I have an elementary question about twists of curves. The reference I am using is Silverman's AEC. In the book, the Galois action on the twisted function field $\bar{K}(C)_\xi$ is defined as $\sigma \cdot f= f^\sigma \xi_\sigma $ where $\xi: G_{\bar{K}/k} \to Isom(E)$ a 1 cocycle. (by $f^\sigma$ I denote the original action and by $\sigma \cdot f$ the twisted one.)
He then proceeds to give an example of a quadratic twist, which is what I am trying to understand. In particular let $L=K(\sqrt d)$ a quadratic extension, $\chi:G_{\bar{K}/K}\to \{\pm 1\}$ by $\chi(\sigma)=\frac{(\sqrt{d})^\sigma}{\sqrt d}$ and $\xi_\sigma=[\chi(\sigma)]$ the 1-cocycle. Let $E/K$ an elliptic curve given by $y^2=f(x)$. We want to see how $G=G_{\bar{K}/K}$ acts on the twisted function field. Note that $[-1](x,y)=(x,-y)$
Here is where Silverman says that $x^\sigma=x$. By that I suppose he means the twisted action, which I denote by $\sigma \cdot x$ (?). If that's the case then $\sigma \cdot x$ by definition should be $x^\sigma \xi_\sigma=x^\sigma$ which is where I am stuck as I get that $\sigma \cdot x= x^\sigma$ instead of $\sigma \cdot x = x$
Similarly I get that $\sigma \cdot y= \chi(\sigma)y^\sigma$ while he gets that $ \sigma \cdot y= \chi(\sigma)y$.
I know I am doing an elementary mistake here, but I would appreciate any hints on what I am doing wrong.