Is $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)\cong\mathrm{Spec}(\overline{\mathbb Q}\otimes\overline{\mathbb Q})$ a group scheme

Question

This question gives a homeomorphism $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)\cong\mathrm{Spec}(\overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q})$, sending $\sigma\in\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ to $\mathfrak p_\sigma:=\ker(\overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q}\to\overline{\mathbb Q}:x\otimes y\mapsto x\sigma(y))$.

Does the group structure on $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ give $\mathrm{Spec}(\overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q})$ the structure of a group scheme?

First of all, the swapping morphism $\overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q}\to \overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q}:x\otimes y\mapsto y\otimes x$ takes $\mathfrak p_\sigma$ to $\mathfrak{p}_{\sigma^{-1}}$, so it should be taken as the inverse map. The problem is, I don't think $\overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q}$ can be given the structure of a $\overline{\mathbb Q}$-algebra structure such that the swapping morphism is a $\overline{\mathbb Q}$-homomorphism. That is, I doubt $\mathrm{Spec}(\overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q})$ can be given the structure of a group scheme over $\overline{\mathbb Q}$.

But over $\mathbb Q$, the fibered product $$\mathrm{Spec}(\overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q})\times_{\mathrm{Spec}\mathbb Q}\mathrm{Spec}(\overline{\mathbb Q}\otimes_\mathbb Q\overline{\mathbb Q})$$ is homeomorphic to $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)^3$, not $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)^2$. Is there any way to remedy this situation?

Answer 1

$\def\Spec{\text{Spec}}\def\QQ{\mathbb{Q}}\def\Qbar{\overline{\QQ}}\def\Gal{\text{Gal}}$This answer is going to be an unsettling "yes". Let $R = \Qbar \otimes_{\QQ} \Qbar$ and consider $R$ as a $\Qbar$-algebra by the action on the first factor. As noted in the linked question, we have natural bijections $\Spec(R) \cong \Spec(R)(\Qbar) \cong \Gal(\Qbar/\QQ)$. I will show that there is a unique $\Qbar$-group scheme structure on $\Spec(R)$ such that the induced group structure on $\Spec(R)(\Qbar)$ is $\Gal(\Qbar/\QQ)$. That said, I really don't understand what this group scheme structure is. In particular, the inverse map can't be $x \otimes y \mapsto y \otimes x$ since, as the OP notes, this is not $\Qbar$-linear.

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Let me start by reviewing different ways of setting up the basic objects of algebraic geometry. Let $k$ be a field and let $A$ be a $k$-algebra. Then

• $\Spec(A)$ is the set of prime ideals of $A$ and
• $\Spec(A)(k)$ is the set of $k$-algebra homeomorphisms $A \to k$.

There is a natural map $\Spec(A)(k) \to \Spec(A)$ sending a homeomorphism $\phi : A \to k$ to its kernel. The linked question checks that, in the case $(A,k) = (R, \Qbar)$, this map is a bijection.

Each $f \in A$ can be thought of as a function $\Spec(A)(k) \to k$. Namely, for $\phi \in \text{Hom}(A,k)$, we put $f(\phi) := \phi(f)$. We'll define $A$ to have enough $k$ points if the map $A \to k^{\Spec(A)(k)}$ is injective. In other words, $A$ has enough $k$-points if, given any nonzero $f \in A$, there is a $k$-algebra hom $\phi : A \to k$ with $\phi(f) \neq 0$.

We call a function $\Spec(A)(k) \to k$ regular if and only it comes from such an element of $A$. So $A$ has enough $k$ points if the ring of regular functions is isomorphic to $A$.

The next result is completely formal manipulations of definitions; please write out the details to convince your self:

Theorem Let $A$ and $B$ be $k$-algebras with enough $k$-points. The $k$-algebra homomorphisms $A \to B$ are in bijection with the set-theoretic maps $\Spec(B)(k) \to \Spec(A)(k)$ under which regular functions pull back to regular functions.

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We claim that $\Spec(R)$ has enough $\Qbar$-points. Indeed, any nonzero $f \in R$ lies in $K \otimes_{\QQ} K$ for some finite Galois extension $K$ of $\QQ$ with Galois group $G$. As discussed in this question, $K \otimes_{\QQ} K$ is isomorphic to the ring of $K$-valued functions on $G$ so, if $f$ is nonzero, then there is some $\sigma \in G$ where $f$ becomes nonzero. Lifting $\sigma$ to some $\widetilde{\sigma}$ in $\Gal(\Qbar/\QQ)$ gives a $\Qbar$ point of $\Spec(R)$ where $f$ is again nonzero.

So, writing $\Gamma = \Spec(R)(\Qbar)$, we have an identification of $R$ with $\Qbar$-valued functions on $\Gamma$ and, by the natural bijection $\Gamma \cong \Gal(\Qbar/\QQ)$, we can identify $R$ with certain $\Qbar$-valued functions on $\Gal(\Qbar/\QQ)$.

Which $\Qbar$-valued functions are they? I claim it is the locally constant functions, for the pro-finite topology on $\Gal(\Qbar/\QQ)$. Indeed, if $K/\QQ$ is a finite Galois extension, then $K \otimes_{\QQ} K$ turns into the $K$-valued functions which are constant on cosets of $\Gal(\Qbar/K)$. Taking the union over all $K/\QQ$, we get all locally constant functions. (Note that, since $\Gal(\Qbar/K)$ is compact, any locally constant function takes only finitely many values, and hence its values are contained in some finite Galois extension and it is constant on the cosets of some finite index subgroup.)

Multiplication gives a set-theoretic map $\Gal(\Qbar/\QQ) \times \Gal(\Qbar/\QQ) \to \Gal(\Qbar/\QQ)$; we want to know if this map is induced by a $\Qbar$-algebra homomorphism $R \to R \otimes_{\Qbar} R$. (And, similarly, inversion gives set-theoretic map $\Gal(\Qbar/\QQ) \to \Gal(\Qbar/\QQ)$ and we want to know if it is induced by a $\Qbar$-algebra homomorphism $R \to R$.)

By the criterion above, what we need to check is whether multiplication and inversion pull regular functions back to regular functions. In other words, we need to check, if $\phi$ is a locally constant function on $\Gal(\Qbar/\QQ)$, whether the functions $\Delta(\phi)(x,y) = \phi(xy)$ and $S(\phi)(x) = \phi(x^{-1})$ will also be locally constant. The answer is clearly yes.

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I am annoyed, therefore, that I can't write down the actual algebra homomorphisms. The one thing I'll note is that all the issues about infinite extensions and topology are distractions; this issue already makes sense for a finite Galois extension. What are the $K$-algebra maps $K \otimes_{\QQ} K \to K \otimes_{\QQ} K \otimes_{\QQ} K$ and $K \otimes_{\QQ} K \to K \otimes_{\QQ} K$ corresponding to multiplication and inversion in the isomorphism $K \otimes_{\QQ} K \cong KG$?

At the moment, I don't know.

To give the tiniest example, let $D \in \QQ$ be a nonsquare and let $K = \QQ(\sqrt{D})$. Note that there are two square roots of $D$ in $K$, but we have chosen one of them to call $\sqrt{D}$. Then $R = K \otimes_{\QQ} K = K[x]/(x^2-D) \cong K \times K$, where the map $K[x]/(x^2-D) \to K \times K$ sends $x$ to $(\sqrt{D}, - \sqrt{D})$.

We then have $R \otimes_K R = K[x_1,x_2]/(x_1^2-D, x_2^2-D) \cong K^4$. Let $\mu$ be the multiplication operation $\Spec R \times_K \Spec R \to \Spec R$. Then $\mu^{\ast}(x)$ is the function which is $\sqrt{D}$ on two of the factors of $K^4$ and $-\sqrt{D}$ on the other two; that function is $\tfrac{x_1 x_2}{\sqrt{D}}$. In short, the group operation comes from the coproduct $\Delta(x) = \tfrac{x_1 x_2}{\sqrt{D}}$.

It is not clear how to generalize this formula.

Answer 2

This is essentially a re-phrasing of David's answer. For any finite Galois extension $K/\mathbb Q$ there is a $\overline{\mathbb Q}$-algebra isomorphism $$\overline{\mathbb Q}\otimes_{\mathbb Q} K\cong \mathrm{Map}(\mathrm{Gal}(K/\mathbb Q),\overline{\mathbb Q}):x\otimes y\mapsto (\sigma\mapsto x\sigma y)$$ such that for extensions $L/K$ the diagram $\require{AMScd}$ \begin{CD} \overline{\mathbb Q}\otimes_{\mathbb Q} K @>{\sim}>> \mathrm{Map}(\mathrm{Gal}(K/\mathbb Q),\overline{\mathbb Q})\\ @V{i}VV @V{j}VV\\ \overline{\mathbb Q}\otimes_{\mathbb Q} L @>{\sim}>> \mathrm{Map}(\mathrm{Gal}(L/\mathbb Q),\overline{\mathbb Q}) \end{CD} commutes, where $i$ is simply the base change of the inclusion $K\hookrightarrow L$ and $j$ is the pre-composition with the surjection $\mathrm{Gal}(L/\mathbb Q)\to \mathrm{Gal}(K/\mathbb Q)$.

Now, taking colimits gives the isomorphism of $\overline{\mathbb Q}$-algebras $$\overline{\mathbb Q}\otimes_{\mathbb Q} \overline{\mathbb Q}\cong \mathrm{colim}\ \mathrm{Map}(\mathrm{Gal}(L/\mathbb Q),\overline{\mathbb Q})\cong\mathrm{Map}_{cont}(\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q),\overline{\mathbb Q})$$ where $\mathrm{Map}_{cont}(\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q),\overline{\mathbb Q})$ is the ring of uniformly continuous functions $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)\to\overline{\mathbb Q}$.

The group structure on $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ gives $\mathrm{Map}_{cont}(\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q),\overline{\mathbb Q})$ a Hopf algebra structure, which translates to a group scheme structure on its spectrum.