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Let $X \subseteq \mathbb A^n$ and $Y \subseteq \mathbb A^m$ be Zariski closed. Then the (Zariski) product $X \times Y \subseteq \mathbb A^{n + m}$ is closed and there is a projection map $p\colon X \times Y \to X$ which is continuous in the Zariski topology.

Question: When is $p$ an open map?

This is always the case when $X \times Y$ is given the product topology but here we give it the Zariski product topology and when we do that the result is not always true. I know it's true when $X$ and $Y$ are irreducible and the field is algebraically closed. Martin gave a very short answer to that effect here (and if someone could point me too a more elementary proof in that case I would appreciate it). The projection $\mathbb R^2 \to \mathbb R$ onto the first coordinate is not open (look at the complement of a circle) so even when $X$ and $Y$ are irreducible we need the field to be algebraically closed. What about when the field is algebraically closed but $X$ and $Y$ are not necessarily irreducible?

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    $\begingroup$ The title and the body ask different questions. At least to the unsuspecting eye of someone unfamiliar with the term Zariski product. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 9, 2013 at 18:43
  • $\begingroup$ Your question has been answered in the linked mathoverflow discussion. So actually you are looking for a more direct proof? $\endgroup$
    – Martin Brandenburg
    Commented Aug 11, 2013 at 11:56
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    $\begingroup$ There is an elementary proof in the case of a projection map. First, it is enough to prove the projection $A^n\times A^1\to A^n$ ($m=1$) is open. It is also enough to prove the image of a principal open subset $D(f)$ is open. Write $f=a_0+a_1T+\cdots a_dT^d$ and let $(x,t)\in D(f)$. Then some $a_i(x)\ne 0$ and $D(a_i)$ is contained in $p(D(f))$: for all $x'\in D(a_i)$, $f(x',t)$ is a non-zero polynomial, hence $f(x',t_0)\ne 0$ for some $t_0$. This means that $(x',t_0)\in D(f)$ and $x'\in p(D(f))$. $\endgroup$
    – user18119
    Commented Aug 11, 2013 at 14:03
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    $\begingroup$ @user18119: This is not a comment, it is an answer. $\endgroup$
    – Martin Brandenburg
    Commented Aug 11, 2013 at 15:53
  • $\begingroup$ I realize now what was confusing me. My example of a circle in $\mathbb R^2$ shows that the projection is not closed, but it is indeed open. $\endgroup$
    – Jim
    Commented Aug 11, 2013 at 18:15

1 Answer 1

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Yes, the projection $p:X\times Y\to X$ is always an open map.
This follows from the following purely algebraic fact:

Theorem
Given two algebras $R,S$ over a field $k$, the projection morphism $Spec(R\otimes_k S)\to Spec(S)$ is open.

Note that there are no finiteness conditions on the algebras.
The proof can be found in De Jong and collaborators' Stacks Project, Lemma 9.38.10.
A more advanced geometric interpretation of this result is that for a scheme $X$ over a field $k$, the structural morphism $X\to Spec(k)$ is universally open.

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