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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