that any other root $\alpha ' \in 1 + I$ of $f(T)$ satisfies $0 = f(\alpha ') - f(\alpha ) = (\alpha ' - \alpha )(1 + i)$ for some $i \in I$, so that, since $1 + i$ is a unit in $A$, we have $\alpha = \alpha '$.

Assume (5). We will show that (2) holds, in other words, that for every étale map $A \to A'$, every section $\sigma : A' \to A/I$ modulo $I$ lifts to a section $A' \to A$. Since $A \to A'$ is étale, the section $\sigma $ determines a decomposition

of $A/I$-algebras. Namely, the surjective ring map $A'/IA' \to A/I$ is étale by Algebra, Lemma 10.143.8 and then we get the desired idempotent by Algebra, Lemma 10.143.9. We will show that this decomposition lifts to a decomposition

of $A$-algebras with $A'_1$ integral over $A$. Then $A \to A'_1$ is integral and étale and $A/I \to A'_1/IA'_1$ is an isomorphism, thus $A \to A'_1$ is an isomorphism by Lemma 15.10.3 (here we also use that an étale ring map is flat and of finite presentation, see Algebra, Lemma 10.143.3).

Let $B'$ be the integral closure of $A$ in $A'$. By Lemma 15.11.5 we may decompose

as $A/I$-algebras compatibly with (15.11.6.1) and we may find $b \in B'$ that lifts $(1, 0)$ such that $B'_ b \to A'_ b$ is an isomorphism. If the decomposition (15.11.6.3) lifts to a decomposition

of $A$-algebras, then the induced decomposition $A' = A'_1 \times A'_2$ will give the desired (15.11.6.2): indeed, since $b$ is a unit in $B'_1$ (details omitted), we will have $B'_1 \cong A'_1$, so that $A'_1$ will be integral over $A$.

Choose a finite $A$-subalgebra $B'' \subset B'$ containing $b$ (observe that any finitely generated $A$-subalgebra of $B'$ is finite over $A$). After enlarging $B''$ we may assume $b$ maps to an idempotent in $B''/IB''$ producing

Since $B'_ b \cong A'_ b$ we see that $B'_ b$ is of finite type over $A$. Say $B'_ b$ is generated by $b_1/b^ n, \ldots , b_ t/b^ n$ over $A$ and enlarge $B''$ so that $b_1, \ldots , b_ t \in B''$. Then $B''_ b \to B'_ b$ is surjective as well as injective, hence an isomorphism. In particular, we see that $C''_1 = A/I$! Therefore $A/I \to C''_1$ is an isomorphism, in particular surjective. By Lemma 15.10.4 we can find an $f(T) \in A[T]$ of the form

with $a_ n, \ldots , a_0 \in I$ and $n \ge 1$ such that $f(b) = 0$. In particular, we find that $B'$ is a $A[T]/(f)$-algebra. By (5) we deduce there is a root $a \in 1 + I$ of $f$. This produces a product decomposition $A[T]/(f) = A[T]/(T - a) \times D$ compatible with the splitting (15.11.6.3) of $B'/IB'$. The induced splitting of $B'$ is then a desired (15.11.6.4). $\square$

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 10.21.3). 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 15.11.6. $\square$

Proof. Immediate from the fourth characterization of henselian pairs in Lemma 15.11.6 and the fact that the composition of integral ring maps is integral. $\square$

Proof. Assume (1). Let $B$ be an integral $A$-algebra. Consider the ring maps

By Lemma 15.11.6 we find that both arrows induce bijections on idempotents. Hence so does the composition. Whence $(A, J)$ is a henselian pair by Lemma 15.11.6.

Conversely, assume (2) holds. Then $(A/I, J/I)$ is a henselian pair by Lemma 15.11.8. Let $B$ be an integral $A$-algebra. Consider the ring maps

By Lemma 15.11.6 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). $\square$

Proof. By Lemma 15.11.8 the pair $(A/I, (I' + I)/I)$ is henselian. Thus we get the conclusion from Lemma 15.11.9. $\square$

Proof. For every $j \in J$, the projection $\prod _{j \in J} A_ j \rightarrow A_ j$ is an integral ring map, so Lemma 15.11.8 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 10.19.1 and Definition 15.11.1. Thus, by Algebra, Lemma 10.19.1 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 15.11.1 $(\prod _{j \in J} A_ j, \prod _{j\in J} I_ j)$ is a Henselian pair. $\square$

Proof. By Categories, Lemma 4.14.11, we only need to consider products and equalizers. For products, the claim follows from Lemma 15.11.11. Thus, consider an equalizer diagram

in which the pairs $(A', I')$ and $(A'', I'')$ are henselian. To check that the pair $(A, I)$ is also henselian, we will use the Gabber's criterion in Lemma 15.11.6. Every element of $1 + I$ is a unit in $A$ because, due to the uniqueness of the inverses of units, this may be checked in $(A', I')$. Thus $I$ is contained in the Jacobson radical of $A$, see Algebra, Lemma 10.19.1. Thus, let

be a polynomial in $A[T]$ with $a_{N - 1}, \dotsc , a_0 \in I$ and $N \ge 1$. The image of $f(T)$ in $A'[T]$ has a unique root $\alpha ' \in 1 + I'$ and likewise for the further image in $A''[T]$. Thus, due to the uniqueness, $\varphi (\alpha ') = \psi (\alpha ')$, to the effect that $\alpha '$ defines a root of $f(T)$ in $1 + I$, as desired. $\square$

Proof. If $u \in 1 + I$ then for some $j \in J$ we see that $u$ is the image of some $u_ j \in 1 + I_ j$. Then $u_ j$ is invertible in $A_ j$ by Algebra, Lemma 10.19.1 and the assumption that $I_ j$ is contained in the Jacobson radical of $A_ j$. Hence $u$ is invertible in $A$. Thus $I$ is contained in the Jacobson radical of $A$ (by the lemma).

Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0 h_0$ be a factorization with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. Write $1 = g_0 g'_0 + h_0 h'_0$ for some $g'_0, h'_0 \in A/I[T]$. Since $A = \mathop{\mathrm{colim}}\nolimits A_ j$ and $A/I = \mathop{\mathrm{colim}}\nolimits A_ j/I_ j$ are filtered colimits we can find a $j \in J$ and $f_ j \in A_ j$ and a factorization $\overline{f}_ j = g_{j, 0} h_{j, 0}$ with $g_{j, 0}, h_{j, 0} \in A_ j/I_ j[T]$ monic and $1 = g_{j, 0} g'_{j, 0} + h_{j, 0} h'_{j, 0}$ for some $g'_{j, 0}, h'_{j, 0} \in A_ j/I_ j[T]$ with $f_ j, g_{j, 0}, h_{j, 0}, g'_{j, 0}, h'_{j, 0}$ mapping to $f, g_0, h_0, g'_0, h'_0$. Since $(A_ j, I_ j)$ is a henselian pair, we can lift $\overline{f}_ j = g_{j, 0} h_{j, 0}$ to a factorization over $A_ j$ and taking the image in $A$ we obtain a corresponding factorization in $A$. Hence $(A, I)$ is henselian. $\square$

Proof. Combine Lemmas 15.11.9, 15.11.10, and 15.11.13. $\square$

Proof. By Lemma 15.11.8 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 10.21.4. By Lemma 15.11.6 this would imply $A$ has a nontrivial idempotent. This is a contradiction. $\square$