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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.

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    with $R = \mathbb{F}_9$, in $(A,\circ)$, $(1,0)$ is of order 9, while there is no element of order 9 in the group $(R \oplus R, +)$2011-02-24
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    @chandok That's right. I found my mistake. In this case, $A\cong C_9\times C_9$ is not isomorphic to $\mathbb{F}_9\oplus \mathbb{F}_9$. So, it seems that all fields of characteristic $3$ are exceptional after all.2011-02-24
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    I confirm that when 3 is invertible, A≅R⊕R. The proof is just that your cocycle is generated by a↦aaa/3, so is a coboundary. It strikes me as unlikely to happen in general (so if there are any other isomorphisms, I would expect them to be coincidences that did not preserve the the normal subgroup {(0,b)}). I haven't checked R=Z yet. It sounded easy to check, but today is crazy.2011-02-24
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    Inspired by Arturo's answer, here is a *slightly* more general type of ring that works: any ring in which the (doubled) binomial coefficient 2*(x choose 3) exists. This includes Z at least.2011-02-24
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    I wonder; clearly, any homomorphism $f\colon R\to S$ induces a homomorphism $\overline{f}\colon(R\times R,\circ)\to(S\times S,\circ)$, and if $f$ is an embedding then so is $\overline{f}$. I wonder if, when $3$ is not a zero divisor, we embed $R$ into $S=R[\frac{1}{3}]$, get the isomorphism for $S$, and somehow get something for $R$ out of it. The problem of course is that the embedding $R\to S$ may not be very compatible with the isomorphism $S\times S\to S\oplus S$.2011-02-24
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    @Anvita: did you make any more progress? Has anyone found any integral domains (characteristic not 3) where it is not isomorphic? It still strikes me as unlikely to happen in general, but I haven't found any such rings where it doesn't.2011-03-01
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    @Jack: No progress so far... The case $R=\mathbb{Z}$, of primary interest to me, was covered in Arturo's answer. The other rings were just a curiosity. Maybe I'll get back to them sometime. (Sorry to reply so late. Being new to this system, I discovered your hidden post only today :)2011-04-05

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