The prime ideals of $\mathbb{C}[x]$ are $(0)$ and $(x-a)$ for $a\in\mathbb{C}$.
Similarly, the prime ideals of $\mathbb{R}[x]$ are $(0)$, $(x-a)$ for $a\in\mathbb{R}$ and $(x^2+ax+b)$ for an irreducible quadratic.
This suggests that we can see $\operatorname{Spec}\mathbb{R}[x]$ as the quotient of $\operatorname{Spec}\mathbb{C}[x]$ by the action of the Galois group $\operatorname{Gal}(\mathbb{C}/\mathbb{R})$.
The same correspondence holds for $\operatorname{Spec}\mathbb{R}[x,y]$, I think.
I would like to know then under what conditions we have a bijection correspondence between $\operatorname{Spec}k[x_1,\dotsc,x_n]$ and $\operatorname{Spec}K[x_1,\dotsc,x_n]/\operatorname{Gal}(K/k)$ for a field extension $K/k$. If we have such a bijection, is it also a homeomorphism for the Zariski topology?
My attempts of proof seem to suggest that we need $K/k$ to be Galois, but I am not sure.
Best Answer
The bijection does indeed require $K/k$ to be Galois. Let's check out our favorite example of a non-Galois extension, $\Bbb Q\subset \Bbb Q[\sqrt[3]{2}]$. The Galois group of this extension is trivial, so the quotient of $\operatorname{Spec} \Bbb Q[\sqrt[3]{2}][x]$ by the Galois action is just $\operatorname{Spec} \Bbb Q[\sqrt[3]{2}][x]$ again. But this is not in bijection with $\operatorname{Spec} \Bbb Q[x]$ in the natural way - for instance, the ideal $(x^3-2)$ has two distinct preimages: $(x-\sqrt[3]{2})$ and $(x^2+\sqrt[3]{2}x+\sqrt[3]{2})$.
When $K/k$ is Galois, this is in fact an isomorphism of schemes and thus automatically a homeomorphism for the Zariski topology. The proof is that $(K[x_1,\cdots,x_n])^{Gal}\cong k[x_1,\cdots,x_n]$ as rings, where $Gal$ is the Galois group, and that $(\operatorname{Spec} A)/G \cong \operatorname{Spec} A^G$ for a nice group $G$ acting on an affine scheme (the Galois group is such a nice group).