TITLE: A commutative group structure on $R\times R$ for a ring $R$
QUESTION [10 upvotes]: Let $R$ be a commutative ring. The Cartesian square $A=R\times R$ is endowed with the operation
$(a_1,b_1)\circ(a_2,b_2)=(a_1+a_2,b_1+b_2+a_1a_2^2+a_1^2a_2)$
which turns $A$ into a commutative group. I have two questions concerning this group.
Question 1: For what $R$ is $(A,\circ)$ isomorphic to $R\oplus R$?
I managed to show that if $3$ is invertible in $R$ then the isomorphism holds.
It also holds for R=F_9, but does not hold for $R=\mathbb{F}_3$, $\mathbb{F}_9$, or $\mathbb{F}_{27}$.
Other rings $R$ in which $3$ is not invertible (including $R=\mathbb{Z}$) remain a mystery to me.
Question 2: When is $(A,\circ)$ generated by the elements of the form $(a,0)$, $a\in R$?
Again, only partial results here. Let $B$ be the subgroup of $A$ generated by $(a,0)$, $a\in R$. Then $B$ contains $(0,a_1a_2^2+a_1^2a_2)$ for all $a_1,a_2\in R$. Hence,
we have $B=A$ if $R$ is additively generated by the elements of the form $a_1a_2^2+a_1^2a_2$. This is so for $R=\mathbb{Z}/p\mathbb{Z}$ with $p$ odd.

REPLY [2 votes]: Ok, so inspired by Arturo's answer, here is another partial answer that includes the Gaussian integers and the 3-adic integers:
Define τ:R→R:n↦2*(n+1 choose 3) and notice the formal equality
$$2\cdot\binom{n+m+1}{3} = 2\cdot\binom{n+1}{3}+2\cdot\binom{m+1}{3} + (mn^2+m^2n)$$
Then define [ a, b ] = ( a, b + τ(a) ) to be a different coordinate system on (R⊕R,∘).  Then [ a1, b1 ]∘[ a2, b2 ] = [ a1+a2, b1+b2 ] is clearly isomorphic to R⊕R.
Suppose R is a domain with field of fractions K.  Then τ:R→K:n↦2*(n+1 choose 3) definitely exists, so one just needs some condition for τ(R) ≤ R.  I thought this was at least sort of common, but I can't think of any real examples other than Z.
Certainly R=Z[x] doesn't work for τ, though both R⊕R and A will be free abelian of countably infinite rank.
I guess the only improvement is that we don't need every element of R to be divisible by 3, only the elements of the form xxx−x.
I believe the ring R=Z[i] works for this, since Z[i]/(3) ≅ Z/3Z × Z/3Z satisfies xxx-x ≡ 0.  This ring is less exciting in some ways though, since the additive group is free abelian.
Rings that are even nicer, where all binomial coefficients exist, are called binomial rings.  For example the p-adic integers for any prime p (and where p=3 is the interesting one for us) also work.  The additive group of p-adic integers is not free abelian, so this is a real gain.