TITLE: $\operatorname{Spec}(\mathbb{C}\otimes_\mathbb{R}\mathbb{C})$ has two points
QUESTION [7 upvotes]: Why does $\operatorname{Spec}(\mathbb{C}\otimes_\mathbb{R}\mathbb{C})$ have two points?
I know that $\operatorname{Spec}(\mathbb{C}\otimes_\mathbb{R}\mathbb{C})=\operatorname{Spec}(\mathbb{C}\oplus\mathbb{C}),$ but then?

REPLY [16 votes]: We have $$\mathbb{C}\otimes_\mathbb{R}\mathbb{C}=\mathbb {R}[T]/(T^2+1)\otimes _\mathbb{R}\mathbb{C}=\mathbb C[T]/(T^2+1)=\mathbb C[T]/((T+i)(T-i))=\mathbb C\times \mathbb C$$ the last equality being,  more precisely, the  isomorphism of $\mathbb C$-algebras:  $$ \mathbb C[T]/((T+i)(T-i))\stackrel {\cong}{\to}\mathbb C\times \mathbb C: \text {class of} \:P(T) \mapsto (P(i),P(-i))     $$ 
We can then conclude by YACP's argument.   
But the correct vision (=  Grothendieck's vision) is that the extension $\mathbb C/\mathbb R$ is separable and 
thus is diagonalized by $\mathbb C$ or, alternatively,  that  $\mathbb C/\mathbb R$ is Galois which means that it diagonalizes itself.
[Recall that a degree $n$  algebra $A$ over the field $K$  is said to be diagonalized (or split) by the field extension $\Omega /K$ if there is  an isomorphism of $\Omega$-algebras  $A\otimes_K\Omega \cong \Omega^n$ ]
Edit
Grothendieck's vision in this toy example could be paraphrased as :
Since $\mathbb C/\mathbb R$ is finite and  separable (= étale) , $Spec(\mathbb C) \to  Spec(\mathbb R)$   is a covering space in the scheme-theoretic sense, hence it  is trivialized by $Spec ( \mathbb C)$, which is a universal covering space of $Spec(\mathbb R)$  because $\mathbb C$ is an algebraic closure of  $\mathbb R$.
How wonderfully  topology illuminates  algebraic geometry   in this vision!
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