we only have to use the exact sequence of localization. The connecting homomorphism yields isomorphisms $H^i(B^3_{x_2+x_3},W)\simeq H^i(C^2)$ for $i=1,2,3$. This shows the assertion.

\[explicit\_b3\] We can give an explicit generator of $H^1(B^3_{x_2+x_3},W)$. Consider the prime ideals $\mathfrak q_1=(x_0^2+1)$ and $\mathfrak q_2=(x_2)$. Taking the Koszul complexes associated to these regular sequences as generators of respectively $\exten 1{B^3_{\mathfrak q_1}}{\kappa({\mathfrak q_1})}{B^3_{\mathfrak q_1}}$ and $\exten 1{B^3_{\mathfrak q_2}}{\kappa({\mathfrak q_2})}{B^3_{\mathfrak q_2}}$, we get symmetric isomorphisms $$\rho_{x_0^2+1}:\kappa({\mathfrak q_1})\to \exten 1{B^3_{\mathfrak q_1}}{\kappa({\mathfrak q_1})}{B^3_{\mathfrak q_1}}$$ and $$\rho_{x_2}:\kappa({\mathfrak q_2})\to \exten 1{B^3_{\mathfrak q_2}}{\kappa({\mathfrak q_2})}{B^3_{\mathfrak q_2}}$$ as in Remark \[explicit\_c2\]. Consider the element $(\kappa({\mathfrak q_1}),(x_0x_1+x_2)\rho_{x_0^2+1})+(\kappa({\mathfrak q_2}), x_1x_3\rho_{x_2})$ in $\bigoplus_{x_1\in \spec {B^3}^{(1)}}W(\kappa(x_1))$. A straightforward computation shows that the element $(\kappa({\mathfrak q_2}), x_1x_3\rho_{x_2})$ is ramified exactly in the prime ideals $(x_1,x_2)$ and $(x_2,x_3)$ (which is the same as $(x_2,x_2+x_3)$). On the other hand, $(\kappa({\mathfrak q_1}),(x_0x_1+x_2)\rho_{x_0^2+1})$ is ramified in both $(x_1,x_2)$ and in the prime ideal $\mathfrak p_1$ of Remark \[explicit\_c2\]. Indeed, the equality in $B^3$ $$(x_0x_1+x_2)(x_0x_1-x_2)-(x_0^2+1)(x_1^2+1)+x_2(x_2+x_3)=0$$ shows that a prime ideal containing $(x_0x_1+x_2)$ and $x_0^2+1$ must contain either $x_2$, either $x_2+x_3$. Such an ideal must therefore contain $(x_1,x_2)$ in the first case, and $(x_2+x_3, x_0^2+1,x_0x_1+x_2)=\mathfrak p_1$ (seen as an ideal in $B^3$) in the second case.

To check that $(\kappa({\mathfrak q_1}),(x_0x_1+x_2)\rho_{x_0^2+1})+(\kappa({\mathfrak q_2}), x_1x_3\rho_{x_2})$ indeed 