3
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $k\cdot 1,\ k\in \Bbb Z$

$\endgroup$
  • $\begingroup$ Why do we need all subgroups to be ideals then, if using the ideal generated by 1 is sufficient? $\endgroup$ – Johnny Apple Aug 22 '13 at 2:10
  • $\begingroup$ @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. $\endgroup$ – Stefan Hamcke Aug 22 '13 at 2:12
  • $\begingroup$ 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. $\endgroup$ – Johnny Apple Aug 22 '13 at 2:14
  • $\begingroup$ @AnthonyVasaturo: Because they wanted you to think about which subgroup you should consider. $\endgroup$ – Stefan Hamcke Aug 22 '13 at 2:15
3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

Hint

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.