MHB Ring of integer p-adic numbers.

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The ring of integer p-adic numbers, denoted as $\mathbb{Z}_p$, is established as a principal ideal domain (PID), meaning all its ideals are principal. The embedding function $\epsilon_p$ maps integers from $\mathbb{Z}$ into $\mathbb{Z}_p$, confirming that $\mathbb{Z}$ is a subset of $\mathbb{Z}_p$. The units in $\mathbb{Z}_p$ are characterized as those elements not divisible by the prime p, represented by the series where the leading coefficient is non-zero. Each non-zero element in $\mathbb{Z}_p$ can be uniquely expressed in the form $x = p^m u$, linking the valuation to the structure of the ideals. The unique maximal ideal of $\mathbb{Z}_p$ is $p\mathbb{Z}_p$, reinforcing the principal nature of its ideals.
evinda
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Hey! (Wave)

Let the ring of the integer $p$-adic numbers $\mathbb{Z}_p$.

Could you explain me the following sentences? (Worried)

  1. It is a principal ideal domain.
    $$$$
  2. The function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding.
    (So, $\mathbb{Z}$ is considered $\subseteq \mathbb{Z}_p$)
    $$$$
  3. The units of the ring $\mathbb{Z}_p$:

    $$\mathbb{Z}^*=\mathbb{Z} \setminus p \mathbb{Z}$$

    so the units are

    $$= \{ \sum_{n=0}^{\infty} a_n p^n | a_0 \neq 0\}$$
  4. Each element $x$ of $\mathbb{Z}_p \setminus \{ 0 \}$ has a unique expression of the form $x=p^m u | m \in \mathbb{N}_0$
  5. $\mathbb{Z}_p$ has exactly these ideals:

    $$0, p^n \mathbb{Z}_p (n \in \mathbb{N}_0)$$

    Furthermore, $\cap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_0=\{0\}$ and $\frac{\mathbb{Z}_p}{p^n \mathbb{Z}_p} \cong \frac{\mathbb{Z}}{p^n \mathbb{Z}}$

    Last but not least, the unique maximal ideal of $\mathbb{Z}_p$ is $p \mathbb{Z}_p$.
 
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  1. Let $\mathfrak{p}$ be a nonzero ideal of $\mathbf{Z}_p$, and a nonzero $x \in \mathfrak{p}$ such that $\nu_p(x) < \infty$ is the smallest order in all of $\mathfrak{p}$. $x = p^{\nu_p(x)} a$ by definition of the valuation, implying $p^{\nu_p(x)} = a^{-1} x$. As the LHS is in $\mathfrak{p}$, $p^{\nu_p(x)} \in \mathfrak{p}$, indicating that $\left \langle p^{\nu_p(x)}\right \rangle$ is sitting inside $\mathfrak{p}$. But then any integer $y$ can be written as $y = p^{\nu_p(y)} a$, and assuming $\nu_p(y) \geq \nu_p(x)$, $y = p^{\nu_p(x)} \cdot (p^{\nu_p(y) - \nu_p(x)} a) \in \left \langle p^{\nu_p(x)} \right \rangle$. Hence $\mathfrak{p}$ sits inside $\left \langle p^{\nu_p(x)} \right \rangle$ in turn, but this is only possible if $\mathfrak{p} = \left \langle p^{\nu_p(x)} \right \rangle = p^{\nu_p(x)} \Bbb Z$, and we prove the first part of $\#5$ more generally. This implies that all of the ideals are principal, hence $\mathbf{Z}_p$ is a PID.
  2. What is your $\epsilon_p$? It's obvious that $\mathbf{Z}_p$ has a copy of $\mathbb{Z}$ sitting inside : take the infinite-tuple $(x_1, x_2, x_3, \cdots)$ where $x_i = n \pmod{p^i}$, $n$ being your given integer.
  3. Left as an exercise.
  4. This is straightforward from $\#4$. $x$ have the $p$-adic representation $$x = \sum_{k \geq \nu_p(x)} a_k p^k = p^{\nu_p(x)} \sum_{k \geq 0} a_k p^k$$ The sum there is a unit from exercise $\#4$, so you have the desired.
 
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