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Let $A$ be a integral domain of finite type over an algebraically closed field $k$. Let $\tilde A$ be the integral closure of $A$. Can we show $\{D(f)|f\in A\}$ is a topological basis of $\operatorname{Spec} \tilde A$? (I actually want to show all open sets of $\operatorname{Spec} \tilde A$ are of the form $\pi^{-1}{V}$ where $V$ is a open set of $\operatorname{Spec} A$ and $\pi$ is the canonical map $\operatorname{Spec} \tilde A\to\operatorname{Spec} A$) Could you provide any help? Thanks!

Richard
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    This is false (consider the normalization of the nodal cubic) - why do you want to prove this result? – KReiser Nov 24 '21 at 22:18
  • @KReiser Well, this is because I have read your answer, but I can't see if you don't have this property how could you strictly prove $\mathscr K^/\pi_\mathcal O_{\tilde{X}}^*=\operatorname{Ca.div}\tilde X$. Could you explain it more explicitly? – Richard Nov 29 '21 at 10:39
  • I've edited my other answer. – KReiser Nov 30 '21 at 02:47

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