Earlier today I had a conversation with a friend about ways of putting topologies on sets of first-order structures; we wound up talking about reducts and expansions from a topological point of view, and this question arose as a special case.


The question has a specific part, and a less specific part. Hopefully these are both appropriate questions for MO; not being an algebraist, I'm not certain these aren't trivial. (Also, in this question I flagrantly abuse notation in the hopes of making things more readable.)


For $G$ a group with underlying set $X$, let $\mathcal{R}\_G$ denote the set of possible ring structures with additive group $G$. We can give $\mathcal{R}\_G$ a topology by taking as a sub-basis all sets of the form $\lbrace R: a\times\_Rb=c\rbrace$ for $a, b, c\in G$. This is (when nonempty) a totally disconnected space, since any two distinct ring structures with the same additive group must disagree on the product of some pair of elements. The automorphism group $Aut(G)$ acts on $\mathcal{R}\_G$ by homeomorphisms in a natural way, and for each $R\in\mathcal{R}\_G$ the group $Aut(R)$ acts on $\mathcal{R}\_G$ by homeomorphisms fixing the point $R$; but in general, I have difficulty saying much of anything about this space. My first question is: what can be said about the spaces $\mathcal{R}\_G$ or the map $G\mapsto\mathcal{R}\_G$, and what is a good source about this?


My vaguer question is this: what is the "right" way of looking at all the compatible ring structures on a given group? I get the sense, just from playing around with it, that the topology $\mathcal{R}\_G$ is too coarse an object (in a completely non-rigorous sense of the word) to get much information out of. Presumably there is a better way of looking at the same set of objects; what is it?



How should one look at the set of compatible ring structures on a given group?
Being a logician, I would look at this from the model theoretic point of view and think about [spaces of types](http://en.wikipedia.org/wiki/Type_%28model_theory%29), which are all rather nice [Stone spaces](http://en.wikipedia.org/wiki/Stone_space).


Here is the setup for your case for $0$-types. Let $\mathcal{L}\_G$ be the language of rings augmented with a constant for each element of the additive group $G$. Let $A\_G$ be the set of purely additive (have no mention of multiplication or the multiplicative identity) sentences of $\mathcal{L}\_G$ that are true in $G$. This is a partial $0$-type and the space $\mathcal{T}\_G^0$ of all (complete) $0$-types extending $A\_G$ is a lot like what you describe. Indeed, if $a,b,c \in G$ then $a\times b = c$ is a sentence of $\mathcal{L}\_G$ which determines a basic clopen set $\lbrace p \in \mathcal{T}\_G^0 : a \times b = c \in p\rbrace$. This is a little finer than the space $\mathcal{R}\_G$ you describe since elements of $\mathcal{R}\_G$ only encode the truth for quantifier-free sentences of $\mathcal{L}\_G$. The advantage of using spaces of types is that they have been extensively studied in model theory. Beyond $0$-types, you can consider the spaces of $k$-types, which instead of sentences use the broader class of formulas of $\mathcal{L}\_G$ with free variables among $v\_1,\dots,v\_k$. Together, these spaces of types give a very nice understanding of the situation you're looking at.


Note that there are some types that are not necessarily compatible with the underlying additive group being precisely $G$. For example, the $1$-types that extend $\lbrace v\_1 \neq a : a \in G \rbrace$ are incompatible with such rings. Nevertheless, if $R$ is a ring with underlying additive group $G$, then $R$ corresponds to the $0$-type $\lbrace \phi : R \vDash \phi\rbrace$, every element $a \in R$ corresponds to the $1$-type $\lbrace \phi(v\_1) : R \vDash \phi(a)\rbrace$, and so on. The Omitting Types Theorem lets you know which types can correspond to rings whose underlying groups are precisely $G$.




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A few more details regarding the last sentence. If $G$ is a countable group and $\tau \in \mathcal{T}^0\_G$ is a complete extension of $A\_G$, then $\tau$ corresponds to a ring with underlying additive group $G$ if and only if there is no single formula $\phi(v\_1)$ such that $\tau \vdash \phi(v\_1) \to v\_1 \neq a$ for every $a \in G$. Of course, this happens exactly when $(\forall v\_1 \lnot\phi(v\_1)) \in \tau$ or $\phi(a) \in \tau$ for some $a \in G$. So the points of $\mathcal{T}^0\_G$ that correspond to types that contain one of these for each formula $\phi(v\_1)$. Since $G$ is countable, this is always a $G\_\delta$ subset of $\mathcal{T}^0\_G$ and therefore it is always a nice Polish subspace of $\mathcal{T}^0\_G$.


Since you're interested in the action of $\mathrm{Aut}(G)$, let me add that $\mathrm{Aut}(G)$ acts on $\mathcal{T}^0\_G$ in a natural way and that the $G\_\delta$ subset described above is invariant under this action. However, since $\mathcal{T}^0\_G$ is compact, there may be much to gain in considering the action on the whole space.


