Corollary 2

Let $(E_i)_{i \in I}$ be a family of subfields of N containing K; put $L = \bigcap_{i \in I} E_i$ and $M = K\left( \bigcup_{i \in I} E_i \right)$. Then g(L) is the smallest closed subgroup of containing $U_{i \in I} g(E_i)$ and we have $g(M) = \bigcap_{i \in I} g(E_i)$.

The first assertion follows from Cor. 1 and the second is immediate.