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Let $A$ be a (not necessarily commutative) algebra over a field $k$. Suppose that for all $a,b\in A$, we have $kab=kba$, i.e. commutativity up to scalar. Show that then $A$ is commutative.

In the assumption, it is important that it holds for all $a,b\in A$, otherwise it would be false. This is a step in Exercise 2.4.8 of Radford's book "Hopf algebras".

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  • $\begingroup$ Why not multiply both sides by $k^{-1}$? $\endgroup$ – G. Chiusole Mar 22 at 11:09
  • $\begingroup$ $k$ is a field. Both sides are not necessarily the same scalar in $k$. $\endgroup$ – user213008 Mar 22 at 11:09
  • $\begingroup$ Ah, $k$ is the field. I misread, sorry $\endgroup$ – G. Chiusole Mar 22 at 11:10
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Suppose $A$ is not commutative; let $x,y\in A$ be such that $xy\neq yx$. Since $kxy=kyx$, we must have $xy\neq 0$ and there is some scalar $c\in k\setminus\{0,1\}$ such that $yx=cxy$. Now compare $$(1+x)y=y+xy$$ and $$y(1+x)=y+yx=y+cxy.$$ Since these are scalar multiples of each other and $c\neq 1$, we conclude that $y$ must be a scalar multiple of $xy$. Swapping the roles of $x$ and $y$, we find that $x$ must also be a scalar multiple of $xy$. But then $x$ and $y$ would be scalar multiples of each other and would commute, which is a contradiction.

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