Assume that all elements under discussion are algebraic over $F$.
Let the notation "$K=F(A)$" mean that $A\subseteq K$ and there is an injective homomorphism $\sigma:F\to K$, and every element of $K$ can be written as a rational function using elements from the image of $\sigma$ and from $A$.
What I'd like is to prove that the union of two finite field extensions of a base field $F$ is also a finite field extension of $F$. That is, if $K=F(\alpha_i)$ and $L=F(\beta_i)$ for some finite sets $\{\alpha_i\}$, $\{\beta_i\}$, then there is a field $M$ such that $M=F(\gamma_i)$ for some finite set $\{\gamma_i\}$ and there are injective homomorphisms from $K,L\to M$. Note in particular that I don't have a splitting field to work over, since this is part of the proof of the existence of a splitting field.
My original idea was to form the set $R$ of polynomials in $K$ with $n$ variables (where there are $n$ different $\beta_i$ variables), then take the set $R/I$ where $I$ is the ideal generated by all polynomials in $F$ (treated as a subset of $L$) whose evaluation under the map $x_i\mapsto\beta_i$ evaluates to $0$, but then I got confused since this always seems to build a larger field, even if you take $K\cup K$. If I take $K=\Bbb Q(\sqrt[3]{2})$ union with itself three times, do I get a field that splits $x^3-2$? If not, how do I get such a splitting field (for this one polynomial)?
Best Answer
It's safe to add one element at a time.
If $K = F(\alpha)$ is a field extension of $F$, then $K \cong F[x] / f(x)$, where $f(x)$ is the minimal polynomial of $\alpha$ over $K$. Thus, if $R$ is any ring over $F$, then the ring homomorphisms $K \to L$ are in one-to-one correspondence with elements $\beta \in R$ satisfying $f(\beta) = 0$.
Now let $K$ and $F(\alpha)$ be field extensions of $F$. You construct the ring $R = K[x] / f(x)$, where $f(x)$ is the minimal polynomial of $\alpha$ (over $F$). It is indeed possible that $R$ is not a field; this happens if and only if $f(x)$ is not irreducible.
So to fix things, what you need is to find an irreducible polynomial $g(x)$ over $K$ such that one of the roots of $g(x)$ is a root of $f(x)$....
An alternative approach, I think, is to recognize that $R$ constructed above is a product of fields.