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The following exercise is taken from ravi vakil's notes on algebraic geometry.

Suppose $X$ is a closed subset of $\mathbb{P}^n_k$ of dimension at least $1$, and $H$ is a nonempty hypersurface in $\mathbb{P}^n_k$. Show that $H\cap X \ne \emptyset$.

The clue suggests to consider the cone over $X$. I'm stuck on this and I realized that i'm at this point again where i'm not sure how a neat formal proof of this should look like.

Thoughts:

Does a hypersurface in projective space mean $H=V_+(f)$, the homogeneous primes not containing $f$?

If $X \hookrightarrow Proj(S_{\bullet})$ is a closed embedding then it corresponds (before taking $Proj$(-)) to a surjection of graded rings $S_{\bullet} \to R_{\bullet}=S_{\bullet}/I_+(X)$ where $I_+(X)$ is the set of all homogeneous elements vanishing on $X$. The cone $C(X)$ over a $X$ is then obtained by taking the $Spec(-)$ of this morphism $C(X) \hookrightarrow Spec(S_{\bullet})$. Is that right? How can this help me prove the theorem above?

I got the feeling so far that there's a very elegant way to describe all hypersurfaces in terms of vanishing of global sections of line bundles. This would help me enourmously since it would enable me to carry my geometric intuition to this setting. In this context the statement would look like:

A global section of a non-trivial line bundle on projective space must have a zero on all zariski closed subsets. This is the same as saying that a nontrivial line bundle on projective space restricts to a nontrivial line bundle on all closed subspaces. And here I have a cohomology problem that feels pretty specific and managable, This all feels much less ambiguous to me than "take the cone over $X$". Clarifying this would help me a lot.

Saal Hardali
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1 Answers1

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Here is a different but most elementary proof that $H\cap X\neq \emptyset$:

The complement $ U=\mathbb P^n\setminus H$ of the hypersurface is an affine variety, an easy by-product of the Veronese embedding: see Theorem 1.1 here.
But it is impossible that $X\subset U$, since an affine variety cannot contain a positive-dimensional dimensional projective variety. Hence $H\cap X$ is non-empty.

Edit
At the OP's request let me remind that given two points on an affine variety $U$ there is a regular function $h\in \mathcal O(U)$ taking different values at them, whereas on a projective varieties $X$ all regular functions are constant.
This is why $U$ cannot contain $X$: consider two points on $X$ and restrict $h$ to $X$ to obtain a contradiction.

Georges Elencwajg
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  • Why can't an affine variety contain a positive dimensional projective variety? Is this a trivial consequence of something I'm missing? – Saal Hardali Jan 04 '16 at 12:54
  • This was very helpful. Thanks! Although I do hope someone will come here to explain this "cone" stuff since Vakil writes there that it is a useful argument to know. Is there a special name for such a scheme whose global sections separates ponts? (this is kind of like hausdorfness) and seems extremely useful... – Saal Hardali Jan 04 '16 at 13:33
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    You are welcome, dear Saal. In complex analysis the corresponding varieties are called "holomorphically separable". I'm not aware of a similar terminology in algebraic geometry. An important class of examples are provided by arbitrary open subsets of affine varieties, the so-called quasi-affine varieties. – Georges Elencwajg Jan 04 '16 at 15:08
  • Thanks again! I asked a different question about this since it seems very interesting and useful. – Saal Hardali Jan 04 '16 at 15:09
  • @Georges Elencwajg The fact $\mathbb P^n \setminus H$ is affine, is it true if $H$ is not irreducible hypersurface, i.e., $H=Z(f)$, $f$ is not irreducible? – user371231 Nov 06 '21 at 08:15