We know that each element $x$ of $\mathbb{Z} \setminus{\{0\}}$ has a unique representation of the form $x=p^m u\mid m \in \mathbb{N}_0, u \in \mathbb{Z}_p^{\star}$.
$$\mathbb{Q}_p=\left \{ \frac{r}{s}: r, s \in \mathbb{Z}_p, s \neq 0 \right \}$$
From the above proposition, we can write $s$ as follows:
$$s=p^mu, m \in \mathbb{N}_0, u \in \mathbb{Z}_p^{\star}$$
$$\Rightarrow \frac{r}{s}=\frac{r}{p^mu}=\frac{ru'}{p^m(uu')}=\frac{ru'}{p^m}$$
$$ru'=\sum_{i=0}^{\infty} a_ip^i$$
Therefore $\mathbb{Q}_p=\{ p^mu\mid u \in \mathbb{Z}_p^{\star}, m \in \mathbb{Z}\} \cup \{0\}$.
Could you exlain me how we deduce that $\mathbb{Q}_p=\{ p^mu\mid u \in \mathbb{Z}_p^{\star}, m \in \mathbb{Z}\} \cup \{0\}$?
Also, knowing that $\mathbb{Z} \hookrightarrow \mathbb{Z}_p$ is an embedding, how can we show that $\mathbb{Q} \hookrightarrow \mathbb{Q}_p$ is also an embedding?
Every nonzero element of the ring of $p$-adic integers $\mathbb{Z}_p$ can be written in a unique way as $p^mu$, where $u$ is invertible in $\mathbb{Z}_p$.
The field $\mathbb{Q}_p$ is the field of fractions of $\mathbb{Z}_p$, so its nonzero elements are of the form $$ \frac{r}{s} $$ with $r,s\in \mathbb{Z}_p$, $r\ne0$. Then $$ r=p^mu,\quad s=p^nv $$ so $$ \frac{r}{s}=\frac{p^mu}{p^nv}=p^{m-n}uv^{-1} $$ Hence every nonzero element of $\mathbb{Q}_p$ can be written as $p^kw$, with $k\in\mathbb{Z}$ and $w$ invertible in $\mathbb{Z}_p$. The converse is obvious, as $\frac{1}{p}=p^{-1}\in\mathbb{Q}_p$.
Since $\mathbb{Z}$ embeds in $\mathbb{Z}_p$, $\mathbb{Q}_p$ has characteristic $0$ and is a field, so the embedding extends to an embedding of the quotient field of $\mathbb{Z}$ into $\mathbb{Q}_p$.