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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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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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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