In page 62 of Shaferevich Basic Algebraic Geometry the statement of Theorem 5 is:

If $f:X\rightarrow Y$ is a regular map of affine varieties and every point $x\in Y$ has an affine neighbourhood $U\ni x$ such that $V=f^{-1}(U)$ is affine and $f:V\rightarrow U$ is finite the $f$ itself is finite.

To prove this theorem he says we can take a neighbourhood $U$ of any point $Y$ such that $U$ is a principal open set and satisfies the assumption of the theorem. How is this so?

I proved that if $U\ni x$ is an principal open set of $Y$ then $f^{-1}(U)$ is also a principal open set of $X$. But then why $f:f^{-1}(U)\rightarrow U$ will also be a finite map?

You may use the results finite maps are closed and surjective.

$\textbf{Edit:}$ I specifically say where my confusion is:

Suppose $x\in Y$ fixed. Then by the hypothesis $\exists V\subseteq Y$, open, $V\ni x$ such that $V$ is affine and corresponding $U=f^{-1}(V)$ is also affine and $f:U\rightarrow V$ is finite, i.e., let $V\cong Z_1\subseteq \mathbb A^n$ and $U\cong Z_2\subseteq\mathbb A^m$, both $Z_1$ and $Z_2$ are closed and the corresponding map $\tilde{f}: Z_2\rightarrow Z_1$ is a finite map. Remember here $V$ and $U$ are given. So we cannot change them. Now, if $V_1\subseteq V$ is principal affine then $V_2=f^{-1}(V_1)\subseteq U$ is also affine. The book says $f:V_2\rightarrow V_1$ is a finite map, i.e., We have to show if $V_2\cong W_2\subseteq\mathbb A^p$ and $V_1\cong W_1\subseteq\mathbb A^q$ then the corresponding map $f':W_2\rightarrow W_1$ is a finite map. Here $p$ and $q$ may be different from $n$ and $m$ right?

Thank you.

  • $\begingroup$ Do you agree that any restriction of a finite map is finite, and that any open set contains a principal open set? $\endgroup$ – Ethan MacBrough Feb 19 '17 at 23:27
  • $\begingroup$ I agree that any open set contains principal open set. But finite maps are defined between affine closed sets in the book. Also I know that restriction of a finite map to any of its closed subset is affine. So I do not know if the restriction on open subset is also finite. The theorem is proved in the book before defining finite maps between quasiprojective sets. $\endgroup$ – user276115 Feb 20 '17 at 13:05
  • $\begingroup$ It's not much harder to define finite maps for open subsets of affine sets than it is to define them for affine sets (in fact, it's the same definition, using the localized coordinate rings). But looking at the theorem in Sharafevich's book, he seems to instead be using the fact that any principal open subset of an affine set is itself affine. I don't know if he proved this beforehand. There's a proof here for when $X$ is $\mathbb{A}^n$ which can easily be adapted to arbitrary affine sets. $\endgroup$ – Ethan MacBrough Feb 20 '17 at 17:25
  • $\begingroup$ @Ethan MacBrugh In my question when I say f from V to U is finite then I mean the corresponding map when we consider them as affine varieties is finite. But when we take principal open set in V and its preimage in U they they may be isomorphic to different closed sets of affine spaces rather than that of V and U. So we cannot take restriction. Or may be I am missing some obvious details. $\endgroup$ – user276115 Feb 20 '17 at 17:34
  • $\begingroup$ They may not be isomorphic as open sets either, if the map is not an injection. They don't need to be isomorphic to make the map between them finite. $\endgroup$ – Ethan MacBrough Feb 20 '17 at 17:37

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