My definitions ($F$, $K$, $L$ denote fields):
A field extension is any field homomorphism. An $F$-extension homomorphism between extensions $\phi\colon F\to K$ and $\psi\colon F\to L$ is a field homomorphism $\xi\colon K\to L$ such that $\xi\circ\phi = \psi$.
A splitting extension of an $f\in F[x]\setminus\{0\}$ is an extension $\phi\colon F\to K$ such that $f_\phi$ splits into linear factors (by $f_\phi$, I mean $f$ with with coefficients replaced with their $\phi$-images), and it satisfies the following "minimality" condition: If $\psi\colon F\to L$ is another extension wherein $f_\psi$ splits, then we have an $F$-extension homomorphism $\xi\colon K\to L$.
I am trying to prove that
Every (nonzero) polynomial of $F$ has a splitting extension unique up to $F$-extension isomorphism.
I am stuck on proving the "uniqueness up to isomorphism" part (having successfully proven the existence part). In particular, the question that has cropped up is essentially this:
Question: Let $\phi\colon F\to K$ be a splitting extension of $f$. Let $f_\phi(\alpha) = 0$ for $\alpha\in K$ so that $f_\phi = (x - \alpha)g$ in $L[x]$, where $L := \phi(F)(\alpha)$. Is the inclusion $L\hookrightarrow K$ a splitting extension of $g$?
Can you help?
Yes. More generally, for all field extensions $F\le L\le K$, if $K$ is a splitting field for some $f(x)\in F[x]$ over $F$, then $K$ is a splitting field for $f(x)$ over $L$.
First, $K$ is by definition an extension of $L$. Let $L\le E\le K$ and suppose $f$ splits over $E$. Now, we have $F\le E$, so $E$ is an intermediate extension of $K/F$ such that $f$ splits over $E$. Since $K$ is a splitting field over $F$, we have $E=K$.
Of course, you can translate all these into mappings (for example, $F\le K$ means there exists a field homomorphism $\phi:F\to K$), but there is really no need to complicate things.