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

1. ## 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

2. Originally Posted by Mollier
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".

3. 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?