# Ring concept … commutativity [duplicate]

Possible Duplicate:
Proving a Ring is commutative

Every element a of some ring $(R,+,\circ)$ satisfy equation $a\circ a=a$.

Is the above ring Commutative...? yes or no ..? please explain

Also please describe the what is Ring. i still am unclear of that term

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## marked as duplicate by rschwieb, Austin Mohr, Henning Makholm, Henry T. Horton, Brett FrankelFeb 2 '13 at 21:36

It's also helpful, if you have very different questions, to ask them separately. For you, I'd advise clearing up what a ring is, before you try to prove things about rings. – rschwieb Feb 2 '13 at 20:48

Such rings are called Boolean and are unusually rigid. Given any set $X$, a typical example of Boolean ring is $(\mathcal{P}(X), \triangle, \cap, \bar{\bullet},\emptyset, X)$, (also known as rings of sets).

For every $a, b \in R$, you have $(a+b)(a+b)=a+b$. The LHS is

$(a+b)(a+b)=a^2 + ab + ba + b^2$.

But, for every $r \in R$, $r^2 = r$. Thus

$(a+b)(a+b) = a +ab +ba + b = a + b$

i.e., $ab + ba = 0$.

Boolean condition also gives us that, for every $r \in R$, $r +r = (r+r)(r+r) = r^2 + r^2 +r^2 +r^2 = r +r +r +r$; thus, $r +r = 0$, that is $r = -r$.

Finally, $ab + ba = ab - ba = 0$, i.e., $ab= ba$. $\square$

Try to prove these funny facts:

1. Rings of sets are actually Boolean unital rings.

2. Boolean rings have characteristic $2$.

3. The only Boolean ring that is an integral domain is $\mathbb{Z}/2\mathbb{Z}$.

4. Every prime ideal is actually maximal and every finitely generated ideal is principal.

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