I'm working on the following problem from Dummit and Foote 15.3 Exercise 15).
Let $V$ be an affine algebraic set over an algebraically closed field $k$. Prove that for some $n$ there is a surjective morphism from $V$ onto $\mathbb{A}^n$ with finite fibers, and that if $V$ is a variety, then $n$ can be taken to be the dimension of $V$.
In particular I'm a little stuck on the second part here. If we assume that $V$ is a variety, then we get that $k[V]$ is a integral domain. It seems like we should be able to use this fact to show that some finite of generators for $k[V]$ are algebraically independent and thus have cardinality equal to that of the dimension of $V$, but I can't quite see how.
The (or one of the) algebraic statement of the Noether Normalization Lemma is the following:
Let $A$ be a finitely generated $k$-algebra and $d$ the Krull dimenstion of $A$. Then there exist algebraically independent elements $x_1, \ldots, x_d \in A$ such that $A$ is an integral extension of $k[x_1, \ldots, x_d]$.
This result has a very beautiful translation in the language of algebraic geometry:
Let $V \subseteq \mathbb{A}^m$ be an affine variety and $n = \text{dim}(V)$. The geometric version of the Noether Normalization Lemma ensures that there exists some finite surjective morphism of affine varieties: $$\phi : V \rightarrow \mathbb{A}^n$$