$\mathbb{C\otimes_QC} $ is not Noether ring, but $\operatorname{Spec}(\mathbb{C\otimes_QC})$ is Noether as topological space

Question

I once heard that $\mathbb{C}\otimes_\mathbb{Q}\mathbb{C} $ is not Noether ring, but $\operatorname{Spec}(\mathbb{C}\otimes_\mathbb{Q}\mathbb{C})$ is Noether as topological space.

This is probably a last in my memory, but I want to check whether this is true or not. Thank you for your help.

Answer 1

Let $$I_n = (\zeta_{3^n}\otimes 1-1\otimes \zeta_{3^n})\subset \mathbb{C}\otimes_\mathbb{Q}\mathbb{C}$$ It is a strictly increasing chain of ideals.

$$\{ P\in Spec(\mathbb{C}\otimes_\mathbb{Q}\mathbb{C}), I_n\subset P\}$$ is a strictly decreasing chain of closed subsets.