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How to show two rings are not isomorphic

by samkolb
Tags: isomorphic, rings
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samkolb
#1
Jan4-09, 12:28 AM
P: 37
I'm just beginning my study of rings, and I'm wondering if there are some standard ways to show that two rings are not isomorphic. I've studied groups quite a bit and I know some of the ways to show that two groups are not isomorphic (G contains an element of order 2 while G' contains no element of order 2, etc...).

In particular, if for two rings R and R' their associated abelian groups A and A' are isomorphic, what are some ways to use the multiplicative properties of R and R' to show that R and R' are not isomprhic?

The problem I'm working on is to show that the rings 2Z and 3Z are not isomorphic.
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quasar987
#2
Jan4-09, 01:26 AM
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Here, 2Z and 3Z are both subrings of Z. You could look at Z/2Z and Z/3Z. If 2Z and 3Z were isomorphic, then Z/2Z and Z/3Z should be isomorphic as well.
lurflurf
#3
Jan4-09, 01:46 AM
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Suppose
2Z~3Z
with f an isomorphism and
f(2)=3k k in Z
2+2=2*2
f(2+2)=f(2*2)
use the properties of f and Z to arrive at a contradiction

joecoz88
#4
Jan4-09, 02:45 AM
P: 14
How to show two rings are not isomorphic

Consider the isomorphism that relates 2Z and 3Z as abelian groups (or nZ for any integer n). As groups, these structures are isomorphic. In order to be isomorphic as rings, the homomorphism property must be satisfied for addition and multiplication. We know that addition works, and we need to check multiplication. Use the same isomorphism you used under addition, and you will see that it fails with multiplication.
adriank
#5
Jan7-09, 02:22 AM
P: 534
Quote Quote by joecoz88 View Post
Consider the isomorphism that relates 2Z and 3Z as abelian groups (or nZ for any integer n). As groups, these structures are isomorphic. In order to be isomorphic as rings, the homomorphism property must be satisfied for addition and multiplication. We know that addition works, and we need to check multiplication. Use the same isomorphism you used under addition, and you will see that it fails with multiplication.
That alone doesn't solve it; you only showed that one particular group isomorphism isn't a ring isomorphism. You need to explain why there is no group isomorphism that is a ring isomorphism. (Then again, there are only two group isomorphisms, and it's easy to check both.)

lurflurf gives a simple solution with a concrete example, and quasar987 gives a more abstract one (but nicer).
joecoz88
#6
Jan7-09, 05:25 AM
P: 14
Yes I see that my solution wasn't general enough, thanks for the correction.

But I am having trouble following quasars logic, maybe someone could elaborate?

I can't see the reason why if 2Z and 3Z are isomorphic, then Z/2Z and Z/3Z should be isomorphic? Is this reasoning valid?

As groups we can see that 2Z and 3Z are isomorphic, but Z/2Z and Z/3Z are not isomorphic.
adriank
#7
Jan7-09, 12:43 PM
P: 534
Well, uh,

good point! :) No, it's not valid, and silly me. And bad quasar. :)
placemat
#8
Apr21-09, 07:15 PM
P: 1
lurflurf's explanation does not describe a specific isomorphism. I agree that picking a function and showing that it is not a ring isomorphism is insufficient to show that two rings are not isomorphic; however, lurflurf's explanation simply made use of the fact that for any function f: A->B, if x belongs to A then f(x) belongs to B. In this example, A=2Z and B=3Z, so if f is an isomorphism from A to B, f(2) belongs to 3Z and thus is of the form 3k for some integer k. lurflurf's other remarks can then be used to show that 2Z and 3Z are not isomorphic, as f was chosen arbitrarily and the description of f(2) came from the description of the codomain of f.
Weensie
#9
Apr21-09, 07:35 PM
P: 2
A first way for that is to look the invertible elements of the rings .
matt grime
#10
Apr22-09, 12:58 PM
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Quote Quote by placemat View Post
lurflurf's other remarks can then be used to show that 2Z and 3Z are not isomorphic

But they are isomorhpic, as rings, with the map 2-->3 (and they're just isomorphic to Z).
shaggymoods
#11
Apr22-09, 11:10 PM
P: 26
The map that matt shows it not a ring isomorphism, for the exact reason that lurflurf gave: f(2 + 2) = f(2*2) which will contradict any isomorphism between the two structures.
matt grime
#12
Apr23-09, 01:16 AM
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Doh, being stupid again, sorry.


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