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Consider an algebraic vector bundle $E$ on a scheme $X$. By definition there is an open cover of $X$ consisting of open subsets on which $E$ is trivial and if $X$ is quasi-compact, a finite cover suffices. The question then is simply: what is the minimum number of open subsets for a cover which trivializes $E$ ? Now this is silly because the answer obviously depends on $E$ ! If $E$ is trivial to begin with, the cover consisting of just $X$ will do, of course, but if you take $\mathcal O(1)$ on $\mathbb P^n_k$ you won't get away with less than $n+1$ trivializing open subsets . Here is why.

Suppose you have $n$ open subsets $U_i\subset \mathbb P^n_k$ over which $\mathcal O(1)$ is trivial. Take regular nonzero sections $s_i\in \Gamma(U_i,\mathcal O(1) )$ and extend them rationally to $\mathbb P^n_k$. Each such extended rational section $\tilde {s_i}$ will have a divisor $D_i$ and the complements $\tilde U_i= X\setminus |D_i|$, $(U_i\subset \tilde U_i)$, of the supports of those divisors will give you a cover of $\mathbb P^n_k$ by $n$ affine open subsets trivializing $\mathcal O(1)$. But this is impossible , because $n$ hypersurfaces in $\mathbb P^n_k$ cannot have empty intersection.

This, conversations with colleagues and some vague considerations/analogies have led me to guess ( I am certainly not calling my rather uninformed musings a conjecture) that the following question might have a positive answer:

Is it true that on a (complete) algebraic variety of dimension $n$ every vector bundle is trivialized by some cover consisting of at most $n+1$ open sets?

For example, the answer is indeed yes for a line bundle on a (not necessarily complete) smooth curve $X$: every line bundle $L$ on $X$ can be trivialized by two open subsets .

Edit Needless to say I'm overjoyed at Angelo's concise and brilliant positive answer. In the other direction ( trivialization with too few opens to be shown impossible) I would like to generalize my observation about projective space. So my second question is:

Consider a (very) ample line bundle $L$ on a complete variety $X$ and a rational section $s \in \Gamma _{rat} (X, L) $. Is it true that its divisor $D= div (s)$ has a support $|D|$ whose complement $X\setminus |D|$ is affine ? Let me emphasize that the divisor $D$ is not assumed to be effective, and that is where I see a difficulty.

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Morally it is true. A vector bundle trivializes away from a divisor and intersection of n+1 divisors is zero. I am afraid that you have to start making some assumptions about the variety to see it through. There may be singular counterexamples... –  Bugs Bunny Dec 15 '10 at 18:42
    
Georges, Angelo's answer is so clean, it seems to prove that every open cover of an n dimensional variety has a refinement by n+1 open sets. So in a way this seems to be a version of one of the open cover definitions of dimension, at least for a space where open sets are dense. Is this sensible? –  roy smith Dec 16 '10 at 22:23
    
Dear roy, first of all I completely agree that "so clean" is an excellent description of Angelo's answer.Secondly,among the "vague considerations"I alluded to in my question, there were indeed analogies with Lebesgue dimension in Topology, about which I learned in §5 of Milnor-Stasheff's book on characteristic classes. However you cannot extract the result you mention about refinements from Angelo's proof, because that result is false. Indeed Roth and Vakil have proved that, given $n$, there is an integral threefold which cannot be covered by less than $n$ affine open sets.(To be continued.) –  Georges Elencwajg Dec 16 '10 at 23:12
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(Continuation) A quasi-projective-scheme of dimension $n$, however, can be covered by $n+1$ open affine subschemes. Be that as it may, in French we have a proverb "Il vaut mieux s'adresser au Bon Dieu qu'à Ses saints". A literal translation would be "It is better to appeal to God than to His saints". However in English you have the less flattering (for me) : "It is better to talk to the organ-grinder than to his monkey".So, ask Angelo... –  Georges Elencwajg Dec 16 '10 at 23:28
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Dear roy: what is the "tin cup"? –  Georges Elencwajg Dec 17 '10 at 1:05
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2 Answers

up vote 25 down vote accepted

This is true if we assume that the vector bundles has constant rank (it is clearly false if we allow vector bundles to have different ranks at different points). Let $U_1$ be an open dense subset of $X$ over which $E$ is trivial, and let $H_1$ be a hypersurface containing the complement of $U_1$. Then $E$ is trivial over $X \smallsetminus H_1$. Now, it is easy to see that there exists an open subset $U_2$ of $X$, containing the generic points of all the components of $H_1$, over which $E$ is trivial (this follows from the fact that a projective module of constant rank over a semi-local ring is free). Let $H_2$ be a hypersurface in $X$ containing the complement of $U_2$, but not containing any component of $H_1$. Then we let $U_3$ be an open subset of $X$ containing the generic points of the components of $H_1 \cap H_2$, and let $H_3$ be a hypersurface containing the complement of $U_3$, but not the generic points of the components of $H_1 \cap H_2$. After we get to $H_{n+1}$, the intersection $H_1 \cap \dots \cap H_{n+1}$ will be empty, and the complements of the $H_i$ will give the desired cover.

[Edit]: now that I think about it, you don't even need the hypersurfaces, just define the $H_i$ to be complement of the $U_i$.

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Dear Angelo, thank you for this fast and perfect answer. Sono impressionato ma, conoscendoti, per nulla stupito! Mille grazie, Angelo. –  Georges Elencwajg Dec 15 '10 at 20:47
    
@Angelo: I don't see the problem with the rank varying. Your proof does not use the rank. –  Qing Liu Dec 15 '10 at 21:06
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To Qing Liu: if the rank varies the statement is false, at least if you interpret "trivialize" as I do. For example, if the rank takes $m$ different values you need at least $m$ different open subsets, and this can also happen if the dimension of $X$ is zero. The problem in the proof arises from the statement that a projective module over a semi-local ring is free, which is false if the rank is not constant. –  Angelo Dec 15 '10 at 21:52
    
To Georges: thank you for your kind words, and congratulations on your impeccable Italian. –  Angelo Dec 15 '10 at 21:53
    
@Angelo: thanks. I see my mistakes. –  Qing Liu Dec 15 '10 at 22:24
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This is an answer to Georges' updated question at the end of his post.

An equivalent formulation of the question is the following:

Question Let $L$ be an ample Cartier divisor on a projective scheme $X$ and suppose there exist effective divisors $D_1, D_2$ such that $L\sim D_1-D_2$. Then is it true that $X\setminus \left({\rm supp}\\,D_1 \cup {\rm supp}\\,D_2\right)$ is affine?

I think this is true in some cases, but not in general.

Claim 1 The answer to the question is YES if $X$ is a projective curve.

Proof Both $D_1$ and $D_2$ are effective and hence ample and similarly so is $A=D_1+D_2$. Clearly $X\setminus \left({\rm supp}\\,D_1 \cup {\rm supp}\\,D_2\right)=X\setminus {\rm supp}\\, A$, which is affine. $\square$

Claim 2 There are many examples for smooth projective varieties for which there exists $L, D_1, D_2$ as above such that $X\setminus \left({\rm supp}\\,D_1 \cup {\rm supp}\\,D_2\right)$ is not affine. In fact, this happens on any smooth projective surface containing a $(-1)$-curve.

Remark I am pretty sure one does not need smoothness and there are also singular examples. (Actually the example below only needs one smooth point.)

Proof Let $Y$ be an arbitrary projective variety (reduced) of dimension at least $2$ and $H$ an effective (very) ample Cartier divisor on $Y$. Let $\sigma : X\to Y$ be the blow up of a smooth point $p\in Y$ that is not contained in $H$ and let the exceptional divisor of $\sigma$ be $E\subset X$.

Then for some $m>0$ positive integer, $L=m\sigma^*H-E$ is ample. (I suspect that most people know this, but if you need a hint for this statement, an explicit estimate on $m$ can be found in Lemma 2 of this answer to another MO question.)

Now let $D_1=m\sigma^*H$ and $D_2=E$. Notice that by the choice of the point that was blown up, $D_1$ and $D_2$ are disjoint. It follows that $X\setminus \left({\rm supp}\\,D_1 \cup {\rm supp}\\,D_2\right)\simeq (Y\setminus {\rm supp}\\, H)\setminus \{p\}$. Furthermore, since $H$ is ample on $Y$, it follows that $Y\setminus {\rm supp}\\, H$ is affine, and hence $(Y\setminus {\rm supp}\\, H)\setminus \{p\}$ is not. $\square$

It is actually true, that for any line bundle there always exists a rational section for which the complement of its divisor is affine.

Claim 3 Let $L$ be an arbitrary Cartier divisor on a projective scheme $X$. Then there exist effective very ample divisors $D_1, D_2$ such that $L\sim D_1-D_2$.

Proof Choose an arbitrary ample Cartier divisor $A$ on $X$. For large enough $r_1\gg 0$ $L+r_1A$ is basepoint-free by the definition (or one of the basic properties depending on what you choose as definition) of ampleness. Then for an even larger $r\gg r_1$ we may assume that $L+rA$ is both basepoint-free and ample and hence very ample and also that $rA$ is very ample as well. Now choose $D_1=L+rA$ and $D_2=rA$. $\square$

And we get as an easy consequence:

Corollary With the notation of Claim 3, we may choose $D_1$ and $D_2$ such that $X\setminus \left({\rm supp}\\,D_1 \cup {\rm supp}\\,D_2\right)$ is affine.

Proof Replace $D_1$ and $D_2$ with general members of their complete linear systems. Then we may assume that they do not have a common component and hence ${\rm supp}\\,(D_1+D_2)={\rm supp}\\,D_1 \cup {\rm supp}\\,D_2$. Since $D_1+D_2$ is also ample, this proves that claim. $\square$

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Dear Sándor, thank you very much for your answer. I really like your ingenious and elegant construction for claim 2. Since I am actually even more interested in your Remark (which is exactly the result I was hoping for), could you be so kind as to briefly elaborate on "very, very ample line bundles" or give a reference? Előre is köszönöm. –  Georges Elencwajg Dec 16 '10 at 8:10
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Hi Georges, thanks. Actually bu "very very" I just meant "very", but I wanted to emphasize that they can be chosen as positive as one wants them to be. The proof is very simple. Choose and arbitrary ample divisor $A$. Then for large enough $r\gg 0$ it follows from the definition of ampleness that $L+rA$ is also ample and it can be made to be very ample with large enough $r$. Then $L=(L+rA)-rA$. I will add this to the answer tomorrow. Kudos for the perfect Hungarian! I am impressed with the correct inclusion of accents!! –  Sándor Kovács Dec 16 '10 at 9:21
    
Dear Sándor, thank you for the quasi-instantaneous answer. I am looking forward to the addition to your answer tomorrow (or earlier !) –  Georges Elencwajg Dec 16 '10 at 10:20
    
Sándor: A Google Translate bárki képes erre. Igazam van? –  Sheikraisinrollbank Dec 16 '10 at 15:54
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Sheikraisinrollbank: good point. And you also proved that Google Translate is only an approximation. Georges' quote was perfect, but yours is missing either a word or a conjugation. –  Sándor Kovács Dec 16 '10 at 17:34
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