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Quick way to tell if two rings are isomorphic?

by Silversonic
Tags: isomorphic, rings
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Silversonic
#1
May15-12, 03:48 PM
P: 129
1. The problem statement, all variables and given/known data

Say if the following rings isomorphic to [itex] \mathbb{Z}_6 [/itex] (no justification needed);

1) [itex] \mathbb {Z}_2 \times \mathbb {Z}_3 [/itex]

2) [itex] \mathbb {Z}_6 \times \mathbb {Z}_6 [/itex]

3) [itex] \mathbb {Z}_{18} / [(0,0) , (2,0)] [/itex]


3. The attempt at a solution

I know how to tell if two rings AREN'T isomorphic - find an essential property that one ring has that another one doesn't. For example part 2), [itex] \mathbb {Z}_6 \times \mathbb {Z}_6 [/itex] has 36 elements, whereas [itex] \mathbb {Z}_6 [/itex] has 6.

But then, how do I show the other two?


My notes say the first one is isomorphic to [itex] \mathbb {Z}_6 [/itex] because 2 and 3 are coprime. But how can it justify that so quickly? Because they are coprime both rings have the same characteristic (i.e. n.1 = 0 for n ≥ 1 in both rings). But is that enough to conclude that they are isomorphic?

And for the last one, where can I begin?


If I can show both rings are generated by their 1, and both rings have the same characteristic, they should be isomorphic right?
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Office_Shredder
#2
May15-12, 03:52 PM
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The first one you can do using the Chinese Remainder Theorem if you've seen it. If not,

If I can show both rings are generated by their 1, and both rings have the same characteristic, they should be isomorphic right?
This is a good way to do it. In particular if something is isomorphic to Z6 it should be really easy to construct the isomorphism and prove that it's an isomorphism

Your third one doesn't seem to make any sense though, since Z18 doesn't contain any elements called (0,0) or (2,0) in any standard notation that I have seen


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