Is the morphism of affine varieties given by Noether Normalisation a closed map?

Question

Let $X\subseteq \mathbb A^n(k)$ be an affine variety over the algebraically closed field $k$. Then by the geometric interpretation of Noether Normalisation Lemma there is a finite and surjective morphism $\varphi:X\to \mathbb A^m(k)$, where $m$ is the dimension of $X$ (see here for the process noether normalization theorem geometric meaning).

Is it true that the morphism $\varphi:X\to \mathbb A^m(k)$ is a closed map?

score 2 · Accepted Answer · answered Jul 07 '17 at 13:48

2

Any finite morphism is proper (hence universally closed).

answered Jul 07 '17 at 13:48

Bernard

175,478

1

But the O.P. spoke of a finite morphism (Normalisation lemma is about a morphism being finite). – Bernard Jul 07 '17 at 13:55
@Bernard I didn't know that fact. Thanks for pointing out! – ugur efem Jul 07 '17 at 13:55
Yes, you are correct. – Nefertiti Jul 07 '17 at 14:03

Is the morphism of affine varieties given by Noether Normalisation a closed map?

1 Answers1