Why $\mathbb{C}\otimes_\mathbb{Q}\mathbb{C} \not\cong \mathbb{C}\times\mathbb{C}$

Question

I am wondering if this is true or not: $\mathbb{C}\otimes_\mathbb{Q}\mathbb{C} \cong \mathbb{C}\times\mathbb{C}$ (as rings). Any help or suggestion would be helpful.

Answer 1

The only proper nontrivial quotients of $\Bbb C\times\Bbb C$ are $\cong\Bbb C$ itself.

But $\Bbb C\otimes_{\Bbb R}\Bbb C\not\cong\Bbb C$ is a proper nontrivial quotient of $\Bbb C\otimes_{\Bbb Q}\Bbb C$.

Answer 2

The equation $x^2=2$ has (at least) $8$ distinct solutions $$\pm \sqrt{2} \otimes 1$$ $$1 \otimes \pm \sqrt{2}$$ $$\pm i \sqrt{2} \otimes i$$ $$i \otimes \pm i\sqrt{2}$$ in $\mathbf C \otimes_{\mathbf Q} \mathbf C$, whereas it only has $4$ solutions $(\pm \sqrt 2, \pm \sqrt 2)$ in $\mathbf C\times \mathbf C$.

Answer 3

$\mathbb{C} \times \mathbb{C}$ is noetherian, but $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ is not noetherian.

Specifically, $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C} \to \mathbb{C} \otimes_{\mathbb{Q}(\sqrt{2})} \mathbb{C} \to \mathbb{C} \otimes_{\mathbb{Q}(\sqrt{2},\sqrt{3})} \mathbb{C} \to \mathbb{C} \otimes_{\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})} \mathbb{C} \dotsc$ is an infinite sequence of surjective ring homomorphisms which are no isomorphisms.