# Any Intersection of affine open subsets is affine

This is an exercise of Shafarevich's Basic Algebraic Geometry 1, Chapter 1 Section 5 Exercise 9 page 66 in the third edition:

Show that any intersection of affine open subsets is affine [Hint: If $$X,Y\subset \mathbb A^n$$ are closed sets, the diagonal $$\Delta\subset\mathbb A^{2n}$$ is a closed subset and there is an isomorphism $$X\cap Y\cong (X\times Y)\cap \Delta$$]

I admit the first thing that's tripping me up is what the author means by "is affine". Does he mean affine closed or open?

Second, I don't see immediately how something like an infinite intersection would work here, especially considering how the hint would suggest looking at an infinite Cartesian product.

Third, I don't see what's wrong with this argument: If $$U_1,U_2\subset \mathbb A^n$$ are open affine subsets, then they can be written as $$U_i=\mathbb A^n\backslash V_i$$ with $$V_i$$ closed affine. Then $$U_1\cap U_2=\mathbb A^n\backslash (V_1\cup V_2)$$ and it is hence an affine open set.

Please avoid mention of schemes or cohomology since this book has not (and will not) introduce these concepts.

• Could you provide a page reference? – bounceback Feb 13 at 17:31
• 3) Doesn't extend to the infinite case, as only finite unions of closed sets are closed – bounceback Feb 13 at 17:33
• @bounceback added it to the question – George Feb 13 at 17:45

This must be a poor translation from the original or just a question that does not sound like what it was supposed to. For example, it would be reasonable to say this question implies that $$\mathbb{Z}=\bigcap_{\lambda\in\mathbb{C}\setminus \mathbb{Z}} (\mathbb{A}_\mathbb{C}^1\setminus\mathbb{V}(x-\lambda))$$ is affine, but the only countable affine varieties over $$\mathbb{C}$$ are finite.
So it's probably just saying that the intersection of finitely many open affines is affine, i.e., isomorphic to a closed algebraic set in some affine space. But your argument doesn't work as is. It's not true that the complement of any closed algebraic set is affine. For example, $$\mathbb{A}^2 \setminus\{(0,0)\}$$ is not affine (i.e., as a quasi-affine variety, it is not isomorphic to any closed subset of any affine space).
Supplemental hint: show that the (open!) subset $$U\times V\subset \mathbb{A}^n\times \mathbb{A}^n$$ (the right hand side is the product as varieties, and the left hand side is just an open subset thereof, so far) with the induced structure of a quasi-affine variety is actually the product of $$U$$ and $$V$$ as varieties. Since the product of affines is affine, and $$\Delta\cap (U\times V)$$ is closed in $$U\times V$$, it too is affine, and it is also isomorphic to $$U\cap V$$.
• Well, it means isomorphic to closed affine. E.g., $\mathbb{A}^1 \setminus \{0\}$ is open inside of the affine $\mathbb{A}^1$, so it's quasi-affine, but it's also isomorphic (using a morphism as a quasi-affine variety) to a closed algebraic set in $\mathbb{A}^2$, namely $\mathbb{V}(xy-1)$. Yeah, I found that book a bit hard to use sometimes, too. I found it helpful to read J.S. Milne's notes alongside it, or just some other 'easy' book, like Reid's or Karen Smith's. – csprun Feb 14 at 1:03