# Proof involving a field and prime numbers

• Jan 16th 2012, 05:54 PM
Proof involving a field and prime numbers
Hi all,
I'm currently in Linear Algebra II and I'm unsure how to proceed with the following proof:

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

We've done proofs on why Zn is a field only when n is prime, which makes me think the question is linked to modulus in some way. I'm simply just not sure how to start this proof - how can I prove adding 1 to a number is different from adding 0? Shouldn't that be defined in the set of numbers we're working with?

Thanks for any help!
• Mar 16th 2012, 11:33 PM
Chris11
Re: Proof involving a field and prime numbers
Hey. First off, you don't prove that for instance, if you're working in Z_p then (p-1)+1=0, right? So, what you want to do is assume that you can't add 1 to its self a bunch of times and end up with 0. Then, show that the first time, the number of summands is prime. Then, you've covered all cases since either it's allways different than zero or it's not.
• Mar 17th 2012, 12:26 AM
Deveno
Re: Proof involving a field and prime numbers
for a field, 1 MUST be different than 0. that is, a field has a minimum of 2 elements. there are two cases:

k1 = 1 + 1 +....+ 1 = 0, for some positive integer k, or:

k1 ≠ 0, for any positive integer k.

in the first case, all we need to do is prove that the smallest integer k for which k1 = 0 is a prime number.

suppose that k is the smallest positive integer for which k1 = 0, where k is composite. then k1 = (mn)1 = (m1)(n1) = 0, for some positive integers 1 < m,n < k.

since F is a field, it is an integral domain, so either m1 = 0, or n1 = 0, contradicting the minimality of k.

so k cannot be composite, leaving us with 2 choices: k = 1, or k is prime. but k = 1 implies 0 = 1, which cannot happen in a field. hence k is prime.
• Mar 17th 2012, 03:40 PM
HallsofIvy
Re: Proof involving a field and prime numbers
I would try an indirect proof. Suppose $1+ 1+ 1+\cdot\cdot\cdot+ 1= 0$ where k, the number of times we used 1, is NOT a prime. Then k= mn for some integers, m and n, neither equal to 1. That means we have $m+ m+ \cdot\cdot\cdot+ m= 0$ n times or mn= 0. And that, of course, is impossible in a ring.