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 q
such 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 q
Set 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\Lambda
We 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 q
Set 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 q
It 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 q
It 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 q
The 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 q
where 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
\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)
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
\text{Ann}_{S^{-1}R/(\gamma _1^ e, \ldots , \gamma _{i - 1}^ e)}(\gamma _ i) = \text{Ann}_{S^{-1}R/(\gamma _1^ e, \ldots , \gamma _{i - 1}^ e)}(\gamma _ i^2).
Thus we get
\text{Ann}_{R/(\gamma _1^ e, \ldots , \gamma _ t^ e)R}(\gamma _{t + 1}) = \text{Ann}_{R/(\gamma _1^ e, \ldots , \gamma _ t^ e)R}(\gamma _{t + 1}^2)
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
\text{Ann}_{S^{-1}\Lambda /({\pi '}_1^ e, \ldots , {\pi '}_ t^ e)}(s\pi _{t + 1}) = \text{Ann}_{S^{-1}\Lambda /({\pi '}_1^ e, \ldots , {\pi '}_ t^ e)}((s\pi _{t + 1})^2).
Hence we may apply Lemma 16.10.1 to find an s' \in S such that
\text{Ann}_{\Lambda /({\pi '}_1^ e, \ldots , {\pi '}_ t^ e)}((s')^ qs\pi _{t + 1}) = \text{Ann}_{\Lambda /({\pi '}_1^ e, \ldots , {\pi '}_ t^ e)}(((s')^ qs\pi _{t + 1})^2).
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
R \to C \otimes _{k[x_1, \ldots , x_ d]} R \to \Lambda \supset \mathfrak q
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
(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 q
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
R_\mathfrak p/\mathfrak p^ nR_\mathfrak p = k[y_1, \ldots , y_ m]_\mathfrak p/ \mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p[t_1, \ldots , t_ d].
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
\xymatrix{ B \ar[r] & D'[t_1, \ldots , t_ d] \\ k[x_1, \ldots , x_ d] \ar[r] \ar[u] & R_\mathfrak p/\mathfrak p^ nR_\mathfrak p \ar[u] }
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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