83
$\begingroup$

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?

$\endgroup$
5
  • 10
    $\begingroup$ Defining $ab=0$ always gives an example, so you might want to mention that you want something else. $\endgroup$
    – Jonas Meyer
    Dec 22, 2011 at 9:23
  • 1
    $\begingroup$ Finitely generated abelian groups are direct sums of cyclic groups, so you may also want to focus on the nonfinitely generated case. $\endgroup$
    – Jonas Meyer
    Dec 22, 2011 at 9:25
  • 6
    $\begingroup$ @Jonas, of course this definition of multiplication will not have an identity element. $\endgroup$
    – Asaf Karagila ♦
    Dec 22, 2011 at 9:34
  • $\begingroup$ I wonder whether infinite dimensional vector spaces over $\mathbb Q$ can be given a multiplication that makes them rings with identity. $\endgroup$
    – Jonas Meyer
    Dec 22, 2011 at 10:10
  • 2
    $\begingroup$ @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 $\kappa$. $\endgroup$
    – Chris Eagle
    Dec 22, 2011 at 11:04

3 Answers 3

Reset to default
96
$\begingroup$

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.

$\endgroup$
3
  • 7
    $\begingroup$ 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. $\endgroup$
    – Jack Schmidt
    Dec 22, 2011 at 16:34
  • 2
    $\begingroup$ 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. $\endgroup$
    – Ehsaan
    Dec 28, 2013 at 5:23
  • 5
    $\begingroup$ And isomorphic, @Ehsaan, to $\Bbb Q/\Bbb Z$. $\endgroup$
    – Lubin
    Nov 7, 2015 at 16:20
18
$\begingroup$

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,

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$
    – Eric Auld
    Mar 15, 2016 at 4:50
  • 1
    $\begingroup$ Oh, now I'm seeing this has to do with the hom-tensor adjunction. That may be a clearer way to look at it. $\endgroup$
    – Eric Auld
    Mar 15, 2016 at 4:51
  • $\begingroup$ @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. $\endgroup$
    – Patrick Da Silva
    Mar 15, 2016 at 5:10
  • $\begingroup$ @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. $\endgroup$
    – Patrick Da Silva
    Mar 15, 2016 at 5:13
  • $\begingroup$ 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 $s \cdot t := \varphi^{-1}( \varphi_t \circ \varphi_s)$. $\endgroup$
    – Eric Auld
    Mar 15, 2016 at 5:14
14
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .