Suppose $X$ is a complete algebraic variety of dimension $n$. Must there exist an affine covering with $n+1$ pieces?

(For a projective variety in $\mathbf{P}^m$, we can always project it to some subspace $\mathbf{P}^n$ with $n$ equals the dimension of X, by a composition of projection from points. Since projection is affine morphism, we are done. For complete varieties, we have Chow lemma, thus $X$ is dominated by some projective variety, but can we find a such a covering or not?)

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  • $\begingroup$ Isn't this proposition 3.6 in these notes by A. Bertram: math.utah.edu/~bertram/6030/Affine.pdf $\endgroup$ – Igor Rivin Apr 30 '15 at 20:51
  • $\begingroup$ @IgorRivin I don't see the connection... $\endgroup$ – YCor Apr 30 '15 at 20:55
  • $\begingroup$ @Ycor Isn't the connection in the following corollary (3.7)? $\endgroup$ – Igor Rivin Apr 30 '15 at 20:57
  • $\begingroup$ @IgorRivin the corollary says "Every quasi-affine variety has an open cover by quasi-affine varieties that are isomorphic to affine varieties." It does not say anything about the number... and anyway it only works with quasi-affine varieties. $\endgroup$ – YCor Apr 30 '15 at 21:01
  • $\begingroup$ @mqx What do you mean by "varieties in $P^m$"? if you mean quasi-projective varieties occurring in $P^m$ the obvious argument yields a covering by quasi-affine varieties (and by the way I don't understand your projection argument). $\endgroup$ – YCor Apr 30 '15 at 21:04

No. Example 4.9 of Roth and Vakil shows that, for any $m$, there is a singular, integral complete $3$-fold which cannot be covered by $m$ open affines. The authors mention as an open problem whether there is a smooth example. If you don't require varieties to be integral, Example 4.8 gives a simpler construction, which the authors credit to Jason Starr.

There was also some discussion in comments about whether this is true for projective or quasi-projective varieties. The answer is yes in both cases. See the proof of Theorem 18.2.6 in Vakil's textbook for the projective case, and the discussion in Section 22.4.15 for the quasi-projective case.

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  • $\begingroup$ The projective case is trivial: $P^m$ is a union of $m+1$ affine open subsets, and hence so is any closed subset of $P^m$. $\endgroup$ – YCor May 1 '15 at 8:22
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    $\begingroup$ @YCor The question is if $X$ is a closed $m$ dimensional subvariety of $\mathbb{P}^n$, with $n>m$, is $X$ a union of $m+1$ affines. The answer is yes (see the proof in Vakil), and it isn't hard, but it is less trivial than that. $\endgroup$ – David E Speyer May 1 '15 at 10:53

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