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.

  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ @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

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 $5\mathbb{Z}$.

  • $\begingroup$ 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

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.