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.