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I aim to prove $P ∩\bigcup_α Q_α =\bigcup_α (P ∩Q_α)$. For $P,Q$ to be open subschemes of an affine scheme $X$.

I have been told that every open scheme is a union of complements of hypersurfaces, an I have attempt write all the $Q_\alpha$'s to be a union of complement $S$ of some hypersurface, Then I have $P ∩\bigcup_\beta S_\beta$. But I am not quite sure how to deal with the right hand side. I know I have $\bigcup_\alpha (P ∩Q_α)=\bigcup (P \cap\bigcup_{\beta}S_\beta)$. But I am not quite sure how to do further. Some reference, answers or guidence, please? Thanks!

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  • $\begingroup$ Only concerning the first line: $P\cap\bigcup_{\alpha}Q_{\alpha} =\bigcup_{\alpha} (P\cap Q_{\alpha})$ is direct, isn't it? Or are $P$ and the $Q_{\alpha}$ not sets here? $\endgroup$ – drhab Sep 26 '17 at 9:23
  • $\begingroup$ As any open subset has a unique structure of open subscheme it's enough to prove the equality as sets. $\endgroup$ – nowhere dense Oct 7 '18 at 20:53
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Apply the following definitions for open subschemes directly

$$\bigcup_{\alpha}D_{I_\alpha} := D_{\sqrt{\sum I_\alpha}}$$ and $$D_I \cap D_J := D_{I\cap J} $$ Assuming you are in my AG class, some facts/theorems you might find useful may also be found on the wattle video W6 L2 Part 2 (you can find the video on the news forum on wattle). Hope that helped.

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  • $\begingroup$ Thanks and anyway I have solved it... I would post a solution latter. $\endgroup$ – PropositionX Sep 26 '17 at 11:55

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