Lemma II 5.8.1 in Silverman "The Arithmetic of Elliptic Curves" I am trying to work through the derivation of the Weierstrass equation where we do not assume that the base field is algebraically closed. This turns on establishing that we have a basis from $K(C)$ for the Riemann-Roch space of functions from from $\bar{K}(C)$. This seems to be more or less Galois descent for an infinite Galois group (the Absolute group of $K$).
Conrad's otherwise excellent treatment rather ducks the issue in his Remark 2.15 by referring to Lemma 5.8.1. So here we are.
If we make the topological assumptions (Krull topology on $G$, discrete on $V$) that the action is continuous and that the stabiliser of $v$ is an open subgroup of $G$, I can make some progress. (I think that my version of Siverman has a typo as he says that the stabiliser has finite indedx in $K$ when I assume he means $G$). However I strugle to prove either of these assumptions.
In due course he defines $w_i$ by a Trace formula and claims that it is clear that $w_i$ is $G$-invariant. Not to me! I have tried to show the $G$-action is equivalent to $Gal(L/K)$ in his notation by factoring through a quotient, but I can't quite see it.
I would be grateful for any suggestions (or another source for infinite Galois descent).
 A: I think that I can see how this goes if we work with the case that I am interested in and accept the topological results. We have the following setup:
$K$ is a field with algebraic closure $\bar{K}$ and absolute Galois group $G:= Gal(\bar{K}/K)$. $C/K$ is a smooth curve with $\mathfrak{D}$ a divisor defined over $K$ so that $G$ acts on the R-R space $\mathcal{L}(\mathfrak{D})$.
Write $V$ for  $\mathcal{L}(\mathfrak{D})\subset \bar{K}(C)$, which is  a finite dimensional $\bar{K}$-vector space. Write
$$
V^G=V\cap K(C):=\{f\in V:f^\sigma=f\,\forall\,\sigma\in G\}.
$$
$V^G$ is clearly a $K$-vector space.
Now for any $f\in V$ we have the stabiliser subgroup of $G$
$$
G_f:=\{\sigma\in G:f^\sigma=f\}.
$$
If we assume that $G_f$ is an open subgroup of $G$ (in the Krull topology), then it has finite index in $G$. If we write $L$ for the Galois closure of the fixed field of $G_f$, we have $[L:K]<\infty$ and $Gal(\bar{K}/L)$ fixes $f$. i.e. $f\in L(C)$.
Suppose $\{\alpha_1,\dots,\alpha_n\}$ is a basis for $L/K$ and $Gal(L/K)=\{\sigma_1,\dots,\sigma_n\}$. For each $1\le i\le n$ set
$$
g_i:=\sum\limits_{j=1}^n(\alpha_i f)^{\sigma_j}\in V.
$$
Since $\alpha_i f\in L(C)$ we have $(\alpha_i f)^{\sigma_j}\in L(C)$ and thus $g_i\in L(C)$.
So $Gal(\bar{K}/L)$ fixes $g_i$ and we can factor the action of $G$ through $G/Gal(\bar{K}/L)=Gal(L/K)$.
But $Gal(L/K)$ just permutes the terms in $g_i$ so it too fixes $g_i$. i.e. $g_i\in V^G$. We can write
$$
(g_1,\dots,g_n)^T=(\alpha_i^{\sigma_j})(f^{\sigma_1},\dots,f^{\sigma_n})^T
$$
and since the $\{\alpha_i\}$ are linearly independent over $K$, we have their discriminant
$D(\alpha_1,\dots,\alpha_n)\ne 0$. Thus $(\alpha_i^{\sigma_j})$ is invertible over $L$ and in particular we can write $f$ as an $L$-linear combination of the $g_i$.
So far $L$ depends on $f$, but if we pick a $K$-basis $h_1,\dots,h_k$ for $V^G$ and express each $g_i$ in terms of this basis, we can express $f$ as a $\bar{K}$-linear combination of  $h_1,\dots,h_k$. i.e. $h_1,\dots,h_k$ is a $\bar{K}$-basis for $V$.
For our purposes, we also want it to be a $K$-basis for $V^G$. To see this, suppose $h\in V^G$ with $h=\beta_1 h_1+\cdots+\beta_kh_k$ and $\beta_i\in \bar{K}$. Since $h$ and the $h_i$  are fixed by $G$ and the $h_i$ are linearly independent over $\bar{K}$, we must have each $\beta_j$ fixed by $G$ and thus $\beta_j\in K$.
I will not accept my answer for a bit in case anyone wants to correct or refute it.
PS
I now have an idea about the stabiliser of $f$ having finite index in $G$ that might be OK:
For any finite subset $S$ of $\bar{K}$ we define the subgroup
$$
G(S):=\{\sigma\in G:\sigma(s)=s\;\forall\;s\in S\}.
$$
(If in addition $S$ is $G$-stable, then by definition $G(S)$ is an open set in $G$ in the Krull topology).
Now suppose $f\in V\subset \bar{K}(C)$ and let $T$ be the (finite) set of coefficients for a pair of polynomials representing $f$.
Then its stabiliser $G_f\supset G(T)$.
Then since any $a\in T$ is algebraic over $K$, its orbit $a^G$ is the finite set of its conjugates so
$$
S:=\bigcup\limits_{a\in T}a^G
$$
is $G$-stable, finite and contains $T$.
$G(S)$ has finite index in $G$ since
$$
G/G(S)\simeq Sym(S).
$$
But $G(S)\subset G(T)\subset G_f$ so $G_f$ has finite index in $G$.
