# On rings isomorphic to the ring of integers...

I have a question that I can't seem to figure out:

Let $R$ be a commutative ring with identity. Note that under addition $R$ is an abelian group. Suppose every subgroup of this group is an ideal. Show that $R$ is isomorphic to the ring of integers or to the integers mod $n$.

I realize $\Bbb Z$ has such a property, and that if $R$ is infinite, we are in case 1, and if it is finite, then case 2, but despite a bunch of manipulating I can't get a sufficient proof working. Any help?

Consider the subgroup of $(R,+)$ generated by the identity $1$. By hypothesis this is an ideal. But an ideal containing $1$ is the entire Ring. That means that each $r\in R$ has the form $n\cdot 1$.

Now there are two cases: Either there is a minimal $n\in \Bbb N$ such that $n\cdot 1=0$ (this is called the characteristic of $R$), or $R$ has infinitely many elements of the form $k\cdot 1,\ k\in \Bbb Z$

• Why do we need all subgroups to be ideals then, if using the ideal generated by 1 is sufficient? Aug 22, 2013 at 2:10
• @AnthonyVasaturo: Because that is also necessary. If $R$ is isomorphic to $\Bbb Z$ or $\Bbb Z_n$, then each subgroup is an ideal. But you are right it would suffice to require that $\langle 1\rangle$ is an ideal. I think they did not want to spoil the solution of this problem. Aug 22, 2013 at 2:12
• What I am saying is, why in the question didn't they just say, suppose the subgroup generated by 1 is an ideal? EDIT: sorry I am just learning how this site works, I didn't see your full answer before I Replied. Aug 22, 2013 at 2:14
• @AnthonyVasaturo: Because they wanted you to think about which subgroup you should consider. Aug 22, 2013 at 2:15

Hint: $\{n\cdot 1_R: n\in\mathbb Z\}$ is an ideal of $R$.

Hint

The group $\langle 1\rangle$ is an ideal so for $a\in A,\ a.1\in \langle 1\rangle$.