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 $k$.

  1. 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?

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

  3. 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?


1 Answer 1


It's probably too late here, but I do not see where one needs "integral domain" in the proof of Noether normalisation. As for projections (3rd question), for infinite fields linear transformations are good enough in Nagata's proof (see Mumford, Red book).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.