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Thread: Prove that adding one to itself is always different from 0

  1. #1
    Member Mollier's Avatar
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    Prove that adding one to itself is always different from 0

    Hi,

    problem:

    $\displaystyle \mathbb{Z}_p=\{0,1,\cdots,p-1\}$.
    If $\displaystyle \alpha$ and $\displaystyle \beta$ are in $\displaystyle \mathbb{Z}_p$, let $\displaystyle \alpha+\beta$ be the least positive remainder obtained by dividing the (ordinary) sum of $\displaystyle \alpha$ and $\displaystyle \beta$ by $\displaystyle p$,
    and similarly, let $\displaystyle \alpha\beta$ be the least positive remainder obtained by dividing the (ordinary) product of $\displaystyle \alpha$ and $\displaystyle \beta$ by $\displaystyle m$

    Prove that if $\displaystyle \mathbb{F}$ is a field, then either the result of repeatedly adding $\displaystyle 1$ to itself is always different from $\displaystyle 0$,
    or else the first time that it is equal to $\displaystyle 0$ occurs when the number of summands is a prime.

    attempt:
    I look at $\displaystyle n\cdot 1$ where $\displaystyle n$ is the number of times 1 is added to itself.

    $\displaystyle n\cdot 1 (mod\;p)$, $\displaystyle p$ is prime since $\displaystyle \mathbb{Z}_p$ is a field.

    $\displaystyle n\cdot 1 (mod\;p)=0$ when $\displaystyle n$ is a multiple of $\displaystyle p$, $\displaystyle n=\alpha p,\;n\in\mathbb{N}$.

    For $\displaystyle n=1$, $\displaystyle 1 (mod\;p)\neq 0$.
    For $\displaystyle n=\alpha p+1$, $\displaystyle \alpha p(mod\;p)+1(mod\;p)\neq 0$

    I am really bad at this so any suggestion is very welcome!
    Thanks
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  2. #2
    Senior Member Shanks's Avatar
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    Quote Originally Posted by Mollier View Post
    Hi,

    problem:

    $\displaystyle \mathbb{Z}_p=\{0,1,\cdots,p-1\}$.
    If $\displaystyle \alpha$ and $\displaystyle \beta$ are in $\displaystyle \mathbb{Z}_p$, let $\displaystyle \alpha+\beta$ be the least positive remainder obtained by dividing the (ordinary) sum of $\displaystyle \alpha$ and $\displaystyle \beta$ by $\displaystyle p$,
    and similarly, let $\displaystyle \alpha\beta$ be the least positive remainder obtained by dividing the (ordinary) product of $\displaystyle \alpha$ and $\displaystyle \beta$ by $\displaystyle m$

    Prove that if $\displaystyle \mathbb{F}$ is a field, then either the result of repeatedly adding $\displaystyle 1$ to itself is always different from $\displaystyle 0$,
    or else the first time that it is equal to $\displaystyle 0$ occurs when the number of summands is a prime.

    attempt:
    I look at $\displaystyle n\cdot 1$ where $\displaystyle n$ is the number of times 1 is added to itself.

    $\displaystyle n\cdot 1 (mod\;p)$, $\displaystyle p$ is prime since $\displaystyle \mathbb{Z}_p$ is a field.

    $\displaystyle n\cdot 1 (mod\;p)=0$ when $\displaystyle n$ is a multiple of $\displaystyle p$, $\displaystyle n=\alpha p,\;n\in\mathbb{N}$.

    For $\displaystyle n=1$, $\displaystyle 1 (mod\;p)\neq 0$.
    For $\displaystyle n=\alpha p+1$, $\displaystyle \alpha p(mod\;p)+1(mod\;p)\neq 0$

    I am really bad at this so any suggestion is very welcome!
    Thanks
    search "Character of field".
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  3. #3
    Member Mollier's Avatar
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    Thanks for the heads up. I did a bit of reading and came up with the following:

    $\displaystyle char(\mathbb{F})$ is the smallest positive number $\displaystyle n$ such that $\displaystyle 1+\cdots+1(n\;summands)=0$.

    If $\displaystyle char(\mathbb{F})=n=ab\;(composite)$,
    then $\displaystyle 1+\cdots+1(ab\;times)=0$,
    which means that $\displaystyle ab=0$.
    Then either $\displaystyle a=0$ or $\displaystyle b=0$, so $\displaystyle char(\mathbb{F})$ is smaller than $\displaystyle ab$ Contradicting the fact that n is the smallest number..(minimality)

    Is this any good?
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