Is it true that if $K/F$ is a splitting field and $\phi \colon K \to K'$ is an isomorphism fixing $F$ then $K = K'$? 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.
 A: It is not true that the fields must be "identical".
Consider the two splitting fields of $x^2+1$ over $\mathbb{R}$, one given by the usual $\mathbb{C}$, and another one, $\mathbb{D}$, given by numbers of the form $a+bj$, with $a,b\in\mathbb{R}$ and $j^2=-1$. As sets, these two fields are not equal, though the two fields are certainly isomorphic over $\mathbb{R}$ (by mapping $a+bi\in\mathbb{C}$ to $a+bj\in\mathbb{D}$). 
If both $K$ and $K'$ are contained in some field $E$, though, then the argument holds: remember that $p(x)$ has at most $\deg(p)$ roots in any integral domain that contains $F$; in particular, since the roots of $p(x)$ in $E$ are $c_1,\ldots,c_n$, and are also $\varphi(c_1),\ldots,\varphi(c_n)$, then the two sets of roots must be equal; i.e., $\varphi$ is a permutation of the roots that are in $E$.
The next step to remember is that part of the definition of "splitting field" is that $K$ is generated over $F$ by the roots; that is, $K = F(c_1,\ldots,c_n)$, and likewise $K'=F(\varphi(c_1),\ldots,\varphi(c_n))$. But if $\{c_1,\ldots,c_n\}=\{\varphi(c_1),\ldots,\varphi(c_n)\}$ as sets, then clearly we have
$$K = F(c_1,\ldots,c_n) = F(\varphi(c_1),\ldots,\varphi(c_n)) = K',$$
giving the desired equality. This is all taking place inside a bigger field $E$.
In this situation, with both $K$ and $K'$ contained in a field $E$, then we do get equality of $K$ and $K'$ as sets, and in particular $\varphi$ is an automorphism of $K$. 
In general, though, you cannot do as you propose. Remember that $F(a,b)$ is only defined contextually. In general, given a field $F$, an extension $E$, and a subset $X$ of $E$, $F(X)$ means "the smallest subfield of $E$ that contains both $F$ and $X$." In order to speak about $F(a,b)$, you need to already have an overfield that contains $F$, contains $a$, and contains $b$; you cannot create it ex nihilo simply by saying "consider $F(a,b)$."
(We can get away with it when we adjoin a single element $\alpha$ for the following reason: there is a theorem that says that given a field $F$ and a polynomial $p$, there is an extension of $F$ that contains a root of $p(x)$. So given $\alpha$ that is to be algebraic, if $p(x)$ is the monic irreducible of $\alpha$, we can consider such an extension and let $F(\alpha)$ be the corresponding field there. Then we have another theorem that tells us any two ways of doing this will result in isomorphic, though not necessarily identical, fields. But there is no parallel result for two or more elements)
Again, consider the situation I give above with $F=\mathbb{R}$, $K=\mathbb{R}(i)$, $K'=\mathbb{R}(j)$. There is no field that contains both $K$ and $K'$ as sets; there is no "$\mathbb{R}(i,j)$" to consider. The assertion is false when you do not assume that both splitting fields are contained in the same field; however, you can prove that any two splitting fields of the same (set of) polynomial(s) over $F$ are isomorphic over $F$.
A: It definitely doesn't make sense to assert that $K=K^\prime$ unless you have already embedded them into some other extension of $F$. What you can say is the following: if $L$ is any extension of $F$ at all, and $i_1,i_2:K\rightarrow L$ are two $F$-monomorphisms of $K$ into $L$, then $i_1(K)=i_2(K)$ (the images are the same). This is because $K/F$ is normal. You can assume $p$ is irreducible over $F$. Let $j$ denote either of $i_1,i_2$. If $\alpha$ is a root of $p$ in $K$, then because $j$ is an $F$-monomorphism, $j(\alpha)$ is a root of $p$ in $L$. Let $\alpha_1,\ldots,\alpha_n$ be the distinct roots of $p$ in $K$. Then $K=F(\alpha_1,\ldots,\alpha_n)$, and it follows immediately that $j(K)=F(j(\alpha_1),\ldots,j(\alpha_n))$. Since $p$ splits over $K$, it also splits over $L$, and its distinct roots are the $j(\alpha_k)$. Now observe that none of this depended on $j$. This means that the sets $\{i_1(\alpha_1),\ldots,i_1(\alpha_n)\}$ and $\{i_2(\alpha_1),\ldots,i_2(\alpha_n)\}$ are the same. Thus the images $i_1(K)$ and $i_2(K)$ are the same (they are generated over $K$ by the same set of elements). 
If we did not assume that $p$ splits in $K$, then this argument would fail. We know if $p$ splits that it has $n$ distinct roots, and it follows that $p$ has $n$ distinct roots in any extension of $F$ in which it splits. Since the elements $i_1(\alpha_1),\ldots,i_1(\alpha_n)$ are $n$ distinct roots of $p$, they give all the roots of $p$ in $L$. The same argument applies for $i_2$. We have two sets of $n$ distinct roots of $p$, and since $p$ has exactly $n$ distinct roots, these sets must be the same. If $p$ didn't split in $K$, say it only had one root, but it split in $L$ with $3$ distinct roots, then by sending the one root to different roots in $L$, we would get two embeddings with different images.
