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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!

Leafar
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1 Answers1

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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.

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Mariano Suárez-Álvarez
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  • Thanks a lot! I have some questions... First, why there is a surjective ring homomorphism $\mathbb{C} \otimes_\mathbb{Z} \mathbb{C} \rightarrow \mathbb{C} \otimes_\mathbb{Z} \mathbb{C}$? Second, how can I prove that $\mathbb{C} \otimes_\mathbb{Q} \mathbb{C} \cong \mathbb{C} \otimes_\mathbb{Z}\mathbb{C}$? Finally, I'm trying to show that $\mathbb{C} \otimes_\mathbb{Q} \mathbb{C}$ is not noetherian: I consider $\mathbb{C}$ as a field extension of $\mathbb{Q}$. Let $F_1 \subsetneq \cdots F_n \subsetneq \cdots$ be a strictly ascending chain of fields between $\mathbb{Q}$ and $\mathbb{C}$. – Leafar May 18 '15 at 14:45
  • Let $I_n = Ker(\mathbb{C} \otimes_\mathbb{Q} \mathbb{C} \rightarrow \mathbb{C} \otimes_{F_n} \mathbb{C})$. For every $n \in \mathbb{N}$ $I_n \subseteq I_{n+1}$. We want to prove that $I_n \subsetneq I_{n+1}$, so that ${I_n}$ is a strictly ascending chain of ideals of $\mathbb{C} \otimes_\mathbb{Q} \mathbb{C}$, and then $\mathbb{C} \otimes_\mathbb{Q} \mathbb{C}$ is not noetherian. So let $f \in F_{n+1} - F_n$. We have $f \otimes 1 - 1 \otimes f \in I_{n+1}$. But $f \otimes 1 - 1 \otimes f \not \in I_n$. So $\mathbb{C} \otimes_\mathbb{Q} \mathbb{C}$ is not noetherian. Is this proof correct? – Leafar May 18 '15 at 15:02
  • If $f:\mathbb C[X]\to\mathbb C$ is a surjective ring morphism, then $f\otimes f:\mathbb C[X]\otimes_\mathbb Z\mathbb C[X]\to\mathbb C\otimes_\mathbb Z\mathbb C$ is a surjective ring morphism. – Mariano Suárez-Álvarez May 18 '15 at 16:17
  • I'll leave $\mathbb C\otimes_\mathbb Q\mathbb C\cong \mathbb C\otimes_\mathbb Z\mathbb C$ as an exercise, which is basic practice on tensor products. – Mariano Suárez-Álvarez May 18 '15 at 16:18
  • But what is the surjective ring morphism $f$ that you consider? – Leafar May 18 '15 at 16:54
  • Any surjective morphism will do. You can surely construct at least one! – Mariano Suárez-Álvarez May 18 '15 at 17:55
  • Oh, of course there is the obvious one: $p(X) \mapsto p(a)$, where $a$ is fixed. Thanks for your help ;) – Leafar May 18 '15 at 18:49
  • Just for understanding better this proof, what is the description of the map $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C} \rightarrow \mathbb{C} \otimes_{\mathbb{F_n}} \mathbb{C}$? – Leafar Jun 25 '15 at 00:01
  • @Leafar, I don't know what youmean by $\mathbb F_n$. – Mariano Suárez-Álvarez Jun 25 '15 at 00:28
  • $F_n$ (sorry, not $\mathbb{F}_n$) is a field between $\mathbb{Q}$ and $\mathbb{C}$ – Leafar Jun 25 '15 at 00:49
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    It is the unique group homomorphism which maps $a\otimes b$ to $a\otimes b$ for all $a$ and all $b$ in $\mathbb Q$. – Mariano Suárez-Álvarez Jun 25 '15 at 00:57