Isomorphic groups but not isomorphic rings

Question

Provide an example of two rings that have the same characteristic, are isomorphic as groups but are not isomorphic as rings. I'm confused with how to being. I know that having the same characteristic means that the concatenation is the same number to receive the zero element.

Answer 1

One easy example is to take a prime $p$ and consider the finite field $\mathbb{F}_{p^2}$ of order $p^2$ and the direct product ring $\mathbb{F}_p \times \mathbb{F}_p$. Both rings have characteristic $p$. As groups, $\mathbb{F}_{p^2}$ and $\mathbb{F}_p \times \mathbb{F}_p$ are both isomorphic to $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$, but they are not isomorphic as rings because the former is a field while the latter is not.

Answer 2

The group isomorphism refers to the additive structure.

Let $R$ be any ring. We can define two ring structures on the set $R\times R$: the addition is the same, so the two additive groups are not only isomorphic, but identical: $$ (a,b)+(c,d)=(a+c,b+d) $$ We can define two different multiplications: $$ (a,b)\cdot(c,d)=(ac,bd) $$ and $$ (a,b)*(c,d)=(ac,ad+bc) $$ It's not difficult to show that $(R\times R,+,\cdot)$ and $(R\times R,+,*)$ are rings. Can you find the characteristic of them and a case where the two rings are not isomorphic?

Answer 3

$\mathbb{R} \times \mathbb{R}$ and $\mathbb{C}$ are isomorphic as additive abelian groups but they have a different multiplicative structure

Answer 4

Somewhat surprisingly, but assuming the axiom of choice $\Bbb R$ and $\Bbb C$ make such example.

Both are $\Bbb Q$-vector spaces of the same dimension, therefore they are isomorphic as $\Bbb Q$-vector spaces, and in particular as groups. However, they are clearly not isomorphic as rings. The same arguments also works with $\Bbb R^n$, and even spaces like $\ell_\infty$ and $\ell_2$.

Interestingly, the existence of such isomorphism(s) relies heavily on the existence of a Hamel basis, and we can prove that it is possible to have a world where the axiom of choice fails, and $\Bbb R$ and $\Bbb C$ are not isomorphic.

In the same spirit, but without relying on the axiom of choice, you can take the [canonical] algebraic closure of $\Bbb Q$, $\overline{\Bbb Q}$, and $\Bbb Q$'s real-closure which is $\overline{\Bbb Q}\cap\Bbb R$. Both are $\Bbb Q$-vector spaces of the same dimension, in this case $\aleph_0$, so they are isomorphic. But they are clearly not isomorphic as fields, since only one of them has a root for $-1$. You can replace one of the fields with any countable but infinite extension of $\Bbb Q$, like $\Bbb Q(\pi)$ for example.

In this case the axiom of choice is avoided, since everything is countable and we can prove the existence of a basis for these vector spaces from this fact alone.

Answer 5

Fro a somewhat general example, consider $\def\Q{\Bbb Q}\Q[X]/(P)$ where $P$ is some monic polynomial. The additive group is isomorphic to $\Q^d$ where $d=\deg P$, so it is unchanged when $P$ varies over polynomials of a fixed degree. However the ring structure depends on other properties of$~P$; for instance it is a field if and only if $P$ is an irreducible polynomial. This allows you to easily make many examples.

Answer 6

Let $G$ and $H$ be non-isomorphic groups of same cardinality and $A$ a ring. Then the additive groups of the group rings $A[G]$ and $A[H]$ are isomorphic (both are just $A^{|G|}$), but the rings themselves are often (at least in concrete examples) easily verified to be non-isomorphic. For example, with $A=\Bbb Z$, $G=(\Bbb Z/2\Bbb Z)^2$, $H=\Bbb Z/4\Bbb Z$, only one of the group rings has a unit of multiplicative order $4$.

As is often the case, answers lead to new questions:

Exercise 1: Are there $A,G,H$ where $G\not\approx H$ and yet $A[G]\approx A[H]$?

Exercise 2: With $A=\Bbb Z/2\Bbb Z$, $G=S_3$, $H=\Bbb Z/6\Bbb Z$, we find an example where the non-isomorphy follows because only one of the rings is commutative. Are there rings $R,S$ with isomorphic additive groups and with less than 64 elements such that only one of them is abelian?

Answer 7

One surprising class of examples comes from the opposite ring construction: given a ring $R$, the opposite ring $R^{op}$ has the same underlying set and same additive structure, but "backwards" multiplication: writing "$\otimes$" for the $R^{op}$ multiplication and "$\times$" for the $R$ multiplication, the rule is $$a\otimes b=b\times a.$$ Even though this looks the same, we need not have $R\cong R^{op}$, even though they are of course isomorphic as groups and extremely similar as rings; see here.