# Local fields. $p$-field $(k,v)$ Proof of lemma.

I study a lemma from 'Local Class Field Theory' and I have difficulties.

To understand this properly we need another lemma

Lemma:1

Let $$\mathfrak{l=o/p}=\mathbb{F}_q$$ for a $$p$$-field $$(k,v)$$. Then, for each $$x \in \mathfrak{o}$$, the limit $$\omega(x) = \lim_{n\to\infty}x^{q^n}$$ exists in $$\mathfrak{o}$$, and the map $$\omega :\mathfrak{o} \to \mathfrak{o}$$ has the following properties: $$\omega(x)\equiv x \ mod \ \mathfrak{p}$$, $$\$$ $$\omega(x)^q = \omega(x) \$$, $$\omega(xy)=\omega(x)\omega(y)$$.

$$\mathfrak{o}=\{x \in k | v(x) \geq 0 \}$$, $$\mathfrak{p} =\{x \in k | v(x) > 0\}$$.

Lemma 2:

Let $$(k,v)$$ be as stated above and let $$V=\{ x\in k | x^{q-1} =1 \}, \ A=V \cup \{0\} = \{x \in k| x^q=x\}.$$ Then $$A$$ is a complete set of representatives of $$\mathfrak{l=o/p}$$ in $$\mathfrak{o}$$,containing $$0$$; $$V$$ is the set of all $$(q-1)$$st roots of unity in $$k$$; and the canonical ring homomorphism $$\mathfrak{o \to l=o/p}$$ induces an isomorphism of multiplicative groups: $$V \cong l^{\times}$$ In particular, $$V$$ is a cyclic group of order $$q-1$$.

Proof of Lemma 2:

Let $$A' = \{\omega(x) | x\in \mathfrak{0} \}$$. As $$\omega(x) \equiv x \ mod \ \mathfrak{p}$$, each reside class of $$\mathfrak{o}$$ mod $$\mathfrak{p}$$ contains at least one element in $$A'$$, and as $$\omega(x)^q =\omega(x)$$, $$A'$$ is a subste of $$A$$. However, the number of elements $$x$$ in $$k$$ satisfying $$x^q-x=0$$ is at most $$q$$, while the number of elemenst in $$\mathfrak{l=o/p}$$ is $$q$$. Hence $$A=A'$$ and $$A$$ is a complete set of representatives of $$\mathfrak{l=o/p}$$ in $$\mathfrak{o}$$. Obviously $$0=\omega(x) \in A$$. Since $$\omega(xy)=\omega(x)\omega(y)$$, the other statements on $$V$$ are clear. QED

My questions:

1) What does it mean that homomorphism induces an isomoprhism?

2) Why we need to consider equation: $$x^q-x=0$$?

3) Why $$\omega(xy)=\omega(x)\omega(y)$$ implies the other statements ? (I understand it implies that map $$\omega: \mathfrak{o} \to \mathfrak{o}$$ is homomorphism)