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In Kollár's book "Lectures on Resolution of Singularities" it is claimed in 3.8 page 125: "Our resolution is strong and functorial with respect to smooth morphisms" I would like to use the statement:

For every variety $X$ there exists a resolution $f: X' \rightarrow X$ which is functorial with respect to smooth morphisms.

Question: what are the assumptions on $X$? i.e. what is a variety in this case?

Sorry for this question, I know very little about about resolutions of singularities.

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  • $\begingroup$ Have you looked in Kollar's book? I think that the definition of a variety is the first sentence of the book. $\endgroup$
    – Matthieu Romagny
    Commented May 2, 2023 at 14:52
  • $\begingroup$ @MatthieuRomagny thanks for your reply, it was confusing to me if that paragraph at the beginning was an informal discussion or if I should take it literally. Also, it would really be helpful to me if an expert or someone who knows what they are talking about just told me: A variety in this book is a "finite type scheme over a field of characteristic zero which is separated and integral" or something along those lines.... $\endgroup$
    – Anette
    Commented May 3, 2023 at 6:48
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    $\begingroup$ you're right, and my answer was not completely fair. In fact it is true that Kollar does not give a full-fledged definition of variety, and presumably is working definition is exactly what you wrote: a scheme of finite type over a field, which is separated and integral. And probably in many places the field will be assumed algebraically closed and of char. 0. $\endgroup$
    – Matthieu Romagny
    Commented May 3, 2023 at 21:07

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