# Can every algebraic variety of dimension $n$ be covered by $n+1$ affine opens?

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?)

• Isn't this proposition 3.6 in these notes by A. Bertram: math.utah.edu/~bertram/6030/Affine.pdf – Igor Rivin Apr 30 '15 at 20:51
• @IgorRivin I don't see the connection... – YCor Apr 30 '15 at 20:55
• @Ycor Isn't the connection in the following corollary (3.7)? – Igor Rivin Apr 30 '15 at 20:57
• @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. – YCor Apr 30 '15 at 21:01
• @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). – 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.
• The projective case is trivial: $P^m$ is a union of $m+1$ affine open subsets, and hence so is any closed subset of $P^m$. – YCor May 1 '15 at 8:22
• @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. – David E Speyer May 1 '15 at 10:53