2
$\begingroup$

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 $+$.

  • 2
    Although we talk about *Abelian groups*, $f$or rin$g$s it's *commutative rings*.2012-12-01

3 Answers 3

5

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 ?

  • 0
    @Alone1990 - I'm glad that helps! :)2012-12-01
2

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 $ab=ba$.

0

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

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