Proof.
Set d = \dim \Lambda _\mathfrak q. Set R = k[x_1, \ldots , x_ d]. Choose n > 0 such that \mathfrak q^ n\Lambda _\mathfrak q \subset \mathfrak h_ A\Lambda _\mathfrak q which is possible as \mathfrak q is minimal over \mathfrak h_ A. Choose generators a_1, \ldots , a_ r of H_{A/R}. Set
B = A[x_1, \ldots , x_ d, z_{ij}]/(x_ i^ n - \sum z_{ij}a_ j)
Each B_{a_ j} is smooth over R it is a polynomial algebra over A_{a_ j}[x_1, \ldots , x_ d] and A_{a_ j} is smooth over k. Hence B_{x_ i} is smooth over R. Let B \to C be the R-algebra map constructed in Lemma 16.3.1 which comes with a R-algebra retraction C \to B. In particular a map C \to \Lambda fitting into the diagram above. By construction C_{x_ i} is a smooth R-algebra with \Omega _{C_{x_ i}/R} free. Hence we can find c > 0 such that x_ i^ c is strictly standard in C/R, see Lemma 16.3.7. Now choose \pi _1, \ldots , \pi _ d \in \Lambda as in Lemma 16.10.2 where n = n, e = 8c, \mathfrak q = \mathfrak q and I = \mathfrak h_ A. Write \pi _ i^ n = \sum \lambda _{ij} a_ j for some \pi _{ij} \in \Lambda . There is a map B \to \Lambda given by x_ i \mapsto \pi _ i and z_{ij} \mapsto \lambda _{ij}. Set R = k[x_1, \ldots , x_ d]. Diagram
\xymatrix{ R \ar[r] & B \ar[rd] \\ k \ar[u] \ar[r] & A \ar[u] \ar[r] & \Lambda }
Now we apply Lemma 16.9.2 to R \to C \to \Lambda \supset \mathfrak q and the sequence of elements x_1^ c, \ldots , x_ d^ c of R. Assumption (2) is clear. Assumption (1) holds for R by inspection and for \Lambda by our choice of \pi _1, \ldots , \pi _ d. (Note that if \text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2), then we have \text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^ c) for all c > 0.) Thus it suffices to resolve
R/(x_1^ e, \ldots , x_ d^ e) \to C/(x_1^ e, \ldots , x_ d^ e) \to \Lambda /(\pi _1^ e, \ldots , \pi _ d^ e) \supset \mathfrak q/(\pi _1^ e, \ldots , \pi _ d^ e)
for e = 8c. By Lemma 16.9.4 it suffices to resolve this after localizing at \mathfrak q. But since x_1, \ldots , x_ d map to a regular sequence in \Lambda _\mathfrak q we see that R_\mathfrak p \to \Lambda _\mathfrak q is flat, see Algebra, Lemma 10.128.2. Hence
R_\mathfrak p/(x_1^ e, \ldots , x_ d^ e) \to \Lambda _\mathfrak q/(\pi _1^ e, \ldots , \pi _ d^ e)
is a flat ring map of Artinian local rings. Moreover, this map induces a separable field extension on residue fields by assumption. Thus this map is a filtered colimit of smooth algebras by Algebra, Lemma 10.158.11 and Proposition 16.5.3. Existence of the desired solution follows from Algebra, Lemma 10.127.4.
\square
Comments (2)
Comment #2760 by Anonymous on
Comment #2868 by Johan on