In a ring $4xy+h^2-2h$ commutes with $x$.

Question

Let $K$ be a ring, and for two elements of the ring we say $$[a,b]=ab-ba.$$

$x,y,h$ are elements of $K$ satisfying: $$[h,x]=2x ,\quad [x,y]=h,\quad [h,y]=-2y.$$

We can already deduce (see here) that $$[h,x^n]=2n\dot{}x^n, \quad [h,y^n]=-2n\dot{}y^n.$$ Now I would like to show that the element $$ 4xy+h^2-2h$$ commute with $x$.

I identify some extra properties such as: $$ [x,y]=-[y,x], \quad[x,y]^2=[y,x]^2$$ but i didn't reach a proof. I'm not sure if i should eliminate h when expanding or bring h in the element to be commute. Can anyone give me a hint?

Answer 1

1) Notice that $[a,b]=[b,a]\iff ab-ba=ba-ab\iff ba=ab\iff[a,b]=0$

2) Then notice also that $[a,bc]=abc-bca=abc-bac+bac-bca=[a,b]c+b[a,c]$, so $[x,xy]=[x,x]y+x[x,y]=x[x,y]$ and $[x,h^2]=[x,h]h+h[x,h]$

3) Finally notice that $[a,n\dot{}b]=[n\dot{}a,b]=n\dot{}[a,b]$ and $[a,b+c]=[a,b]+[a,c]$

So for previously computing $$\begin{align}[x,4\dot{}xy+h^2-2\dot{}h]&=4\dot{}[x,xy]+[x,h^2]-2\dot{}[x,h]\\&=4\dot{}x[x,y]+[x,h]h+h[x,h]-2\dot{}[x,h]\\&=4\dot{}xh-2\dot{}xh-2\dot{}hx-2\dot{}[x,h]\\&=2\dot{}(xh-hx-[x,h])\\&=2\dot{}([x,h]-[x,h])\\&=0\end{align}$$ So for 1) we conclude.

Answer 2

What you want to show is that $[x,4xy+h^2-2h]=0$.

Note that the linked answer includes the identity $[a,bc] = [a,b]c+b[a,c]$.

This can be used to compute $[x,xy]$ and $[x,h^2]$, which should be enough to reach the answer (using linearity in the sense that $[x,a+b] = [x,a] + [x,b]$, for instance).