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:



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> 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? 
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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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An algebraic vector bundle is trivialized by open sets. How many does one need?
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 $U\_i$.


