Proposition 2

*Let E and $E'$ be two fields and u an isomorphism of a subfield K of E onto a subfield $K'$ of $E'$. Let $\mathbf{x} = (x_i)_{i \in I}$ (resp. $\mathbf{x}' = (x'_i)_{i \in I}$) be a family of elements of E (resp. $E'$) which is algebraically free over K (resp. $K'$). Then there exists a unique isomorphism $v$ of $L = K(x_i)_{i \in I}$ onto $L' = K'(x'_i)_{i \in I}$, which induces $u$ on $K$ and maps $x_i$ to $x'_i$ for each $i \in I$.

The uniqueness of $v$ is clear. Let us put $A = K[x_i]_{i \in I}$ and $A' = K'[x'_i]_{i \in I}$. By hypothesis the monomials $x'' = \prod_{i \in I} x_i^{\alpha_i}$ (for $\alpha = (\alpha_i)_{i \in I}$, in $\mathbf{N}^{(I)}$) form a basis of the vector K-space $A$ and there is a similar property of $A'$. Hence there exists a ring isomorphism $w : A \to A'$ mapping each element $\sum_{\alpha \in \mathbf{N}^{(I)}} c_\alpha x^\alpha$ to $\sum_{\alpha \in \mathbf{N}^{(I)}} u(c_\alpha) {x'}^\alpha$. Since $L$ is the field of fractions of $A$ and $L'$ that of $A'$, the isomorphism $w$ extends to an isomorphism $v$ of the field $L$ onto the field $L'$.