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Let \( \displaystyle R \) be an associative ring with unitary element \( \displaystyle 1_{R} \) such that \[ \displaystyle \left( a \cdot b \right)^{2} = a^{2} \cdot b^{2} \; , \; \forall a,b \in R \] Show that \( \displaystyle R \) is commutative.

\(\displaystyle{... \left(a\cdot b\right)^2=a^2\cdot b^2\,\,\forall\,a\,,b\in R\,\,(I)...}\) .

Let \(\displaystyle{a\,,b\in R}\). We'll prove that \(\displaystyle{a\cdot b=b\cdot a}\) .

\(\displaystyle{\begin{aligned}\left[\left(a+1_{R}\right)\cdot b\right]^2\stackrel{(I)}{=}\left(a+1_{R}\right)^2\cdot b^2&\implies \left(a\cdot b+b\right)^2=\left(a^2+2\,a+1_{R}\right)\cdot b^2\\&\implies \left(a\cdot b+b\right)\,\left(a\cdot b+b\right)=a^2\cdot b^2+2\,a\cdot b^2+b^2\\&\implies \left(a\cdot b\right)^2+a\cdot b^2+b\cdot a\cdot b+b^2=a^2\cdot b^2+2\,a\cdot b^2+b^2\\&\stackrel{(I)}{\implies} b\cdot a\cdot b=a\cdot b^2\,(1)\end{aligned}}\)

Also,

\(\displaystyle{\begin{aligned}\left[a\cdot \left(b+1_{R}\right)\right]^2\stackrel{(I)}{=}a^2\cdot \left(b+1_{R}\right)^2&\implies \left(a\cdot b+a\right)^2=a^2\,\left(b^2+2\,b+1_{R}\right)\\&\implies \left(a\cdot b+a\right)\,\left(a\cdot b+a\right)=a^2\cdot b^2+2\,a^2\cdot b+a^2\\&\implies \left(a\cdot b\right)^2+a\cdot b\cdot a+a^2\cdot b+a^2=a^2\cdot b^2+2\,a^2\cdot b+a^2\\&\stackrel{(I)}{\implies} a\cdot b\cdot a=a^2\cdot b\,(2)\end{aligned}}\)

By using the relations \(\displaystyle{(1)\,,(2)}\) and

\(\displaystyle{\left[\left(a+1_{R}\right)\,\left(b+1_{R}\right)\right]^2=\left(a+1_{R}\right)^2\,\left(b+1_{R}\right)^2}\) ,

we get : \(\displaystyle{a\cdot b=b\cdot a}\) .

So, the ring \(\displaystyle{R=\left(R,+,\cdot\right)}\) is commutative.