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I'm trying to show that the cardinality of any variety of positive dimension is $ |k |$ where $k $ is the field being considered. This is part of exercise I.4.8 in Hartshorne's Algebraic Geometry:

Show that any variety of positive dimension over $k $ has the same cardinality as $ k $. Hints: Do $\mathbb{A}^{n} $ and $\mathbb{P}^{n}$ first. Then for any $ X $, use induction on the dimension $ n $. Use (4.9) to make $ X $ birational to a hypersurface $ H \subset\mathbb{P}^{n+1 }$. Use (Ex. 3.7) to show that the projection of $ H$ to $\mathbb{P}^{n}$ from a point not on $ H$ is finite-to-one and surjective.

So far I successfully showed this result for $ \mathbb{A}^{ n} $ and $ \mathbb{P}^{ n} $. For the general case, since $ X$ sits in a projective space, we have $|X | \le | k| $. To show the opposite inquality, $ X $ has an affine open subset $ U $of positive dimension. Hence there is a nonconstant polynomial as a regular function on $ U $. If I show that this polynomial is surjective, I'm done. I'm unable to show this so far.

I'm interested in completing my approach. If it's hopeless or too difficult, I'm fine with a solution following the hint of the book.

Thank you

PeterM
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  • Rereading their sketch, induction isn't used at all in the part they describe: it is simply reducing the general case to the case of projective space. The need for induction will appear when working through the details of using the reduction to compute the cardinality. –  Jul 16 '14 at 17:05
  • @Hurkyl Thanks for your comment. Even ignoring induction, I don't know how birationality with a hypersurface in the projective space helps. Is it somehow possible to show that an open subset of the hypersurface has cardinality $ |k |$? This seems to me the same question I'm trying to prove, and no reduction of dimension happens. I would appreciate a more detailed answer, if possible. – PeterM Jul 16 '14 at 17:34
  • The proof sketch wants you to compare the cardinality of a variety with the cardinality of a projective space, so you can use what you've already proven about projective space. –  Jul 16 '14 at 17:46
  • @Hurkyl The birational map allows me to compare the cardinality of an open set in the given variety with an open set in the projective space. I don't know (yet) what the cardinality of the latter open set is. However, I think I have a better idea: The variety $X$ contains an open affine set $U$. $U$ has a surjective regular function (for example : projection on first coordinate). Hence $|X| \ge |k|$. Does this make sense? – PeterM Jul 16 '14 at 20:12
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    What about using the Noether Normalization Lemma to obtain a finite surjective map to some affine space from your open set $U$? – rfauffar Aug 16 '14 at 17:23
  • @RobertAuffarth I didn't know about the geometric statement of this lemma. It looks like it can be used indeed. – PeterM Aug 16 '14 at 20:12

1 Answers1

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Your approach runs into the difficulty of showing surjectivity of any nonconstant regular function. However, it can be made to work after some tweaks:

It suffices to show that an affine variety $U$, with $\dim U > 0$, has $|U| \ge |k|$. A morphism $U \to \mathbb{A}^1_k$ is the same as an element of the coordinate ring $A(U) = k[x_1, \ldots, x_n]/I(U)$. Since $k = \overline{k}$ and $\dim U = \dim A(U) > 0$, there exists $f \in A(U)$ transcendental over $k$, i.e. $\phi^* : k[x] \to A(U), x \mapsto f$, is an injection. Then $\phi : U \to \mathbb{A}^1_k$ is dominant (in fact, $\phi(U)$ contains a nonempty open set), so $|\mathbb{A}^1_k \setminus \phi(U)| < \infty \implies |U| \ge |\phi(U)| = |\mathbb{A}^1_k| = |k|$.

zcn
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  • @AlexYoucis: Yes, as mentioned in the comments. I wanted to avoid the full strength though :) – zcn Aug 24 '14 at 05:13
  • Oops! I missed that conversation. Apologies. :) – Alex Youcis Aug 24 '14 at 05:14
  • +1 Thank you for your answer. I see why $\varphi$ is dominant since $\varphi^*$ is injective. but why is $\varphi(U)$ an open set? The density of $\varphi(U)$ alone isn't sufficient to conclude that $|\mathbb{A}^{1} \setminus \varphi(U)| < \infty$, right? – PeterM Sep 02 '14 at 09:07
  • @PeterM: One (admittedly high-powered) way to see that $\phi(U)$ contains a nonempty open set is by Chevalley's Theorem, which guarantees that $\phi(U)$ is constructible. Now any dense constructible subset of an irreducible Zariski space (which $\mathbb{A}^1_k$ is) contains a nonempty open set – zcn Sep 02 '14 at 20:21
  • Sorry, but i dont understand the part of the density of $\phi(U)$ can you explain a little more? – Dimitri Feb 22 '15 at 07:15