Prove that adding one to itself is always different from 0

• January 11th 2010, 09:23 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, 09:49 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, 04:09 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?