computation by Definition .

\[cor:gdz\] Denote by $\zz$ the $\Q$-linear span of the $y_{s,0}$ ($s\in
\Z_{>0}$). Then for $z\in \zz$ the map $\gd_z$ is a derivation on $\fA^1$ which preserves $\fA^0$. Moreover, $\gd_z$ can be extended to a derivation on $\fA$ by Prop. \[prop:gdz\](ii).

Define $\gd'$ as in Prop. \[prop:gdz\](ii). For any $z=y_{s,0}\in\zz$ and $y_{t,i}$ a generator of $\fA^1$ we have $$\gd'_z(y_{t,i})=a^{t-1}\gd'_z(b_i)=a^{t-1}(a+b_i)y_{s,i}
 =y_{s,0}*y_{t,i}-y_{s,0}y_{t,i}=\gd_z(y_{t,i}).$$ Note that for $z\in \zz$ and $w, w'\in\fA$ we have $\tau_w\Big(\gd_z\big(\tau_{-w}(w')\big)\Big)=\gd_z(w')$. So indeed $\gd_z$ can be extended to a derivation on $\fA$. It’s obvious that $\gd_z$ fixes both $\fA^0$ and $\fA^1$.

We can define another operation on $\caS$ by $y_{s,i}\circ y_{t,j}=y_{s+t,i+j}.$ We can then restrict this to $\{y_{s,0}: s\in\Z_{>0}\}$ then extend linearly to $\zz$. The following result is then straight-forward.

The vector space $\zz$ becomes a commutative and associative algebra with respect to the multiplication $\circ$ defined by $$z\circ z'=z*z'-zz'-z'z.$$

The following proposition is one of the keys to the proof of Theorem \[thm:RDS\] (cf. [@IKZ Prop. 4]).

\[circ\] Let $u$ be a formal variable. For $z\in \caS$ we have $$\exp(zu\tau_z) (\bfone) =(2-\exp_{\circ}(z u\tau_z) )^{-1}  (\bfone).$$ (The inverse on the right is with respect to the concatenation product.)

Define power series $$f(u \tau_z)=\exp_\circ(z u\tau_z )-1=z u\tau_z +z\circ 