I am reading the article Noncommutative quadric surfaces (here) by S. Paul Smith and M. Van den Bergh. There is a sentence in the introduction to the article:
$S$ is a not-necessarily-commutative connected graded ring, $z\in S_2$ is a central regular element. (If $z$ were a normal regular element we could replace $S$ by a suitable Zhang twist in which $z$ becomes central, so there is no loss of generality in assuming $z$ is central.)
I want to prove this property: If $z$ were a normal regular element in $S_2$, we could replace $S$ by a suitable Zhang twist in which $z$ becomes central.
I computed some examples but I can't prove it. Here is my idea:
$z$ is normal and regular, then $\forall a\in S,az=z\sigma(a)$. It is easy to verify that $\sigma$ is an automorphism of $S$.I want to find a square root of $\sigma$ which fixs $z$.
And I wonder if this property is still true if $z\in S_d$.