- #1
evinda
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MHB
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Hello! (Smile)
We have the canonical function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p, x \mapsto (\overline{x})_{k \in \mathbb{N}_0}=(\overline{x}, \overline{x}, \overline{x}, \dots )$.
The function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding. We interpret with it $\mathbb{Z}$ as a subset of $\mathbb{Z}_p$.
Proof:
Let $x \in \mathbb{Z}$ with $\epsilon_p(x)=(\overline{x})_k=0$, i.e. $x \equiv 0 \mod{p^{k+1}}$ for all $k \in \mathbb{N}_0$. The only integer number that is divisible by any high $p$-power is zero, so it holds $x=0$. Thus, $\epsilon_p$ is an embedding.
Could you explain me the above proof? (Thinking)
We have the canonical function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p, x \mapsto (\overline{x})_{k \in \mathbb{N}_0}=(\overline{x}, \overline{x}, \overline{x}, \dots )$.
The function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding. We interpret with it $\mathbb{Z}$ as a subset of $\mathbb{Z}_p$.
Proof:
Let $x \in \mathbb{Z}$ with $\epsilon_p(x)=(\overline{x})_k=0$, i.e. $x \equiv 0 \mod{p^{k+1}}$ for all $k \in \mathbb{N}_0$. The only integer number that is divisible by any high $p$-power is zero, so it holds $x=0$. Thus, $\epsilon_p$ is an embedding.
Could you explain me the above proof? (Thinking)