Understanding a step in the proof of Silverman's Proposition 5.4 about twists of elliptic curves On page 343 in the book Arithmetic of Elliptic Curves by Silverman, we have the following result:

Proposition 5.4. Assume that $char(K)\neq 2,3$, and let
$$
n = \begin{cases}
 2 & \text{if }j(E) \neq 0,1728, \\
 4 & \text{if }j(E) = 1728, \\
6 & \text{if }j(E) = 0.
\end{cases}
$$
Then $Twist((E,O)/K)$ is canonically isomorphic to $K^*/(K^*)^n$. 
More precisely, choose a Weierstrass equation $$E: y^2 = x^3 + Ax + B$$
for $E/K$ and let $D \in K^*$. Then the elliptic curve $E_D \in Twist((E,O)/K)$ corresponding to $D (\mod (K^*)^n)$ has Weierstrass equation  
(i) $E_D: y^2 = x^3 + D^2Ax + D^3 B$ if $j(E) \neq 0,1728$, 
(ii) $E_D: y^2 = x^3 + DAx$ if $j(E) = 1728$ (so $B=0$),  (iii)
$E_D: y^2 = x^3 + DB$ if $j(E) = 0$ (so $A=0$).

In the proof of this result, there is the following section which I have trouble understanding:

The action of $G_{\bar{K}/K}$ on the twisted field $\bar{K}(E)_\xi$ is given by
$$\delta^\sigma = \xi_\sigma \delta, \quad x^\sigma = \xi^2_\sigma x, \quad y^\sigma = \xi_\sigma y$$
(own note: here, for $D \in K^*$ and a forth root $\delta \in \bar{K}$ of $D$, $\xi$ is the cocycle defined by $$\xi: G_{\bar{K}/K} \to \mu_4, \quad \xi_\sigma = \delta^\sigma/\delta)$$
The subfield fixed by $G_{\bar{K}/K}$ thus contains the functions
$$X=\delta^{-2} x \quad and \quad Y = \delta^{-1} y,$$
and these functions satisfy the equation $$Y^2 = DX^3 + AX.$$

In particular, I do not understand what

*

*this field $\bar{K}(E)_\xi$ is,

*how to obtain the equations with $x^\sigma$ and $y^\sigma$,

*and what this subfield fixed by $G_{\bar{K}/K}$ is.

I guess obtaining $Y^2 = DX^3 + AX$ is just a matter of direct computation.
I also think that the following isomorphism in the proof may play a role as well:
$$
[]: \mu_4 \to Aut(E), \quad [\zeta](x,y) = (\zeta^2 x, \zeta y)
$$
(I suppose that $\zeta$ is a generator f $\mu_4$).
Could you please help me with these mentioned problems?
 A: For the first question, the twisted field is defined on page 319. Namely, for a cocycle $\xi: G_K\rightarrow \operatorname{Isom} C$, $\bar{K}(C)_\xi$ has the same underlying field with $\bar{K}(C)$ but with a different action of $G_K$, which is $f^\sigma:=f^\sigma\circ\xi_\sigma$.
Now, on page 344, we have a cocycle $\xi: G_K\rightarrow\mu_4$ and $[]: \mu_4\rightarrow\operatorname{Aut}(E)$, and the twisted field $\bar{K}(C)_\xi$ is actually $\bar{K}(C)_{[]\circ\xi}$.
For the second question, $x$ is the projection onto the first factor, note that $E$ is defined over $k$, so $x^\sigma=x\circ(([]\circ\xi)(\sigma))=x\circ([\xi_\sigma])=x\circ(x(\xi_\sigma)^2,y\xi_\sigma)=x(\xi_\sigma)^2$, and you can do this for $y$ similarly.
For the third question, you can also check that $(\delta^{-2}x)^\sigma=((\delta^{-2})^\sigma x)\circ(x(\xi_\sigma)^2,y\xi_\sigma)=(\delta^{-2})^\sigma x(\xi_\sigma)^2=\delta^{-2}x$ and $(\delta^{-1}y)^\sigma=\delta^{-1}y$, hence $\delta^{-2}x,\delta^{-1}y\in(\bar{K}(C)_\xi)^{G_K}$. I guess that from this, you can get $(\bar{K}(C)_\xi)^{G_K}=K(\delta^{-2}x,\delta^{-1}y)$.
