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    #1
    Hello!!!

    I want to prove that at the field $\mathbb{Q}_p$, where $p$ is a prime, it stands:

    $$-1=\sum_{0}^{\infty} (p-1)p^i$$

    That's what I have tried so far:


    $$-1=\sum_{i=0}^{\infty}(p-1)p^i =(p-1)+(p-1)p+(p-1)p^2+\dots \\ \Rightarrow 1-1=1+p-1+p^2-p+p^2-p^2+\dots \\ \Rightarrow 0=(1-1)+(p-p)+(p^2-p^2)+\dots$$

    How can I continue?

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    #2
    Quote Originally Posted by evinda View Post
    Hello!!!

    I want to prove that at the field $\mathbb{Q}_p$, where $p$ is a prime, it stands:

    $$-1=\sum_{0}^{\infty} (p-1)p^i$$

    That's what I have tried so far:


    $$-1=\sum_{i=0}^{\infty}(p-1)p^i =(p-1)+(p-1)p+(p-1)p^2+\dots \\ \Rightarrow 1-1=1+p-1+p^2-p+p^2-p^2+\dots \\ \Rightarrow 0=(1-1)+(p-p)+(p^2-p^2)+\dots$$

    How can I continue?
    Hi evinda,

    I think you're misunderstanding the meaning of the equation $-1 = \sum_0^\infty (p-1)p^i$. This is an equation in $\Bbb Q_p$, so to show that the formula holds, you need to prove

    $ \displaystyle \left|\sum_0^N (p-1)p^i - (-1)\right|_p \to 0$ as $ \displaystyle N \to \infty.$

    Here, $|\cdot|_p$ denotes the $p$-adic norm on $\Bbb Q_p$. For all nonnegative integers $N$,

    $ \displaystyle \sum_0^N (p - 1)p^i = \sum_0^N (p^{i+1} - p^i) = p^{N+1} - 1.$

    Hence,

    $ \displaystyle \left|\sum_0^N (p-1)p^i - (-1)\right|_p = |p^{N+1}|_p = p^{-(N+1)} \to 0$ as $ \displaystyle N \to \infty.$

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    #3 Thread Author
    Quote Originally Posted by Euge View Post
    Hi evinda,

    I think you're misunderstanding the meaning of the equation $-1 = \sum_0^\infty (p-1)p^i$. This is an equation in $\Bbb Q_p$, so to show that the formula holds, you need to prove

    $ \displaystyle \left|\sum_0^N (p-1)p^i - (-1)\right|_p \to 0$ as $ \displaystyle N \to \infty.$

    Here, $|\cdot|_p$ denotes the $p$-adic norm on $\Bbb Q_p$. For all nonnegative integers $N$,

    $ \displaystyle \sum_0^N (p - 1)p^i = \sum_0^N (p^{i+1} - p^i) = p^{N+1} - 1.$

    Hence,

    $ \displaystyle \left|\sum_0^N (p-1)p^i - (-1)\right|_p = |p^{N+1}|_p = p^{-(N+1)} \to 0$ as $ \displaystyle N \to \infty.$
    So, do we use this definition?

    A $p-$ norm of $\mathbb{Q}_p$ is a function $||_p: \mathbb{Q}_p \to \mathbb{R}$

    $$x \neq 0, x=p^{w_p(x)}u \mapsto p^{-w_p(x)}$$

    $$\text{ For } x=0 \Leftrightarrow w_p(x)=\infty \\ p^{-\infty}=:0$$

    Is so, is it $w_p(x)=N+1$ and $u=1$ ?

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    #4
    Quote Originally Posted by evinda View Post
    So, do we use this definition?

    A $p-$ norm of $\mathbb{Q}_p$ is a function $||_p: \mathbb{Q}_p \to \mathbb{R}$

    $$x \neq 0, x=p^{w_p(x)}u \mapsto p^{-w_p(x)}$$

    $$\text{ For } x=0 \Leftrightarrow w_p(x)=\infty \\ p^{-\infty}=:0$$

    Is so, is it $w_p(x)=N+1$ and $u=1$ ?
    The $p$-adic norm on $\Bbb Q_p$ is defined by a map $|\cdot|_p : \Bbb Q \to \{p^n\, |\, n\in \Bbb Z\} \cup \{0\}$ such that $|0|_p = 0$ and $|x|_p = p^{-\text{ord}_p(x)}$ for $x \neq 0$. When $x = p^{N+1}$, $\text{ord}_p(x) = N+1$ and thus $|x|_p = p^{-(N+1)}$.

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    #5 Thread Author
    Quote Originally Posted by Euge View Post
    The $p$-adic norm on $\Bbb Q_p$ is defined by a map $|\cdot|_p : \Bbb Q \to \{p^n\, |\, n\in \Bbb Z\} \cup \{0\}$ such that $|0|_p = 0$ and $|x|_p = p^{-\text{ord}_p(x)}$ for $x \neq 0$. When $x = p^{N+1}$, $\text{ord}_p(x) = N+1$ and thus $|x|_p = p^{-(N+1)}$.
    I understand!!! Could you also give me an other example, with an other $x$, at which I can use the difinition of the $p-$adic norm?

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    #6
    Quote Originally Posted by evinda View Post
    I understand!!! Could you also give me an other example, with an other $x$, at which I can use the difinition of the $p-$adic norm?
    Sure. Let $p$ be a prime and let $n$ be an integer not divisible by $p$. Set $x = \frac{n}{p^2}$. Then

    $ \displaystyle \text{ord}_p(x) = \text{ord}_p(n) - \text{ord}_p(p^2) = 0 - 2 = -2.$

    Hence, $|x|_p = p^{-\text{ord}_p(x)} = p^2$.

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    Quote Originally Posted by Euge View Post
    Sure. Let $p$ be a prime and let $n$ be an integer not divisible by $p$. Set $x = \frac{n}{p^2}$. Then

    $ \displaystyle \text{ord}_p(x) = \text{ord}_p(n) - \text{ord}_p(p^2) = 0 - 2 = -2.$

    Hence, $|x|_p = p^{-\text{ord}_p(x)} = p^2$.
    Why do we take the difference $\text{ord}_p(n) - \text{ord}_p(p^2)$ ?

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    #8
    Quote Originally Posted by evinda View Post
    Why do we take the difference $\text{ord}_p(n) - \text{ord}_p(p^2)$ ?
    Recall that $\text{ord}_p(x)$ is by definition the highest power of $p$ which divides $x$ when $x\in \Bbb Z$ and $\text{ord}_p(a) - \text{ord}_p(b)$ when $x$ is of the form $\frac{a}{b}$, $a, b\in \Bbb Z$, $b \neq 0$. So in the case $x = \frac{n}{p^2}$, we have by definition $\text{ord}_p(x) = \text{ord}_p(n) - \text{ord}_p(p^2)$.

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