Showing that rings kZ and lZ, where l≠k are not isomorphic.

  • Thread starter jmjlt88
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  • #1
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I have shown that Z is not isomorphic to 2Z and that 2Z is not isomorphic to 3Z. I need now to generalize this. Thus, to prove that rings kZ and lZ, where l≠k are not isomorphic, I need to define an arbitrary isomorphism, and reach a contradiction. So here's what I am thinking. I let f: kZ->lZ be a isomorphism. Then, f(k)=ln , where n is some integer. This is true since k in kZ has to go to some multiple of l. From here, I have just been tinkering around trying to get k to map to 0 so that f is not injective. This worked for the first two. Is this the right approach? Not really looking for a hint, but just an "okay" to keep trying working with this line of reasoning.
 

Answers and Replies

  • #2
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Yeah, I think that's a valid approach. Go on, you'll get there.

Ask anytime if you want a hint :tongue:
 
  • #3
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Okay, I give!! =) A little hint would be great. When working with specific numbers, it was so easy to show that an isomorphism cannot exist. Now, I cannot get anything to work (or not work, I guess). A nudge in the right direction would be nice. Or else, this problem will keep me up all night!
 
  • #4
22,129
3,297
Okay, I give!! =) A little hint would be great. When working with specific numbers, it was so easy to show that an isomorphism cannot exist. Now, I cannot get anything to work (or not work, I guess). A nudge in the right direction would be nice. Or else, this problem will keep me up all night!

Think about [itex]f(k^2)[/itex].
 
  • #5
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Hmmmm...

I worked with that for while. Perhaps, I couldn't see the forest through the trees (or maybe I made an error). Here's what I had.

f(k2)=f(k)f(k)=l2n2

f(k k)=f(k)+f(k)+....+f(k) [k-times]=k(ln)

Now, I know that k≠l. But perhaps there would be a problem if k=ln?
 
  • #6
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OK, so we can deduce that k=ln.

Now think about the inverse map [itex]f^{-1}[/itex] and do something similar.
 

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