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From yesterday I'm with the following exercise and I'd appreciate If someone could check what I have so far. Thanks:

Let $R$ be a commutative ring, not necessarily with identity, and whose only ideals are $R$ and $\{0\}$, and $R$ is not the trivial ring. Show that $R$ is either a field or $(R,+)$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ (which I denote for simplicity $Z_p$) for $p$ a prime number and the product of any two elements is always zero.

The first part is completely trivial, because if it has a $1$, it follows immediately that is a field (as is commutative and doesn't have proper ideals). But if not have a $1$. We shall show that the product of any two elements is always $0$. If not, let $rs\not=0$ then $(r)$ (the ideal generated by $r$) it has to be entire Ring, i.e., $Rr=R$ and so there must be some $a\in R$, s.t., $ar=r$ and is easy to see that this $a$ must be the $1$, contradiction.

Since the product of all the elements is zero and the Ring is not the zero ring, for a non zero $r$, $Rr=\{0\}$. Let $\langle r \rangle$ the subgroup of $(R, +)$ generated by $r$. It's clearly an ideal an so there must be the entire ring $R$. So there is a $n$, s.t., $nr=0$.

We claim that $n\not=0$. Suppose not, we have that $2r\not= 0$, in fact $2rn\not=0$ for all $n$. Thus, by the same argument as above we have that $\langle 2r \rangle$ is an ideal and also is not the zero ideal, but it cannot be $R$, since $r\notin \langle 2r \rangle$, therefore it has to be proper, contradiction. For instances, either $n$ or $-n$ is positive and we can pick the least such value, say $n$.

Now we have to show that $n$ is prime. If $n$ were composite, so $n=ab$, and also the subgroup generated by $ar$ is an ideal and has to be the entire ring. But this is a contradiction, since the order of $ar$ is $b<n$ where $n$ is the order of $\langle r \rangle=R$. Then $n$ has to be a prime number. Thus, defining the surjective homomorphism $\mathbb{Z}\to (R,+);\,\langle r \rangle$, $n\mapsto nr$ is clear that $n\mathbb{Z}$ is the kernel and by the isomorphism thm, it follows that $Z_n$ and $(R,+)$ are isomorphic,

as was to be shown.

Jose Antonio
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  • There are a few details you should really elaborate on (i.e. you don't explain why $r$ has $\mathbb{Z}$-torsion), but overall it looks good. – Andrew Dudzik Oct 04 '14 at 16:24
  • Well, since $\langle r \rangle= R$ and $0\in R$ so there must be a $nr =0$, and we can choose the least $n$ s.t., this happens. What do you think @Slade ? and thanks for take your time and check the argument I really appreciate it. – Jose Antonio Oct 04 '14 at 16:31
  • This argument doesn't show that we may choose $n\neq 0$. – Andrew Dudzik Oct 04 '14 at 16:34
  • And no problem, it's kind of a nice result and I don't get to think about rings without unity very often... – Andrew Dudzik Oct 04 '14 at 16:35
  • @Slade Ok I know what is the flaw, mmm let me think how to show that $n\not=0$ – Jose Antonio Oct 04 '14 at 16:42
  • @Slade If $nr=0$ just when $n=0$, it follows that $\langle 2r \rangle$ is by the same argument as above an ideal. But it cannot be $(0)$ but also it cannot be $R$ since $r\notin \langle 2r \rangle$, contradiction. Hence there must be a $n\in \mathbb{N}\setminus {0}$ s.t. $nr=0$ – Jose Antonio Oct 04 '14 at 17:46
  • Right. The point as I see it is that $\mathbb{Z}$ contains nontrivial ideals. – Andrew Dudzik Oct 04 '14 at 18:12
  • Thanks @Slade for your help :) – Jose Antonio Oct 04 '14 at 18:17

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