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Usual approaches to the inverse Galois problem start with realizations of a group $G$ over a larger field, and then try to specialize to ${\Bbb Q}$.

One could also start by building suitable objects over ${\Bbb Q}$ and then trying to locate among them a field.

Writing $n=|G|$, we could start with $A=<G>_{\Bbb Q}$, the $n$-dimensional ${\Bbb Q}$-vector space with basis $G$. Then equipping $A$ with a product $\ast$ (not to be confused with the group operation of $G$) making it a (not necessarily associative) algebra just means choosing structure constants $c_{g,h,j}\in {\Bbb Q}$ so that $$ g \ast h = \sum_{j\in G} c_{g,h,j} j\ .$$

Next we want the natural action of $G$ on the basis of $A$ to preserve $\ast$, so for $k\in G$, $$ kg \ast kh = \sum_{j\in G} c_{g,h,j} kj = \sum_{j\in G} c_{g,h,k^{-1}j} j\ .$$ This means $c_{kg,kh,j} = c_{g,h,k^{-1}j}$. Making $\ast$ commutative also forces $c_{g,h,k}=c_{h,g,k}$.

With no loss of generality, we can make the group identity $e\in G$ serve also as the identity for $\ast$, with $c_{e,h,j}=\delta_{h,j}$.

By linearity, imposing associativity on $A$ merely requires $(g\ast h)\ast j = g\ast (h\ast j)$ for basis elements. Since the action of $G$ preserves $\ast$, we can even fix $g$, say as $e$, the identity of $G$, a further reduction. So associativity imposes a family of quadratic relations on the structure constants. In any event, we get an affine variety $V_G$ that parametrizes commutative, associative algebras with a $G$-action that looks like the regular representation of $G$. (Is anything known in general about these varieties?)

It seems to me that the real difficulty here lies in requiring $\ast$ to form a field. I suppose it suffices to make $A$ a domain, and thus make multiplication by each element $\alpha=\sum_{g\in G} a_g g\in A$ injective. For a fixed $\alpha$ this comes down the the non-vanishing of a determinant, but then we must quantify over all non-zero $\alpha\in A$, so getting an algebraic condition on the structure constants involves some elimination of quantifiers or classical elimination theory.

But the difficulty here reminds me of the classical story about classifying division algebras over ${\Bbb R}$, with the related story about the existence of orthogonal families of vector fields on spheres and links to $K$-theory. I'm not expert in those matters, but I wonder whether analogous ideas might have some bearing here.

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The only question I can see in your post is in parentheses, and it is also somewhat vague. Can you enunciate a more precise question? –  Pete L. Clark Dec 11 '10 at 8:04
    
@Peter I do consider my final sentence a kind of question. I'd like to know if anyone has pursued a similar line of thought. If this line of thought is necessarily fruitless, I'd like to know technical reasons why. I'd love help from someone with expertise concerning the matters in my final paragraph on pushing through some sort of analogy, or explaining why the analogy breaks down. And frankly I'd like to hear any informed free association that my technical suggestion inspires. So I didn't want to lead the witnesses too much. –  David Feldman Dec 11 '10 at 8:24
    
I don't get the connection between the title and the (which?) question. –  Martin Brandenburg Dec 11 '10 at 10:16
    
Since you want to consider fields that are commutative and associative, rather than division algebras in general, the connection you suggest with parallelizability of spheres doesn't seem particularly strong. The commutative associative real division algebras were basically classified by Gauss - long before the development of K-theory. –  S. Carnahan Dec 11 '10 at 11:56
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@David Feldman (Peter who?): In your last comment I see a precise question: "Does every finite group $G$ occur as a group of automorphisms of a (not necessarily associative or commutative) division algebra $D/\mathbb{Q}$ with $[D:\mathbb{Q}] = |G|$?" In my opinion this is a much more MO-worthy question than writing down some observations about $G$-Galois algebras, saying that it reminds you of K-theory and asking if anyone has any comments. I encourage you to edit it into the question (or, perhaps, ask it as a new question). –  Pete L. Clark Dec 11 '10 at 22:51

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