$\DeclareMathOperator\Ker{Ker}\DeclareMathOperator\U{U}$I'm reading Serre's text, "A course in arithmetic". I have some problems with this argument.
Let $\U=\mathbb{Z}_p^*$ the group of $p$-adic units. For every $n \le 1$ let $\U_n=1+p^n\mathbb{Z}_p$. I don't understand how to prove that $\U_n=\Ker(f_n)$ with $f_n:\U \to (\mathbb{Z}/p^n\mathbb{Z})^*$ morphism. In particular, I see that $\U_n \subseteq\Ker(f_n)$ but not the inverse direction.
Then Serre says that the map $(1+p^nx) \to x \pmod p$ from $\U_n/\U_{n+1}$ to $\mathbb{Z}/p\mathbb{Z}$ is an isomorphism because$(1+p^nx)(1+p^ny)=1+p^n(x+y)$ modulo $p^{n+1}$. Also this point isn't clear to me. Is there anybody who has some suggestions? Thanks!
$\DeclareMathOperator\Ker{Ker}\DeclareMathOperator\U{U}$To prove $\Ker f_n=1+p^n\Bbb Z_p$, recall that the group homomorphism $f_n:\U\to(\Bbb Z/p^n\Bbb Z)^\times$ is induced by the ring homomorphism $\varepsilon_n:\Bbb Z_p\to\Bbb Z/p^n\Bbb Z$ whose kernel is $\Ker\varepsilon_n=p^n\Bbb Z_p$ (cfr. proposition 1). Consequently, \begin{align} \Ker f_n &=\{x\in\U:f_n(x)=1\}\\ &=\U\cap\{x\in\Bbb Z_p:\varepsilon_n(x)=1\}\\ &=\U\cap (1+\Ker\varepsilon_n)\\ &=(\Bbb Z_p\setminus p\Bbb Z_p)\cap (1+p^n\Bbb Z_p)\\ &=1+p^n\Bbb Z_p \end{align} where note that $\U=\Bbb Z_p\setminus p\Bbb Z_p$ as stated in proposition 2.
For the second question, consider the function \begin{align} &\varphi:\Bbb Z_p\to\U_n/\U_{n+1}& &x\mapsto(1+p^nx)\U_{n+1} \end{align} Then the formula $(1+p^nx)(1+p^ny)\equiv 1+p^n(x+y)\pmod{p^{n+1}}$ proves that $\varphi$ is a group homomorphism from the additive group $\Bbb Z_p$ to the multiplicative group $\U_n/\U_{n+1}$. Indeed, there exists some $w\in\Bbb Z_p$ such that $$(1+p^nx)(1+p^ny)=(1+p^n(x+y))+p^{n+1}w$$ hence $$z=\frac w{1+p^n(x+y)}$$ belongs to $\Bbb Z_p$ because $1+p^n(x+y)$ is invertible and satisfy $$(1+p^nx)(1+p^ny)=(1+p^n(x+y))(1+p^{n+1}z)$$ Since $1+p^{n+1}z\in\U_{n+1}$, it follows that $1+p^n(x+y)$ and $(1+p^nx)(1+p^ny)$ determine the same coset in $\U_n/\U_{n+1}$, hence $\varphi(x+y)=\varphi(x)\varphi(y)$.
Clearly $\varphi$ is surjective and $\Ker\varphi=p\Bbb Z_p$ thus giving rise to a group isomorphism $\bar\varphi:\Bbb Z_p/p\Bbb Z_p\to\U_n/\U_{n+1}$. Finally, the ring homomorphism $\varepsilon_1:\Bbb Z_p\to\Bbb Z/p\Bbb Z$ is surjective and induce a (ring, hence) group isomorphism $\bar\varepsilon_1:\Bbb Z_p/p\Bbb Z_p\to\Bbb Z/p\Bbb Z$. Then the composition $$\U_n/\U_{n+1}\xrightarrow[\sim]{\bar\varphi^{-1}}\Bbb Z_p/p\Bbb Z_p\xrightarrow[\sim]{\bar\varepsilon_1}\Bbb Z/p\Bbb Z$$ gives the required group isomorphism.