# Ring structures on abelian groups [duplicate]

My question is: given an abelian group $G$ with addition $+$, is there some natural multiplicative structure that arises so that we can define a ring $(G, +, \cdot)$. For instance, multiplication on $\mathbb{Z}$ and $\mathbb{Z}_n$ are entirely determined by addition, since it must be that $ma = a + \ldots + a$, where the addition is $m$ times.

For finitely generated abelian groups $G$, we know that its representation according to the Fundamental Theorem of Finitely Generated Abelian Groups is

$$G = \mathbb{Z}_{p_1^{r_1}} \times \ldots \times \mathbb{Z}_{p_n^{r_n}} \times \mathbb{Z} \times \ldots \times \mathbb{Z}$$

Since each of those factors has a natural ring structure, we can define a ring structure on $G$ as the product of these ring structures. That leaves the question: can we define a "natural" ring structure on infinitely generated abelian groups $G$?

• $ma$ is not actually the same as $ba$ were $b\in\mathbb Z_n$. In fact, in $\mathbb Z_n$, any generator could be used as the identity of the ring.There is no "natural" choice for the multiplicative identity. (Also, bad to talk about $na$ in the context of $\mathbb Z_n$.) Nov 5, 2014 at 0:41
• Also, while the Fundamental Theorem is true, there isn't a "natural" representation of the group in this form. There are plenty of representations in this form... Nov 5, 2014 at 0:45
• Is it not true, however, that no matter the choice of generator as multiplicative identity on $\mathbb{Z}_n$, that the resulting ring structures will be isomorphic? Thus can we assume $1$ is unity? Nov 5, 2014 at 0:46
• Let's say someone gives you a group $\{0,a,b\}$ with $a+a=b$, $b+b=a$, $a+b=0$. This is isomorphic to $(\mathbb Z_3,+)$ in two different ways, one with $a$ sent to $1$, one with $b$ set to $1$. So there is no "natural" way to create a ring out of this group, only if we choose arbitrarily from the isomorphisms with $\mathbb Z_3$ is there an isomorphism. Nov 5, 2014 at 0:47
• There's no ring structure on the Prüfer groups, IIRC. Nov 5, 2014 at 0:47

The Prüfer group for $p$ has the property that all elements are of finite order, and there is an element of order $p^k$ for every $k$.
Now, in a ring, every element must be of additive order equal to a divisor of the additive order of $1$, because if $n1=0$ then $nr=n(1r)=(n1)r=0$ for any $r\in R$. So the Prüfer group cannot be made a ring.
• Every abelian group can be made a ring without unity by defining $a\cdot b = 0$ for all $a,b$. Nov 5, 2014 at 2:26
• No, if $a,b$ are in the group, let $p^k$ be the additive order of $a$. Find $b'$ so that $p^kb'=b$. Then $ab=(a(p^kb'))=(p^ka)b'=0$. Nov 5, 2014 at 13:45