The question is the following:

A ring is actually an abelian group under addition, along with an extra operation called multiplication.

Is is possible for two rings to be isomorphic as abelian groups but not isomorphic as rings??

Thanks!(Evilgrin)

- May 6th 2009, 09:57 PMxxieIs is possible for two rings to be isomorphic as abelian groups but not.....
The question is the following:

A ring is actually an abelian group under addition, along with an extra operation called multiplication.

Is is possible for two rings to be isomorphic as abelian groups but not isomorphic as rings??

Thanks!(Evilgrin) - May 6th 2009, 10:40 PMGamma
Sure, the integers form a ring under the standard operations of addition and multiplication.

But you can also form a trivial ring by defining a*b=0 for all a,b. It is kind of lame, but it is still a ring. With these two types of multiplication imposed on the integers it is clear there is no ring isomorphism between the two.

The standard integer operations have a multiplicative identity, 1. Unity. This other multiplication clearly has no unity because for any nonzero element a.

$\displaystyle ab=0 \not = a$ for all b, so there is no identity for the nonzero integers and the unity must give back the element for multiplication on everything in the ring. - May 6th 2009, 11:18 PMxxie
sorry, i don't quite understand what u mean by "a trivial ring by defining a*b=0 for all a,b"

what's this ring?? - May 6th 2009, 11:31 PMGamma
It is still a ring of the integers, I have just defined multiplication differently than the way you learned in school. I am telling you that for every pair of integers a and b, define multiplication $\displaystyle a\circ b=0$. Leave the addition alone.

So the result of multiplying any two elements of this ring call it $\displaystyle \mathbb{Z}^{\circ}$ is 0.

These rings are better than isomorphic as abelian groups under addition, they are the same group! The integers.

However, as rings they are not isomorphic because one has unity and one does not. - May 6th 2009, 11:38 PMxxie
all right! got cha! Thanks!(Evilgrin)