Multiplicative group of $p$-adic numbers $\mathbb{Q}_p$ $\DeclareMathOperator\Ker{Ker}\DeclareMathOperator\U{U}$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!
 A: $\DeclareMathOperator\Ker{Ker}\DeclareMathOperator\U{U}$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.
