Multiplicative type groups and Galois character modules

If $M$ is finitely generated and the action is continuous, intersect the open stabilizers of a finite generating list. That intersection fixes every element of $M$, so is the kernel of the action and is open and normal. A Krull-open subgroup contains $\mathrm{Gal}(k_s\mathrm{/}L)$ for a finite Galois $L\mathrm{/}k$: take the normal closure of the finite extension defining a basic open neighborhood. Thus the action factors through the finite group $\mathrm{Gal}(L\mathrm{/}k)$. Conversely an action factoring through this group is continuous. For $G$, F3 and F2 show $X^{\ast }(G)$ is finitely generated, and its basis characters under a splitting isomorphism over $L$ are fixed by $\mathrm{Gal}(k_s\mathrm{/}L)$, so its action is continuous.

Given $M$, choose $L$ as in step 1.1. The action on $L[M]$ respects multiplication and all Hopf maps. By F4, $A=L[M{]}^{\mathrm{Gal}(L\mathrm{/}k)}$ is a finite-type Hopf algebra with $L\otimes A\cong L[M]$. Hence $D(M)=\mathrm{Spec}A$ is of multiplicative type. Its character module identifies with $M$ by F2, and the identification is equivariant because $e_m$ transforms to $e_{\sigma m}$. The fixed algebra equals $(k_s[M]{)}^{{\mathrm{\Gamma }}_k}$: invariance under $\mathrm{Gal}(k_s\mathrm{/}L)$ means every coefficient lies in $L$, since this subgroup fixes every monomial; taking the remaining finite-group invariants gives $A$. Thus the construction is independent of the chosen $L$.

For $G$, the evaluation map $k_s[X^{\ast }(G)]\to \mathcal{O}(G)\otimes k_s$ sends $e_{\chi }$ to its character function and is an equivariant Hopf isomorphism by F2 and F3. The finitely many images of a Hopf generating set of $k_s[X^{\ast }(G)]$ and of their antipodes involve finitely many coefficients in $k_s$, hence lie over a common finite Galois extension. Restricting the evaluation isomorphism to ${\mathrm{\Gamma }}_k$-fixed elements identifies its source with $(k_s[X^{\ast }(G)]{)}^{{\mathrm{\Gamma }}_k}=\mathcal{O}(D(X^{\ast }(G)))$ and its target with $(\mathcal{O}(G)\otimes k_s{)}^{{\mathrm{\Gamma }}_k}=\mathcal{O}(G)$, the latter because a fixed element is defined over one finite Galois subextension, where F4's canonical fixed-algebra clause computes the invariants. Hence $\mathcal{O}(D(X^{\ast }(G)))\cong \mathcal{O}(G)$, and so $G\cong D(X^{\ast }(G))$. Given $G,H$, take a common finite Galois splitting extension. F2 identifies every group map there with a reversed character-module map; this identification respects the actions. F4 says precisely the equivariant maps descend uniquely, giving the displayed bijection. The evaluation identifications commute with maps because both send a monomial to the corresponding pulled-back character. They are therefore natural and prove the anti-equivalence.