Suppose $A$ is subvariety of an irreducible complex space(analytic variety) $X$. Is there an analytic hypersurface of $X$ containing $A$?
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Take an open $U\subset X$ such that $W=U\cap A$ is non-empty (thus it is dense), let $P$ be a prime ideal of codimension $1$ in $O_X(U)_{I(W)}$, let $H$ be the closure of the vanishing set of $P\cap O_X(U)$. – reuns Dec 17 '20 at 20:24
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1@reuns this doesn't work in the analytic setting, see the crosspost at MO. To the OP: cross-posting so quickly is frowned upon, and cross-posting without cross-linking is extremely frowned upon. – KReiser Dec 18 '20 at 02:48
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@KReiser Tks I overlooked the term analytic variety. This answer looks fine https://math.stackexchange.com/a/36658/276986 – reuns Dec 18 '20 at 07:02
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What is your definition of an analytic hypersurface? Is it required to be closed? – Moishe Kohan Dec 20 '20 at 20:06
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@MoisheKohan https://link.springer.com/content/pdf/10.1007/s40306-017-0214-3.pdf – Ng Chikeung Dec 22 '20 at 13:02
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@MoisheKohan I am reading above article. The hypersurface in Prop2.7 I think is not closed. But in general it should be closed. – Ng Chikeung Dec 22 '20 at 13:08
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@NgChikeung: The answer really depends on this. – Moishe Kohan Dec 22 '20 at 14:32