I was in a class recently where we were trying to roughly count the dimensions of certain spaces of rational maps from algebraic curves into closed subschemes $Z \subseteq \mathbb{A}^n$. One way to simplify the proof is to find a map from $Z$ to $\mathbb{A}^{n-1}$ with finite fibers, but we couldn't figure out whether you could always do this. I'd like to know when and how badly this can fail.

Let's assume that $Z$ is a reduced closed subscheme of $\mathbb{A}^n$ (and not equal to all of $\mathbb{A}^n$), and that everything is over an infinite field $k$.

Noether normalization says that if $Z$ is irreducible, then there's a finite map $Z \to \mathbb{A}^{n-1}$. Are there examples of reducible $Z$ of pure dimension $n-1$ such that no such map exists?

Are there examples of reducible $Z$ (no restrictions on dimension) such that no map $Z \to \mathbb{A}^{n-1}$ with finite fibers exists?

We could restrict attention to the linear projections $\pi:\mathbb{A}^n \to \mathbb{A}^{n-1}$. In this case, $\pi|_Z$ has a positive-dimensional fiber iff $Z$ contains a line in the direction of the kernel of $\pi$. So: are there examples of $Z$ that contain a line in every direction?