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I have to prove that the ring $(B,+,*)$ is abelian only when for every $(a,b) \in B^2$, $(a+b)^2=a^2+2ab+b^2$.

I don't know where to start, and also I can relate $*$ to the ring, not $+$.

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You were missing a $2$ in the formula - it is certainly not the case in all abelian rings that $(a+b)^2=a^2+ab+b^2$. – Thomas Andrews Dec 1 '12 at 17:45
Although we talk about Abelian groups, for rings it's commutative rings. – lhf Dec 1 '12 at 19:35
up vote 5 down vote accepted

Hint: Write out $$(a+b)^{2}=(a+b)(a+b)=a^{2}+ab+ba+b^{2}$$

note that in general you have to multiply this way since in some rings $ab\neq ba$ .

Now assume $$(a+b)^{2}=a^{2}+2ab+b^{2}$$ and compare both calculations. What do you get ?

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So,since we have * in the ring, I can write (a+b)*(a+b),which is = a^2+ab+ba+b^2 ,and compared to (a+b)^=a^2+2ab+b^2...i do the math and I have ab=ba so,yes this is an abelian ring,thanks! – Alone1990 Dec 1 '12 at 17:59
@Alone1990 - I'm glad that helps! :) – Belgi Dec 1 '12 at 18:24

Let $a,b\in B$. Then your hypothesis is that $(a+b)^2=a^2+2ab+b^2$. So $$ a^2+2ab+b^2=(a+b)^2=a^2+ab+ba+b^2. $$ Since we can cancel additive terms, we get $$ 2ab=ab+ba; $$ subtracting $ab$ from both sides, we get $ab=ba$.

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Similarly, and more simply, the $ $ difference of squares $ $ formula is equivalent to commutativity

$$\rm\begin{eqnarray} a^2-b^2 &=&\,\rm (a-b)\,(a+b) \\ &=&\,\rm ab-ba\, +\, a^2-b^2 \\ \iff\ 0\, &=&\,\rm ab-ba \end{eqnarray}$$

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