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Given an algebraic variety $X$, I'm asking about the existence of a variety $X^{aff}$ and an "affinization" morphism

$f:X\rightarrow X^{aff}$

such that:

  • (a1) $f$ is injective when restricted to closed connected affine subvarieties of $X$,
  • (a2) Complete connected subvarieties of $X$ get shrinked by $f$ to points.

Condition (a2) may be strenghtened to

  • (a2') $f(x)=f(y)$ iff $x$ and $y$ are not separated by a regular function on $X$.

Then we may require that

  • (a3) whenever $f':X\rightarrow X'$ enjoys the above properties, then there is a closed embedding $j:X^{aff} \rightarrow X'$such that $f'=j\circ f$.

Q1: Does such an "affinization" morphism exist? Feel free to change the requirements (a1), (a2), (a3) in your aswer (i.e. answer a different question!), if it helps to get a better notion of what ought to be an "affinization" of $X$ (as my requirements may not be the best).

Q2: If it exists, is it unique?

I have something like a candidate for $f$, but I'm not sure the following makes sense. Consider the map

$X\rightarrow X^{aff}:=\operatorname{Spec} \mathcal{O}(X)=\operatorname{Spec}(H^{0}(X,\mathcal{O}_X))$

$x \mapsto \mathfrak{m}_x$

where $\mathfrak{m}_x$ is the ideal of functions in $\mathcal{O}(X)$ vanishing at $x$.

Q1': Is it even a morphism? Does this work as an "affinization" morphism?

An analogous question would involve a hypothetical "properization morphism"

$g:X\rightarrow X^{prop}$

such that:

  • (p1) $g$ is injective when restricted to complete connected subvarieties of $X$
  • (p2) closed connected affine subvarieties of $X$ get shrinked to points by $g$
  • (p3) an analogous "universal property" holds (if you like).

Condition (p2) maybe might be stregthened as:

  • (p2') closed quasi-affine varieties get shrinked to points.

Q3: Does such a "properization" morphism exist? Again, feel free to change my requirements in such a way that they meet a good heuristic definition of what a "properization morphism" should be, if it should exist at all.

Q4: In case it exists, what about uniqueness?

Then I could ask, in case of existence, if it would be the case that $f$ (resp. $g$) factors through a proper (resp. affine) morphism to $X^{aff}$ (resp. X^{prop}), but I feel that the above questions Qi are already sloppy enough! So, first I'm waiting for some answers or remarks that may point out some obvious things that I may have been missing.

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  • $\begingroup$ $H^0(X,\mathcal{O}_X)$ need not be a finitely generated algebra. $\endgroup$
    – naf
    Commented Apr 23, 2010 at 10:46
  • $\begingroup$ @unknown(google): could you give me an example in which H^0(X,O) is not fin gen ? $\endgroup$
    – Qfwfq
    Commented Apr 23, 2010 at 11:04
  • $\begingroup$ (continued) Wait...I'm going to post it as a separate question: it'd be worth having this example easily searchable. $\endgroup$
    – Qfwfq
    Commented Apr 23, 2010 at 11:06
  • $\begingroup$ (continued) Oh, it seems there's already a question like this (about finite gen. of O(X)) on MO. $\endgroup$
    – Qfwfq
    Commented Apr 23, 2010 at 11:08
  • $\begingroup$ Also have a look here for an example: math.stanford.edu/~vakil/files/nonfg.pdf $\endgroup$
    – Lars
    Commented Apr 25, 2013 at 12:01

2 Answers 2

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I believe that as you stated it, an affinization cannot exist. Consider $\mathbb{P}^2$ with a point removed p. Any line through p is affine and should "survive" affinization. Every curve not through p should be contracted. Clearly this cannot happen!

You might try considering the spectrum of the global regular functions: this might be a substitute for what you want...

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  • $\begingroup$ I've considered the spectrum of global regular functions in my question Q1'. - Does it work in some sense? Surely your example shows that there's something wrong with my "axioms" for an affinization... $\endgroup$
    – Qfwfq
    Commented Apr 23, 2010 at 9:55
  • $\begingroup$ By the way, the total space of O(1) of a projective variety $X$ should be "affinizable" to the affine cone of $X$, right? $\endgroup$
    – Qfwfq
    Commented Apr 23, 2010 at 10:01
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    $\begingroup$ The spectrum of the regular functions is obviously an affine scheme and it is universal with respect to morphisms from X to affine varieties. Since a proper subvariety of an affine variety has dimension zero, clearly proper subvarieties are contracted by the morphism in Q1'. On the other hand, injectivity is very difficult to achieve. For instance, if X admits an embedding in a proper variety with complement of codimension at least two, then the only global regular functions on X are constants. $\endgroup$
    – damiano
    Commented Apr 23, 2010 at 10:05
  • $\begingroup$ I am not sure what it is that you really want, but whatever it is, see if it survives the following check. Assume that X is irreducible, quasi-projective and not projective and that Y is a projective closure of X. Facts: 1) given any finite number of points in Y there is an irreducible curve containing them; 2) an integral curve without a point is affine. In particular there are closed affine subvarieties in X containing any finite subset of X. Moreover, if the complement of X in Y has codimension at least two, then there are also plenty of projective curves contained in X. $\endgroup$
    – damiano
    Commented Apr 23, 2010 at 10:15
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    $\begingroup$ The affine cone over X comes from the total space of $O(-1)$, not of $O(1)$! $\endgroup$
    – Sasha
    Commented Apr 23, 2010 at 20:28
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Ok, you can take a ring A, the global sections of the scheme. You have a morphism from X to Spec A, mapping each x to the maximal ideal of the local ring O_{X,x}. This is a locally ringed morphism.

If you like this case, you have some nice properties, for example, every morphism to another affine scheme will factorize by your f.

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