= \langle 0 \rangle$, and the possible values for $U_{\sigma_1} \cap (\cap_{i \in \tau_1} U^c_i)$ are $U_1 \cap U_1^c = \emptyset$ (when $\sigma_1 = \tau_1 = 1$), $U_1 \cap X = U_1$ (when $\sigma_1=1$ and $\tau_1 = \emptyset$), $X \cap U_1^c = U_1^c$ (when $\sigma_1 = \emptyset$ and $\tau_1 = 1$), and $X \cap X = X$ (when $\sigma_1=\tau_1 = \emptyset$). Since the possible values of $ U_{\sigma_2} \cap (\cap_{j \in \tau_2} U^c_j)$ are also $\emptyset$, $U_1$, $U_1^c$, and $X$, and since none of these four sets are equal, $U_{\sigma_1} \cap (\cap_{i \in \tau_1} U^c_i) = U_{\sigma_2} \cap (\cap_{j \in \tau_2} U^c_j)$ only when $\sigma_1 = \sigma_2$ and $\tau_1 = \tau_2$. This implies that $x_{\sigma_1} \prod_{i \in \tau_1} (1+x_i) + x_{\sigma_2} \prod_{j \in \tau_2} (1+x_j) = x_{\sigma_1} \prod_{i \in \tau_1} (1+x_i) + x_{\sigma_1} \prod_{j \in \tau_1} (1+x_j)   = 0 \in J_C$.

In the case of $C_2$, $J_C  = \langle 1+x_1 \rangle$, $U_1 = X$, and $U_1^c = \emptyset$. This means that there are two choices of $\sigma_1$ and $\tau_1$ for which $U_{\sigma_1} \cap (\cap_{i \in \tau_1} U^c_i) = \emptyset$: $\sigma_1 = \tau_1=1$, or $\sigma_1 = \emptyset$ and $\tau_1 = 1$. Similarly, $U_{\sigma_1} \cap (\cap_{i \in \tau_1} U^c_i) = X$ when 