Field extensions and monomorphism

Question

Suppose $[E_1:F]=m<\infty$ and $E_1$ is algebraic extension of $F$. If $K$ is any extension of $F$ then the number of monomorphism of $E_1/F$ into $K/F$ is at most $m$.

I am trying to prove this by induction on $[E_1:F]$. If I pick an irreducible polynomial $g(x)$ with degree $m$ over $F$ and look at the solutions for $g(x)$ in $E_1$, I can construct a sort of intermediate field between $E_1$ and the field generated by adjoining the root of $g(x)$ to $F$. However how do I translate this information and know something about $K$ as it can bigger or smaller field?

Answer 1

Show that E is a splitting field over F of some polynomial in F[x] using the fact that [E:F] is finite. Then m is the maximum number of automorphisms of E/F. If K/F contains no monomorphic images of E/F we are done. Otherwise any two monomorphic images of E in K/F are isomorphic. So we can regard all the monomorphic images as being a single subset of K. If z is one of the monomorphisms let it correspond to the identity automorphism. If w is another monomorphism let it correspond to the automorphism (z^-1)w. We are done.

Answer 2

Here is perhaps the induction proof you had in mind.

First, if $E=F$, the proposition is trivial. Now suppose that we have proved the proposition for every subfield $F \subseteq E' \subsetneq E$. Taking $E'$ to be maximal, and $a\in E\setminus E'$, we have $E = E'[a]$. Let $f(x)$ be the minimal polynomial of $a$ over $E'$.

Now we count $F$-morphisms $E=E'[a]\to K$. By the inductive hypothesis, there are at most $[E':F]$ $F$-morphisms $E'\to K$. If $\varphi: E\to K$ is an $F$-morphism, then $a$ must be mapped to a root of $\varphi(f)$, a polynomial with at most $[E:E']$ roots.

Since $\varphi$ is determined by what it does to $E'$ and what it does to $a$, it follows that there are at most $[E':F][E:E']=[E:F]$ morphisms $E\to K$ that fix $F$.