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?
Best Answer
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.