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Show that the ring $3\mathbb Z$ is not isomorphic to the ring $5\mathbb Z$.

I see that they are not but I am not sure how to go about proving it. We went over a similar problem, disproving it by using that the number of units in the rings were not the same but that doesn't seem to apply in this case.

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    $\begingroup$ What are $3\Bbb Z$ and $5\Bbb Z$? $\endgroup$ – Parcly Taxel Oct 24 '16 at 9:16
  • $\begingroup$ They would be the sets that are the multiples of 3 {...-3, 0, 3, 6,..} and the multiples of 5 {...-5, 0, 5, 10, 15...} $\endgroup$ – Chrizi Oct 24 '16 at 9:17
  • $\begingroup$ They're not even... rings? They're called pseudo-rings. $\endgroup$ – Parcly Taxel Oct 24 '16 at 9:20
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    $\begingroup$ @ParclyTaxel Several authors don't require a unity in their rings. $\endgroup$ – egreg Oct 24 '16 at 9:38
  • $\begingroup$ @egreg Maybe I was mistaken... $\endgroup$ – Parcly Taxel Oct 24 '16 at 9:39
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They are not isomorphic, to see this, note that $3+3+3=3^2$, thus $3\mathbb{Z}$ has a non-zero element $x$ such that $3x=x^2$. There is no such element in $5\mathbb{Z}$.

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  • $\begingroup$ Which can be easily generalized to $a\mathbb{Z}$ and $b\mathbb{Z}$, with distinct positive $a$ and $b$. Without loss of generality, $a\nmid b$: in $a\mathbb{Z}$ there exists $x$ with $ax=x^2$ (precisely $a$), but no such element exists in $b\mathbb{Z}$. $\endgroup$ – egreg Oct 24 '16 at 9:44
  • $\begingroup$ Indeed, I enjoyed this question, hadn't given this any thought before and thought for a second that they could be isomorphic. It's interesting one really has to use the detailed structures of the rings to see that they are not isomorphic. I wonder whether there is a more general property (less detailed) to distinguish these rings. $\endgroup$ – Mathematician 42 Oct 24 '16 at 9:55
  • $\begingroup$ Thank you so much for your help!! I really appreciated it. I knew that worked between 2Z and 3Z with 2+2=2*2 but I hadn't thought about it in 3Z. Thank you again! $\endgroup$ – Chrizi Oct 24 '16 at 9:58
  • $\begingroup$ @Mathematician42 You're really using the property that they are subrings (without a unit) of the integral domain $\Bbb Z$ right? Since the solution to $3x=x^2$ means that $x(3-x)=0$ must be true in $\Bbb Z$ then either $x=0$ or $3-x=0$ by the i.d. cancellation property. So really when excluding $0$, there's no other non-zero element that has this property. $\endgroup$ – snulty Oct 24 '16 at 10:04

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