I am having trouble to understand automorphism in the article "Galois for beginners" by John Stillwell, like my friend Leo's post and Alberto's post. The main question is related to the automorphism described in theorem 2, but instead of one question, I am breaking it down so the it becomes easy to answer.
First I state the questions, the complete proof is given below.
a. What does $\sigma'\sigma|_{B(\alpha)}=\sigma'|_{B(\alpha)}\sigma|_{B(\alpha)}$ mean in below proof?
b. Does $\sigma'\sigma|_{B(\alpha)}(\beta)=\sigma'|_{B(\alpha)}(\beta)\sigma|_{B(\alpha)}(\beta)$?
c. Why we need $\sigma'\sigma|_{B(\alpha)}=\sigma'|_{B(\alpha)}\sigma|_{B(\alpha)}$ in the proof?
d. In the proof, it is written that, if $\alpha$ is a pth root of unity $\zeta$ then $\sigma(\alpha)=\alpha^i$ but according to the fact that $\sigma \in \text{Gal}(E/B(\alpha))$, it should be $\sigma(\alpha)=\alpha$, so why $\sigma$ is not fixing $\alpha$?
Note, $\sigma \in \text{Gal}(E/B(\alpha)) \implies \sigma \in \text{Gal}(E/B)$ because $B(\alpha) \supset B$, if $\sigma$ fixes $B(\alpha)$ it also fixes $B$, since it fixes $B(\alpha)$, so $\sigma(\alpha)=\alpha$, by the definition. This also makes the below line found in the proof redundant-
$\sigma|_{B(\alpha)}$ is completely determined by the value $\sigma(\alpha)$.
because by the definition, $\sigma(\alpha)=\alpha$.
e. If $\alpha$ is a pth root of unity $\zeta$ then $\sigma(\alpha)=\alpha^i$ according to author, but why not $\sigma(\alpha)=a\zeta^i+b$ where $a,b \in B$, note, $a\zeta^i+b$ still $ \in B(\alpha)$, so $\sigma$ is remains an automorphism of $E$.
f. If $\alpha$ is not a pth root of unity, why $\sigma(\alpha)=\alpha \zeta^i$ for some pth root of unity $\zeta$?
g. Using the similar argument in e., why not $\sigma(\alpha)=a\zeta^i+b$ instead of $\sigma(\alpha)=\alpha \zeta^i$ for some pth root of unity $\zeta$, when $\alpha$ is not a pth root of unity,?
See the proof given below-
Edit:
a. It means you take the restrictions of $\sigma, \sigma'$ and of their composite. The composite of the restrictions is claimed (by that equation) to be the same as the restriction of the composite.
b. No, the "product" $\sigma'\sigma$ means their composite, not the function $x\mapsto\sigma'(x)\cdot\sigma(x)$.
c. As Thomas said, to get the homomorphism.
d. It is not assumed that $\sigma$ fixes $B(\alpha)$.
e. If $b\neq0$ and $\sigma(\alpha^p)=\sigma(\alpha)^p=(a\zeta^i+b)^p=1$, then $b$ will actually turn out to be of the form $c\zeta^j$ for suitable $c, j$ that will make $a\zeta^i+b=a\xi^k$ for a root $\xi$ of unity.
f. $\alpha^p\in B$, so $\sigma(\alpha^p)=\alpha^p$. Since we may choose $b=0$, $\sigma(\alpha)=\alpha\cdot z$ for some $z$ with the property $z^p=1$.