# Example of two non-isomorphic, non-trivial rings with the same underlying group

I'm looking for two rings that have the same additive group, but the multiplication is defined differently. For instance, if we have $(\mathbb{Z}_6,+)$ as the additive group, we naturally have the ring $(\mathbb{Z}_6,+,\cdot)$. But can we define multiplication differently so that it still satisfies the ring axioms with some other operation? This new ring, call it $(\mathbb{Z}_6,+,*)$ should not be isomorphic to $(\mathbb{Z}_6,+,\cdot)$.

I found two trivial examples. The identity ring works out, but again, it's trivial. The other one I found was the opposite ring, which in the non-Abelian case is just defined as switching the order of a given ring. Except that's isomorphic to the original ring, so it's not really what I'm looking for.

Are there any two rings that satisfy this property? Any finite rings? Given an arbitrary ring, can we find another non-isomorphic ring with the same group, or does the first ring need to satisfy specific properties to allow this?

• If $n$ is square-free (e.g. $n=6$), then there is a unique unital ring of cardinality $n$ (see here). – Watson Sep 26 '16 at 18:05

For an example with finite rings, $\mathbb{F}_4$ (the field with four elements) and $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ have the same additive group (the Klein four-group), but are not isomorphic as rings, because $\mathbb{F}_4$ is a field while $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ has zero divisors.
$\Bbb{C}$ and $\Bbb{R}$ are isomorphic as $\Bbb{Q}$-vector spaces (they have the same dimension), hence as groups. But obviously they are not isomorphic as rings.
I think we can show that two rings $$3\mathbb{Z},~~~2\mathbb{Z}$$ are no isomorphic from ring theory point of view. Just assume that there are an isomorphis and then get a contradiction!
For a nontrivial finite example, $\mathbb{Z}_2[x]/(1+x+x^2)$ (the finite field with 4 elements) and $\mathbb{Z}_2 \times \mathbb{Z}_2$ are nonisomorphic rings with isomorphic additive groups.
If you allow rings without $1$, you can redefine the multiplication on any ring by $xy = 0$ and still have a ring (with the same additive group).