The Stacks project

88.4 The completed base category

The following “completion” of the category $\mathcal{C}_\Lambda $ will serve as the base category of the completion of a category cofibered in groupoids over $\mathcal{C}_\Lambda $ (Section 88.7).

Definition 88.4.1. Let $\Lambda $ be a Noetherian ring and let $\Lambda \to k$ be a finite ring map where $k$ is a field. We define $\widehat{\mathcal{C}}_\Lambda $ to be the category with

  1. objects are pairs $(R, \varphi )$ where $R$ is a Noetherian complete local $\Lambda $-algebra and where $\varphi : R/\mathfrak m_ R \to k$ is a $\Lambda $-algebra isomorphism, and

  2. morphisms $f : (S, \psi ) \to (R, \varphi )$ are local $\Lambda $-algebra homomorphisms such that $\varphi \circ (f \bmod \mathfrak m) = \psi $.

As in the discussion following Definition 88.3.1 we will usually denote an object of $\widehat{\mathcal{C}}_\Lambda $ simply $R$, with the identification $R/\mathfrak m_ R = k$ understood. In this section we discuss some basic properties of objects and morphisms of the category $\widehat{\mathcal{C}}_\Lambda $ paralleling our discussion of the category $\mathcal{C}_\Lambda $ in the previous section.

Our first observation is that any object $A \in \mathcal{C}_\Lambda $ is an object of $\widehat{\mathcal{C}}_\Lambda $ as an Artinian local ring is always Noetherian and complete with respect to its maximal ideal (which is after all a nilpotent ideal). Moreover, it is clear from the definitions that $\mathcal{C}_\Lambda \subset \widehat{\mathcal{C}}_\Lambda $ is the strictly full subcategory consisting of all Artinian rings. As it turns out, conversely every object of $\widehat{\mathcal{C}}_\Lambda $ is a limit of objects of $\mathcal{C}_\Lambda $.

Suppose that $R$ is an object of $\widehat{\mathcal{C}}_\Lambda $. Consider the rings $R_ n = R/\mathfrak m_ R^ n$ for $n \in \mathbf{N}$. These are Noetherian local rings with a unique nilpotent prime ideal, hence Artinian, see Algebra, Proposition 10.60.7. The ring maps

\[ \ldots \to R_{n + 1} \to R_ n \to \ldots \to R_2 \to R_1 = k \]

are all surjective. Completeness of $R$ by definition means that $R = \mathop{\mathrm{lim}}\nolimits R_ n$. If $f : R \to S$ is a ring map in $\widehat{\mathcal{C}}_\Lambda $ then we obtain a system of ring maps $f_ n : R_ n \to S_ n$ whose limit is the given map.

Lemma 88.4.2. Let $f: R \to S$ be a ring map in $\widehat{\mathcal{C}}_\Lambda $. The following are equivalent

  1. $f$ is surjective,

  2. the map $\mathfrak m_ R/\mathfrak m_ R^2 \to \mathfrak m_ S/\mathfrak m_ S^2$ is surjective, and

  3. the map $\mathfrak m_ R/(\mathfrak m_\Lambda R + \mathfrak m_ R^2) \to \mathfrak m_ S/(\mathfrak m_\Lambda S + \mathfrak m_ S^2)$ is surjective.

Proof. Note that for $n \geq 2$ we have the equality of relative cotangent spaces

\[ \mathfrak m_ R/(\mathfrak m_\Lambda R + \mathfrak m_ R^2) = \mathfrak m_{R_ n}/(\mathfrak m_\Lambda R_ n + \mathfrak m_{R_ n}^2) \]

and similarly for $S$. Hence by Lemma 88.3.5 we see that $R_ n \to S_ n$ is surjective for all $n$. Now let $K_ n$ be the kernel of $R_ n \to S_ n$. Then the sequences

\[ 0 \to K_ n \to R_ n \to S_ n \to 0 \]

form an exact sequence of directed inverse systems. The system $(K_ n)$ is Mittag-Leffler since each $K_ n$ is Artinian. Hence by Algebra, Lemma 10.86.4 taking limits preserves exactness. So $\mathop{\mathrm{lim}}\nolimits R_ n \to \mathop{\mathrm{lim}}\nolimits S_ n$ is surjective, i.e., $f$ is surjective. $\square$

Lemma 88.4.3. The category $\widehat{\mathcal{C}}_\Lambda $ admits pushouts.

Proof. Let $R \to S_1$ and $R \to S_2$ be morphisms of $\widehat{\mathcal{C}}_\Lambda $. Consider the ring $C = S_1 \otimes _ R S_2$. This ring has a finitely generated maximal ideal $\mathfrak m = \mathfrak m_{S_1} \otimes S_2 + S_1 \otimes \mathfrak m_{S_2}$ with residue field $k$. Set $C^\wedge $ equal to the completion of $C$ with respect to $\mathfrak m$. Then $C^\wedge $ is a Noetherian ring complete with respect to the maximal ideal $\mathfrak m^\wedge = \mathfrak mC^\wedge $ whose residue field is identified with $k$, see Algebra, Lemma 10.97.5. Hence $C^\wedge $ is an object of $\widehat{\mathcal{C}}_\Lambda $. Then $S_1 \to C^\wedge $ and $S_2 \to C^\wedge $ turn $C^\wedge $ into a pushout over $R$ in $\widehat{\mathcal{C}}_\Lambda $ (details omitted). $\square$

We will not need the following lemma.

Lemma 88.4.4. The category $\widehat{\mathcal{C}}_\Lambda $ admits coproducts of pairs of objects.

Proof. Let $R$ and $S$ be objects of $\widehat{\mathcal{C}}_\Lambda $. Consider the ring $C = R \otimes _\Lambda S$. There is a canonical surjective map $C \to R \otimes _\Lambda S \to k \otimes _\Lambda k \to k$ where the last map is the multiplication map. The kernel of $C \to k$ is a maximal ideal $\mathfrak m$. Note that $\mathfrak m$ is generated by $\mathfrak m_ R C$, $\mathfrak m_ S C$ and finitely many elements of $C$ which map to generators of the kernel of $k \otimes _\Lambda k \to k$. Hence $\mathfrak m$ is a finitely generated ideal. Set $C^\wedge $ equal to the completion of $C$ with respect to $\mathfrak m$. Then $C^\wedge $ is a Noetherian ring complete with respect to the maximal ideal $\mathfrak m^\wedge = \mathfrak mC^\wedge $ with residue field $k$, see Algebra, Lemma 10.97.5. Hence $C^\wedge $ is an object of $\widehat{\mathcal{C}}_\Lambda $. Then $R \to C^\wedge $ and $S \to C^\wedge $ turn $C^\wedge $ into a coproduct in $\widehat{\mathcal{C}}_\Lambda $ (details omitted). $\square$

An empty coproduct in a category is an initial object of the category. In the classical case $\widehat{\mathcal{C}}_\Lambda $ has an initial object, namely $\Lambda $ itself. More generally, if $k' = k$, then the completion $\Lambda ^\wedge $ of $\Lambda $ with respect to $\mathfrak m_\Lambda $ is an initial object. More generally still, if $k' \subset k$ is separable, then $\widehat{\mathcal{C}}_\Lambda $ has an initial object too. Namely, choose a monic polynomial $P \in \Lambda [T]$ such that $k \cong k'[T]/(P')$ where $p' \in k'[T]$ is the image of $P$. Then $R = \Lambda ^\wedge [T]/(P)$ is an initial object, see proof of Lemma 88.3.8.

If $R$ is an initial object as above, then we have $\mathcal{C}_\Lambda = \mathcal{C}_ R$ and $\widehat{\mathcal{C}}_\Lambda = \widehat{\mathcal{C}}_ R$ which effectively brings the whole discussion in this chapter back to the classical case. But, if $k' \subset k$ is inseparable, then an initial object does not exist.

Lemma 88.4.5. Let $S$ be an object of $\widehat{\mathcal{C}}_\Lambda $. Then $\dim _ k \text{Der}_\Lambda (S, k) < \infty $.

Proof. Let $x_1, \ldots , x_ n \in \mathfrak m_ S$ map to a $k$-basis for the relative cotangent space $\mathfrak m_ S/(\mathfrak m_\Lambda S + \mathfrak m_ S^2)$. Choose $y_1, \ldots , y_ m \in S$ whose images in $k$ generate $k$ over $k'$. We claim that $\dim _ k \text{Der}_\Lambda (S, k) \leq n + m$. To see this it suffices to prove that if $D(x_ i) = 0$ and $D(y_ j) = 0$, then $D = 0$. Let $a \in S$. We can find a polynomial $P = \sum \lambda _ J y^ J$ with $\lambda _ J \in \Lambda $ whose image in $k$ is the same as the image of $a$ in $k$. Then we see that $D(a - P) = D(a) - D(P) = D(a)$ by our assumption that $D(y_ j) = 0$ for all $j$. Thus we may assume $a \in \mathfrak m_ S$. Write $a = \sum a_ i x_ i$ with $a_ i \in S$. By the Leibniz rule

\[ D(a) = \sum x_ iD(a_ i) + \sum a_ iD(x_ i) = \sum x_ iD(a_ i) \]

as we assumed $D(x_ i) = 0$. We have $\sum x_ iD(a_ i) = 0$ as multiplication by $x_ i$ is zero on $k$. $\square$

Lemma 88.4.6. Let $f : R \to S$ be a morphism of $\widehat{\mathcal{C}}_\Lambda $. If $\text{Der}_\Lambda (S, k) \to \text{Der}_\Lambda (R, k)$ is injective, then $f$ is surjective.

Proof. If $f$ is not surjective, then $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2)$ is nonzero by Lemma 88.4.2. Then also $Q = S/(f(R) + \mathfrak m_ R S + \mathfrak m_ S^2)$ is nonzero. Note that $Q$ is a $k = R/\mathfrak m_ R$-vector space via $f$. We turn $Q$ into an $S$-module via $S \to k$. The quotient map $D : S \to Q$ is an $R$-derivation: if $a_1, a_2 \in S$, we can write $a_1 = f(b_1) + a_1'$ and $a_2 = f(b_2) + a_2'$ for some $b_1, b_2 \in R$ and $a_1', a_2' \in \mathfrak m_ S$. Then $b_ i$ and $a_ i$ have the same image in $k$ for $i = 1, 2$ and

\begin{align*} a_1a_2 & = (f(b_1) + a_1')(f(b_2) + a_2') \\ & = f(b_1)a_2' + f(b_2)a_1' \\ & = f(b_1)(f(b_2) + a_2') + f(b_2)(f(b_1) + a_1') \\ & = f(b_1)a_2 + f(b_2)a_1 \end{align*}

in $Q$ which proves the Leibniz rule. Hence $D : S \to Q$ is a $\Lambda $-derivation which is zero on composing with $R \to S$. Since $Q \not= 0$ there also exist derivations $D : S \to k$ which are zero on composing with $R \to S$, i.e., $\text{Der}_\Lambda (S, k) \to \text{Der}_\Lambda (R, k)$ is not injective. $\square$

Lemma 88.4.7. Let $R$ be an object of $\widehat{\mathcal{C}}_\Lambda $. Let $(J_ n)$ be a decreasing sequence of ideals such that $\mathfrak m_ R^ n \subset J_ n$. Set $J = \bigcap J_ n$. Then the sequence $(J_ n/J)$ defines the $\mathfrak m_{R/J}$-adic topology on $R/J$.

Proof. It is clear that $\mathfrak m_{R/J}^ n \subset J_ n/J$. Thus it suffices to show that for every $n$ there exists an $N$ such that $J_ N/J \subset \mathfrak m_{R/J}^ n$. This is equivalent to $J_ N \subset \mathfrak m_ R^ n + J$. For each $n$ the ring $R/\mathfrak m_ R^ n$ is Artinian, hence there exists a $N_ n$ such that

\[ J_{N_ n} + \mathfrak m_ R^ n = J_{N_ n + 1} + \mathfrak m_ R^ n = \ldots \]

Set $E_ n = (J_{N_ n} + \mathfrak m_ R^ n)/\mathfrak m_ R^ n$. Set $E = \mathop{\mathrm{lim}}\nolimits E_ n \subset \mathop{\mathrm{lim}}\nolimits R/\mathfrak m_ R^ n = R$. Note that $E \subset J$ as for any $f \in E$ and any $m$ we have $f \in J_ m + \mathfrak m_ R^ n$ for all $n \gg 0$, so $f \in J_ m$ by Artin-Rees, see Algebra, Lemma 10.51.4. Since the transition maps $E_ n \to E_{n - 1}$ are all surjective, we see that $J$ surjects onto $E_ n$. Hence for $N = N_ n$ works. $\square$

Lemma 88.4.8. Let $\ldots \to A_3 \to A_2 \to A_1$ be a sequence of surjective ring maps in $\mathcal{C}_\Lambda $. If $\dim _ k (\mathfrak m_{A_ n}/\mathfrak m_{A_ n}^2)$ is bounded, then $S = \mathop{\mathrm{lim}}\nolimits A_ n$ is an object in $\widehat{\mathcal{C}}_\Lambda $ and the ideals $I_ n = \mathop{\mathrm{Ker}}(S \to A_ n)$ define the $\mathfrak m_ S$-adic topology on $S$.

Proof. We will use freely that the maps $S \to A_ n$ are surjective for all $n$. Note that the maps $\mathfrak m_{A_{n + 1}}/\mathfrak m_{A_{n + 1}}^2 \to \mathfrak m_{A_ n}/\mathfrak m_{A_ n}^2$ are surjective, see Lemma 88.4.2. Hence for $n$ sufficiently large the dimension $\dim _ k (\mathfrak m_{A_ n}/\mathfrak m_{A_ n}^2)$ stabilizes to an integer, say $r$. Thus we can find $x_1, \ldots , x_ r \in \mathfrak m_ S$ whose images in $A_ n$ generate $\mathfrak m_{A_ n}$. Moreover, pick $y_1, \ldots , y_ t \in S$ whose images in $k$ generate $k$ over $\Lambda $. Then we get a ring map $P = \Lambda [z_1, \ldots , z_{r + t}] \to S$, $z_ i \mapsto x_ i$ and $z_{r + j} \mapsto y_ j$ such that the composition $P \to S \to A_ n$ is surjective for all $n$. Let $\mathfrak m \subset P$ be the kernel of $P \to k$. Let $R = P^\wedge $ be the $\mathfrak m$-adic completion of $P$; this is an object of $\widehat{\mathcal{C}}_\Lambda $. Since we still have the compatible system of (surjective) maps $R \to A_ n$ we get a map $R \to S$. Set $J_ n = \mathop{\mathrm{Ker}}(R \to A_ n)$. Set $J = \bigcap J_ n$. By Lemma 88.4.7 we see that $R/J = \mathop{\mathrm{lim}}\nolimits R/J_ n = \mathop{\mathrm{lim}}\nolimits A_ n = S$ and that the ideals $J_ n/J = I_ n$ define the $\mathfrak m$-adic topology. (Note that for each $n$ we have $\mathfrak m_ R^{N_ n} \subset J_ n$ for some $N_ n$ and not necessarily $N_ n = n$, so a renumbering of the ideals $J_ n$ may be necessary before applying the lemma.) $\square$

Lemma 88.4.9. Let $R', R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. Suppose that $R = R' \oplus I$ for some ideal $I$ of $R$. Let $x_1, \ldots , x_ r \in I$ map to a basis of $I/\mathfrak m_ R I$. Set $S = R'[[X_1, \ldots , X_ r]]$ and consider the $R'$-algebra map $S \to R$ mapping $X_ i$ to $x_ i$. Assume that for every $n \gg 0$ the map $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ has a left inverse in $\mathcal{C}_\Lambda $. Then $S \to R$ is an isomorphism.

Proof. As $R = R' \oplus I$ we have

\[ \mathfrak m_ R/\mathfrak m_ R^2 = \mathfrak m_{R'}/\mathfrak m_{R'}^2 \oplus I/\mathfrak m_ RI \]

and similarly

\[ \mathfrak m_ S/\mathfrak m_ S^2 = \mathfrak m_{R'}/\mathfrak m_{R'}^2 \oplus \bigoplus kX_ i \]

Hence for $n > 1$ the map $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ induces an isomorphism on cotangent spaces. Thus a left inverse $h_ n : R/\mathfrak m_ R^ n \to S/\mathfrak m_ S^ n$ is surjective by Lemma 88.4.2. Since $h_ n$ is injective as a left inverse it is an isomorphism. Thus the canonical surjections $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ are all isomorphisms and we win. $\square$


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 06GV. Beware of the difference between the letter 'O' and the digit '0'.