Let $C/S$ be a semistable curve, and $G$ a finite group acting faithfully on $C/S$, whose order is invertible on $S$.


Let $P\in C$ be a geometric point with image $s\in S$, such that $P$ is a node in $C\_s$. For short, let $A := \widehat{\mathcal{O}}\_s$, and $R := \widehat{\mathcal{O}}\_P$. Then, at $P$, we may fix "local coordinates" - ie, we fix an identification
$$R = A[[X,Y]]/(XY-a)$$
where $a\in m\_A$ (the maximal ideal of $A$).


Let $x,y$ be the images of $X,Y$, then relative to these coordinates, any $A$-linear automorphism $\sigma$ which fixes the ideals $(x),(y)\subset R$ (which are called the "branches" of the node) is determined uniquely by a pair $(u,\gamma)\in R^\times\times A^\times$ with $\gamma a = a$, where
$$\sigma(x) = ux,\qquad \sigma(y) = \gamma u^{-1}y$$
Now suppose that the stabilizer $G\_P\le G$ of $P$ is cyclic and fixes the branches, so any $g\in G\_P$ is described by $(u,\gamma)$ as above.


On the bottom of page 44 of ["Champs de Hurwitz"](https://perso.univ-rennes1.fr/matthieu.romagny/articles/champs_de_Hurwitz.pdf), the author deduces that "this description, together with theoreme 90, allows us to assume that the action of $G\_P$ is given by a pair of opposite characters $\chi,\chi^{-1}$ of order $e > 1$, via:
$$\sigma(x) = \chi(\sigma)x,\qquad \sigma(y) = \chi(\sigma)^{-1}y$$
"


It is not at all obvious to me why this should be true, or even how "theoreme 90" (presumably Hilbert's thm 90) would come in.


EDIT: I should add that the action of $G\_P$ on the cotangent space of $C$ at $P$ has determinant 1.



local picture of ramification at a node
Okay! Finally I see how to prove this.


First note that any $r\in R$ has a unique expansion as:
$$r = r\_0 + \sum\_{i\ge 1}a\_ix^i + \sum\_{i\ge 1}b\_iy^i\qquad r\_0,a\_i,b\_i\in A$$
Since $R$ is a local ring (with maximal ideal $m\_R := (x,y,m\_A)$), the quotient $\varphi : R\rightarrow R/m\_R \cong k$ is a local ring homomorphism, and hence $r\in R$ is a unit if and only if $\varphi(r)\in k$ is a unit - ie, if and only if $r\_0\in A^\times$.


Note that $k = R/m\_R = A/m\_A$ is algebraically closed, since we are looking at geometric points.


Under our presentation of $R$, we find that for any $\sigma\in G\_P$, there is a unique unit $u\_\sigma\in R^\times$ and $\gamma\_\sigma\in A^\times$ such that $\sigma(x) = u\_\sigma x$ and $\sigma(y) = u\_\sigma^{-1}\gamma\_\sigma y$. Further, there is a unique $\chi(\sigma)\in k^\times$ such that we may write the action as


$$\begin{array}{ll} x\mapsto \chi(\sigma) u\_\sigma x \\ 
y \mapsto \chi(\sigma)^{-1} u\_\sigma^{-1}\gamma\_\sigma y
\end{array}$$
where now $u\_\sigma\equiv 1\mod m\_R$. Since $\sigma$ is $A$-linear, $\chi(\sigma)$ defines a character $\chi : G\_P\rightarrow k^\times$.


Now the map $\sigma\mapsto u\_\sigma$ defines a 1-cocycle in $Z^1(G\_P,R^\times)$. Ie, it satisfies:
$$u\_{\sigma\tau} = u\_\sigma\cdot\sigma(u\_\tau)$$


Let $g$ be a generator of $G\_P$, so that $G\_P = \{1,g,g^2,\ldots,g^{e-1}\}$. For any $\sigma\in G\_P$, consider the function $f\_{u\_\sigma} : R\rightarrow R$ given by


$$f\_{u}(r) = \sum\_{\sigma\in G\_P}u\_\sigma\sigma(r) = \sum\_{i=0}^{e-1}u\_{g^i}g^i(r) = r + u\_gg(r) + g(u\_g)u\_gg(r) + \cdots + \left(g^{e-2}(u\_g)\cdots g(u\_g)u\_gg(r)\right)$$
The critical observation is that here, each term in the above sum fixes the residue class of $r$ mod $m\_R$ (ie, fixes $r\_0$ in it the $x,y$-expansion of $r$), and hence $f\_{u}(r) \equiv er\_0\mod m\_R$. Since $e$ is invertible in $A$ by hypothesis, $r\_0\in A^\times\iff er\_0\in A^\times$, so $f\_{u}$ sends units to units. Then, the usual computation in the proof of Hilbert's theorem 90 shows:


Let $b\in R^\times$, then for all $\tau\in G\_P$,
$$\tau(f\_u(b)) = \sum\_{\sigma\in G\_P}\tau(u\_\sigma)\tau(\sigma(b)) = \sum\_{\sigma\in G\_P}u\_\tau^{-1}u\_{\tau\sigma}\tau(\sigma(b)) = u\_\tau^{-1}\sum\_{\sigma\in G\_P}u\_{\tau\sigma}(\tau\sigma)(b) = u\_{\tau}^{-1}f\_u(b)$$
Thus, $u\_\tau = \frac{f\_u(b)}{\tau(f\_u(b))} = \frac{\tau(f\_u(b)^{-1})}{f\_u(b)^{-1}}$. Let $\beta\_u := f\_u(b)$, then
$$\sigma(\beta\_u x) = \frac{\sigma(\beta\_u)}{\beta\_u}\chi(\sigma)u\_\sigma \beta\_u x = \frac{\sigma(\beta\_u)}{\beta\_u}\chi(\sigma)u\_\sigma \beta\_u x = \frac{\sigma(\beta\_u)}{\beta\_u}\chi(\sigma)\frac{\beta\_u}{\sigma(\beta\_u)} \beta\_u x = \chi(\sigma)\beta\_u x$$


Thus, replacing $x$ by $\beta\_u x$ will get us the desired statement for the action of $G\_P$ on $x$.


For the action of $G\_P$ on $y$, observe that on the geometric fiber above $s$, the cotangent space at $P$ (ie, the cotangent space of $R/m\_AR$) is generated by $x,y$, upon which $\sigma$ acts via the matrix
$$\begin{bmatrix}
\chi(\sigma) & 0 \\
0 & \chi(\sigma)^{-1}\gamma\_\sigma
\end{bmatrix}\mod m\_AR$$
The requirement that the action have determinant one then implies that $\gamma\_\sigma = 1\mod m\_A$, and hence viewed in $R$, $\gamma\_\sigma = 1\mod m\_R$. This implies that the 1-cocycle $\sigma\mapsto u\_\sigma^{-1}\gamma\_\sigma$ also has the property that the function $f\_{u^{-1}\gamma} : R\rightarrow R$ given by
$$f\_{u^{-1}\gamma} := \sum\_{\sigma\in G\_P}u\_\sigma^{-1}\gamma\_\sigma\sigma $$
also takes units to units, and thus we may apply the same argument above to find a unit $\beta\_{u^{-1}\gamma}\in R^\times$ such that
$$\sigma(\beta\_{u^{-1}\gamma}y) = \chi(\sigma)\beta\_{u^{-1}\gamma} y$$
which completes the proof.


