Lemma 59.88.7. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume for some $q \geq 1$ we have that $BC(f, n, q - 1)$ is true, but $BC(f, n, q)$ is not. Then there exist a commutative diagram

\[ \xymatrix{ X \ar[d]_ f & X' \ar[d] \ar[l] & Y \ar[l]^ h \ar[d] \\ S & S' \ar[l] & \mathop{\mathrm{Spec}}(K) \ar[l] } \]

with both squares cartesian, where

$S'$ is affine, integral, and normal with algebraically closed function field,

$K$ is algebraically closed and $\mathop{\mathrm{Spec}}(K) \to S'$ is dominant (in other words $K$ is an extension of the function field of $S'$)

and there exists an integer $d | n$ such that $R^ qh_*(\mathbf{Z}/d\mathbf{Z})$ is nonzero.

**Proof.**
First choose a diagram and $\mathcal{F}$ as in Lemma 59.88.6. We may and do assume $S'$ is affine (this is obvious, but see proof of the lemma in case of doubt). By Lemma 59.88.3 we may assume $K$ is algebraically closed. Then $\mathcal{F}$ corresponds to a $\mathbf{Z}/n\mathbf{Z}$-module. Such a modules is a direct sum of copies of $\mathbf{Z}/d\mathbf{Z}$ for varying $d | n$ hence we may assume $\mathcal{F}$ is constant with value $\mathbf{Z}/d\mathbf{Z}$. By Lemma 59.88.4 we may replace $S'$ by the normalization of $S'$ in $\mathop{\mathrm{Spec}}(K)$ which finishes the proof.
$\square$

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