Lemma 16.11.1. Let k be a field of characteristic p > 0. Let (\Lambda , \mathfrak m, K) be an Artinian local k-algebra. Assume that \dim H_1(L_{K/k}) < \infty . Then \Lambda is a filtered colimit of Artinian local k-algebras A with each map A \to \Lambda flat, with \mathfrak m_ A \Lambda = \mathfrak m, and with A essentially of finite type over k.
16.11 Inseparable residue fields
In this section we explain how to solve a local problem in the case of an inseparable residue field extension.
Proof. Note that the flatness of A \to \Lambda implies that A \to \Lambda is injective, so the lemma really tells us that \Lambda is a directed union of these types of subrings A \subset \Lambda . Let n be the minimal integer such that \mathfrak m^ n = 0. We will prove this lemma by induction on n. The case n = 1 is clear as a field extension is a union of finitely generated field extensions.
Pick \lambda _1, \ldots , \lambda _ d \in \mathfrak m which generate \mathfrak m. As K is formally smooth over \mathbf{F}_ p (see Algebra, Lemma 10.158.7) we can find a ring map \sigma : K \to \Lambda which is a section of the quotient map \Lambda \to K. In general \sigma is not a k-algebra map. Given \sigma we define
using \sigma on elements of K and mapping x_ i to \lambda _ i. Claim: there exists a \sigma : K \to \Lambda and a subfield k \subset F \subset K finitely generated over k such that the image of k in \Lambda is contained in \Psi _\sigma (F[x_1, \ldots , x_ d]).
We will prove the claim by induction on the least integer n such that \mathfrak m^ n = 0. It is clear for n = 1. If n > 1 set I = \mathfrak m^{n - 1} and \Lambda ' = \Lambda /I. By induction we may assume given \sigma ' : K \to \Lambda ' and k \subset F' \subset K finitely generated such that the image of k \to \Lambda \to \Lambda ' is contained in A' = \Psi _{\sigma '}(F'[x_1, \ldots , x_ d]). Denote \tau ' : k \to A' the induced map. Choose a lift \sigma : K \to \Lambda of \sigma ' (this is possible by the formal smoothness of K/\mathbf{F}_ p we mentioned above). For later reference we note that we can change \sigma to \sigma + D for some derivation D : K \to I. Set A = F[x_1, \ldots , x_ d]/(x_1, \ldots , x_ d)^ n. Then \Psi _\sigma induces a ring map \Psi _\sigma : A \to \Lambda . The composition with the quotient map \Lambda \to \Lambda ' induces a surjective map A \to A' with nilpotent kernel. Choose a lift \tau : k \to A of \tau ' (possible as k/\mathbf{F}_ p is formally smooth). Thus we obtain two maps k \to \Lambda , namely \Psi _\sigma \circ \tau : k \to \Lambda and the given map i : k \to \Lambda . These maps agree modulo I, whence the difference is a derivation \theta = i - \Psi _\sigma \circ \tau : k \to I. Note that if we change \sigma into \sigma + D then we change \theta into \theta - D|_ k.
Choose a set of elements \{ y_ j\} _{j \in J} of k whose differentials \text{d}y_ j form a basis of \Omega _{k/\mathbf{F}_ p}. The Jacobi-Zariski sequence for \mathbf{F}_ p \subset k \subset K is
As \dim H_1(L_{K/k}) < \infty we can find a finite subset J_0 \subset J such that the image of the first map is contained in \bigoplus _{j \in J_0} K\text{d}y_ j. Hence the elements \text{d}y_ j, j \in J \setminus J_0 map to K-linearly independent elements of \Omega _{K/\mathbf{F}_ p}. Therefore we can choose a D : K \to I such that \theta - D|_ k = \xi \circ \text{d} where \xi is a composition
Let f_ j = \xi (\text{d}y_ j) \in I for j \in J_0. Change \sigma into \sigma + D as above. Then we see that \theta (a) = \sum _{j \in J_0} a_ j f_ j for a \in k where \text{d}a = \sum a_ j \text{d}y_ j in \Omega _{k/\mathbf{F}_ p}. Note that I is generated by the monomials \lambda ^ E = \lambda _1^{e_1} \ldots \lambda _ d^{e_ d} of total degree |E| = \sum e_ i = n - 1 in \lambda _1, \ldots , \lambda _ d. Write f_ j = \sum _ E c_{j, E} \lambda ^ E with c_{j, E} \in K. Replace F' by F = F'(c_{j, E}). Then the claim holds.
Choose \sigma and F as in the claim. The kernel of \Psi _\sigma is generated by finitely many polynomials g_1, \ldots , g_ t \in K[x_1, \ldots , x_ d] and we may assume their coefficients are in F after enlarging F by adjoining finitely many elements. In this case it is clear that the map A = F[x_1, \ldots , x_ d]/(g_1, \ldots , g_ t) \to K[x_1, \ldots , x_ d]/(g_1, \ldots , g_ t) = \Lambda is flat. By the claim A is a k-subalgebra of \Lambda . It is clear that \Lambda is the filtered colimit of these algebras, as K is the filtered union of the subfields F. Finally, these algebras are essentially of finite type over k by Algebra, Lemma 10.54.4. \square
Lemma 16.11.2. Let k be a field of characteristic p > 0. Let \Lambda be a Noetherian geometrically regular k-algebra. Let \mathfrak q \subset \Lambda be a prime ideal. Let n \geq 1 be an integer and let E \subset \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q be a finite subset. Then we can find m \geq 0 and \varphi : k[y_1, \ldots , y_ m] \to \Lambda with the following properties
setting \mathfrak p = \varphi ^{-1}(\mathfrak q) we have \mathfrak q\Lambda _\mathfrak q = \mathfrak p \Lambda _\mathfrak q and k[y_1, \ldots , y_ m]_\mathfrak p \to \Lambda _\mathfrak q is flat,
there is a factorization by homomorphisms of local Artinian rings
k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to D \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak qwhere the first arrow is essentially smooth and the second is flat,
E is contained in D modulo \mathfrak q^ n\Lambda _\mathfrak q.
Proof. Set \bar\Lambda = \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q. Note that \dim H_1(L_{\kappa (\mathfrak q)/k}) < \infty by More on Algebra, Proposition 15.35.1. Pick A \subset \bar\Lambda containing E such that A is local Artinian, essentially of finite type over k, the map A \to \bar\Lambda is flat, and \mathfrak m_ A generates the maximal ideal of \bar\Lambda , see Lemma 16.11.1. Denote F = A/\mathfrak m_ A the residue field so that k \subset F \subset K. Pick \lambda _1, \ldots , \lambda _ t \in \Lambda which map to elements of A in \bar\Lambda such that moreover the images of \text{d}\lambda _1, \ldots , \text{d}\lambda _ t form a basis of \Omega _{F/k}. Consider the map \varphi ' : k[y_1, \ldots , y_ t] \to \Lambda sending y_ j to \lambda _ j. Set \mathfrak p' = (\varphi ')^{-1}(\mathfrak q). By More on Algebra, Lemma 15.35.2 the ring map k[y_1, \ldots , y_ t]_{\mathfrak p'} \to \Lambda _\mathfrak q is flat and \Lambda _\mathfrak q/\mathfrak p' \Lambda _\mathfrak q is regular. Thus we can choose further elements \lambda _{t + 1}, \ldots , \lambda _ m \in \Lambda which map into A \subset \bar\Lambda and which map to a regular system of parameters of \Lambda _\mathfrak q/\mathfrak p' \Lambda _\mathfrak q. We obtain \varphi : k[y_1, \ldots , y_ m] \to \Lambda having property (1) such that k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to \bar\Lambda factors through A. Thus k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to A is flat by Algebra, Lemma 10.39.9. By construction the residue field extension F/\kappa (\mathfrak p) is finitely generated and \Omega _{F/\kappa (\mathfrak p)} = 0. Hence it is finite separable by More on Algebra, Lemma 15.34.1. Thus k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to A is finite by Algebra, Lemma 10.54.4. Finally, we conclude that it is étale by Algebra, Lemma 10.143.7. Since an étale ring map is certainly essentially smooth we win. \square
Lemma 16.11.3. Let \varphi : k[y_1, \ldots , y_ m] \to \Lambda , n, \mathfrak q, \mathfrak p and
be as in Lemma 16.11.2. Then for any \lambda \in \Lambda \setminus \mathfrak q there exists an integer q > 0 and a factorization
such that D \to D' is an essentially smooth map of local Artinian rings, the last arrow is flat, and \lambda ^ q is in D'.
Proof. Set \bar\Lambda = \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q. Let \bar\lambda be the image of \lambda in \bar\Lambda . Let \alpha \in \kappa (\mathfrak q) be the image of \lambda in the residue field. Let k \subset F \subset \kappa (\mathfrak q) be the residue field of D. If \alpha is in F then we can find an x \in D such that x \bar\lambda = 1 \bmod \mathfrak q. Hence (x \bar\lambda )^ q = 1 \bmod (\mathfrak q)^ q if q is divisible by p. Hence \bar\lambda ^ q is in D. If \alpha is transcendental over F, then we can take D' = (D[\bar\lambda ])_\mathfrak m equal to the subring generated by D and \bar\lambda localized at \mathfrak m = D[\bar\lambda ] \cap \mathfrak q \bar\Lambda . This works because D[\bar\lambda ] is in fact a polynomial algebra over D in this case. Finally, if \lambda \bmod \mathfrak q is algebraic over F, then we can find a p-power q such that \alpha ^ q is separable algebraic over F, see Fields, Section 9.28. Note that D and \bar\Lambda are henselian local rings, see Algebra, Lemma 10.153.10. Let D \to D' be a finite étale extension whose residue field extension is F(\alpha ^ q)/F, see Algebra, Lemma 10.153.7. Since \bar\Lambda is henselian and F(\alpha ^ q) is contained in its residue field we can find a factorization D' \to \bar\Lambda . By the first part of the argument we see that \bar\lambda ^{qq'} \in D' for some q' > 0. \square
Lemma 16.11.4. Let k \to A \to \Lambda \supset \mathfrak q be as in Situation 16.9.1 where
k is a field of characteristic p > 0,
\Lambda is Noetherian and geometrically regular over k,
\mathfrak q is minimal over \mathfrak h_ A.
Then k \to A \to \Lambda \supset \mathfrak q can be resolved.
Proof. The lemma is proven by the following steps in the given order. We will justify each of these steps below.
Pick an integer N > 0 such that \mathfrak q^ N\Lambda _\mathfrak q \subset H_{A/k}\Lambda _\mathfrak q.
Pick generators a_1, \ldots , a_ t \in A of the ideal H_{A/R}.
Set d = \dim (\Lambda _\mathfrak q).
Set B = A[x_1, \ldots , x_ d, z_{ij}]/(x_ i^{2N} - \sum z_{ij}a_ j).
Consider B as a k[x_1, \ldots , x_ d]-algebra and let B \to C be as in Lemma 16.3.1. We also obtain a section C \to B.
Choose c > 0 such that each x_ i^ c is strictly standard in C over k[x_1, \ldots , x_ d].
Set n = N + dc and e = 8c.
Let E \subset \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q be the images of generators of A as a k-algebra.
Choose an integer m and a k-algebra map \varphi : k[y_1, \ldots , y_ m] \to \Lambda and a factorization by local Artinian rings
k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to D \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak qsuch that the first arrow is essentially smooth, the second is flat, E is contained in D, with \mathfrak p = \varphi ^{-1}(\mathfrak q) the map k[y_1, \ldots , y_ m]_\mathfrak p \to \Lambda _\mathfrak q is flat, and \mathfrak p \Lambda _\mathfrak q = \mathfrak q \Lambda _\mathfrak q.
Choose \pi _1, \ldots , \pi _ d \in \mathfrak p which map to a regular system of parameters of k[y_1, \ldots , y_ m]_\mathfrak p.
Let R = k[y_1, \ldots , y_ m, t_1, \ldots , t_ m] and \gamma _ i = \pi _ i t_ i.
If necessary modify the choice of \pi _ i such that for i = 1, \ldots , d we have
\text{Ann}_{R/(\gamma _1^ e, \ldots , \gamma _{i - 1}^ e)R}(\gamma _ i) = \text{Ann}_{R/(\gamma _1^ e, \ldots , \gamma _{i - 1}^ e)R}(\gamma _ i^2)There exist \delta _1, \ldots , \delta _ d \in \Lambda , \delta _ i \not\in \mathfrak q and a factorization D \to D' \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q with D' local Artinian, D \to D' essentially smooth, the map D' \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q flat such that, with \pi _ i' = \delta _ i \pi _ i, we have for i = 1, \ldots , d
(\pi _ i')^{2N} = \sum a_ j\lambda _{ij} in \Lambda where \lambda _{ij} \bmod \mathfrak q^ n\Lambda _\mathfrak q is an element of D',
\text{Ann}_{\Lambda /({\pi '}_1^ e, \ldots , {\pi '}_{i - 1}^ e)}({\pi '}_ i) = \text{Ann}_{\Lambda /({\pi '}_1^ e, \ldots , {\pi '}_{i - 1}^ e)}({\pi '}_ i^2),
\delta _ i \bmod \mathfrak q^ n\Lambda _\mathfrak q is an element of D'.
Define B \to \Lambda by sending x_ i to \pi '_ i and z_{ij} to \lambda _{ij} found above. Define C \to \Lambda by composing the map B \to \Lambda with the retraction C \to B.
Map R \to \Lambda by \varphi on k[y_1, \ldots , y_ m] and by sending t_ i to \delta _ i. Further introduce a map
k[x_1, \ldots , x_ d] \longrightarrow R = k[y_1, \ldots , y_ m, t_1, \ldots , t_ d]by sending x_ i to \gamma _ i = \pi _ i t_ i.
It suffices to resolve
R \to C \otimes _{k[x_1, \ldots , x_ d]} R \to \Lambda \supset \mathfrak qSet I = (\gamma _1^ e, \ldots , \gamma _ d^ e) \subset R.
It suffices to resolve
R/I \to C \otimes _{k[x_1, \ldots , x_ d]} R/I \to \Lambda /I\Lambda \supset \mathfrak q/I\LambdaWe denote \mathfrak r \subset R = k[y_1, \ldots , y_ m, t_1, \ldots , t_ d] the inverse image of \mathfrak q.
It suffices to resolve
(R/I)_\mathfrak r \to C \otimes _{k[x_1, \ldots , x_ d]} (R/I)_\mathfrak r \to \Lambda _\mathfrak q/I\Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q/I\Lambda _\mathfrak qSet J = (\pi _1^ e, \ldots , \pi _ d^ e) in k[y_1, \ldots , y_ m].
It suffices to resolve
(R/JR)_\mathfrak p \to C \otimes _{k[x_1, \ldots , x_ d]} (R/JR)_\mathfrak p \to \Lambda _\mathfrak q/J\Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q/J\Lambda _\mathfrak qIt suffices to resolve
(R/\mathfrak p^ nR)_\mathfrak p \to C \otimes _{k[x_1, \ldots , x_ d]} (R/\mathfrak p^ nR)_\mathfrak p \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak qIt suffices to resolve
(R/\mathfrak p^ nR)_\mathfrak p \to B \otimes _{k[x_1, \ldots , x_ d]} (R/\mathfrak p^ nR)_\mathfrak p \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak qThe ring D'[t_1, \ldots , t_ d] is given the structure of an R_\mathfrak p/\mathfrak p^ nR_\mathfrak p-algebra by the given map k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to D' and by sending t_ i to t_ i. It suffices to find a factorization
B \otimes _{k[x_1, \ldots , x_ d]} (R/\mathfrak p^ nR)_\mathfrak p \to D'[t_1, \ldots , t_ d] \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak qwhere the second arrow sends t_ i to \delta _ i and induces the given homomorphism D' \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q.
Such a factorization exists by our choice of D' above.
We now give the justification for each of the steps, except that we skip justifying the steps which just introduce notation.
Ad (1). This is possible as \mathfrak q is minimal over \mathfrak h_ A = \sqrt{H_{A/k}\Lambda }.
Ad (6). Note that A_{a_ i} is smooth over k. Hence B_{a_ j}, which is isomorphic to a polynomial algebra over A_{a_ j}[x_1, \ldots , x_ d], is smooth over k[x_1, \ldots , x_ d]. Thus B_{x_ i} is smooth over k[x_1, \ldots , x_ d]. By Lemma 16.3.1 we see that C_{x_ i} is smooth over k[x_1, \ldots , x_ d] with finite free module of differentials. Hence some power of x_ i is strictly standard in C over k[x_1, \ldots , x_ n] by Lemma 16.3.7.
Ad (9). This follows by applying Lemma 16.11.2.
Ad (10). Since k[y_1, \ldots , y_ m]_\mathfrak p \to \Lambda _\mathfrak q is flat and \mathfrak p \Lambda _\mathfrak q = \mathfrak q \Lambda _\mathfrak q by construction we see that \dim (k[y_1, \ldots , y_ m]_\mathfrak p) = d by Algebra, Lemma 10.112.7. Thus we can find \pi _1, \ldots , \pi _ d \in \Lambda which map to a regular system of parameters in \Lambda _\mathfrak q.
Ad (12). By Algebra, Lemma 10.106.3 any permutation of the sequence \pi _1, \ldots , \pi _ d is a regular sequence in k[y_1, \ldots , y_ m]_\mathfrak p. Hence \gamma _1 = \pi _1 t_1, \ldots , \gamma _ d = \pi _ d t_ d is a regular sequence in R_\mathfrak p = k[y_1, \ldots , y_ m]_\mathfrak p[t_1, \ldots , t_ d], see Algebra, Lemma 10.68.10. Let S = k[y_1, \ldots , y_ m] \setminus \mathfrak p so that R_\mathfrak p = S^{-1}R. Note that \pi _1, \ldots , \pi _ d and \gamma _1, \ldots , \gamma _ d remain regular sequences if we multiply our \pi _ i by elements of S. Suppose that
holds for i = 1, \ldots , t for some t \in \{ 0, \ldots , d\} . Note that \gamma _1^ e, \ldots , \gamma _ t^ e, \gamma _{t + 1} is a regular sequence in S^{-1}R by Algebra, Lemma 10.68.9. Hence we see that
Thus we get
after replacing \pi _{t + 1} by s\pi _{t + 1} for some s \in S by Lemma 16.10.1. By induction on t this produces the desired sequence.
Ad (13). Let S = \Lambda \setminus \mathfrak q so that \Lambda _\mathfrak q = S^{-1}\Lambda . Set \bar\Lambda = \Lambda _\mathfrak q/\mathfrak q^ n \Lambda _\mathfrak q. Suppose that we have a t \in \{ 0, \ldots , d\} and \delta _1, \ldots , \delta _ t \in S and a factorization D \to D' \to \bar\Lambda as in (13) such that (a), (b), (c) hold for i = 1, \ldots , t. We have \pi _{t + 1}^ N \in H_{A/k}\Lambda _\mathfrak q as \mathfrak q^ N \Lambda _\mathfrak q \subset H_{A/k}\Lambda _\mathfrak q by (1). Hence \pi _{t + 1}^ N \in H_{A/k} \bar\Lambda . Hence \pi _{t + 1}^ N \in H_{A/k}D' as D' \to \bar\Lambda is faithfully flat, see Algebra, Lemma 10.82.11. Recall that H_{A/k} = (a_1, \ldots , a_ t). Say \pi _{t + 1}^ N = \sum a_ j d_ j in D' and choose c_ j \in \Lambda _\mathfrak q lifting d_ j \in D'. Then \pi _{t + 1}^ N = \sum c_ j a_ j + \epsilon with \epsilon \in \mathfrak q^ n\Lambda _\mathfrak q \subset \mathfrak q^{n - N}H_{A/k}\Lambda _\mathfrak q. Write \epsilon = \sum a_ j c'_ j for some c'_ j \in \mathfrak q^{n - N}\Lambda _\mathfrak q. Hence \pi _{t + 1}^{2N} = \sum (\pi _{t + 1}^ N c_ j + \pi _{t + 1}^ N c'_ j) a_ j. Note that \pi _{t + 1}^ Nc'_ j maps to zero in \bar\Lambda ; this trivial but key observation will ensure later that (a) holds. Now we choose s \in S such that there exist \mu _{t + 1j} \in \Lambda such that on the one hand \pi _{t + 1}^ N c_ j + \pi _{t + 1}^ N c'_ j = \mu _{t + 1j}/s^{2N} in S^{-1}\Lambda and on the other (s \pi _{t + 1})^{2N} = \sum \mu _{t + 1j}a_ j in \Lambda (minor detail omitted). We may further replace s by a power and enlarge D' such that s maps to an element of D'. With these choices \mu _{t + 1j} maps to s^{2N}d_ j which is an element of D'. Note that \pi _1, \ldots , \pi _ d are a regular sequence of parameters in S^{-1}\Lambda by our choice of \varphi . Hence \pi _1, \ldots , \pi _ d forms a regular sequence in \Lambda _\mathfrak q by Algebra, Lemma 10.106.3. It follows that {\pi '}_1^ e, \ldots , {\pi '}_ t^ e, s\pi _{t + 1} is a regular sequence in S^{-1}\Lambda by Algebra, Lemma 10.68.9. Thus we get
Hence we may apply Lemma 16.10.1 to find an s' \in S such that
for any q > 0. By Lemma 16.11.3 we can choose q and enlarge D' such that (s')^ q maps to an element of D'. Setting \delta _{t + 1} = (s')^ qs and we conclude that (a), (b), (c) hold for i = 1, \ldots , t + 1. For (a) note that \lambda _{t + 1j} = (s')^{2Nq}\mu _{t + 1j} works. By induction on t we win.
Ad (16). By construction the radical of H_{(C \otimes _{k[x_1, \ldots , x_ d]} R)/R} \Lambda contains \mathfrak h_ A. Namely, the elements a_ j \in H_{A/k} map to elements of H_{B/k[x_1, \ldots , x_ n]}, hence map to elements of H_{C/k[x_1, \ldots , x_ n]}, hence a_ j \otimes 1 map to elements of H_{C \otimes _{k[x_1, \ldots , x_ d]} R/R}. Moreover, if we have a solution C \otimes _{k[x_1, \ldots , x_ n]} R \to T \to \Lambda of
then H_{T/R} \subset H_{T/k} as R is smooth over k. Hence T will also be a solution for the original situation k \to A \to \Lambda \supset \mathfrak q.
Ad (18). Follows on applying Lemma 16.9.2 to R \to C \otimes _{k[x_1, \ldots , x_ d]} R \to \Lambda \supset \mathfrak q and the sequence of elements \gamma _1^ c, \ldots , \gamma _ d^ c. We note that since x_ i^ c are strictly standard in C over k[x_1, \ldots , x_ d] the elements \gamma _ i^ c are strictly standard in C \otimes _{k[x_1, \ldots , x_ d]} R over R by Lemma 16.2.7. The other assumption of Lemma 16.9.2 holds by steps (12) and (13).
Ad (20). Apply Lemma 16.9.4 to the situation in (18). In the rest of the arguments the target ring is local Artinian, hence we are looking for a factorization by a smooth algebra T over the source ring.
Ad (22). Suppose that C \otimes _{k[x_1, \ldots , x_ d]} (R/JR)_\mathfrak p \to T \to \Lambda _\mathfrak q/J\Lambda _\mathfrak q is a solution to
Then C \otimes _{k[x_1, \ldots , x_ d]} (R/I)_\mathfrak r \to T_\mathfrak r \to \Lambda _\mathfrak q/I\Lambda _\mathfrak q is a solution to the situation in (20).
Ad (23). Our n = N + dc is large enough so that \mathfrak p^ nk[y_1, \ldots , y_ m]_\mathfrak p \subset J_\mathfrak p and \mathfrak q^ n \Lambda _\mathfrak q \subset J\Lambda _\mathfrak q. Hence if we have a solution C \otimes _{k[x_1, \ldots , x_ d]} (R/\mathfrak p^ nR)_\mathfrak p \to T \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q of (22 then we can take T/JT as the solution for (23).
Ad (24). This is true because we have a section C \to B in the category of R-algebras.
Ad (25). This is true because D' is essentially smooth over the local Artinian ring k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p and
Hence D'[t_1, \ldots , t_ d] is a filtered colimit of smooth R_\mathfrak p/\mathfrak p^ nR_\mathfrak p-algebras and B \otimes _{k[x_1, \ldots , x_ d]} (R_\mathfrak p/\mathfrak p^ nR_\mathfrak p) factors through one of these.
Ad (26). The final twist of the proof is that we cannot just use the map B \to D' which maps x_ i to the image of \pi _ i' in D' and z_{ij} to the image of \lambda _{ij} in D' because we need the diagram
to commute and we need the composition B \to D'[t_1, \ldots , t_ d] \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q to be the map of (14). This requires us to map x_ i to the image of \pi _ i t_ i in D'[t_1, \ldots , t_ d]. Hence we map z_{ij} to the image of \lambda _{ij} t_ i^{2N} / \delta _ i^{2N} in D'[t_1, \ldots , t_ d] and everything is clear. \square
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