Lemma 37.48.5. 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 37.48.2. By Lemma 37.48.3 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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