Questions about page 108-109 of Janusz's book Algebraic Number Fields

Question

I'm reading Janusz's book Algebraic Number Fields (second edition) and I'm confused by the proofs presented on page 108-109. My questions are:

1. In the proof of lemma 4.2 on page 108, why does $f(\alpha^n)=f(\alpha)^n$ imply $f(\alpha) \leq 1$?
2. Second line on page 109 : he proved that the minimum value $m$ of $|z-\alpha|$ occurs at some $\alpha_0 \in \mathbb{C}$ by showing that the function $g(\beta)=|z-\beta|$ is continuous on the closed disk $B_{\varepsilon}=\{\beta \in \mathbb{C} \mid |\beta| \leq m+\varepsilon+|z|\}$. I can see the function is continuous. If he is using the extreme value theorem, then the extreme value theorem says that $\max_{\alpha \in B_{\varepsilon}} |z-\alpha|=|z-\alpha_{\varepsilon}|$ for some $\alpha_{\varepsilon} \in B_{\varepsilon}$. Why does this imply $m=\min_{\alpha \in \mathbb{C}}=|z-\alpha|=|z-\alpha_0|$ for some $\alpha_0 \in \mathbb{C}$?
3. 7th line on paeg 109: he used the following formula $$z^n-\alpha^n=(z-\alpha)(z-\omega \alpha)\cdots (z-\omega^{n-1}\alpha)$$ What is this formula? How is it derived? Can anyone point me to a reference?

Thank you!

Answer 1

(1) Actually it is $f(\alpha^n)=f(\alpha)^n$ together with $f(a)$ being bounded by $\sqrt 2$ that implies $|f(\alpha)|\leq1$. Otherwise, suppose you have $|f(\alpha)|>1$ for some $\alpha$, then $|f(\alpha^n)|=|f(\alpha)|^n\to\infty$ as $n\to\infty$, contradicting the boundedness.

(2) For notational simplicity let $$A=\{\alpha\in\mathbb{C}:|z-\alpha|\leq m+\epsilon\}\\ B_\epsilon=\{\beta\in\mathbb{C}:|\beta|\leq m+\epsilon+|z|\}$$ Then $A\subset B$. Now you are asking why $$m=\inf_{\alpha\in\mathbb{C}}|z-\alpha|=\inf_{\alpha\in A}|z-\alpha|=\min_{\beta\in B_\epsilon}|z-\beta|$$ First by the definition of $A$ we have $\alpha\notin A\implies |z-\alpha|>m+\epsilon$. So we can remove the compliment of $A$ from consideration. Hence $$\inf_{\alpha\in\mathbb{C}}|z-\alpha|=\inf_{\alpha\in A}|z-\alpha|$$ And from the set inclusions $A\subset B_\epsilon\subset\mathbb{C}$ we have $$\inf_{\alpha\in A}|z-\alpha|\geq\inf_{\beta\in B_\epsilon}|z-\beta|\geq\inf_{\alpha\in\mathbb{C}}|z-\alpha|$$ thus finishing the equation.

(3) You can see this as a generalization of the equation $x^n-1=(x-1)(x-\omega)\cdots(x-\omega^{n-1})$.