The Stacks project

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. $\square$