A fundamental insight of Michael Artin is that you can approximate objects of a limit preserving stack. Namely, given an object $x$ of the stack over a Noetherian complete local ring, you can find an object $x_ A$ over an algebraic ring which is “close to” $x$. Here an algebraic ring means a finite type $S$-algebra and close means adically close. In this section we present this in a simple, yet general form.
To formulate the result we need to pull together some definitions from different places in the Stacks project. First, in Criteria for Representability, Section 97.5 we introduced limit preserving on objects for $1$-morphisms of categories fibred in groupoids over the category of schemes. In More on Algebra, Definition 15.50.1 we defined the notion of a G-ring. Let $S$ be a locally Noetherian scheme. Let $A$ be an $S$-algebra. We say that $A$ is of finite type over $S$ or is a finite type $S$-algebra if $\mathop{\mathrm{Spec}}(A) \to S$ is of finite type. In this case $A$ is a Noetherian ring. Finally, given a ring $A$ and ideal $I$ we denote $\text{Gr}_ I(A) = \bigoplus I^ n/I^{n + 1}$.
Proof.
Choose an affine open $\mathop{\mathrm{Spec}}(\Lambda ) \subset S$ such that $k$ is a finite $\Lambda $-algebra, see Morphisms, Lemma 29.16.1. We may and do replace $S$ by $\mathop{\mathrm{Spec}}(\Lambda )$.
We may write $R$ as a directed colimit $R = \mathop{\mathrm{colim}}\nolimits C_ j$ where each $C_ j$ is a finite type $\Lambda $-algebra (see Algebra, Lemma 10.127.2). By assumption (b) the object $x$ is isomorphic to the restriction of an object over one of the $C_ j$. Hence we may choose a finite type $\Lambda $-algebra $C$, a $\Lambda $-algebra map $C \to R$, and an object $x_ C$ of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(C)$ such that $x = x_ C|_{\mathop{\mathrm{Spec}}(R)}$. The choice of $C$ is a bookkeeping device and could be avoided. For later use, let us write $C = \Lambda [y_1, \ldots , y_ u]/(f_1, \ldots , f_ v)$ and we denote $\overline{a}_ i \in R$ the image of $y_ i$ under the map $C \to R$. Set $\mathfrak m_ C = C \cap \mathfrak m_ R$.
Choose a $\Lambda $-algebra surjection $\Lambda [x_1, \ldots , x_ s] \to k$ and denote $\mathfrak m'$ the kernel. By the universal property of polynomial rings we may lift this to a $\Lambda $-algebra map $\Lambda [x_1, \ldots , x_ s] \to R$. We add some variables (i.e., we increase $s$ a bit) mapping to generators of $\mathfrak m_ R$. Having done this we see that $\Lambda [x_1, \ldots , x_ s] \to R/\mathfrak m_ R^2$ is surjective. Then we see that
98.10.1.1
\begin{equation} \label{artin-equation-surjection} P = \Lambda [x_1, \ldots , x_ s]_{\mathfrak m'}^\wedge \longrightarrow R \end{equation}
is a surjective map of Noetherian complete local rings, see for example Formal Deformation Theory, Lemma 90.4.2.
Choose lifts $a_ i \in P$ of $\overline{a}_ i$ we found above. Choose generators $b_1, \ldots , b_ r \in P$ for the kernel of (98.10.1.1). Choose $c_{ji} \in P$ such that
\[ f_ j(a_1, \ldots , a_ u) = \sum c_{ji} b_ i \]
in $P$ which is possible by the choices made so far. Choose generators
\[ k_1, \ldots , k_ t \in \mathop{\mathrm{Ker}}(P^{\oplus r} \xrightarrow {(b_1, \ldots , b_ r)} P) \]
and write $k_ i = (k_{i1}, \ldots , k_{ir})$ and $K = (k_{ij})$ so that
\[ P^{\oplus t} \xrightarrow {K} P^{\oplus r} \xrightarrow {(b_1, \ldots , b_ r)} P \to R \to 0 \]
is an exact sequence of $P$-modules. In particular we have $\sum k_{ij} b_ j = 0$. After possibly increasing $N$ we may assume $N - 1$ works in the Artin-Rees lemma for the first two maps of this exact sequence (see More on Algebra, Section 15.4 for terminology).
By assumption $\mathcal{O}_{S, s} = \Lambda _{\Lambda \cap \mathfrak m'}$ is a G-ring. Hence by More on Algebra, Proposition 15.50.10 the ring $\Lambda [x_1, \ldots , x_ s]_{\mathfrak m'}$ is a $G$-ring. Hence by Smoothing Ring Maps, Theorem 16.13.2 there exist an étale ring map
\[ \Lambda [x_1, \ldots , x_ s]_{\mathfrak m'} \to B, \]
a maximal ideal $\mathfrak m_ B$ of $B$ lying over $\mathfrak m'$, and elements $a'_ i, b'_ i, c'_{ij}, k'_{ij} \in B'$ such that
$\kappa (\mathfrak m') = \kappa (\mathfrak m_ B)$ which implies that $\Lambda [x_1, \ldots , x_ s]_{\mathfrak m'} \subset B_{\mathfrak m_ B} \subset P$ and $P$ is identified with the completion of $B$ at $\mathfrak m_ B$, see remark preceding Smoothing Ring Maps, Theorem 16.13.2,
$a_ i - a'_ i, b_ i - b'_ i, c_{ij} - c'_{ij}, k_{ij} - k'_{ij} \in (\mathfrak m')^ N P$, and
$f_ j(a'_1, \ldots , a'_ u) = \sum c'_{ji} b'_ i$ and $\sum k'_{ij}b'_ j = 0$.
Set $A = B/(b'_1, \ldots , b'_ r)$ and denote $\mathfrak m_ A$ the image of $\mathfrak m_ B$ in $A$. (Note that $A$ is essentially of finite type over $\Lambda $; at the end of the proof we will show how to obtain an $A$ which is of finite type over $\Lambda $.) There is a ring map $C \to A$ sending $y_ i \mapsto a'_ i$ because the $a'_ i$ satisfy the desired equations modulo $(b'_1, \ldots , b'_ r)$. Note that $A/\mathfrak m_ A^ N = R/\mathfrak m_ R^ N$ as quotients of $P = B^\wedge $ by property (2) above. Set $x_ A = x_ C|_{\mathop{\mathrm{Spec}}(A)}$. Since the maps
\[ C \to A \to A/\mathfrak m_ A^ N \cong R/\mathfrak m_ R^ N \quad \text{and}\quad C \to R \to R/\mathfrak m_ R^ N \]
are equal we see that $x_ A$ and $x$ agree modulo $\mathfrak m_ R^ N$ via the isomorphism $A/\mathfrak m_ A^ N = R/\mathfrak m_ R^ N$. At this point we have shown properties (1) – (5) of the statement of the lemma. To see (6) note that
\[ P^{\oplus t} \xrightarrow {K} P^{\oplus r} \xrightarrow {(b_1, \ldots , b_ r)} P \quad \text{and}\quad P^{\oplus t} \xrightarrow {K'} P^{\oplus r} \xrightarrow {(b'_1, \ldots , b'_ r)} P \]
are two complexes of $P$-modules which are congruent modulo $(\mathfrak m')^ N$ with the first one being exact. By our choice of $N$ above we see from More on Algebra, Lemma 15.4.2 that $R = P/(b_1, \ldots , b_ r)$ and $P/(b'_1, \ldots , b'_ r) = B^\wedge /(b'_1, \ldots , b'_ r) = A^\wedge $ have isomorphic associated graded algebras, which is what we wanted to show.
This last paragraph of the proof serves to clean up the issue that $A$ is essentially of finite type over $S$ and not yet of finite type. The construction above gives $A = B/(b'_1, \ldots , b'_ r)$ and $\mathfrak m_ A \subset A$ with $B$ étale over $\Lambda [x_1, \ldots , x_ s]_{\mathfrak m'}$. Hence $A$ is of finite type over the Noetherian ring $\Lambda [x_1, \ldots , x_ s]_{\mathfrak m'}$. Thus we can write $A = (A_0)_{\mathfrak m'}$ for some finite type $\Lambda [x_1, \ldots , x_ n]$ algebra $A_0$. Then $A = \mathop{\mathrm{colim}}\nolimits (A_0)_ f$ where $f \in \Lambda [x_1, \ldots , x_ n] \setminus \mathfrak m'$, see Algebra, Lemma 10.9.9. Because $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ is limit preserving on objects, we see that $x_ A$ comes from some object $x_{(A_0)_ f}$ over $\mathop{\mathrm{Spec}}((A_0)_ f)$ for an $f$ as above. After replacing $A$ by $(A_0)_ f$ and $x_ A$ by $x_{(A_0)_ f}$ and $\mathfrak m_ A$ by $(A_0)_ f \cap \mathfrak m_ A$ the proof is finished.
$\square$
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