Theorem of primitive element 
I don understand why we have at least 2 embeddings of $E$ coincide when restricted. Also, there is this sentence ' If we select $c$ different from each of these values, we reach a contradiction of our assumption that $F(\alpha +c \beta) \neq E$.' What is the contradiction here?
Can someone enlighten me?
 A: Well, (3.5.2) probably says that there are exactly $m = [F(\alpha + c \beta) : F]$ distinct $F$-monomorphisms from $F(\alpha + c \beta)$ to $C$. 
Then we have two sets: $A$ is the set of all $F$-monomorphisms from $E$ to $C$, and $B$ is the set of all $F$-monomorphisms from $F(\alpha + c\beta)$ to $C$. And there is the "restriction map" from $A$ to $B$ that takes monomorphisms and restricts their domain from $E$ to $F(\alpha + c\beta)$. Since $|A| = n$ is strictly larger that $|B|=m$, at least two monomorphisms must get squashed together when restricted. This is the pigeonhole principle.
As for the second part, think of it this way. Imagine that we select $c$ in the beginning of the proof rather than in the end. Right before the words "Now by (3.5.2)..." we insert this prhase: "$F$ is an infinite field, so let us pick $c$ in such a way that it is different from $\frac{\alpha' - \alpha''}{\beta' - \beta''}$, for any $\alpha', \alpha''$ conjugate to $\alpha$ and any $\beta', \beta''$ conjugate to $\beta$". Then the proof proceeds as before, and when we arrive at the equality $c = \frac{\sigma(\alpha) - \tau(\alpha)}{\tau(\alpha) - \sigma(\alpha)}$, we immediately have our contradiction. And from this contradiction it follows that the assumption $F(\alpha + c\beta) \neq E$ is wrong, as required.
