Extension of isomorphism of fields

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!

Solution 1:

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 $M$.