The Stacks project

Lemma 86.10.2. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $B$ be an object of (86.2.0.2) which is rig-étale over $(A, I)$. Then there exists a finite type $A$-algebra $C$ and an isomorphism $B \cong C^\wedge $.

Proof. We prove this lemma by induction on the number of generators of $I$. Say $I = (a_1, \ldots , a_ t)$. If $t = 0$, then $I = 0$ and there is nothing to prove. If $t = 1$, then the lemma follows from Lemma 86.10.1. Assume $t > 1$.

For any $m \geq 1$ set $\bar A_ m = A/(a_ t^ m)$. Consider the ideal $\bar I_ m = (\bar a_1, \ldots , \bar a_{t - 1})$ in $\bar A_ m$. Observe that $V(I \bar A_ m) = V(\bar I_ m)$. Let $B_ m = B/(a_ t^ m)$ be the base change of $B$ for the map $(A, I) \to (\bar A_ m, \bar I_ m)$, see Remark 86.2.4. By Lemma 86.8.6 we find that $B_ m$ is rig-étale over $(\bar A_ m, \bar I_ m)$.

By induction hypothesis (on $t$) we can find a finite type $\bar A_ m$-algebra $C_ m$ and a map $C_ m \to B_ m$ which induces an isomorphism $C_ m^\wedge \cong B_ m$ where the completion is with respect to $\bar I_ m$. By Lemma 86.8.4 we may assume that $\mathop{\mathrm{Spec}}(C_ m) \to \mathop{\mathrm{Spec}}(\bar A_ m)$ is étale over $\mathop{\mathrm{Spec}}(\bar A_ m) \setminus V(\bar I_ m)$.

We claim that we may choose $A_ m \to C_ m \to B_ m$ as in the previous paragraph such that moreover there are isomorphisms $C_ m/(a_ t^{m - 1}) \to C_{m - 1}$ compatible with the given $A$-algebra structure and the maps to $B_{m - 1} = B_ m/(a_ t^{m - 1})$. Namely, first fix a choice of $A_1 \to C_1 \to B_1$. Suppose we have found $C_{m - 1} \to C_{m - 2} \to \ldots \to C_1$ with the desired properties. Note that $C_ m/(a_ t^{m - 1})$ is étale over $\mathop{\mathrm{Spec}}(\bar A_{m - 1}) \setminus V(\bar I_{m - 1})$. Hence by Lemma 86.8.7 there exists an étale extension $C_{m - 1} \to C'_{m - 1}$ which induces an isomorphism modulo $\bar I_{m - 1}$ and an $\bar A_{m - 1}$-algebra map $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ inducing the isomorphism $B_ m/(a_ t^{m - 1}) \to B_{m - 1}$ on completions. Note that $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is étale over the complement of $V(\bar I_{m - 1})$ by Morphisms, Lemma 29.36.18 and over $V(\bar I_{m - 1})$ induces an isomorphism on completions hence is étale there too (for example by More on Morphisms, Lemma 37.12.3). Thus $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is étale. By the topological invariance of étale morphisms (Étale Morphisms, Theorem 41.15.2) there exists an étale ring map $C_ m \to C'_ m$ such that $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is isomorphic to $C_ m/(a_ t^{m - 1}) \to C'_ m/(a_ t^{m - 1})$. Observe that the $\bar I_ m$-adic completion of $C'_ m$ is equal to the $\bar I_ m$-adic completion of $C_ m$, i.e., to $B_ m$ (details omitted). We apply Lemma 86.9.1 to the diagram

\[ \xymatrix{ & C'_ m \ar[r] & C'_ m/(a_ t^{m - 1}) \\ C''_ m \ar@{..>}[ru] \ar@{..>}[rr] & & C_{m - 1} \ar[u] \\ & \bar A_ m \ar[r] \ar[uu] \ar@{..>}[lu] & \bar A_{m - 1} \ar[u] } \]

to see that there exists a “lift” of $C''_ m$ of $C_{m - 1}$ to an algebra over $\bar A_ m$ with all the desired properties.

By construction $(C_ m)$ is an object of the category (86.2.0.1) for the principal ideal $(a_ t)$. Thus the inverse limit $B' = \mathop{\mathrm{lim}}\nolimits C_ m$ is an $(a_ t)$-adically complete $A$-algebra such that $B'/a_ t B'$ is of finite type over $A/(a_ t)$, see Lemma 86.2.1. By construction the $I$-adic completion of $B'$ is isomorphic to $B$ (details omitted). Consider the complex $\mathop{N\! L}\nolimits _{B'/A}^\wedge $ constructed using the $(a_ t)$-adic topology. Choosing a presentation for $B'$ (which induces a similar presentation for $B$) the reader immediately sees that $\mathop{N\! L}\nolimits _{B'/A}^\wedge \otimes _{B'} B = \mathop{N\! L}\nolimits _{B/A}^\wedge $. Since $a_ t \in I$ and since the cohomology modules of $\mathop{N\! L}\nolimits _{B'/A}^\wedge $ are finite $B'$-modules (hence complete for the $a_ t$-adic topology), we conclude that $a_ t^ c$ acts as zero on these cohomologies as the same thing is true by assumption for $\mathop{N\! L}\nolimits _{B/A}^\wedge $. Thus $B'$ is rig-étale over $(A, (a_ t))$ by Lemma 86.8.2. Hence finally, we may apply Lemma 86.10.1 to $B'$ over $(A, (a_ t))$ to finish the proof. $\square$

Proof of Lemma 86.7.3 in case $A$ is a G-ring. This proof is easier in that it does not depend on the somewhat delicate deformation theory argument given in the proof of Lemma 86.7.2, but of course it requires a very strong assumption on the Noetherian ring $A$.

Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Choose generators $g_1, \ldots , g_ m \in J$. Choose generators $k_1, \ldots , k_ t$ of the module of relations between $g_1, \ldots , g_ m$, i.e., such that

\[ (A[x_1, \ldots , x_ r]^\wedge )^{\oplus t} \xrightarrow {k_1, \ldots , k_ t} (A[x_1, \ldots , x_ r]^\wedge )^{\oplus m} \xrightarrow {g_1, \ldots , g_ m} A[x_1, \ldots , x_ r]^\wedge \]

is exact in the middle. Write $k_ i = (k_{i1}, \ldots , k_{im})$ so that we have

86.8.0.1
\begin{equation} \label{restricted-equation-relations-straight-up} \sum k_{ij}g_ j = 0 \end{equation}

for $i = 1, \ldots , t$. Let $I^ c = (a_1, \ldots , a_ s)$. For each $l \in \{ 1, \ldots , s\} $ we know that multiplication by $a_ l$ on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is zero in $D(B)$. By Lemma 86.3.4 we can find a map $\alpha _ l : \bigoplus B\text{d}x_ i \to J/J^2$ such that $\text{d} \circ \alpha _ l$ and $\alpha _ l \circ \text{d}$ are both multiplication by $a_ l$. Pick an element $f_{l, i} \in J$ whose class modulo $J^2$ is equal to $\alpha _ l(\text{d}x_ i)$. Then we have for all $l = 1, \ldots , s$ and $i = 1, \ldots , r$ that

86.8.0.2
\begin{equation} \label{restricted-equation-derivatives} \sum \nolimits _{i'} (\partial f_{l, i}/ \partial x_{i'}) \text{d}x_{i'} = a_ l \text{d}x_ i + \sum h_{l, i}^{j', i'} g_{j'} \text{d}x_{i'} \end{equation}

for some $h_{l, i}^{j', i'} \in A[x_1, \ldots , x_ r]^\wedge $. We also have for $j = 1, \ldots , m$ and $l = 1, \ldots , s$ that

86.8.0.3
\begin{equation} \label{restricted-equation-ci} a_ l g_ j = \sum h_{l, j}^ if_{l, i} + \sum h_{l, j}^{j', j''}g_{j'} g_{j''} \end{equation}

for some $h_{l, j}^ i$ and $h_{l, j}^{j', j''}$ in $A[x_1, \ldots , x_ r]^\wedge $. Of course, since $f_{l, i} \in J$ we can write for $l = 1, \ldots , s$ and $i = 1, \ldots , r$

86.8.0.4
\begin{equation} \label{restricted-equation-in-ideal} f_{l, i} = \sum h_{l, i}^ jg_ j \end{equation}

for some $h_{l, i}^ j$ in $A[x_1, \ldots , x_ r]^\wedge $.

Let $A[x_1, \ldots , x_ r]^ h$ be the henselization of the pair $(A[x_1, \ldots , x_ r], IA[x_1, \ldots , x_ r])$, see More on Algebra, Lemma 15.12.1. Since $A$ is a Noetherian G-ring, so is $A[x_1, \ldots , x_ r]$, see More on Algebra, Proposition 15.50.10. Hence we have approximation for the map $A[x_1, \ldots , x_ r]^ h \to A[x_1, \ldots , x_ r]^\wedge $ with respect to the ideal generated by $I$, see Smoothing Ring Maps, Lemma 16.14.1. Choose a large integer $M$. Choose

\[ G_ j, K_{ij}, F_{l, i}, H_{l, j}^ i, H_{l, j}^{j', j''}, H_{l, i}^ j \in A[x_1, \ldots , x_ r]^ h \]

such that analogues of equations (86.8.0.1), (86.8.0.3), and (86.8.0.4) hold for these elements in $A[x_1, \ldots , x_ r]^ h$, i.e.,

\[ \sum K_{ij}G_ j = 0,\quad a_ l G_ j = \sum H_{l, j}^ iF_{l, i} + \sum H_{l, j}^{j', j''} G_{j'} G_{j''},\quad F_{l, i} = \sum H_{l, i}^ j G_ j \]

and such that we have

\[ G_ j - g_ j, K_{ij} - k_{ij}, F_{l, i} - f_{l, i}, H_{l, j}^ i - h_{l, j}^ i, H_{l, j}^{j', j''} - h_{l, j}^{j', j''}, H_{l, i}^ j - h_{l, i}^ j \in I^ M A[x_1, \ldots , x_ r]^ h \]

where we take liberty of thinking of $A[x_1, \ldots , x_ r]^ h$ as a subring of $A[x_1, \ldots , x_ r]^\wedge $. Note that we cannot guarantee that the analogue of (86.8.0.2) holds in $A[x_1, \ldots , x_ r]^ h$, because it is not a polynomial equation. But since taking partial derivatives is $A$-linear, we do get the analogue modulo $I^ M$. More precisely, we see that

86.8.0.5
\begin{equation} \label{restricted-equation-derivatives-analogue} \sum \nolimits _{i'} (\partial F_{l, i}/ \partial x_{i'}) \text{d}x_{i'} - a_ l \text{d}x_ i - \sum h_{l, i}^{j', i'} G_{j'} \text{d}x_{i'} \in I^ MA[x_1, \ldots , x_ r]^\wedge \end{equation}

for $l = 1, \ldots , s$ and $i = 1, \ldots , r$.

With these choices, consider the ring

\[ C^ h = A[x_1, \ldots , x_ r]^ h/(G_1, \ldots , G_ r) \]

and denote $C^\wedge $ its $I$-adic completion, namely

\[ C^\wedge = A[x_1, \ldots , x_ r]^\wedge /J',\quad J' = (G_1, \ldots , G_ r)A[x_1, \ldots , x_ r]^\wedge \]

In the following paragraphs we establish the fact that $C^\wedge $ is isomorphic to $B$. Then in the final paragraph we deal with show that $C^ h$ comes from a finite type algebra over $A$ as in the statement of the lemma.

First consider the cokernel

\[ \Omega = \mathop{\mathrm{Coker}}(J'/(J')^2 \longrightarrow \bigoplus C^\wedge \text{d}x_ i) \]

This $C^\wedge $ module is generated by the images of the elements $\text{d}x_ i$. Since $F_{l, i} \in J'$ by the analogue of (86.8.0.4) we see from (86.8.0.5) we see that $a_ l \text{d}x_ i \in I^ M\Omega $. As $I^ c = (a_ l)$ we see that $I^ c \Omega \subset I^ M \Omega $. Since $M > c$ we conclude that $I^ c \Omega = 0$ by Algebra, Lemma 10.19.1.

Next, consider the kernel

\[ H_1 = \mathop{\mathrm{Ker}}(J'/(J')^2 \longrightarrow \bigoplus C^\wedge \text{d}x_ i) \]

By the analogue of (86.8.0.3) we see that $a_ l J' \subset (F_{l, i}) + (J')^2$. On the other hand, the determinant $\Delta _ l$ of the matrix $(\partial F_{l, i}/ \partial x_{i'})$ satisfies $\Delta _ l = a_ l^ r \bmod I^ M C^\wedge $ by (86.8.0.5). It follows that $a_ l^{r + 1} H_1 \subset I^ M H_1$ (some details omitted; use Algebra, Lemma 10.14.5). Now $(a_1^{r + 1}, \ldots , a_ s^{r + 1}) \supset I^{(sr + 1)c}$. Hence $I^{(sr + 1)c}H_1 \subset I^ M H_1$ and since $M > (sr + 1)c$ we conclude that $I^{(sr + 1)c}H_1 = 0$.

By Lemma 86.3.5 we conclude that multiplication by an element of $I^{2(sr + 1)c}$ on $\mathop{N\! L}\nolimits ^\wedge _{C^\wedge /A}$ is zero (note that the bound does not depend on $M$ or the choice of the approximation, as long as $M$ is large enough). Since $G_ j - g_ j$ is in the ideal generated by $I^ M$ we see that there is an isomorphism

\[ \psi _ M : C^\wedge /I^ MC^\wedge \to B/I^ MB \]

As $M$ is large enough we can use Lemma 86.7.1 with $d = d(I \subset A \to B)$, with $C^\wedge $ playing the role of $B$, with $2(rs + 1)c$ instead of $c$, to find a morphism

\[ \psi : C^\wedge \longrightarrow B \]

which agrees with $\psi _ M$ modulo $I^{q - 2(rs + 1)c}$ where $q$ is the quotient of $M$ by the number of generators of $I$. We claim $\psi $ is an isomorphism. Since $C^\wedge $ and $B$ are $I$-adically complete the map $\psi $ is surjective because it is surjective modulo $I$ (see Algebra, Lemma 10.95.1). On the other hand, as $M$ is large enough we see that

\[ \text{Gr}_ I(C^\wedge ) \cong \text{Gr}_ I(B) \]

as graded $\text{Gr}_ I(A[x_1, \ldots , x_ r]^\wedge )$-modules by More on Algebra, Lemma 15.4.2. Since $\psi $ is compatible with this isomorphism as it agrees with $\psi _ M$ modulo $I$, this means that $\text{Gr}_ I(\psi )$ is an isomorphism. As $C^\wedge $ and $B$ are $I$-adically complete, it follows that $\psi $ is an isomorphism.

This paragraph serves to deal with the issue that $C^ h$ is not of finite type over $A$. Namely, the ring $A[x_1, \ldots , x_ r]^ h$ is a filtered colimit of étale $A[x_1, \ldots , x_ r]$ algebras $A'$ such that $A/I[x_1, \ldots , x_ r] \to A'/IA'$ is an isomorphism (see proof of More on Algebra, Lemma 15.12.1). Pick an $A'$ such that $G_1, \ldots , G_ m$ are the images of $G'_1, \ldots , G'_ m \in A'$. Setting $C = A'/(G'_1, \ldots , G'_ m)$ we get the finite type algebra we were looking for. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AKA. Beware of the difference between the letter 'O' and the digit '0'.