Fundamental question about field extensions Let $k$ be an algebraically closed field. Let $K\subset k$ be a subfield, and suppose that it has an algebraic  extension $K\subset L$. Is it true (or at least well defined, because I'm not even sure about this) that $L\subseteq k$? One slightly different point of view is: given injections $i:K\to  k$, $j:K\to  L$, the last one being algebraic, is  there an injection $l:L\to  k$ such that $l\circ j=i$ (i.e. $l$ is a homomorphism of $K$-algebras)? The doubt came from this other question of mine (Exercise 5.2, Atiyah-MacDonald), where I assumed (without thinking of it) that the answer to this one question is positive. Thanks for any clarify.
 A: You cannot prove that $L\subseteq k$, but you're right that there exists an embedding $L\to k$ that's the identity o $K$.
Consider the set $\mathscr{E}$ of pairs $(F,j_F)$, where $K\subseteq F\subseteq L$ (as subfields) and $j_F\colon F\to k$ is the identity on $K$.
We can partially order $\mathscr{E}$ by declaring $(F_1,j_{F_1})\le(F_2,j_{F_2})$ if (and only if) $F_1\subseteq F_2$ and the restriction of $j_{F_2}$ to $F_1$ is $j_{F_1}$.
Clearly $\mathscr{E}$ is not empty, because $(K,i)\in\mathscr{E}$. Moreover any chain in $\mathscr{E}$ has an upper bound: take the union of the first components of the pair and define the homomorphism accordingly.
By Zorn's lemma, there is a maximal element $(F,j_{F})$ and we want to prove that $F=L$. Here the fact that $L$ is algebraic over $K$ is needed (up to now we didn't use it).
If $a\in L\setminus F$, the element $a$ is algebraic over $K$ and all it takes is to prove that we can extend $j_F$ to a homomorphism $j\colon F(a)\to k$. This is standard. But now $(F(a),j)$ contradicts the maximality of $(F,j_{F})$ and we're done.
