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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?

Andrew
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Gabriel Soranzo
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2 Answers2

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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!

Community
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Georges Elencwajg
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If you know that, use the following elementary description of the prime ideals of a finite product of commutative rings: $P$ is a prime ideal of $R_1\times\cdots\times R_n$ iff there is $p_i$ a prime ideal of $R_i$ such that $P=R_1\times\cdots R_{i-1}\times p_i\times R_{i+1}\times\cdots\times R_n$. Then you get that there are only two prime (=maximal) ideals: $\{0\}\times\mathbb{C}$ and $\mathbb{C}\times\{0\}$.