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
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 $|\mathbb{A}^1_k \setminus \phi(U)| < \infty \implies |U| \ge |\phi(U)| = |\mathbb{A}^1_k| = |k|$.