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In this question, I denote $\mathbb{P}^n$ for the standard $n$-dimensional projective space over $\mathbb{C}$. Also, with a variety I always mean a Zariski closed subset.

I am currently following an introductory course in algebraic geometry (I am 5 chapters in 'An invitation to algebraic geometry') and stumbled on the following problem:

Let $g \in \mathbb{C}[x_0 , ... , x_n]$ be a homogeneous polynomial of with degree at least 1. Suppose $Y \subset \mathbb{P}^n$ is a projective algebraic variety such that $Y \cap \mathbb{V}(g) = \emptyset$. Prove that $Y$ is affine.

I'm not sure I have a good understanding of the concept of a projective variety being affine. In my course we defined $U_i$ to be the set of $[x_0 : ... : x_n] \in \mathbb{P}^n$ where $x_i \neq 0$ and $\psi_i : U_i \to \mathbb{A}^n : [x_0 : ... : x_n] \mapsto [\frac{x_0}{x_i} : ... : \frac{x_n}{x_i}]$, which is a homeomorphism with respect to the Zariski topologies. Therefore we can "identify $\mathbb{A}^n$ as $U_i$, being a subset of $\mathbb{P}^n$", and we can thus "see every affine algebraic variety $V \subset \mathbb{A}^n$ as a subset of $\mathbb{P}^n$". What's meant by this (according to me) is that we can identify an affine algebraic variety $V \subset \mathbb{A}^n$ with $\psi_i^{-1}(V)$ as a subset of $\mathbb{P}^n$. Then we call a projective variety $W \subset \mathbb{P}^n$ affine if we can find an $i$ and an affine variety $V \subset \mathbb{A}^n$ such that $W = \psi_i^{-1}(V)$ (or that's how I interpreted this).

I can prove that this is equivalent to $W \subset U_i$ for some $i$. But $U_i = \mathbb{P}^n \setminus \mathbb{V}(x_i)$ is just the whole projective space minus a hypersurface, so shouldn't every projective variety $W$ for which we can find a hypersurface $H$ such that $W \subset \mathbb{P}^n \setminus H$ be called affine? Such $W$ is definetly isomorphic to an affine variety. And what about projective varieties $W \subset \mathbb{P}^n \setminus \mathbb{V}(g)$ for some homogeneous $g \in \mathbb{C}[x_0 , ... , x_n]$, like in the above problem? (Maybe we could differ the case where $g$ is irreducible from the case where it is not?)

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so shouldn't every projective variety $W$ for which we can find a hypersurface $H$ such that $W \subset \mathbb{P}^n \setminus H$ be called affine?

Yes! The trick is that most of the time, you can't find such a hypersurface $H$. If $W\subset\Bbb P^n$ is a projective variety of dimension $d > 0$, then every hypersurface intersects $W$: going to the affine cone, $C(W\cap V(g)) = C(W)\cap C(V(g))\subset\Bbb A^{n+1}$, an intersection of something of dimension $d+1$ and $n$ which meet at the origin. Every irreducible component will be of dimension at least $d$, and an irreducible component exists (since the origin is in the intersection). Therefore $W\cap V(g)$ has dimension $\geq d-1$, and so in particular, when $d\geq 1$, it must be nonempty. Therefore the only projective varieties which can be missed by a hypersurface (of any sort) are finite.

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  • $\begingroup$ Not everything is very clear to me. Could you give some clarifications to the following questions please? Please keep in mind I haven't yet seen much dimension theory at all, and also not much commutative algebra. 1) I'm guessing that the affine cone has the same dimension as the original variety, and follows from the statement that $W$ is irreducible iff $C(W)$ is. I had trouble proving the 'only if'-statement, since if we write $C(W)$ as the union of 2 affine varieties, the polynomials defining these 2 varieties need not be homogeneous. $\endgroup$
    – Steve
    Commented Dec 13, 2023 at 21:07
  • $\begingroup$ 2) I don't see why every irreducible component of $C(W) \cap C(V(g))$ has dimension at least $d$. Also if I understand this correctly, we do can take an irreducible component $X$ of $C(W)$ and $Y$ of $C(V(g))$ and use the link you provided to see $\dim (X \cap Y) \geq \dim (X) + \dim (Y) - n = d+1 + n - n = d+1$. But doesn't that mean that $W \cap V(g)$ has dimension at least $d$? 3) How can be proven that a variety that has dimension zero is finite? 4) Why is a finite set in $\mathbb{P}^n$ affine? $\endgroup$
    – Steve
    Commented Dec 13, 2023 at 21:22
  • $\begingroup$ 1) The affine cone has dim $d+1$ - if $V\subset\Bbb P^n$ is cut out by $I\subset k[x_0,\cdots,x_n]$, then $C(V)\subset \Bbb A^{n+1}$ is cut out by $I\subset k[x_0,\cdots,x_n]$ too. Irreducibility follows from the projection map $C(V)\setminus 0\to V$: the fibers are irreducible and of the same dimension, while the target is irreducible. You can find this on MSE. 2) By the linked proof, $\dim C(Y)\cap C(V(g)) \ge d+1+(n-1)+1-(n+1) = d$. 3) From the definition of dimension as length of chains of irreducible subvarieties. $\endgroup$
    – KReiser
    Commented Dec 13, 2023 at 22:14
  • $\begingroup$ 4) Find a hyperplane missing it, use that $\Bbb P^n\setminus H$ is affine, and closed subvarieties of affines are affine. $\endgroup$
    – KReiser
    Commented Dec 13, 2023 at 22:14

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