I have seen here: Silverman *Arithmetic of Elliptic Curves* Problem 1.12 (a) that $H^1(G_{\overline{K} / K}, \overline{K}^+) = 0$ implies that $H^1(G_{\overline{K} / K}, I(V)) = 0$, where $I(V)$ is a prime ideal in $\overline{K}[X]$ (based on the comments it looks like the assumption that $I(V)$ is prime is unnecessary), but I'm not able to follow the reasoning for why this is the case.
Could someone provide additional explanation for how we can write an element of $Z^1(G_{\overline{K}/ K}, I(V))$ as a linear combination of elements in $Z^1(G_{\overline{K} / K}, \overline{K}^+)$?
Since $V$ is defined over $K$, there exists $b_1,\dots,\in K[X]$ defined over $K$ that forms a basis for $I(V)$. This is also a $\bar{K}$-basis for $I(V)\bar{K}[X]$.
Now, by hypothesis $F^\sigma - F \in I(V)\bar{K}[X]$. So, we have $F^\sigma -F = \sum_j c_j(\sigma)b_j$, with almost all $c_j(\sigma)=0$. The claim is that $c_j(\sigma)$ are cocycles of $\bar{K}^+$. We use the fact that $F^\sigma - F$ is a cocyle:
$$F^{\sigma\tau} - F =\sum_j c_j(\sigma\tau)b_j $$ and \begin{eqnarray*} F^{\sigma\tau} - F &=& (F^\sigma-F)^\tau + (F^\tau-F)\\ &=&\left(\sum_j c_j(\sigma)b_j\right)^\tau + \sum_j c_j(\tau)b_j \\ &=&\left(\sum_j c_j(\sigma)^\tau b_j^\tau \right) + \sum_j c_j(\tau)b_j \\ &=&\sum_j (c_j(\sigma)^\tau + c_j(\tau))b_j \end{eqnarray*} Comparing coefficients gives $$c_j(\sigma)^\tau + c_j(\tau)=c_j(\sigma\tau) $$ and so each $c_j$ is a cocycle of $\bar{K}^+$