# Commutativity up to scalar implies commutativity in an algebra

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".

• Why not multiply both sides by $k^{-1}$? – G. Chiusole Mar 22 '20 at 11:09
• $k$ is a field. Both sides are not necessarily the same scalar in $k$. – user213008 Mar 22 '20 at 11:09
• Ah, $k$ is the field. I misread, sorry – G. Chiusole Mar 22 '20 at 11:10

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.