Silverman *Arithmetic of Elliptic Curves* Problem 1.12 (a)

Question

I am trying to understand Problem 1.12 (a) in Silverman's Arithmetic of Elliptic Curves. Here is the problem

Let $K$ be a perfect field, $V/K$ be an affine variety, and let $G_K = \operatorname{Gal}(\overline{K}/K)$. Prove that $K[V] = \{f\in\overline{K}[V]:f^\sigma=f\;\;\forall\sigma\in G_K\}$.

Here is the hint: the $\subset$ direction is clear. Conversely, if $F\in\overline{K}[X]$ is a representative of $f$, the map $G_K\to\mathcal{I}(V)$ via $\sigma\mapsto F^\sigma-F$ is a 1-cocycle (indeed, if $f^\sigma=f$ in $\overline{K}[V]$, then $F^\sigma-F\in\mathcal{I}(V)$ and $F^{\sigma\tau}-F = (F^{\sigma}-F)^\tau+ (F^\tau-F)$). It's the next point that I do not understand. Using $H^1(G_K,\overline{K}^+) = 0$, I'm supposed to deduce that there is some $G\in\mathcal{I}(V)$ such that $F+G\in K[X]$. I see that this would solve the problem, since in that case $f = (F+G)+\mathcal{I}(V)\in K[V]$.

Here's what I don't understand. Since $H^1(G_K,\overline{K}^+) = \frac{Z^1(G_K,\overline{K}^+)}{B^1(G_K,\overline{K}^+)}$, so if $H^1 = 0$, this means $Z^1(G_K,\overline{K}^+) = B^1(G_K,\overline{K}^+)$, so the function $\sigma\mapsto F^\sigma-F$ out to come from a boundary.

Question 1: It seems that $\sigma\mapsto F^\sigma-F$ is already in $B^1(G_K,\overline{K}^+)$ by definition, so how am I supposed to use the condition that $H^1 = 0$?

Question 2: In addition, it seems to me that $H^1(G_K,\overline{K}^+)$ is also the wrong group to be looking at. We're treating $\overline{K}[X]$ as an additive $G_K$-module, not $\overline{K}^+$, so shouldn't I be looking at $H^1(G_K,\overline{K}[X])$ or $H^1(G_K,\overline{K}[V])$?

Any elucidation would be much appreciated.

Answer 1

$f\in \overline{K}[X]$

$ c(\sigma)=f^\sigma-f$ is in $Z^1(G_K,\overline{K}[X])$

We are told that $f^\sigma \equiv f\in \overline{K}[V]$ ie. $c\in Z^1(G_K,I(V)\overline{K}[X])$.

We are told that $H^1(G_K,\overline{K})=0$.

This implies that $H^1(G_K,I(V)\overline{K}[X])=0$. This is because a $K$-vector space basis of $I(V)$ gives a $G_K$-invariant $\overline{K}$-vector space basis of $I(V)\overline{K}[X]$.

This gives that $c(\sigma)=g^\sigma-g$ with $g\in I(V)\overline{K}[X]$

so that $(f-g)^\sigma-(f-g)=0\in \overline{K}[X]$ ie. $f-g\in K[X] $ and $f\equiv f-g\in K[V]$