Extending field homomorphisms in the general setting If we have arbitrary field extensions $\tilde{K}/K$, $\tilde{L}/L$ and a homomorphism $\sigma:K\rightarrow L$. Under what conditions there exists a homomorphism $\tilde{\sigma}:\tilde{K}\rightarrow \tilde{L}$ that extends $\sigma$? and if such homomorphism exist, is there a way to count the number of possible $\tilde{\sigma}$ in function of the separability degree/degree of the extensions (possibly $[\tilde{L}/\sigma(K)]_s$)?
 A: If $\tilde{K}/K$ is algebraic, $\sigma$ can be extended to $\tilde{K}$ if and only if it can be extended to every subfield of $\tilde{K}$ which is of finite degree over $K$.
In particular, if $\tilde{K}/K$ is separable, an extension exists iff for every $\alpha\in\tilde{K}$ the image $\sigma(f^{\alpha}_{K})\in L[X]$ of the minimum polynomial $f^{\alpha}_{K}$ of $\alpha$ over $K$ has a root in $\tilde{L}$. For when $K\subseteq F \subseteq \tilde{K}$ and $[F:K]$ is finite, $F$ is of the form $K(\alpha)=K[X]/(f^{\alpha}_{K})$, for some $\alpha\in\tilde{K}$.
Another special case often seen is when $\tilde{K}/K$ is algebraic but not necessarily separable, and for each $\alpha\in\tilde{K}$ the polynomial $\sigma(f^{\alpha}_{K})$ completely splits into linear factors in $\tilde{L}[X]$. An extension for $\sigma$ is then easily found by means of Zorn's Lemma.
Zorn's Lemma does not appear to lend itself to the general situation, and we proceed as follows. Let $\mathcal{C}=\{F\mid K\subseteq F \subseteq \tilde{K},\,[F:K]<\infty\}$ be the set of finite subextensions. For every $F\in\mathcal{C}$, let $\sigma_{F}:F\to\tilde{L}$ be an extension of $\sigma$, and put $\mathcal{C}_{F}=\{C\in\mathcal{C}\mid F\subseteq C\}$. Finite intersections of the $\mathcal{C}_{F}$ are non-empty, hence there exists an ultrafilter $\mathcal{F}$ over $\mathcal{C}$ that contains each of the sets $\mathcal{C}_{F}$.
$\tilde{K}$ maps into the ultraproduct $(\prod_{F\in\mathcal{C}}F)/\mathcal{F}$ by $\alpha\mapsto\tilde{\alpha}\text{ mod }\mathcal{F}$, where for $F\in\mathcal{C}$ the $F$-th component of $\tilde{\alpha}$ is given by $\alpha$ when $\alpha\in F$ and is $0$ otherwise. This works because $\mathcal{C}_{K(\alpha)}\in\mathcal{F}$. In turn, the ultraproduct maps into the ultrapower $M:=\tilde{L}^{\mathcal{C}}/\mathcal{F}$, component-wise via the $\sigma_{F}$. This gives a homomorphism $\tilde{\sigma}:\tilde{K}\to M$.
Let $\tilde{L}_{1}$ be the (canonical) image of $\tilde{L}$ in $M$, containing the image  $K_{1}$, say, of $\sigma(K)$. Then $\tilde{\sigma}(\tilde{K})$ is algebraic over $\tilde{\sigma}(K)=K_{1}$. It follows that $\tilde{\sigma}(\tilde{K})\subseteq\tilde{L}_{1}$. For if an element $(\beta_{F})_{F\in\mathcal{C}}\text{ mod }\mathcal{F}$ of $M$ is algebraic over $K_{1}$, there is a non-zero polynomial $f\in K[X]$ such that $\{F\in\mathcal{C}\mid \sigma(f)(\beta_{F})=0\}$ is in $\mathcal{F}$. But if $\gamma_{1},\cdots,\gamma_{n}$ are the zeroes of $\sigma(f)$ in $\tilde{L}$, this set is the union for $1\leq i\leq n$ of the sets $\{F\in\mathcal{C}\mid \beta_{F}=\gamma_{i}\}$, so that one of these must be in $\mathcal{F}$.
Thus $\tilde{\sigma}:\tilde{K}\to \tilde{L}_{1}$, and using the canonical isomorphism $\tilde{L}_{1}\to\tilde{L}$, this yields the desired extension of $\sigma$.
As an application, if $K$ is perfect, $\tilde{K}/K$ is algebraic, and every irreducible polynomial over $K$ has a root in $\tilde{K}$, then $\tilde{K}=\overline{K}$, the algebraic closure of $K$. For, by the above, $\overline{K}$ maps into $\tilde{K}$. (This result is well-known, and it holds for non-perfect $K$ as well.)
