Permuting roots in splitting fields Currently, I've just started to study Field and Galois theory. In one of my textbooks, I have found the following (probably important) theorem:
If $K/F$ is a splitting field for the irreducible polynomial $f \in F[X]$ and $a,b \in K$ are two roots of $f$, then there is an automorphism in $\text{Aut}(K/F)$ which sends $a$ to $b$.
The proof is omitted in my textbook, so I try to find it by myself. I know how to show that there is an isomorphism $\tau:F(a) \to F(b)$ sending $a$ to $b$ and being the identitiy when restricted to $F$. To complete the proof, it seems to be necessary to show that $\tau$ can be extended to an isomorphism $\sigma:K \to K$. My problem: I have no idea how to show that a suitable $\sigma$ exists. My question: How to show this?
The textbook is "Basic Abstract Algebra" by Schilling (1975) and the theorem is called "Proposition" and can be found in chaper 8, section 8.5.
 A: I might as well make my comments into an answer. Let $ S(n) $ denote the following statement:

For any two fields $ F $ and $ F' $, any polynomial $ f \in F[X] $ of degree $ n $ and any splitting field $ E $ of $ f $ over $ F $, if there exists a field isomorphism $ \varphi: F \to F' $, then for any splitting field $ E' $ of $ \varphi(f) $ over $ F' $ (where $ \varphi(f) $ is the polynomial in $ F'[X] $ obtained by replacing each coefficient of $ f $ by its image under $ \varphi $), there exists an extension of $ \varphi $ to an isomorphism $ \tilde{\varphi}: E \to E' $.

We implicitly assume in the statement above that with a splitting field comes a field embedding of the base field into it.
Clearly, $ S(1) $ is true, so let us assume that $ S(n) $ is true for some $ n \in \Bbb{N} $.
Let $ f \in F[X] $ have degree $ n + 1 $. Let $ p $ be an irreducible factor (over $ F $) of $ f $. Then extend $ \varphi $ to a field isomorphism $ \tilde{\varphi}: F(\alpha) \to F'(\beta) $, where $ \alpha $ is a root of $ p $ in $ E $ and $ \beta $ is a root of $ \varphi(p) $ in $ E' $. Hence, there now exists a $ g \in F(\alpha)[X] $ such that
$$
f                  = (X - \alpha) \cdot g \quad \text{and} \quad
\tilde{\varphi}(f) = (X - \beta)  \cdot \tilde{\varphi}(g).
$$
Note that $ E $ and $ E' $ remain splitting fields of $ g $ and $ \tilde{\varphi}(g) $ over $ F(\alpha) $ and $ F'(\beta) $ respectively. The induction hypotheses thus hold, and as $ \deg(g) = n $, we can extend $ \tilde{\varphi} $ to a field isomorphism
$$
\tilde{\tilde{\varphi}}: E \to E'.
$$
As $ \tilde{\tilde{\varphi}} $ obviously extends $ \varphi $, we deduce that $ S(n + 1) $ is true.
By mathematical induction, $ S(n) $ is true for all $ n \in \Bbb{N} $.
A: A useful result for you might be the Isomorphism Extension theorem.  You can find a detailed proof, IIRC, in Dummit & Foote (or by Googling).  This theorem states the following:
Let $F$ be a field and $E$ a finite extension of $F$.  Suppose we have a field isomorphism $\phi:F \rightarrow F'$.  Letting $\overline{F'}$ denote an algebraic closure of $F'$, we can extend $\phi$ to an isomorphism $\tau$ from $E$ to a subfield of $\overline{F'}$ wherein $\tau|_F = \phi$.  
