Characteristic or a finite field is a prime number?

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SUMMARY

The characteristic of a finite field is definitively a prime number. This conclusion arises from the fact that if the characteristic were composite, it would lead to the existence of nontrivial zero-divisors, which contradicts the properties of a field. Specifically, if p = nm (where n and m are both greater than 1), then the equation 0 = 1 + 1 + ... + 1 (p times) implies that either a or b must equal zero, which is not possible if p is minimal. Thus, the characteristic must be prime.

PREREQUISITES
  • Understanding of finite fields and their properties
  • Familiarity with the concept of zero-divisors in algebra
  • Knowledge of basic number theory, particularly prime and composite numbers
  • Experience with abstract algebra, specifically field theory
NEXT STEPS
  • Study the properties of finite fields in detail, focusing on their characteristics
  • Learn about zero-divisors and their implications in algebraic structures
  • Explore the relationship between integers and fields, particularly through mappings like Z to R
  • Read Dummit and Foote's "Abstract Algebra" for deeper insights into field operations
USEFUL FOR

Mathematicians, students of abstract algebra, and anyone interested in the foundational properties of finite fields and their characteristics.

el3orian
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Why is the characteristic of a finite field a prime number?!
 
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A finite field clearly has a characteristic (among the elements 1, 1 + 1, 1 + 1 + 1, ... there must be two that equal one another, since we have only finitely many elements in the field). Let p be the least number of ones we need to add up in order to get 0. Suppose p = nm with 1 < n, m < p (i.e. p is not prime). Then

0 = 1 + 1 ... + 1 (p times) = p = nm = (1 + ... + 1)(1 + ... + 1) := ab

where a is the first paranthesis (containing n ones) and b is the second paranthesis (containing m ones). But since we're in a field, this implies that either a or b is 0, contradicting the fact that p minimal.
 
Last edited:
Short (but same) answer: char(F) is clearly not zero. If it were composite, then it's easy to find a nontrivial zero-divisor.
 
map the integers Z to R by sending 1 to 1. if n goes to zero, this induces an injection fron Z/n to R. but since R is a domain, so is Z/n, hence n is prime.
 
Hurkyl said:
Short (but same) answer: char(F) is clearly not zero. If it were composite, then it's easy to find a nontrivial zero-divisor.

Would you mind expanding on this explanation a bit? What is the significance of a nontrivial zero-divisor? Thanks!
 
subGiambi said:
Would you mind expanding on this explanation a bit? What is the significance of a nontrivial zero-divisor? Thanks!
How can a field have a nontrivial zero-divisor?
 
if 1+1+...+1, n times =0, and n is a product of a and b, then ab = 0 in your field, so one of a or b is already zero, so some smaller sum of 1's is already zero.
 
this is the same as my response that if under the map Z---R sending 1 to 1, ab goes to zero, then look at what a and b go to. the product of their images is zero, so one of them is.

(the point is that in a field if AB=0 then either A=0 or B=0.)
 
in dummit and foote's abstract algebra the proof is not very clear i guess. he did not define the binary operation between positive integers and members of the field F.A mapping should be defined to make it clear.Also (1+1+1...ntimes).(1+1+...mtimes) can be (1+1+...mn times) clearly due to properties of the field so it is evident that this step answers all the questions asked above,is'nt it??
 

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