Let $p$ be a prime, $x \in \mathbb{Z}_p$. I say that if $x \in \mathbb{Z}_p$ then $\text{val}_p(x) \ge 0$ as follows:

Let $x = a_0 + a_1 p + a_2 p^2 + \dots$. Then, $\text{val}_p(x) \ge \text{min}(\text{val}_p(a_0), \text{val}_p(a_1), \dots ) \ge 0 $.

Now, I think that $\text{val}_p(x) \ge 0 \implies x \in \mathbb{Z}_p$ but could not prove it. I start by assuming $\text{val}_p(x) \ge 0 $ and $x\notin \mathbb{Z}_p$. Since $x\in \mathbb{Q}_p $, say $$x= \dots + a_{-k} \cfrac{1}{p^k} + \dots$$ for some $a_{-k} \in \{1,\dots,p-1\} $. Then, we can write

$$x = \cfrac{1}{p^k}(\dots + a_0 + \dots)$$ and we have $\text{val}_p(x) = \text{val}_p(\cfrac{1}{p^k}) \text{val}_p(\dots + a_0 + \dots) = -k + \text{val}_p(\dots + a_0 + \dots) \ge 0$.

However, I stuck here. So, can we say

$\text{val}_p(x) \ge 0 \implies x \in \mathbb{Z}_p$

or not?

  • 3
    $\begingroup$ What’s your definition of $\Bbb Z_p$ ? $\endgroup$ – Lubin Jun 29 at 23:52
  • $\begingroup$ Series of the form $\displaystyle \sum_{k=0}^{\infty} a_k p^k $ where $0 \le a_k < p$. $\endgroup$ – Ninja Jun 30 at 0:05
  • $\begingroup$ What is your definition of the valuation then? Because the way I define it you do not have to prove anything. $\endgroup$ – ThorWittich Jun 30 at 0:08
  • $\begingroup$ Given $x \in \mathbb{Q}$, one can write $x = p^k \frac{a}{b}$ where $k$ is an integer, $p,a,b$ are relatively prime integers. Then, $val_p(x) = k$. $\endgroup$ – Ninja Jun 30 at 0:11
  • 1
    $\begingroup$ Don’t try for a proof by contradiction. If $v_p(z)\ge0$, is not $z$ of form $\sum_0^\infty a_kp^p$? $\endgroup$ – Lubin Jun 30 at 0:11

Let $x \in \mathbb{Q}_p$. That means we can write $$x = \sum_{k \geq -m} a_kp^k,$$ where $\text{val}_p(x) = -m$. Now assume that $\text{val}_p(x) \geq 0$. That means $x$ has to be of the form $$x = \sum_{k \geq 0} a_kp^k,$$ which just means it is an element of $\mathbb{Z}_p$.

  • $\begingroup$ Why does it have to be in that form? $\endgroup$ – Ninja Jun 30 at 1:10
  • $\begingroup$ Because of the valuation. The valuation is the smallest index such that the coefficient corresponding to the index is non-zero. $\endgroup$ – ThorWittich Jun 30 at 1:12
  • $\begingroup$ The valuation of $x$ - the valuation of the sum - is greater than or equal to $0$. So it must be greater than something negative, then how can we say your argument then? $\endgroup$ – Ninja Jun 30 at 1:28
  • $\begingroup$ I am not sure if understand your comment. If the valuation is bigger or equal to zero we know that the smallest index having s non-zero coefficient is $0$. Therefore we can start the series there and these elements are exactly the elements of the $p$-adic integers. $\endgroup$ – ThorWittich Jun 30 at 13:09

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