Lemma 37.47.5. Let $S$ be a scheme. Let $\{ S_ i \to S\} _{i \in I}$ be an fppf covering. Then there exists a surjective integral morphism $S' \to S$ and an open covering $S' = \bigcup U'_\alpha$ such that for each $\alpha$ the morphism $U'_\alpha \to S$ factors through $S_ i \to S$ for some $i$.

Proof. Choose $S = \bigcup U_ j$, $W_ j \to U_ j$, $W_ j = \bigcup W_{j, k}$, and $T_{j, k} \to W_{j, k}$ as in Lemma 37.47.2. By Lemma 37.47.3 we can extend $W_ j \to U_ j$ to a surjective integral morphism $\overline{W}_ j \to S$. After this we can extend $T_{j, k} \to W_{j, k}$ to a surjective integral morphism $\overline{T}_{j, k} \to \overline{W}_ j$. We set $\overline{T}_ j$ equal to the product of all the schemes $\overline{T}_{j, k}$ over $\overline{W}_ j$ (Limits, Lemma 32.3.1). Then we set $S'$ equal to the product of all the schemes $\overline{T}_ j$ over $S$. If $x \in S'$, then there is a $j$ such that the image of $x$ in $S$ lies in $U_ j$. Hence there is a $k$ such that the image of $x$ under the projection $S' \to \overline{W}_ j$ lies in $W_{j, k}$. Hence under the projection $S' \to \overline{T}_ j \to \overline{T}_{j, k}$ the point $x$ ends up in $T_{j, k}$. And $T_{j, k} \to S$ factors through $S_ i$ for some $i$. Finally, the morphism $S' \to S$ is integral and surjective by Limits, Lemmas 32.3.3 and 32.3.2. $\square$

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