From Wikipedia on Quasi Projective Varieties, it is stated that, "Quasi-projective varieties are locally affine in the same sense that a manifold is locally Euclidean : every point of a quasi-projective variety has a neighborhood which is an affine variety. This yields a basis of affine sets for the Zariski topology on a quasi-projective variety."

My question is, how can we explicitly show the above statement? The projective space $P^n$ can be covered by n affine space $A^n$ so do we just intersect these sets? And does this give us an open neighborhood that is still in the projective variety?

  • 3
    $\begingroup$ Yes, that's the right idea. Why don't you try to work through it and see where you get stuck? $\endgroup$
    – KReiser
    Commented Dec 7, 2022 at 18:23
  • $\begingroup$ We need (n+1) A^n to cover P^n. $\endgroup$
    – Yos
    Commented Dec 7, 2022 at 18:24

1 Answer 1


You can justify this in two steps, which I'll leave you to try and show. We may write a quasiprojective variety as $U \cap Z \subset \mathbb{P}^n$ where $U$ is open and $Z$ is closed.

Step 0: $\mathbb{A}^n$ has a basis of affine open sets.

Step 1: $\mathbb{P}^n$ has a basis of affine open sets, so we may write $U$ as a union of affine open sets.

Step 2: A closed subset of an affine open subset is affine.

Now we can conclude every element of $U \cap Z$ has an affine open neighborhood.

Challenging exercise: If $U, U'$ are affine open subsets of a quasiprojective variety then $U \cap U'$ is affine.


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