Show that $\mathbb{A}_\mathbb{C}^2 \ncong \mathbb{A}_\mathbb{C}^1 \times_{Spec(\mathbb{Z})} \mathbb{A}_\mathbb{C}^1$

Question

Show that $\mathbb{A}_\mathbb{C}^2 \ncong \mathbb{A}_\mathbb{C}^1 \times_{Spec(\mathbb{Z})} \mathbb{A}_\mathbb{C}^1$

Honestly I don't know where to begin...

It's the same as proving that $Spec(\mathbb{C}[X,Y]) \ncong Spec(\mathbb{C}[X]) \times_{Spec(\mathbb{Z})} Spec(\mathbb{C}[Y])$.

If we had $Spec(\mathbb{C})$ instead of $Spec(\mathbb{Z})$, then clearly we had an isomorphism.

How do I begin?

Thanks!

Answer 1

One way of showing that $\def\ZZ{\mathbb Z}\def\CC{\mathbb C}\CC[X,Y]$ and $\CC[X]\otimes_\ZZ\CC[Y]$ are not isomorphic as rings is to notice that the first one is a noetherian ring while the second is not.

Indeed, there is a surjective ring homomorphism $\CC[X]\otimes_\ZZ\CC[Y]\to\CC\otimes_\ZZ\CC$, so it is enough to show that $\CC\otimes_\ZZ\CC$ is not noetherian, and this follows from Martin's answer here: it is easy to see that $\CC\otimes_\ZZ\CC$ is isomorphic to $\CC\otimes_\mathbb Q\CC$, and since $\operatorname{tr.deg.}\mathbb C/\mathbb Q$ is infinite, his fifth bullet point does what we want.