The Stacks project

86.8 Rig-étale algebras

In view of our definition of rig-smooth algebras (Definition 86.4.1), the following definition should not come as a surprise.

Definition 86.8.1. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Let $B$ be an object of (86.2.0.2). We say $B$ is rig-étale over $(A, I)$ if there exists an integer $c \geq 0$ such that for all $a \in I^ c$ multiplication by $a$ on $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is zero in $D(B)$.

Condition (7) in the next lemma is one of the conditions used in [ArtinII] to define formal modifications. We have added it to the list of conditions to facilitate comparison with our conditions later on.

Lemma 86.8.2. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Let $B$ be an object of (86.2.0.2). Write $B = A[x_1, \ldots , x_ r]^\wedge /J$ (Lemma 86.2.2) and let $\mathop{N\! L}\nolimits _{B/A}^\wedge = (J/J^2 \to \bigoplus B\text{d}x_ i)$ be its naive cotangent complex (86.3.0.1). The following are equivalent

  1. $B$ is rig-étale over $(A, I)$,

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

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

  4. 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 $n \geq 1$ where $A_ n = A/I^ n$ and $B_ n = B/I^ nB$,

  5. for every $a \in I$ there exists a $c \geq 0$ such that

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

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

  6. for every $a \in I$ there exist $f_1, \ldots , f_ r \in J$ and $c \geq 0$ such that

    1. $\det _{1 \leq i, j \leq r}(\partial f_ j/\partial x_ i)$ divides $a^ c$ in $B$, and

    2. $a^ c J \subset (f_1, \ldots , f_ r) + J^2$.

  7. choosing generaters $f_1, \ldots , f_ t$ for $J$ we have

    1. the Jacobian ideal of $B$ over $A$, namely the ideal in $B$ generated by the $r \times r$ minors of the matrx $(\partial f_ j/\partial x_ i)_{1 \leq i \leq r, 1 \leq j \leq t}$, contains the ideal $I^ cB$ for some $c$, and

    2. the Cramer ideal of $B$ over $A$, namely the ideal in $B$ generated by the image in $B$ of the $r$th Fitting ideal of $J$ as an $A[x_1, \ldots , x_ r]^\wedge $-module, contains $I^ cB$ for some $c$.

Proof. The equivalence of (1) and (2) is a restatement of Definition 86.8.1.

The equivalence of (2) and (3) follows from More on Algebra, Lemma 15.83.11.

The equivalence of (3) and (4) follows from the fact that the systems $\{ \mathop{N\! L}\nolimits _{B_ n/A_ n}\} $ and $\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B_ n$ are strictly isomorphic, see Lemma 86.3.3. Some details omitted.

Assume (2). Let $a \in I$. Let $c$ be such that multiplication by $a^ c$ is zero on $\mathop{N\! L}\nolimits _{B/A}^\wedge $. By More on Algebra, Lemma 15.83.4 part (1) 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 $\alpha (\text{d}x_ i)$. A simple calculation gives that (6)(a), (b) hold.

We omit the verification that (6) implies (5); it is just a statement on two term complexes over $B$ of the form $M \to B^{\oplus r}$.

Assume (5) holds. Say $I = (a_1, \ldots , a_ t)$. Let $c_ i \geq 0$ be the integer such that (5)(a), (b) hold for $a_ i^{c_ i}$. Then we see that $I^{\sum c_ i}$ annihilates $H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge )$. Let $f_{i, 1}, \ldots , f_{i, r} \in J$ be as in (5)(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 (5)(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 10.16.4). Thus the kernel of $B^{\oplus r} \to \bigoplus B\text{d}x_ i$ is $a_ i$-power torsion, and hence $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge ) = \mathop{\mathrm{Ker}}(J/J^2 \to \bigoplus B\text{d}x_ i)$ is $a_ i$-power torsion. Since $B$ is Noetherian (Lemma 86.2.2), all modules including $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ are finite. Thus $a_ i^{d_ i}$ annihilates $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ for some $d_ i \geq 0$. It follows that $I^{\sum d_ i}$ annihilates $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ and we see that (3) holds.

Thus conditions (2), (3), (4), (5), and (6) are equivalent. Thus it remains to show that these conditions are equivalent with (7). Observe that the Cramer ideal $\text{Fit}_ r(J) B$ is equal to $\text{Fit}_ r(J/J^2)$ as $J/J^2 = J \otimes _{A[x_1, \ldots , x_ r]^\wedge } B$, see More on Algebra, Lemma 15.8.4 part (3). Also, observe that the Jacobian ideal is just $\text{Fit}_0(H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge ))$. Thus we see that the equivalence of (3) and (7) is a purely algebraic question which we discuss in the next paragraph.

Let $R$ be a Noetherian ring and let $I \subset R$ be an ideal. Let $M \xrightarrow {d} R^{\oplus r}$ be a two term complex. We have to show that the following are equivalent

  1. the cohomology of $M \to R^{\oplus r}$ is annihilated by a power of $I$, and

  2. the ideals $\text{Fit}_ r(M)$ and $\text{Fit}_0(\text{Coker}(d))$ contain a power of $I$.

Since $R$ is Noetherian, we can reformulate part (2) as an inclusion of the corresponding closed subschemes, see Algebra, Lemmas 10.17.2 and 10.32.5. On the other hand, over the complement of $V(\text{Fit}_0(\mathop{\mathrm{Coker}}(d)))$ the cokernel of $d$ vanishes and over the complement of $V(\text{Fit}_ r(M))$ the module $M$ is locally generated by $r$ elements, see More on Algebra, Lemma 15.8.6. Thus (B) is equivalent to

  1. away from $V(I)$ the cokernel of $d$ vanishes and the module $M$ is locally generated by $\leq r$ elements.

Of course this is equivalent to the condition that $M \to R^{\oplus r}$ has vanishing cohomology over $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ which in turn is equivalent to (A). This finishes the proof. $\square$

Lemma 86.8.3. Let $A$ be a Noetherian ring and let $I$ be an ideal. Let $B$ be an object of (86.2.0.2). If $B$ is rig-étale over $(A, I)$, then $B$ is rig-smooth over $(A, I)$.

Proof. Immediate from Definitions 86.4.1 and 86.8.1. $\square$

Lemma 86.8.4. 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 86.8.2.

  2. If $B^\wedge $ satisfies the equivalent conditions of Lemma 86.8.2, 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 86.8.2. 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 10.97.2) and $\mathop{N\! L}\nolimits _{B^\wedge /A}^\wedge $ 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 10.20.1) 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 10.143.1.

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 _{B^\wedge /A}^\wedge $ is the derived completion of $\mathop{N\! L}\nolimits _{B/A}$ (see Lemma 86.3.3) it follows that $B^\wedge $ satisfies the equivalent conditions of Lemma 86.8.2. $\square$

Lemma 86.8.5. Let $(A_1, I_1) \to (A_2, I_2)$ be as in Remark 86.2.3 with $A_1$ and $A_2$ Noetherian. Let $B_1$ be in (86.2.0.2) 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. By Lemma 86.3.4 there is a map

\[ \mathop{N\! L}\nolimits _{B_1/A_1} \otimes _{B_2} B_1 \to \mathop{N\! L}\nolimits _{B_2/A_2} \]

which induces and isomorphism on $H^0$ and a surjection on $H^{-1}$. Thus the result by More on Algebra, Lemma 15.83.8. $\square$

Lemma 86.8.6. Let $A_1 \to A_2$ be a map of Noetherian rings. Let $I_ i \subset A_ i$ be an ideal such that $V(I_1A_2) = V(I_2)$. Let $B_1$ be in (86.2.0.2) for $(A_1, I_1)$. Let $B_2$ be the base change of $B_1$ as in Remark 86.2.3. If $B_1$ is rig-étale over $(A_1, I_1)$, then $B_2$ is rig-étale over $(A_2, I_2)$.

Proof. Follows from Lemma 86.8.5 and Definition 86.8.1 and the fact that $I_2^ c \subset I_1A_2$ for some $c \geq 0$ as $A_2$ is Noetherian. $\square$

Lemma 86.8.7. 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 86.8.4 and 86.3.5 we see that the cohomology groups of $\mathop{N\! L}\nolimits _{A^\wedge /\! _\varphi B^\wedge }^\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 15.106.4. Hence $A \to D$ is a finite type algebra map which induces an isomorphism $A^\wedge \to D^\wedge $. By Lemma 86.8.4 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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