# Number of affines needed to cover a variety

Hello,

I am aware of the related question "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

• Since any scheme is locally affine, for your last question it suffices to take any affine neighbourhood of any point in $Y$, no? Then we obtain, that one can take $C=D=1$ (at least) for irreducible varieties. – Mikhail Bondarko Jan 23 '12 at 18:55
• If $Y$ is irreducible then it is OK, but otherwise problematic. And even if I want a result only for irreducible varieties, I can't continue by induction since I am not sure that the complement $Y-U$ will be irreducible. – Sasha Jan 23 '12 at 18:58
• Sasha, you can take an affine on each irreducible component that is disjoint from all the other components. Then the union of all of these is a disjoint union of irreducible affines, hence affine itself. This you can repeat. – Sándor Kovács Jan 23 '12 at 19:23
• Maybe I don't understand, but how can I repeat this? at the second step, this procedure will find an open affine inside the closed complement of the first open, not a global open affine. How will I be sure that I can extend it to an open subset which is affine? – Sasha Jan 23 '12 at 21:42

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 $X$.

• Certainly more elementary than my answer! – Laurent Moret-Bailly Jan 24 '12 at 9:30

I am not sure what you mean by "hard", but here is an answer to (2). I claim that if $X$ is quasiprojective of dimension $d$, there is an affine morphism $X\to\mathbb{P}^d$. This obviously implies (2): cover $X$ by the preimages of the $d+1$ standard affine charts.

Proof of claim: take a dense open immersion $j:X\hookrightarrow\overline{X}$, with $\overline{X}$ projective. Blowing up if necessary, you can assume that $\overline{X}\setminus X$ is a Cartier divisor. Then $j$ is an affine morphism (locally defined by inverting one function). On the other hand, by Noether normalization, there is a finite (hence affine) morphism $p:\overline{X}\to\mathbb{P}^d$, so the composite map $p\circ j$ is affine.

(In this argument, the only subtle point is the notin of affine morphism, in particular the fact that it is a local condition on the target.)

About the last question: for given $X$, a positive answer is equivalent to the "Chevalley property" that every finite subset of $X$ has an affine neighborhood. This clearly holds for quasiprojective $X$. If $X$ is smooth and complete, this property is equivalent to projectivity (Kleiman).

Here is a more geometric but equally elementary argument.

Lemma. Let $U$ be an $n$-dimensional quasi-projective scheme (over any field $k$). Then there exists an open cover of $U$ by $n+1$ affines.

Proof. Let $X$ be a projective closure of $U$, and $Z = X \setminus U$. Blowing up in $Z$, we may assume that $Z$ is a Cartier divisor. If $\mathscr L$ is ample, then for some $d \gg 0$ both $\mathscr L^{\otimes d}$ and $\mathscr L^{\otimes d} + Z$ are ample. Write $\mathscr L^{\otimes d} =: \mathcal O(1)$, and consider the embedding $X \to \mathbb P^N$ it defines.

There exists a section $H$ of $\mathcal O_{\mathbb P^N}(1)$ not containing $X$, since $\bigcap H^0(\mathcal O_{\mathbb P^N}(1)) = \varnothing$. An easy induction then shows that there exist $n+1$ sections $H_1, \ldots, H_{n+1}$ of $\mathcal O_{\mathbb P^N}(1)$ satisfying \begin{align*} \dim(H_1 \cap \ldots \cap H_r \cap X) = n - r & & \text{ for } & 0 \leq r \leq n+1. \end{align*} Here we say that $\dim Y = -1$ iff $Y = \varnothing$. Letting $H'_i = (H_i \cap X) + Z$, we get $$H'_1 \cap \ldots \cap H'_{n+1} = Z.\label{Eq 1}\tag{1}$$ Moreover, the $H'_i$ are all ample divisors, since $H'_i \in |\mathscr L^{\otimes d} + Z|$. Thus, their complements $U_i$ are affine, and (\ref{Eq 1}) shows that their union is $U$. $\square$

Remark. Instead of there exists a section, I could have written a general section. But over finite fields, that is not enough to prove existence.

Remark. Throughout the argument, I only care about divisors set-theoretically. For example, in the blow-up step, we really should say that (the underlying set of) $Z$ is the support of a divisor. In (\ref{Eq 1}) we only have set-theoretic equality, which is good enough for the argument.