Let $K$ and $K'$ be finite extensions over a field $F$ such that $K$ is a splitting field for a polynomial $p(x)$ over $F$, and let $\varphi \colon K \to K'$ be an isomorphism which fixes $F$. Is it true that $K = K'$?
This is stated as a theorem in Pinter but I am having some difficulty following the (sparse) proof.
The proof begins with the fact that any such isomorphism takes a root of a polynomial to some other root of the polynomial. So, if $c_1,\ldots,c_n$ are the roots of $p(x)$ in $K$ then $\varphi(c_1),\ldots,\varphi(c_n)$ are the roots of $p(x)$ in $K'$. But Pinter asserts that $\varphi$ permutes the roots, by which I assume he means that $\varphi(c_k) = c_j$ for some $j$ for each $k$. But it's not clear to me why such a statement of equality would even make sense, since $\varphi(c_k) \in K'$ and $c_j \in K$.
I came across this answer by Qiaochu Yuan which seems to hint at where some of my confusion is coming from. It seems like I need to first assume that $K$ and $K'$ are embedded in another field $E$ before even using the symbol "$=$". I would certainly appreciate some clarification on this point. This is why I asked "Is it true?" in the title.
Then, if I understand correctly, we can use the fact that $K$ and $K'$ are simple extensions, say $K = F(a)$ and $K' = F(b)$, and embed them in $F(a,b)$.
I think the next step would be to extend $\varphi$ to an automorphism on $F(a,b)$. Then, since the roots of $p(x)$ are already in $F(a,b)$ and $\varphi$ sends roots to roots, it must permute the roots. But I don't see how to construct such an extension.
Any help would be greatly appreciated.