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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 Frankel Feb 2 '13 at 21:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

@JaySatishTeli: Welcome to MSE! It really helps readability if you format your questions using MathJax. Also, if this is homework, please tag it as such. Lastly, what have you tried or why are you confused would help the MSE Community. Regards – Amzoti Feb 2 '13 at 20:47
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
up vote 2 down vote accepted

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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