$I(\kappa(x_1))=0$ for any closed point $x_1$. Therefore, $H^1(C^1,I)\simeq H^1(C^1,\overline I)$. The latter is just $Pic(C)/2$, which is trivial by [@Sw1].

\[comput\_c2\] We have $$H^i(C^2,W)=\left\{ \begin{array}{cl} 
\Z/2 & \mbox{if $i=1$,}\\
0 & \mbox{if $i\geq 2$.}
\end{array}\right.$$

Since $C^2$ is a surface, we have $H^i(C^2,W)=0$ if $i\geq 3$. Because $C^2$ has no real point, we have $H^2(C^2,I^2)=0$ by [@Fa4 Proposition 5.1] (observe that $C^2$ has a trivial canonical bundle). This shows that $H^2(C^2,W)=H^2(C^2,I)=0$.

Lemma \[deg\_1\] shows that $H^1(C^2,I)=H^1(C^2,W)$. We then use the sequence comparing $I$ and $I^2$: $$\xymatrix@C=1.4em{\ldots\ar[r] & H^1(C^2,I^2)\ar[r] & H^1(C^2,I)\ar[r] & H^1(C^2,\overline I)\ar[r] & H^2(C^2,I^2)\ar[r] & H^2(C^2,I)\ar[r] & 0.}$$ which reduces to $$\xymatrix@C=1.4em{\ldots\ar[r] & H^1(C^2,I^2)\ar[r] & H^1(C^2,I)\ar[r] & H^1(C^2,\overline I)\ar[r]& 0}$$ by the above discussion.

Since none of the residue fields of $\spec {C^2}$ admits an ordering, we obtain $H^i(C^2,I^2)=H^i(C^2,\overline I^2)$ (argue as in [@Suj]). Now $H^1(C^2,\overline I^2)=0$ by [@CS Corollary 3.3, Remark 3.3.1]. This implies that the map $H^1(C^2,I)\to H^1(C^2,\overline I)$ is an isomorphism. Since $Pic(C^2)/2=\Z/2$ by [@Sw1 Theorem 9.2(b)], we get $H^1(C^2,W)=\Z/2$.

\[explicit\_c2\] The generator of $H^1(C^2,W)$ can be explicitly obtained. Using [@Sw1 Remark, p 471], we find that the generator of $Pic(C^2)$ is the class of the prime ideal $\mathfrak p_1=(x_0^2+1, x_0x_1+x_2)$. Since $(x_0^2+1)(x_1^2+1)=(x_0x_1+x_2)(x_0x_1-x_2)$ and $(x_0x_1-x_2)\not\in\mathfrak p_1$ 