Theorem $:$ Let $L|K$ be a finite field extension and $E|K$ be any field extension. Let $M$ be an intermediary field between $K$ and $L.$ Then for any $\sigma \in \text {Emb}_K (M,E)$ there exist a finite field extension $E'|E$ and a $\sigma' \in \text {Emb}_K (L,E')$ such that $\sigma'\big |_M = \sigma.$
I observed that it is enough to prove the above result when $L|K$ is a finite simple extension. So what essentially I need to prove is the following lemma $:$
Lemma $:$ Let $L|K$ be a finite field extension and $E|K$ be any field extension. Then for any $x \in L$ and for any $K$-embedding $\sigma : M \longrightarrow E$ there exist a finite field extension $E'|E$ and a $K$-embedding $\sigma' : M(x) \longrightarrow E'$ such that $\sigma' \big |_M = \sigma.$
I read a proposition couple of days back which is as follows $:$
Proposition $:$ Let $K$ be a field. Let $K(x)|K$ be a simple algebraic field extension and $E|K$ be any field extension. Then there is a natural bijective correspondence between $\text {Emb}_K (K(x),E)$ and the set $V_E(\mu_{x,K})$ of all zeros of $\mu_{x,K}$ lying inside $E,$ where $\mu_{x,K}$ denotes the minimal polynomial of $x$ over $K.$
Is the above proposition anyway helpful in proving the above lemma? If so how? Any suggestion regarding this will be highly appreciated.
Thank you very much for reading.
Since $L|K$ is a finite field extension then so is $L|M$ where $M$ is any intermediary field between $K$ and $L;$ so that any $x \in L$ is algebraic over $M.$ Let $\mu_{x,M}$ be the minimal polynomial of $x$ over $M.$ Then clearly $\mu_{x,M}$ is irreducible over $M.$ Consider the polynomial $$\overline {f} = \sigma (a_0) + \sigma (a_1) X + \cdots + \sigma (a_n) X^n \in \overline {M} [X] \subseteq E[X]$$ for any $f = a_0 + a_1 X + \cdots + a_n X^n \in M[X],$ where $\overline {M} = \sigma (M).$ Consider the $K$-algebra homomorphism $\varphi : M[X] \longrightarrow \overline {M} [X]$ defined by $f \mapsto \overline {f}.$ Then it is easy to verify that $\varphi$ is a $K$-algebra isomorphism since $\sigma \in \text {End}_K (M,E).$ Now let $y$ be a zero of $\overline {\mu_{x,M}}$ in some finite field extension $E'|E$ (This is possible due to Kronecker's theorem). Let us consider another $K$-algebra homomorphism $\psi : M[X] \longrightarrow E'$ defined by $f \mapsto \overline {f} (y).$ Let $\overline {\mu_{y, \overline {M}}}$ be the minimal polynomial of $y$ over $\overline {M}.$ Let $\mu_{y,\overline {M}}$ be the pullback of $\overline {\mu_{y, \overline {M}}}$ in $M[X]$ under $\varphi.$ Clearly $\mu_{y,\overline {M}}$ is a monic polynomial in $M[X]$ since $\overline {\mu_{y, \overline {M}}}$ is a monic polynomial in $\overline {M}[X].$ Then it can be checked that $\text {Ker} (\psi) = \langle \mu_{y, \overline {M}} \rangle.$ We claim that $\mu_{x,M} = \mu_{y,\overline {M}}.$
Since $y$ is a zero of $\overline {\mu_{x,M}} \in \overline {M} [X]$ and $\overline {\mu_{y, \overline {M}}}$ is the minimal polynomial for $y$ over $\overline {M}$ so we must have $\overline {\mu_{y, \overline {M}}}\ \bigg |\ \overline {\mu_{x, M}}$ in $\overline {M} [X].$ So $\exists$ $\overline {g} \in \overline {M} [X]$ such that $\overline {\mu_{x, M}} = \overline {\mu_{y, \overline {M}}}\ \overline {g}.$ Let $g$ be the pullback of $\overline {g}$ in $M[X]$ under $\varphi.$ Then since $\varphi$ is a $K$-algebra isomorphism we find that $\mu_{x,M} = \mu_{y, \overline {M}} g.$ Now since $\mu_{x,M}$ is irreducible it follows that $g = a \in M^{\times}.$ So we have $\mu_{x,M} = a\ \mu_{y,\overline {M}}.$ But since both the polynomials $\mu_{x,M}, \mu_{y, \overline {M}} \in M[X]$ are monic it follows that $a=1$ and hence $\mu_{x,M} = \mu_{y, \overline {M}},$ as claimed.
So by the first isomorphism theorem we get a injective $K$-algebra homomorphism $\sigma' : M(x) = M[X] / \langle \mu_{x,M} \rangle \longrightarrow E'$ with the property that $\sigma' \big |_M = \sigma$ i.e. we get a $\sigma' \in \text {End}_K(M(x),E')$ such that $\sigma' \big |_M = \sigma,$ as required.