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?

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    $\begingroup$ Even if the point of view is completely different, in the second volume (if I remember correctly) of "Infinite Abelian Groups" by Laszlo Fuchs, there is a nice discussion about the possible ring structures on a given Abelian group $G$. (probably you are aware of this, but it is not always possible to find such a ring structure on a given group) $\endgroup$ – Simone Virili Dec 12 '12 at 10:19
  • $\begingroup$ Thanks! I was aware that it's not always possible (hence the "when nonempty" in my question), but I wasn't aware of the source you mention. I'll give it a look. $\endgroup$ – Noah Schweber Dec 23 '12 at 18:21
  • $\begingroup$ @SimoneVirili this claim is highly sensitive on the definition of "ring". Every abelian group has the identically zero multiplication, which is associative and commutative. For the group $C_{p^\infty}$ or $\mathbf{Q}/\mathbf{Z}$ it's indeed the only one and in particular there's no unital ring structure. $\endgroup$ – YCor Apr 8 '20 at 1:21

Being a logician, I would look at this from the model theoretic point of view and think about spaces of types, which are all rather nice Stone spaces.

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 $G$.

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.

  • $\begingroup$ Sorry for the delay in my response - I've been away from my computer. This is a nice answer! $\endgroup$ – Noah Schweber Dec 23 '12 at 18:27
  • $\begingroup$ No problem, Noah. This was a nice question! $\endgroup$ – François G. Dorais Dec 23 '12 at 23:25

In the case of a countable group, this kind of thing often arises in the subject of Borel equivalence relation theory, which has been considered in a few MO questions (see also links in this answer). One may regard the space of all rings extending a given countable group as a Polish space, just along the lines you suggest. This is important when comparing the complexity of corresponding classification problems, since one can sometimes say that one isomorphism relation does or does not reduce to another, by a Borel reduction, when they live on Polish spaces.


I think the answer depends on what you want to accomplish.

At this writing, two other answers have given a (primarily in my opinion) point of view from a mathematical logic perspective. Let me suggest a general algebraic or computer science perspective, but also note that with some imagination one could find some useful applications from a perspective based in dynamical systems or statistics or number theory or combinatorics.

I may need to design an information representation system which will help with data storage or transmission. In fact, I may need several such representations which have certain algebraic relationships. Assuming I have the group G as a required form of addition, I might view the possible extensions as a family of clones on the set which are generated by the addition and one of the choices for multiplication. I may find that several of the choices give the same term functions or operations, in which case it may be useful to mod out by an appropriate equivalence to organize the analysis. If you ask, I can spin out some other requirements with algebraic consequences for this generalized scenario.

Gerhard "Which Hammers Will You Use?" Paseman, 2012.12.12


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