Prove that adding one to itself is always different from 0

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January 11th 2010, 08:23 PM
Mollier

Prove that adding one to itself is always different from 0

Hi,

problem:

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

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

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

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

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

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

I am really bad at this so any suggestion is very welcome!
Thanks
January 11th 2010, 08:49 PM
Shanks

Quote:

Originally Posted by Mollier

Hi,

problem:

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

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

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

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

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

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

I am really bad at this so any suggestion is very welcome!
Thanks

search "Character of field".
January 12th 2010, 03:09 AM
Mollier

Thanks for the heads up. I did a bit of reading and came up with the following:

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

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

Is this any good?