The Stacks project

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. $\square$

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. $\square$

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 $q' > 0$. $\square$

Proof. The lemma is proven by the following steps in the given order. We will justify each of these steps below.

1. Pick an integer $N > 0$ such that $\mathfrak q^ N\Lambda _\mathfrak q \subset H_{A/k}\Lambda _\mathfrak q$.

2. Pick generators $a_1, \ldots , a_ t \in A$ of the ideal $H_{A/R}$.

3. Set $d = \dim (\Lambda _\mathfrak q)$.

4. Set $B = A[x_1, \ldots , x_ d, z_{ij}]/(x_ i^{2N} - \sum z_{ij}a_ j)$.

5. 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 $C \to B$.

6. Choose $c > 0$ such that each $x_ i^ c$ is strictly standard in $C$ over $k[x_1, \ldots , x_ d]$.

7. Set $n = N + dc$ and $e = 8c$.

8. Let $E \subset \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q$ be the images of generators of $A$ as a $k$-algebra.

9. 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 $\mathfrak p \Lambda _\mathfrak q = \mathfrak q \Lambda _\mathfrak q$.

10. Choose $\pi _1, \ldots , \pi _ d \in \mathfrak p$ which map to a regular system of parameters of $k[y_1, \ldots , y_ m]_\mathfrak p$.

11. Let $R = k[y_1, \ldots , y_ m, t_1, \ldots , t_ m]$ and $\gamma _ i = \pi _ i t_ i$.

12. If necessary modify the choice of $\pi _ i$ such that for $i = 1, \ldots , d$ we have

    \[ \text{Ann}_{R/(\gamma _1^ e, \ldots , \gamma _{i - 1}^ e)R}(\gamma _ i) = \text{Ann}_{R/(\gamma _1^ e, \ldots , \gamma _{i - 1}^ e)R}(\gamma _ i^2) \]

13. 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 $i = 1, \ldots , d$

    1. $(\pi _ i')^{2N} = \sum a_ j\lambda _{ij}$ in $\Lambda $ where $\lambda _{ij} \bmod \mathfrak q^ n\Lambda _\mathfrak q$ is an element of $D'$,

    2. $\text{Ann}_{\Lambda /({\pi '}_1^ e, \ldots , {\pi '}_{i - 1}^ e)}({\pi '}_ i) = \text{Ann}_{\Lambda /({\pi '}_1^ e, \ldots , {\pi '}_{i - 1}^ e)}({\pi '}_ i^2)$,

    3. $\delta _ i \bmod \mathfrak q^ n\Lambda _\mathfrak q$ is an element of $D'$.

14. 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 $C \to B$.

15. 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 $\gamma _ i = \pi _ i t_ i$.

16. It suffices to resolve

    \[ R \to C \otimes _{k[x_1, \ldots , x_ d]} R \to \Lambda \supset \mathfrak q \]

17. Set $I = (\gamma _1^ e, \ldots , \gamma _ d^ e) \subset R$.

18. It suffices to resolve

    \[ R/I \to C \otimes _{k[x_1, \ldots , x_ d]} R/I \to \Lambda /I\Lambda \supset \mathfrak q/I\Lambda \]

19. We denote $\mathfrak r \subset R = k[y_1, \ldots , y_ m, t_1, \ldots , t_ d]$ the inverse image of $\mathfrak q$.

20. It suffices to resolve

    \[ (R/I)_\mathfrak r \to C \otimes _{k[x_1, \ldots , x_ d]} (R/I)_\mathfrak r \to \Lambda _\mathfrak q/I\Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q/I\Lambda _\mathfrak q \]

21. Set $J = (\pi _1^ e, \ldots , \pi _ d^ e)$ in $k[y_1, \ldots , y_ m]$.

22. It suffices to resolve

    \[ (R/JR)_\mathfrak p \to C \otimes _{k[x_1, \ldots , x_ d]} (R/JR)_\mathfrak p \to \Lambda _\mathfrak q/J\Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q/J\Lambda _\mathfrak q \]

23. It suffices to resolve

    \[ (R/\mathfrak p^ nR)_\mathfrak p \to C \otimes _{k[x_1, \ldots , x_ d]} (R/\mathfrak p^ nR)_\mathfrak p \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q \]

24. It suffices to resolve

    \[ (R/\mathfrak p^ nR)_\mathfrak p \to B \otimes _{k[x_1, \ldots , x_ d]} (R/\mathfrak p^ nR)_\mathfrak p \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q \]

25. 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 $D' \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q$.

26. 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. $\square$