Show that two rings are not isomorphic

  • Thread starter Mr Davis 97
  • Start date
  • Tags
    Rings
In summary, to show that the rings ##2 \mathbb{Z}## and ##3 \mathbb{Z}## are not isomorphic, we can assume they are isomorphic and then find a contradiction by considering the consequences of this assumption. One way to do this is by trying to find a ring homomorphism between the two rings and see if we run into contradictions.
  • #1
1,462
44

Homework Statement


Show that the rings ##2 \mathbb{Z}## and ##3 \mathbb{Z}## are not isomorphic.

Homework Equations




The Attempt at a Solution


I know how to show that two structures are isomorphic: find an isomorphism. However, I am not quite sure how to show that there exists no isomorphism at all
 
Physics news on Phys.org
  • #2
Mr Davis 97 said:

Homework Statement


Show that the rings ##2 \mathbb{Z}## and ##3 \mathbb{Z}## are not isomorphic.

Homework Equations




The Attempt at a Solution


I know how to show that two structures are isomorphic: find an isomorphism. However, I am not quite sure how to show that there exists no isomorphism at all
I don't have any help for you but I have to comment that your post gave me a chuckle because my son and I were just recently discussing a linguistic phenomenon that I noticed some time ago. I make no representation that I know what YOU mean, but I know what I mean, and what most people mean with the following construct:

"I don't exactly know how to ... " or "I don't quite know how to ... " generally means "I don't have even the tiniest clue how to and in fact I'm not even sure how to spell it" :biggrin:
 
  • Like
Likes nuuskur and Mr Davis 97
  • #3
Assume they were isomorphic and consider the consequences, or try to find a ring homomorphism by defining ##\varphi (2) = 3x## for some integer ##x## and see if you run into contradictions.
 
  • #4
Suppose, for a contradiction, a ring isomorphism [itex]f :2\mathbb{Z}\to3\mathbb{Z}[/itex] existed. Then [itex]f(2) = 3m[/itex] for some [itex]m\in\mathbb{Z}[/itex].
Since [itex]f[/itex] respects addition and multiplication, then [itex]f(2)+f(2) =f(2+2) =f(4)= f(2\cdot 2)=f(2)\cdot f(2)[/itex]. But this is a problem. Can you explain, why?
 
Last edited:
  • Like
Likes Mr Davis 97

Suggested for: Show that two rings are not isomorphic

Back
Top