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 90.7).
Definition 90.4.1. 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
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
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 90.3.1 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 90.4.2. Let $f: R \to S$ be a ring map in $\widehat{\mathcal{C}}_\Lambda $. The following are equivalent
$f$ is surjective,
the map $\mathfrak m_ R/\mathfrak m_ R^2 \to \mathfrak m_ S/\mathfrak m_ S^2$ is surjective, and
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 90.3.5 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 90.4.3. 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 90.4.4. 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 90.3.8.
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 90.4.5. 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 90.4.6. 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 90.4.2. 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 90.4.7. 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 Krull's intersection theorem, 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 90.4.8. 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 90.4.2. 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 90.4.7 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 90.4.9. 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 90.4.2. 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)