The Stacks project

85.4 Rig-étale homomorphisms

In this and some of the later sections we will study ring maps as in Lemma 85.4.1. Condition (4) is one of the conditions used in [ArtinII] to define modifications. Ring maps like this are sometimes called rig-étale or rigid-étale ring maps in the literature. These and the analogously defined rig-smooth ring maps were studied in [Elkik]. A detailed exposition can also be found in [Abbes]. Our main goal will be to show that rig-étale ring maps are completions of finite type algebras, a result very similar to results found in Elkik's paper [Elkik].

Lemma 85.4.1. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Let $B$ be an object of (85.2.0.2). The following are equivalent

  1. there exists a $c \geq 0$ such that multiplication by $a$ on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is zero in $D(B)$ for all $a \in I^ c$,

  2. there exits a $c \geq 0$ such that $H^ i(\mathop{N\! L}\nolimits ^\wedge _{B/A})$, $i = -1, 0$ is annihilated by $I^ c$,

  3. there exists a $c \geq 0$ such that $H^ i(\mathop{N\! L}\nolimits _{B_ n/A_ n})$, $i = -1, 0$ is annihilated by $I^ c$ for all $n \geq 1$,

  4. $B = A[x_1, \ldots , x_ r]^\wedge /J$ and for every $a \in I$ there exists a $c \geq 0$ such that

    1. $a^ c$ annihilates $H^0(\mathop{N\! L}\nolimits ^\wedge _{B/A})$, and

    2. there exist $f_1, \ldots , f_ r \in J$ such that $a^ c J \subset (f_1, \ldots , f_ r) + J^2$.

Proof. The equivalence of (1) and (2) follows from Lemma 85.3.5. The equivalence of (1) $+$ (2) and (3) follows from Lemma 85.3.1. Some details omitted.

Assume the equivalent conditions (1), (2), (3) holds and let $B = A[x_1, \ldots , x_ r]^\wedge /J$ be a presentation (see Lemma 85.2.1). Let $a \in I$. Let $c$ be such that multiplication by $a^ c$ is zero on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ which exists by (1). By Lemma 85.3.4 there exists a map $\alpha : \bigoplus B\text{d}x_ i \to J/J^2$ such that $\text{d} \circ \alpha $ and $\alpha \circ \text{d}$ are both multiplication by $a^ c$. Let $f_ i \in J$ be an element whose class modulo $J^2$ is equal to $\alpha (\text{d}x_ i)$. Then we see that (4)(a), (b) hold.

Assume (4) holds. Say $I = (a_1, \ldots , a_ t)$. Let $c_ i \geq 0$ be the integer such that (4)(a), (b) hold for $a_ i^{c_ i}$. Then we see that $I^{\sum c_ i}$ annihilates $H^0(\mathop{N\! L}\nolimits ^\wedge _{B/A})$. Let $f_{i, 1}, \ldots , f_{i, r} \in J$ be as in (4)(b) for $a_ i$. Consider the composition

\[ B^{\oplus r} \to J/J^2 \to \bigoplus B\text{d}x_ i \]

where the $j$th basis vector is mapped to the class of $f_{i, j}$ in $J/J^2$. By (4)(a) and (b) the cokernel of the composition is annihilated by $a_ i^{2c_ i}$. Thus this map is surjective after inverting $a_ i^{c_ i}$, and hence an isomorphism (Algebra, Lemma 10.15.4). Thus the kernel of $B^{\oplus r} \to \bigoplus B\text{d}x_ i$ is $a_ i$-power torsion, and hence $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A}) = \mathop{\mathrm{Ker}}(J/J^2 \to \bigoplus B\text{d}x_ i)$ is $a_ i$-power torsion. Since $B$ is Noetherian (Lemma 85.2.2), all modules including $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A})$ are finite. Thus $a_ i^{d_ i}$ annihilates $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A})$ for some $d_ i \geq 0$. It follows that $I^{\sum d_ i}$ annihilates $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A})$ and we see that (2) holds. $\square$

Lemma 85.4.2. Let $A$ be a Noetherian ring and let $I$ be an ideal. Let $B$ be a finite type $A$-algebra.

  1. If $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$, then $B^\wedge $ satisfies the equivalent conditions of Lemma 85.4.1.

  2. If $B^\wedge $ satisfies the equivalent conditions of Lemma 85.4.1, then there exists $g \in 1 + IB$ such that $\mathop{\mathrm{Spec}}(B_ g)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$.

Proof. Assume $B^\wedge $ satisfies the equivalent conditions of Lemma 85.4.1. The naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}$ is a complex of finite type $B$-modules and hence $H^{-1}$ and $H^0$ are finite $B$-modules. Completion is an exact functor on finite $B$-modules (Algebra, Lemma 10.96.2) and $\mathop{N\! L}\nolimits ^\wedge _{B^\wedge /A}$ is the completion of the complex $\mathop{N\! L}\nolimits _{B/A}$ (this is easy to see by choosing presentations). Hence the assumption implies there exists a $c \geq 0$ such that $H^{-1}/I^ nH^{-1}$ and $H^0/I^ nH^0$ are annihilated by $I^ c$ for all $n$. By Nakayama's lemma (Algebra, Lemma 10.19.1) this means that $I^ cH^{-1}$ and $I^ cH^0$ are annihilated by an element of the form $g = 1 + x$ with $x \in IB$. After inverting $g$ (which does not change the quotients $B/I^ nB$) we see that $\mathop{N\! L}\nolimits _{B/A}$ has cohomology annihilated by $I^ c$. Thus $A \to B$ is étale at any prime of $B$ not lying over $V(I)$ by the definition of étale ring maps, see Algebra, Definition 10.142.1.

Conversely, assume that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$. Then for every $a \in I$ there exists a $c \geq 0$ such that multiplication by $a^ c$ is zero $\mathop{N\! L}\nolimits _{B/A}$. Since $\mathop{N\! L}\nolimits ^\wedge _{B^\wedge /A}$ is the derived completion of $\mathop{N\! L}\nolimits _{B/A}$ (see Lemma 85.3.1) it follows that $B^\wedge $ satisfies the equivalent conditions of Lemma 85.4.1. $\square$

Lemma 85.4.3. Assume the map $(A_1, I_1) \to (A_2, I_2)$ is as in Remark 85.2.3 with $A_1$ and $A_2$ Noetherian. Let $B_1$ be in (85.2.0.2) for $(A_1, I_1)$. Let $B_2$ be the base change of $B_1$. If multiplication by $f_1 \in B_1$ on $\mathop{N\! L}\nolimits ^\wedge _{B_1/A_1}$ is zero in $D(B_1)$, then multiplication by the image $f_2 \in B_2$ on $\mathop{N\! L}\nolimits ^\wedge _{B_2/A_2}$ is zero in $D(B_2)$.

Proof. Choose a presentation $B_1 = A_1[x_1, \ldots , x_ r]^\wedge /J_1$. Since $A_2/I_2^ n[x_1, \ldots , x_ r] = A_1/I_1^{cn}[x_1, \ldots , x_ r] \otimes _{A_1/I_1^{cn}} A_2/I_2^ n$ we have

\[ A_2[x_1, \ldots , x_ r]^\wedge = (A_1[x_1, \ldots , x_ r]^\wedge \otimes _{A_1} A_2)^\wedge \]

where we use $I_2$-adic completion on both sides (but of course $I_1$-adic completion for $A_1[x_1, \ldots , x_ r]^\wedge $). Set $J_2 = J_1 A_2[x_1, \ldots , x_ r]^\wedge $. Arguing similarly we get the presentation

\begin{align*} B_2 & = (B_1 \otimes _{A_1} A_2)^\wedge \\ & = \mathop{\mathrm{lim}}\nolimits \frac{A_1/I_1^{cn}[x_1, \ldots , x_ r]}{J_1(A_1/I_1^{cn}[x_1, \ldots , x_ r])} \otimes _{A_1/I_1^{cn}} A_2/I_2^ n \\ & = \mathop{\mathrm{lim}}\nolimits \frac{A_2/I_2^ n[x_1, \ldots , x_ r]}{J_2(A_2/I_2^ n[x_1, \ldots , x_ r])} \\ & = A_2[x_1, \ldots , x_ r]^\wedge /J_2 \end{align*}

for $B_2$ over $A_2$. Consider the commutative diagram

\[ \xymatrix{ \mathop{N\! L}\nolimits ^\wedge _{B_1/A_1} : \ar[d] & J_1/J_1^2 \ar[r]_-{\text{d}} \ar[d] & \bigoplus B_1\text{d}x_ i \ar[d] \\ \mathop{N\! L}\nolimits ^\wedge _{B_2/A_2} : & J_2/J_2^2 \ar[r] & \bigoplus B_2\text{d}x_ i } \]

The induced arrow $J_1/J_1^2 \otimes _{B_1} B_2 \to J_2/J_2^2$ is surjective because $J_2$ is generated by the image of $J_1$. By Lemma 85.3.4 there is a map $\alpha _1 : \bigoplus B\text{d}x_ i \to J_1/J_1^2$ such that $f_1 \text{id}_{\bigoplus B_1\text{d}x_ i} = \text{d} \circ \alpha _1$ and $f_1 \text{id}_{J_1/J_1^2} = \alpha _1 \circ \text{d}$. We define $\alpha _2 : \bigoplus B_1\text{d}x_ i \to J_2/J_2^2$ by mapping $\text{d}x_ i$ to the image of $\alpha _1(\text{d}x_ i)$ in $J_2/J_2^2$. Because the image of the vertical arrows contains generators of the modules $J_2/J_2^2$ and $\bigoplus B_2 \text{d}x_ i$ it follows that $\alpha _2$ also defines a homotopy between multiplication by $f_2$ and the zero map. $\square$

Lemma 85.4.4. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $B$ be a finite type $A$-algebra such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$. Let $C$ be a Noetherian $A$-algebra. Then any $A$-algebra map $B^\wedge \to C^\wedge $ of $I$-adic completions comes from a unique $A$-algebra map

\[ B \longrightarrow C^ h \]

where $C^ h$ is the henselization of the pair $(C, IC)$ as in More on Algebra, Lemma 15.12.1. Moreover, any $A$-algebra homomorphism $B \to C^ h$ factors through some étale $C$-algebra $C'$ such that $C/IC \to C'/IC'$ is an isomorphism.

Proof. Uniqueness follows from the fact that $C^ h$ is a subring of $C^\wedge $, see for example More on Algebra, Lemma 15.12.4. The final assertion follows from the fact that $C^ h$ is the filtered colimit of these $C$-algebras $C'$, see proof of More on Algebra, Lemma 15.12.1. Having said this we now turn to the proof of existence.

Let $\varphi : B^\wedge \to C^\wedge $ be the given map. This defines a section

\[ \sigma : (B \otimes _ A C)^\wedge \longrightarrow C^\wedge \]

of the completion of the map $C \to B \otimes _ A C$. We may replace $(A, I, B, C, \varphi )$ by $(C, IC, B \otimes _ A C, C, \sigma )$. In this way we see that we may assume that $A = C$.

Proof of existence in the case $A = C$. In this case the map $\varphi : B^\wedge \to A^\wedge $ is necessarily surjective. By Lemmas 85.4.2 and 85.3.2 we see that the cohomology groups of $\mathop{N\! L}\nolimits ^\wedge _{A^\wedge /\! _\varphi B^\wedge }$ are annihilated by a power of $I$. Since $\varphi $ is surjective, this implies that $\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Ker}}(\varphi )^2$ is annihilated by a power of $I$. Hence $\varphi : B^\wedge \to A^\wedge $ is the completion of a finite type $B$-algebra $B \to D$, see More on Algebra, Lemma 15.98.4. Hence $A \to D$ is a finite type algebra map which induces an isomorphism $A^\wedge \to D^\wedge $. By Lemma 85.4.2 we may replace $D$ by a localization and assume that $A \to D$ is étale away from $V(I)$. Since $A^\wedge \to D^\wedge $ is an isomorphism, we see that $\mathop{\mathrm{Spec}}(D) \to \mathop{\mathrm{Spec}}(A)$ is also étale in a neighbourhood of $V(ID)$ (for example by More on Morphisms, Lemma 37.12.3). Thus $\mathop{\mathrm{Spec}}(D) \to \mathop{\mathrm{Spec}}(A)$ is étale. Therefore $D$ maps to $A^ h$ and the lemma is proved. $\square$


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