I'm reading the book of Razmyslov "Identities of Algebras and their representations" and he uses some "supposedly known fact" from field theory. As I could understand, a more or less general statement of the claim can be stated as follows (probably there are superfluous hypothesis).
Theorem. Let $K$ be any infinite field (of any characteristic) and suppose $L\mid K$ an arbitrary field extension (not necessarily finite). Let $K_0=K(x_i\mid i\in\mathbb{N})$ be the field of fractions of the polynomial ring over $K$ with countable infinite number of variables. Suppose $\delta_1:L\to K_0$ and $\delta_2:L\to K_0$ two $K$-monomorphisms. Then there exist a field extension $K_0\subset M$ and a $K$-automorphism $\sigma$ of $M$ such that $\sigma\circ\delta_1=\delta_2$.
In his book, Razmyslov says that $M$ is algebraically closed and has infinite transcendence degree over $K_0$. He also references the book of Lang, "Algebra" (1965); but a quick search could not reveal me anything similar to above statement (most of Lang's book deals with finite extensions).
Since I'm not specialist in field theory, I would like to ask if someone can either prove the Theorem, disprove it, or give reference for the result.
Thank you in advance!
This is essentially just the uniqueness of algebraic closures (or equivalently, the fact that any isomorphism between fields extends to an isomorphism between their algebraic closures). Let $L_1=\delta_1(L)$ and $L_2=\delta_2(L)$. We have an isomorphism $f:L_1\to L_2$ given by $\delta_2\circ\delta_1^{-1}$.
Now let $M$ be any algebraically closed extension of $K_0$ which has infinite transcendence degree over $K_0$. Let $B_1$ be a transcendence basis for $M$ over $L_1$ and let $B_2$ be a transcendence basis for $M$ over $L_2$. Note that $B_1$ and $B_2$ have the same cardinality (namely, the transcendence degree of $M$ over $K_0$), since $K_0$ has countable transcendence degree over both $L_1$ and $L_2$. So choosing a bijection between $B_1$ and $B_2$, we get an isomorphism $g:L_1(B_1)\to L_2(B_2)$ extending $f$. But now $M$ is an algebraic closure of both $L_1(B_1)$ and $L_2(B_2)$, so $g$ extends to an isomorphism $\sigma:M\to M$. Since $\sigma$ extends $f$, it satisfies $\sigma\circ\delta_1=\delta_2$.
More generally, this argument shows that if $K_0$ is a field with subfields $L_1$ and $L_2$ and an isomorphism $f:L_1\to L_2$ and $M$ is an algebraically closed extension of $K_0$ whose transcendence degree is infinite and greater than or equal to the transcendence degrees of $K_0$ over $L_1$ and $L_2$, then $f$ extends to an automorphism of $M$.