Why is the characteristic of a finite field a prime number???!
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.
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.
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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