Filtered Colimit of Subalgebras

Question

Let $K/k$ be a finitely generated field extension of characteristic $0$, and let $\mathcal{S}$ be the set of field extensions of $k$ which have finite index in $K$. Is it true that $\text{Hom}_{k \text{-linear}} (K, K) \cong \text{colimit}_{F \in \mathcal{S}} \text{Hom}_{F} (K, K)$ as $k$-algebras under composition?

Answer 1

No. For instance, if $k=\mathbb{Q}$ and $K=\mathbb{Q}(x)$, then $\mathcal{S}$ is countable and $\operatorname{Hom}_F(K,K)$ is countable for any $F\in S$, so the colimit is countable. However, $\operatorname{Hom}_k(K,K)$ is uncountable since $K$ has infinite dimension over $k$.

Answer 2

As stated, the statement would certainly not be true, because the only element in the set on the right hand side is the identity. Thus, I assume that there has been a typographical mistake, and $S$ are in fact the finite field extensions of $k$ which are subfields of $K$.

In this case, observe, in the case of field towers $k \le S \le T \le K$ the existence of a forgetful functor $$ \mathcal C_T \to \mathcal C_S, $$ where $\mathcal C_T$ is the category whose only object is $K$, together with the $T$-linear morphisms. This then induces the morphisms of algebrae $$ \operatorname{Aut}_T(K) \to \operatorname{Aut}_S(K) $$ which then form the inductive system over which the colimit is taken.

Note that in this case, $S = k$ is a minimum (or maximum, whichever way you define inductive systems), and it is quite a general theorem that when a minimum exists, this minimum is precisely the inductive limit.