Hello,


I am aware of the related question ["Minimal size of an open affine cover"](https://mathoverflow.net/questions/13478/minimal-size-of-an-open-affine-cover), but would like to ask more specifically:


Do you have some elementary (i.e. not using hard things like compactification and such) proof for one of the following (here "variety" is separated over alg. closed field):



> 
> (1) Let $X$ be a variety; Can you show that $X$ can be covered by $C \cdot dim(X) + D$ open affines, where $C,D$ are universal constants?
> 
> 
> (2) Let $X$ be quasi-projective; Can you show (1) for it with $C=1,D=1$?
> 
> 
> (3) Let $X$ be smooth quasi-projective, and char. = 0; Can you show (2) for it?
> 
> 
> 


It is easy for a variety $X$ to find an open affine whose complement is of smaller dimension than $X$. But I don't see how given $Y$ closed in $X$, to find an affine open $U$ in $X$ such that $Y-U$ is of smaller dimension than $Y$.


Sasha



Number of affines needed to cover a variety
Here is a way to do (2) (and hence (3)):


Let $X$ be a quasi-projective variety, i.e., $X=Y\setminus W$, where $Y,W\subseteq \mathbb P^n$ are (closed) projective varieties. Consider the irreducible decomposition $Y=\cup\_i Y\_i$ and observe that $I\_W\not\subseteq \cup I\_{Y\_i}$ where $I\_T\subseteq k[x\_o,\dots,x\_N]$ denotes the ideal of the set $T\subseteq \mathbb P^N$. Pick a homogenous polynomial $f$ of degree $d$ such that $f\in I\_W\setminus (\cup\_i I\_{Y\_i})$. Let $H=Z(f)$. Then $\mathbb P^n\setminus H$ is affine and hence so is $Y\setminus H$. 


By construction $Y\setminus H\subseteq Y\setminus W=X$ and $H\not\supseteq Y\_i$ for any $i$
by the choice of $f$. Therefore $\dim (Y\_i\cap H)<\dim Y\_i$ so we may use induction on $\dim X$. Notice that the affine subset is obtained as an affine subset of the ambient projective space intersected with our variety, so the affine varieties obtained subsequently are restrictions of affine subvarieties of the original $X$.


