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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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With $R=\mathbb{Z}$, you get an isomorphism. Note that $(\mathbb{Z}\times\mathbb{Z},\circ)$ is torsionfree abelian: given $(a,b)\in\mathbb{Z}\times\mathbb{Z}$, if $a\neq 0$, then $n(a,b) = (na,x)$ for some $x$, so $n(a,b)=(0,0)$ requires $n=0$. If $a=0$, then $n(0,b) = (0,nb)$, so again $n(0,b)=0$ requires $n=0$ or $b=0$. (I'm using $m(x,y)$ to mean $(x,y)$ $\circ$-added to itself $m$ times if $m\gt 0$, and the $\circ$-inverse if $m\lt 0$, as usual for an abelian group).

Also, $(\mathbb{Z}\times\mathbb{Z},\circ)$ is $2$-generated: $(1,0)$ and $(0,1)$ certainly generate: to get an arbitrary $(a,b)$ using $(1,0)$ and $(0,1)$, just take $a(1,0)$, which will give an element of the form $(a,x)$ for some $x$, and then take $(a,x)\circ(0,b-x)$.

So $(\mathbb{Z}\times\mathbb{Z},\circ)$ is free abelian and $2$-generated, hence either cyclic or isomorphic to $\mathbb{Z}\times\mathbb{Z}$. Moding out by the infinite cyclic normal subgroup $\{(0,b)\mid b\in\mathbb{Z}\}$ we get a group isomorphic to $\mathbb{Z}$, so the group $(\mathbb{Z}\times\mathbb{Z},\circ)$ cannot be infinite cyclic (because the quotient of an infinite cyclic group by a nontrivial subgroup is finite). Thus, with $R=\mathbb{Z}$ you get an isomorphism.

More generally, if you have an isomorphism for $R$ and one for $S$, then you also get an isomorphism for $R\oplus S$, since the addition will just be "coordinate wise", so you can take the isomorphism $f\colon (R\times R,\circ)\to R\oplus R$ and $g\colon (S\times S,\circ)\to S\oplus S$, and obtain an isomorphism $f\times g\colon (R\oplus S\times R\oplus S,\circ)\to (R\oplus R)\times (S\oplus S)\cong (R\oplus S)\times (R\oplus S)$.

This will give you isomorphisms for all rings of the form $$\mathbb{Z}^t \times\frac{\mathbb{Z}}{p_1^{a_1}\mathbb{Z}}\times\cdots\times\frac{\mathbb{Z}}{p_r^{a_r}\mathbb{Z}}$$ where $p_1,\ldots,p_r$ are primes, different from $3$, $a_i\gt 0$, $t,r\geq 0$, and the ring structure is the obvious one.

Added. (Well that was silly). For rings of characteristic $3$, you get an isomorphism if and only if $a^3=0$ for all $a\in R$.

To see that this is necessary, notice that \begin{align*} 3(a,0) &= (a,0)\circ(a,0)\circ(a,0) = (2a,2a^3)\circ(a,0)\\ &= (3a,2x^3 + 4a^3+2a^3) = (0,8a^3) = (0,-a^3). \end{align*} Thus, $(R\times R,\circ)$ has characteristic $3$ if and only if $a^3=0$ for all $a$.

To see that the condition is sufficient, if $R$ is of characteristic $3$ and $a^3=0$ for all $a\in R$,then $(R\times R,\circ)$ is an abelian $3$-group, hence a vector space over $\mathbb{F}_3$. Since it has the same cardinality as $R\oplus R$, which is also a vector space over $\mathbb{F}_3$, they are isomorphic as $\mathbb{F}_3$-vector spaces, and hence as abelian groups.

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    Thanks! A crazy day indeed. (ZxZ,o) is an extension of Z by Z, so of course it splits. Your first two paragraphs even describes a complement: the subgroup generated by (1,0) is not only torsion-free, but it intersects the subgroup generated by (0,1) trivially.2011-02-24
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    I think your idea might work for characteristic 3^n too, at least for unital rings. 3^n(a,0) = (3^na, 2*binomial(3^n+1,3)*a^3), and so as long as a^3 is not annihilated by 3 (so a=1), 3^n(a,0)≠0, and so the exponent of A is strictly larger than that of R×R.2011-02-25
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    Is there an example of a ring $R$ of charactristic 3 such that $a^3=0$ for all $a\in R$?2011-02-25
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    Hmm.. I guess the additive cyclic group $C_3$ with zero multiplication will do.2011-02-25
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    @Anvita: Obviously, no such ring can have an identity. There are rings that do not have zero multiplication, but have $A^3=0$, so such a ring with characteristic $3$ would do as well.2011-02-25
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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.

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    Well, don't we get the $p$-adics for $p\neq 3$ from the fact that $3$ being invertible in $\mathbb{Z}_p$? We do seem to gain $\mathbb{Z}_3$, though...2011-02-24
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    Yup, this is a *tiny* improvement, one measly group.2011-02-25
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    A *big* group, though.... (-:2011-02-25
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    If $R=\mathbb{Z}[x]$ then $A$ is certainly torsion-free. But why is it free?2011-02-25
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    R is free abelian, A is an extension of R by R, and all extensions of free abelian groups by free abelian groups are free abelian.2011-02-25