The Stacks project

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

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

in $P$ which is possible by the choices made so far. Choose generators

and write $k_ i = (k_{i1}, \ldots , k_{ir})$ and $K = (k_{ij})$ so that

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

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

1. $\kappa (\mathfrak m') = \kappa (\mathfrak m_ B)$ 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,

2. $a_ i - a'_ i, b_ i - b'_ i, c_{ij} - c'_{ij}, k_{ij} - k'_{ij} \in (\mathfrak m')^ N P$, and

3. $f_ j(a'_1, \ldots , a'_ u) = \sum c'_{ji} b'_ i$ 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

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

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. $\square$