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I'm currently reading Olivier Debarre's book “Higher-Dimensional Algebraic Geometry” and I'm stuck on a small remark made in the definition of Cartier divisors (it's on page 3 if you have the book). Note that here, all schemes are Noetherian and separated (this is an assumption made earlier). To be precise, I can't prove the passage in bold :

A Cartier divisor on a scheme $X$ is a collection of pairs $(U_i, f_i)$, where $(U_i)$ is an open cover of $X$ and $f_i$ an invertible element of $\mathcal K_X(U_i)$ (the sheaf of total quotient rings of $\mathcal O_X$), such that $f_i / f_j$ is in $\mathcal O_X^*(U_i \cap U_j)$. When $X$ is reduced, we may take integral open sets $U_i$ (...).

I am not able to build these integral open sets. I've tried to “play” with the finitely many irreducible components of X and the assumption of $X$ being reduced, but it doesn't seem obvious to me. All $U_i$ are reduced and Noetherian too, so I tried to look at the irreducible components of these open sets to cobble something together, but it came to nothing. Do you have any ideas?

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  • $\begingroup$ You can try proving the following statement: $X$ is an integral scheme if and only if $X$ is irreducible and reduced. $\endgroup$ Commented Dec 4, 2024 at 12:16
  • $\begingroup$ @jhzg Hey, thanks for your answer. I do know this property but I'm not sure how to apply it in this case. If $X$ is integral, of course, it is straightforward. Could you please tell me a little more about it? $\endgroup$ Commented Dec 4, 2024 at 13:06
  • $\begingroup$ $X$ is reduced, so every irreducible component of $X$ is also reduced. Hence, each irreducible component of $X$ is an integral scheme. Naturally, we can consider Cartier divisors on each irreducible component. $\endgroup$ Commented Dec 5, 2024 at 0:24
  • $\begingroup$ The components of X are always integral since, usually, we equip them with the reduced structure (even when X is a non-reduced scheme). I don't get your point. $\endgroup$ Commented Dec 5, 2024 at 6:01
  • $\begingroup$ Sorry, I made a mistake. My teacher said that more conditions are needed here; otherwise, the conclusion is incorrect. $\endgroup$ Commented Dec 5, 2024 at 11:19

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