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

notassumed to be effective, and that is where I see a difficulty.

canbe 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... $\endgroup$ – Georges Elencwajg Dec 16 '10 at 23:284more comments