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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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