Let $R$ be a non-zero commutative ring with identity. The following is well known:

If $x,y \in R$ are idempotents then $x+y-2xy$ is also an idempotent and more than that by defining $x*y = x+y-2xy$, the set of idempotent elemetns becomes a 2-group.

Now Let $T$ be the set of all elements $x$ with $x^3 =x$. Is there any general binary operation on $T$ that makes $T$ a (semi)group ?

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    $\begingroup$ I think you've asked too many questions too quickly. This is like 3 or 4 already today. I'd normally restrict myself to 1 per day, so I could actually respond to comments people leave and think through the answers they give. Also, if you post too many questions too quickly you'll have lower chances of all being answered. $\endgroup$ – David White Dec 29 '12 at 23:26
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    $\begingroup$ Are you certain the operation that you named $\ast$ is associative? $\endgroup$ – Todd Trimble Dec 29 '12 at 23:41
  • $\begingroup$ Note that for such an x, xx is an idempotent, so you already have such an operation on a subset of T. Gerhard "Ask Me About System Design" Paseman, 2012.12.29 $\endgroup$ – Gerhard Paseman Dec 30 '12 at 0:07
  • $\begingroup$ Don't you mean x+y-xy? You are supposed to think of + as union and product as intersection and you are trying to construct the symmetric difference. $\endgroup$ – Benjamin Steinberg Dec 30 '12 at 3:08
  • $\begingroup$ I think the answer to your question is that there is no "natural operation" a commutative semigroup of idempotents is automatically a meet-semilattice and inside of a ring forms a boolean algebra. You are trying to take the additive part of the corresponding boolean ring. $\endgroup$ – Benjamin Steinberg Dec 30 '12 at 3:10

Any set $T$ has a binary operation on it that makes the set into a semigroup! Namely $x\ast y=x$, the "choose the left" operator. (It is easy to check that this operation is associative.)

Only slightly less trivially, any set $T$ has a binary operation on it that makes the set into a group! Fix a group $G$ with the same cardinality as $T$, and fix a bijection $G\cong T$. This endows $T$ with a group structure.

That said, let $R$ be a ring and let $T=\{x\in R\ :\ x^3=x\}$. You might wonder whether there exists a binary function $\ast:T\times T\to T$ which makes $T$ into a (semi)group, is symmetric in the two coordinates, and is defined equationally. For the semigroup question, the answer is also trivially yes, by just taking $\ast$ to be multiplication in $R$!

For the group question, assume $x\ast y=f(x,y)$ where $f(a,b)\in \mathbb{Z}[a,b]$. Since $x^3=x$ and $y^3=y$, and $f$ is symmetrtic, we may write

$$f(a,b)=s_0+s_1(a+b)+s_2(a^2b+ab^2)+s_3(a^2+b^2)+s_4ab+s_5a^2b^2,\qquad s_i\in \mathbb{Z}.$$

In the ring $\mathbb{Z}$ the only elements in $T$ are $0,1$. So one of the two is the group identity, hence either $f(1,1)=1$ or $f(0,0)=0$. Thus, the same holds in any other ring.

Case 1: First, assume $f(0,0)=0$. This implies $s_0=0$. Now, since $0$ is the group identity, $f(x,0)=x$. This implies $s_1=1$ and $s_3=0$ (seen by working over the ring $\mathbb{Z}[x]/(x^3=x)$). Next, we must have $f(1,1)=0$ since in $R=\mathbb{Z}$ the set $T=\{0,1\}$ can only have one group structure with $0$ the identity. Hence $s_5=-2-2s_2-s_4$.

Also $f(x,x)^3=f(x,x)$. But $f(x,x)=(2+2s_2)(x-x^2)$. In the ring $\mathbb{Z}[x]/(x^3=x)$, this cubes to itself only if $s_2=-1$. So we now have $f(a,b)=a+b-a^2b-ab^2+tab-ta^2b^2)$. If we compute $f(x,y)^3-f(x,y)$ in $\mathbb{Z}[x,y]/(x^3=x,y^3=y)$, the coefficient on $xy$ is $4t^3-t$, which implies $t=0$.

With $f(a,b)=a+b-a^2b-ab^2$, we find that $f$ does not give an associative operation.

Case 2: Now assume $f(1,1)=1$. We can reduce to case 1 by using the new polynomial $g(a,b)=1-f(1-a,1-b)$.


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