I am reading the lecture notes [INTRODUCTION TO DONALDSON–THOMAS INVARIANTS](https://www.maths.tcd.ie/~mozgovoy/data/dt_lecture.pdf). I have a question in the end of page 1 about the proof of a map is an automorphism.


Let $m>0$ be an integer. Let $\overline{A} = Q[x\_1, x\_2]$ be an algebra with multiplication:
\begin{align}
x\_1^ax\_2^b \cdot x\_1^c x\_2^d = (-1)^{m(ad-bc)} x\_1^{a+c} x\_2^{b+d}.
\end{align}


For any $a,b \in \mathbb{N}^2\backslash \{0\}$, define a homomorphism $T\_{a,b}: \overline{A} \to \overline{A}$ by 
\begin{align}
x\_1 \mapsto x\_1 \cdot (1-x\_1^a x\_2^b)^{-mb}, \\
x\_2 \mapsto x\_2 \cdot (1-x\_1^a x\_2^b)^{ma}.
\end{align}
It is said that this map is an automorphism. How to show that $T\_{a,b}$ is a bijection? 


It is said that the algebra $\overline{A}$ is commutative. But according to the definition of the multiplication, this algebra is not commutative. Am I correct? 


Thank you very much.



How to show that a map which relates to Donaldson–Thomas invariants is an automorphism?
The fact that the algebra is commutative has been discussed in the comments. So, I will address the bijectivity of $T\_{a,b}$. I will assume we believe that we have a well defined algebra homomorphism since it is only asked why $T\_{a,b}$ is a bijection. Here is a sketch of bijectivity.


Our multiplication respects the usual (multi)degree of monomials since $x\_1^ax\_2^b \cdot x\_1^cx\_2^d = \pm x\_1^{a+c}x\_2^{b+d}$. We will consider some [term order](https://en.wikipedia.org/wiki/Monomial_order). Notice that $T\_{a,b}(x\_1^cx\_2^d)$ is a power series with lowest term $x\_1^cx\_2^d$. More generally given any formal power series $f$ with lowest term $x\_1^cx\_2^d$, the power series $T\_{a,b}(f)$ has lowest term $x\_1^cx\_2^d$. It follows that $T\_{a,b}(f) = 0$ only if $f = 0$ and so $T\_{a,b}$ is injective. To show $T\_{a,b}$ is surjective we will show any monomial $x\_1^cx\_2^d$ is in the image. To do this start with $f\_0 = x\_1^cx\_2^d$, then $T\_{a,b}(f\_0)$ is a power series with lowest term $x\_1^cx\_2^d$. Now define $f\_1$ so that $T\_{a,b}(f\_1)$ cancels the second lowest term of $T\_{a,b}(f\_0)$. Consider $T\_{a,b}(f\_0 + f\_1)$ and repeat.


