Lemma 16.9.4. Let R \to A \to \Lambda \supset \mathfrak q be as in Situation 16.9.1. Let \mathfrak p = R \cap \mathfrak q. Assume
\mathfrak q is minimal over \mathfrak h_ A,
R_\mathfrak p \to A_\mathfrak p \to \Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q can be resolved, and
\dim (\Lambda _\mathfrak q) = 0.
Then R \to A \to \Lambda \supset \mathfrak q can be resolved.
Proof.
By (3) the ring \Lambda _\mathfrak q is Artinian local hence \mathfrak q\Lambda _\mathfrak q is nilpotent. Thus (\mathfrak h_ A)^ N \Lambda _\mathfrak q = 0 for some N > 0. Thus there exists a \lambda \in \Lambda , \lambda \not\in \mathfrak q such that \lambda (\mathfrak h_ A)^ N = 0 in \Lambda . Say H_{A/R} = (a_1, \ldots , a_ r) so that \lambda a_ i^ N = 0 in \Lambda . By Lemma 16.9.3 we can find a factorization A \to C \to \Lambda with C of finite presentation such that \mathfrak h_ C \not\subset \mathfrak q. Write C = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m). Set
B = A[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z, t_{ij}]/ (f_ j - \sum y_ i t_{ij}, zy_ i)
where t_{ij} is a set of rm variables. Note that there is a map B \to C[y_ i, z]/(y_ iz) given by setting t_{ij} equal to zero. The map B \to \Lambda is the composition B \to C[y_ i, z]/(y_ iz) \to \Lambda where C[y_ i, z]/(y_ iz) \to \Lambda is the given map C \to \Lambda , maps z to \lambda , and maps y_ i to the image of a_ i^ N in \Lambda .
We claim that B is a solution for R \to A \to \Lambda \supset \mathfrak q. First note that B_ z is isomorphic to C[y_1, \ldots , y_ r, z, z^{-1}] and hence is smooth. On the other hand, B_{y_\ell } \cong A[x_ i, y_ i, y_\ell ^{-1}, t_{ij}, i \not= \ell ] which is smooth over A. Thus we see that z and a_\ell y_\ell (compositions of smooth maps are smooth) are all elements of H_{B/R}. This proves the lemma.
\square
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