Lemma 59.90.3. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume $n$ is invertible on $S$ and that 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 $K$ 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 59.88.7. We may and do assume $S'$ is affine (this is obvious, but see proof of the lemma in case of doubt). Let $K'$ be the function field of $S'$ and let $Y' = X' \times _{S'} \mathop{\mathrm{Spec}}(K')$ to get the diagram

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

By Lemma 59.90.2 the total direct image $R(Y \to Y')_*\mathbf{Z}/d\mathbf{Z}$ is isomorphic to $\mathbf{Z}/d\mathbf{Z}$ in $D(Y'_{\acute{e}tale})$; here we use that $n$ is invertible on $S$. Thus $Rh'_*\mathbf{Z}/d\mathbf{Z} = Rh_*\mathbf{Z}/d\mathbf{Z}$ by the relative Leray spectral sequence. This finishes the proof.
$\square$

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