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Given some Abelian group $(G, +)$, does there always exist a binary operation $*$ such that $(G, +, *)$ is a ring? That is, $*$ is associative and distributive:

\begin{align*} &a * (b * c) = (a*b) * c \\ &a * (b + c) = a * b + a * c \\ &(a + b) * c = a * c + b * c \\ \end{align*}

We also might have multiplicative identity $1 \in G$, with $a * 1 = 1 * a = a$ for any $a \in G$. Multiplication may or may not be commutative.

Depending on the definition, the answer could be no in the case of the group with one element: then $1 = 0$. But the trivial ring is not a very interesting case. For cyclic groups the statement is certainly true, since $(\mathbb{Z}_n, +, \cdot)$ and $(\mathbb{Z}, +, \cdot)$ are both rings. What about in general? Is there some procedure to give arbitrary abelian groups ring structure?

Mikko Korhonen
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    Defining $ab=0$ always gives an example, so you might want to mention that you want something else. – Jonas Meyer Dec 22 '11 at 09:23
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    Finitely generated abelian groups are direct sums of cyclic groups, so you may also want to focus on the nonfinitely generated case. – Jonas Meyer Dec 22 '11 at 09:25
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    @Jonas, of course this definition of multiplication will not have an identity element. – Asaf Karagila Dec 22 '11 at 09:34
  • I wonder whether infinite dimensional vector spaces over $\mathbb Q$ can be given a multiplication that makes them rings with identity. – Jonas Meyer Dec 22 '11 at 10:10
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    @Jonas Meyer: They can. For $\kappa$ an infinite cardinal, any characteristic $0$ field of transcendence degree $\kappa$ is also a $\mathbb{Q}$-vector space of dimension $\kappa$. – Chris Eagle Dec 22 '11 at 11:04

4 Answers4

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If your group has the property that every element has finite order, but there is no upper bound on the orders of the elements, then it is not the additive abelian group of a ring with identity. The reason is that if there were such a ring structure with an identity $1$, then $1$ would have finite additive order $k$, and then for all $a$ in your group, $k\cdot a=(k\cdot1)a=0a=0$, which forces $a$ to have order at most $k$.

For each prime $p$, the Prüfer $p$-group $\mathbb Z(p^\infty)$ is an example of such a group. The quotient group $\mathbb Q/\mathbb Z$ is another. Direct sums (but not direct products) of infinitely many finite cyclic groups of unbounded order would also be examples.

Jonas Meyer
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    Even if you do not require a unit, then things still go wrong for a torsion divisible group G like the Prüfer groups. G has the property that Hom(G⊗G,G) = 0, so the only distributive (but not necessarily unital, associative, or commutative) multiplication is ab = 0. Unbounded DSC groups are not covered by this tensor product argument, but are covered by your use of 1. – Jack Schmidt Dec 22 '11 at 16:34
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    A very simple example of such an abelian group is the multiplicative group of complex roots of unity. It's an infinite abelian torsion group with unbounded exponent. – Ehsaan Dec 28 '13 at 05:23
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    And isomorphic, @Ehsaan, to $\Bbb Q/\Bbb Z$. – Lubin Nov 07 '15 at 16:20
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There are several canonical group structures that can be equipped naturally on a group ; I am going to list the ones that I know that are interesting, perhaps other people can add some to the list.

Finite abelian groups are isomorphic to product of finite cyclic groups so you can naturally form a ring from a direct product of finite rings. The same goes for finitely generated abelian groups ; you can deduce a ring structure from the products of the underlying natural rings.

One ring structure that is always there will be the trivial ring structure, i.e. just add a multiplication that means nothing ; it sends everything to the identity (of $G$). Clearly it respects all the ring axioms and it gives you a ring (without a ring identity, though). It is not very interesting in such, because it means "we do have a ring structure on $G$, but it is not telling us anything at all about $G$".

Another ring structure which does not always have its underlying abelian group isomorphic to $G$ but is canonically defined by $G$ is the endomorphism ring of $G$, i.e. let $$ R = \{ \phi : G \to G \, | \, \phi \text{ is a group endormorphism of } G \text{ to } G \}. $$ Equip a ring structure on $R$ by saying that $(R,+, \circ)$ is a ring by defining the following : the endomorphism $\phi_1 + \phi_2$ is defined as $$ (\phi_1 + \phi_2)(g) = \phi_1(g) + \phi_2(g) $$ and $$ (\phi_1 \circ \phi_2)(g) = \phi_1(\phi_2(g)). $$ If $G$ is abelian, then you can readily see that the operation $+$ gives you an abelian group on $R$, because the inverse of the endomorphism $\phi$ will just be the endomorphism that sends $g$ to $-\phi(g)$ instead of $\phi(g)$, so that $\phi + (-\phi) = 0_R$. Commutativity follows from the fact that $G$ is abelian and the other properties follow naturally. The fact that the $\circ$ operation gives you a ring structure comes from the fact that endomorphisms are in particular homomorphisms, so that $$ (\phi \circ (\psi_1 + \psi_2))(g) = \phi( \psi_1(g) + \psi_2(g)) = (\phi \circ \psi_1)(g) + (\phi \circ \psi_2)(g) $$ In particular, the identity of the ring $R$ would be the identity endomorphism of $G$.

I know it is not exactly what you asked for (in this last part), but I didn't know much about it at first when I saw it and I believed it was interesting, so I just thought you might wanna take a look.

Hope that helps,

Patrick Da Silva
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  • It seems that a unital ring with underlying abelian group $G$ is equivalent to an injective group homomorphisim $G \to \operatorname{End}_{\mathbb{Z}}G$ with $\text{Id}_G$ in its image. Would you agree? Can we tighten this up at all? – Eric Auld Mar 15 '16 at 04:50
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    Oh, now I'm seeing this has to do with the hom-tensor adjunction. That may be a clearer way to look at it. – Eric Auld Mar 15 '16 at 04:51
  • @EricAuld : A group homomorphism $G \to \mathrm{End}{\mathbb Z}(G)$ corresponds to a morphism $G \otimes{\mathbb Z} G \to G$ under the tensor-hom adjunction, but such a morphism is just a $\mathbb Z$-bilinear map ; associativity does not come automatically. The $\mathbb Z$-bilinearity gives you distributivity, but this is not enough to give you a ring. – Patrick Da Silva Mar 15 '16 at 05:10
  • @EricAuld : Actually I thought a bit more ; you need an extra property of your map $\varphi : G \to \mathrm{End}{\mathbb Z}(G)$ which is that for $s,t \in G$, you have $\varphi_s \circ \varphi_t = \varphi{\varphi_s(t)}$. This is what gives you associativity of multiplication, since you want to define $s \cdot t \overset{def}= \varphi_s(t)$. So a map $\varphi$ as above with $\varphi_s \circ \varphi_t = \varphi_{\varphi_s(t)}$ and an element $1_G$ with $\varphi_{1_G} = \mathrm{id}_G$ gives you a unital ring. Forget my group structure comment. – Patrick Da Silva Mar 15 '16 at 05:13
  • You mean that condition gives the associativity without assuming the map $G \to \operatorname{End}_{\mathbb{Z}}(G)$ is injective, or that hypothesis is still required even if we assume the map is injective? My thought was to define $s \cdot t := \varphi^{-1}( \varphi_t \circ \varphi_s)$. – Eric Auld Mar 15 '16 at 05:14
  • I posted this as a question here. http://math.stackexchange.com/questions/1698186/ring-structure-with-underlying-abelian-group-g – Eric Auld Mar 15 '16 at 05:17
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Every finite Abelian group is a product of finite cyclic groups, so you get a ring for free. Similarly, every finitely generated abelian group is isomorphic to some copies of $\mathbb{Z}$ times a finite Abelian group, so you get a ring for free there as well. The only interesting case remaining would be a non-finitely-generated Abelian group. There are a few more steps that one can probably go, but I don't know how we can get a ring in general.

Alexander Vlasev
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Jonas' nice answer leaves open the question of whether every torsion-free abelian group $A$ admits a ring structure, since his argument does not apply to this case. So here is a torsion-free counterexample.

Consider the abelian group $A \subset \mathbb{Q}$ consisting of rational numbers of the form $\frac{a}{b}$ where $b$ is squarefree; equivalently, this is the additive subgroup of $\mathbb{Q}$ generated by $\{ \frac{1}{p} \}$ where $p$ runs over all primes. Since we use infinitely many primes, any finitely generated subgroup can only consist of rational numbers with denominators divisible by a finite set of primes, so $A$ is not finitely generated (we will need this later).

Now, if $\ast : A \times A \to A$ is a ring structure on $A$ (for me all ring structures are unital) then it defines by currying a map

$$A \ni a \mapsto (a \ast (-)) \in \text{End}(A)$$

sending an element $a \in A$ to the operation of left multiplication by $a$, which is an additive homomorphism (by distributivity) and injective (since if $a \neq 0$ then $a \ast 1_A \neq 0$). So:

Corollary: If $A$ is an abelian group such that $\text{End}(A)$ does not admit an additive subgroup isomorphic to $A$, then $A$ cannot have a ring structure.

Now we in fact have that

$$\boxed{ \text{End}(A) \cong \mathbb{Z} }.$$

To see this, let $f : A \to A$ be an endomorphism. Then $f$ acts by multiplication by $n = f(1) \in \mathbb{Q}$ on $\mathbb{Z} \subset A$. To extend this action to all of $A$ we consider

$$b f \left( \frac{a}{b} \right) = f(a).$$

All divisions in $A$ are unique when they exist (since this is true of $\mathbb{Q}$), so if $b \neq 0$ then $f \left( \frac{a}{b} \right) = \frac{f(a)}{b}$, and it follows that $f$ acts by multiplication by $n \in \mathbb{Q}$ on all of $A$. But we must have $n \in \mathbb{Z}$ or else multiplication by $n$ is not defined on some fraction $\frac{1}{p}$ (for some prime $p$ dividing the denominator). So $n \in \mathbb{Z}$ and $\text{End}(A) \cong \mathbb{Z}$ as desired, and $\mathbb{Z}$ does not contain $A$ as a subgroup (e.g. because $A$ is not finitely generated).

There is a dual argument that can be used to rule out groups like $\mathbb{Q}/\mathbb{Z}$ as mentioned elsewhere in the discussion here: another necessary condition for $A$ to have a ring structure is that the multiplication $\ast : A \times A \to A$ induces a map $A \otimes A \to A$ which is surjective (again because of the identity). So if $A \otimes A$ does not admit a quotient isomorphic to $A$ we can again rule out the existence of a ring structure on $A$, and for $A = \mathbb{Q}/\mathbb{Z}$ or the Prüfer $p$-group $\mathbb{Q}_p/\mathbb{Z}_p$ this tensor product vanishes.

Qiaochu Yuan
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  • Nice. Another way to write your $A$ is as $\colim\left(\mathbb{Z}\overset{2\cdot}{\to}\mathbb{Z}\overset{3\cdot}{\to}\mathbb{Z}\overset{5\cdot}{\to}\cdots\right)$. Then one can readily verify that $\left[A,A\right]\simeq\mathbb{Z}$ or even (dually, as you suggest) that $\left[A\otimes A,A\right]\simeq 0$ via that $\left[,\right]$ is (contravariantly) continuous in its first argument. – Rafi Jan 27 '25 at 20:05