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Notation: $k$ is an algebraically closed field. By a variety I mean a separated ringed space $(X,O_X)$ that is locally isomorphic to $(Z,\mathcal O_Z)$ where $Z\subseteq\mathbb A^n_k$ is a closed Zariski subset and $\mathcal O_U$ is the structural sheaf of regular functions.


Let $(X,\mathcal O_X)$ be an irreducible variety; I've found two definitions for a prime divisor of $X$:

  1. It is a closed irreducible subvariety $Y$ of codimension $1$.
  2. It is an irreducible subvariety $Y$ of codimension $1$.

Which of them do you prefer? why?

Thanks in advance.

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    $\begingroup$ Where have you seen the definition of prime divisor without it being required to be closed? $\endgroup$
    – rfauffar
    Nov 4, 2013 at 23:51
  • $\begingroup$ For example on Milne's notes at page 195, but also on wikipedia at the voice "Weil divisor". $\endgroup$
    – Dubious
    Nov 4, 2013 at 23:58
  • $\begingroup$ Sometimes subvarieties are defined to be closed. I've been to a course on complex manifolds where submanifolds were assumed closed... $\endgroup$
    – Zhen Lin
    Nov 5, 2013 at 0:30
  • $\begingroup$ I would think one would always want it to be closed. For example, you would like for a prime divisor on an affine scheme $\mbox{Spec}(A)$ to correspond to a prime ideal of $A$; for this you need it to be closed. $\endgroup$
    – rfauffar
    Nov 5, 2013 at 0:45

1 Answer 1

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Prime divisors are closed by definition. See any introduction to algebraic geometry. If some texts doesn't write "closed", this is probably included in the definition of a "subvariety".

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