This is problem 9.2.E in Vakil's notes:
Show that the points of Spec$(\overline{\mathbb{Q}}\otimes\overline{\mathbb{Q}})$ and are in natural bijection with Gal$(\overline{\mathbb{Q}}/\mathbb{Q})$, and the Zariski topology on the former agrees with the profinite topology on the latter.
I have worked out the case for finite Galois extensions.
If $K/\mathbb{Q}$ is a Galois extension of degree $n<\infty$ with minimal polynomial $f$, then $\overline{\mathbb{Q}}\otimes_\mathbb{Q} K\cong \overline{\mathbb{Q}}[x]/(f)\cong \overline{\mathbb{Q}}^n$, so
$$\textrm{Spec}(\overline{\mathbb{Q}}\otimes_\mathbb{Q} K) \cong \bigsqcup \textrm{Spec}(\overline{\mathbb{Q}}).$$
This is the disjoint union of $n$ points, which is in bijection with $\textrm{Gal}(K/\mathbb{Q})$ since $|\textrm{Gal}(K/\mathbb{Q})|=\deg f = n$.
I want to be able to conclude by just taking the limit, but I'm not sure whether or not this will preserve the bijection, especially because we are taking limits in distinct categories (affine schemes and groups) both different from the category of sets. I'm also not sure how I would show that the topologies agree afterwards.
Best Answer
The idea of taking limits is the right one.
Let $A = \overline{\mathbb{Q}}\otimes_{\mathbb{Q}}\overline{\mathbb{Q}}$.
Fact 1. All prime ideals of $A$ are maximal and all its residue fields are canonically isomorphic with $\overline{\mathbb{Q}}$.
Proof. For the proof note that $\mathbb{Q}\hookrightarrow \overline{\mathbb{Q}}$ is an integral morphism. Hence $\overline{\mathbb{Q}} \hookrightarrow A$ is an integral morphism (as a base change of the previously considered morphism). Pick a prime ideal $\mathfrak{p}\subseteq A$, then the induced map $\overline{\mathbb{Q}} \hookrightarrow A/\mathfrak{p}$ is integral. Now if $K\hookrightarrow D$ is an integral morphism, where $K$ is a field and $D$ is an integral domain, then $D$ is a field. Thus $A/\mathfrak{p}$ is a field algebraic over $\overline{\mathbb{Q}}$. Hence $\mathfrak{p}$ is maximal and $A/\mathfrak{p} \cong \overline{\mathbb{Q}}$.
Fact 2. Let $R$ be a commutative ring such that all prime ideals of $R$ are maximal. Then $\mathrm{Spec}\,R$ is a Hausdorff topological space. In particular, $\mathrm{Spec}\,R$ is compact.
Proof. By dividing $R$ by nilradical, we may assume that $R$ has no nilpotent elements. Since every prime ideal $\mathfrak{p}$ of $R$ is maximal, we deduce that $R_{\mathfrak{p}}$ is a reduced local ring with single prime ideal. Hence $R_{\mathfrak{p}}$ is a field. Pick now distinct prime ideals $\mathfrak{p}$, $\mathfrak{q}$ of $R$. Since they are maximal, there exists $f\in \mathfrak{p}\setminus \mathfrak{q}$. Now $\frac{f}{1}\in R_{\mathfrak{p}}$ is zero (ideal $\mathfrak{p}R_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$ is zero as $R_{\mathfrak{p}}$ is a field). Thus there exists $g\not \in \mathfrak{p}$ such that $g\cdot f =0$. Since $g\cdot f = 0 \in \mathfrak{q}$ and $f\not \in \mathfrak{q}$, we deduce that $g\in \mathfrak{q}$. Thus $g\in \mathfrak{q}\setminus \mathfrak{p}$ and $f\cdot g = 0$. Therefore, we have (for distinguished open subsets of $\mathrm{Spec}\,R$) $$\mathfrak{p}\in D(g),\,\mathfrak{q}\in D(f),\,D(f)\cap D(g) = \emptyset$$ and thus $\mathrm{Spec}\,R$ is a Hausdorff space. Since it is quasi-compact, we deduce that it is compact.
Now we are ready to prove that $\mathrm{Spec}\,A$ is homeomorphic to $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Let $\mathcal{K}$ be a diagram consisting of all finite extensions of $\mathbb{Q}$ and inclusions between them. Consider the limit $X$ in the category of topological spaces of the induced diagram $$\big\{\mathrm{Spec}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K\right)\big\}_{K\in \mathcal{K}}$$ Clearly $X$ is homeomorphic with $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ (this is just the definiton of a profinite topology on a Galois group). It suffices to check that $X$ is homeomorphic with $\mathrm{Spec}\,A$. Note that $$A=\mathrm{colim}_{K\in \mathcal{K}}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K\right)$$ in the category of $\overline{\mathbb{Q}}$-algebras. Hence $\mathrm{Spec}\,A$ gives rise to a cone over topological diagram $\big\{\mathrm{Spec}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K\right)\big\}_{K\in \mathcal{K}}$ and thus there exists a unique comparison map $f:\mathrm{Spec}\,A\rightarrow X$. Clearly $f$ is continuous. We have
$$\mbox{Points of }\mathrm{Spec}\,A\,=\mathrm{Mor}_{\overline{\mathbb{Q}}}\left(A, \overline{\mathbb{Q}}\right) = \mathrm{Mor}_{\overline{\mathbb{Q}}}\left(\mathrm{colim}_{K\in \mathcal{K}}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K\right), \overline{\mathbb{Q}}\right) =$$ $$= \lim_{K\in \mathcal{K}}\mathrm{Mor}_{\overline{\mathbb{Q}}}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K, \overline{\mathbb{Q}}\right) = \mbox{Points of }X$$ where the first identification is a consequence of Fact 1. Moreover, (as you can check carefully) this identification is induced by $f$. Thus $f$ is bijective. Therefore, $f$ is a continuous bijection with compact source (by Fact 2) and Hausdorff target. Hence $f$ is a homeomorphism.