Sorry David if you can't prove this you don't understand finite fields. It's a 2 line proof. Go back and review what it means to be characteristic of a field and what being a field says about divisors.
It's still hard to understand for me though, maybe we can create a finite field with 6 objects... I mean we did with 4 (0,1,a,b) I mean the char was 2 but still. That is why it's hard for me to understand this susbject, I am not fully convinced, which means there is something I am not getting here..
David,
In any field F, two additive laws of "exponents":
1. for n an integer and x, y in F, n(xy) = (nx)y = x(ny)
2. for x in F and m,n integers, (mn)x=m(nx)
Apply this to your problem.
OK, let's start with something basic:
In any field, we have 1. Since we can add in a field, we can form:
1+1, 1+1+1, etc.
Now it could be that we NEVER get:
1+1+...+1 = 0, no matter how many 1's we add. If this is so, we say our field is of characteristic 0. In that case, the collection of all "sums of 1's" together with 0, forms a structure isomorphic to the natural numbers, and the minimal field we can create from that is isomorphic to the rational numbers, like so:
natural numbers ---> add "negatives" ---> integers ---> add "fractions" ---> rational numbers.
However, it might be the case that there is some n, for which:
1 + 1 +...+ 1 = 0 (where we have n 1's), in which case our field is said to be of characteristic n.
First, let's prove something more basic:
Define k1 as: 1 + 1 +...+ 1 (k times).
I claim: S = {k1: k in Z} forms a subgroup of (F,+).
Associativity is clear, the sums in S are also sums in F, and + is associative in F.
Also, since n1 = 0, we have 0 in S, so S possesses an additive identity.
Finally, for any k1 in S, we also have (n-k)1 in S, which means that:
k1 + (n-k)1 = (1 + 1 +...+ 1) + (1 + 1 +...+ 1) (where the first term has k 1's, and the second has n-k 1's)
= n1 = 0, so we have additive inverses.
We can similarly show that (k1)(m1) = (km)1, which means that multiplication (of F) is closed in S (I urge you to write this out in full detail). Thus (S,+,*) is a RING (since the associative law, and distributivity are inherited from F).
In fact, it is a commutative ring with unity. The mapping:
k + nZ ---> k1 is, in fact, a ring-isomorphism of the ring Z/nZ with S.
Now, let's pause for a moment and work back inside the larger field F.
Consider the mapping F - {0}-->F - {0} given by x-->sx, for s = k1 in S - {0}.
Since F is a field, if sx = sy, and s is non-zero, we have: (1/s)(sx) = (1/s)(sy), that is: x = y.
This shows that the mapping s --> sx is injective, so in particular, when we restrict it to a mapping S - {0} -->S - {0} (by only taking x in S - {0}), it is STILL injective.
Now S - {0} is finite, and since s-->sx is injective for any s in S - {0}, it must also be surjective on S - {0}. In particular, we must have:
sx = 1, for any s in S - {0}, and "some" x (which will depend on s, of course) in S - {0}.
But this means that every s in S - {0} has a multiplicative inverse, that is: U(S) = S - {0}, so S is a subfield of F.
So we may as well ask: for which n is Z/nZ a field?
Suppose that n is composite: so n = kd, for 1 < k,d < n.
Clearly, we have in Z/nZ: [k][d] = [n] = [0], which means in Z/nZ, [k] is a zero-divisor. Zero-divisors cannot have inverses, for if they did:
say [k][u] = [1], then:
[d] = [1][d] = [k][u][d] = [u][k][d] = [u][kd] = [u][n] = [u][0] = [0], which is impossible since 1 < d < n.
So that leaves two possibilities: n = 1, or n = p, a prime.
If n = 1, then Z/1Z = {[0]}, which cannot be a field, since in a field the additive identity, and the multiplicative identity must be different.
So, the only possible candidates for fields are Z/pZ, for a prime p.
The question remains: are these ACTUALLY fields?
If we can show every non-zero element in Z/pZ is a unit, then the answer is YES.
Well consider Z/pZ = {[0],[1],[2],...,[p-1]}.
Clearly all the integers 1,2,...,p-1 are co-prime to p, that is gcd(k,p) = 1 for k = 1,2,...,p-1.
This means that there exists integers a,b such that:
ak + bp = 1.
Taking this equation mod p, we get:
[ak + bp] = [1]
[ak] + [bp] = [1]
[a][k] + [b][p] = [1]
[a][k] + [b][0] = [1]
[a][k] = [1]
so [a] (that is: a (mod p)) is an inverse for [k]. This settles the matter.