Question

24. Show that (x ^p) − x has p distinct zeros in Zp, for any prime p. Conclude that (x ^p) − x = x(x − 1)(x − 2)· · ·(x − (p − 1)).

(this is not as simple as showing that each element in Z_p is a root -- after all, we've

seen that in Z₆[x], the polynomial x^2-5x has 4 roots, 0, 5, 2, and 3, but x^2-5 is not equal to (x-0)(x-5)(x-2)(x-3))

Answer 1

Recall Fermat's little theorem states that if p is a prime number, then for any integer a, .

Then choose any element , then we can find a unique integer , be such that the image of a in is . Then gives us the fact , in the ring (rather field) . Thus this proves the fact that every element of is a root of the polynomial . Since the ring has p many distinct elements then we have has p many distinct roots.

Now note that since is a field this implies is an Euclidean domain. Also recall in Euclidean domain we can divide.

We will prove the fact that if is a root of a non constant polynomial  , then divides .

Note that since we are in an Euclidean domain, we can find (by division algo) and , with , (that is deg(r)=0 gives us r is a constant) such that , now putting , (and recall r is a constant ) gives us the fact r=0 . Thus we get .

Using this fact and the result we have proved earlier that , we can say for any , divides , (Now note that a, b divides c and both of a and b are prime implies ab divides c). Here for any , is a prime (as a domain) , thus we have   divides , this gives us there exists some , bu such that . Now comparing the the degree on the both side we can say deg()=0, gives is a constant. Again note that as we are in a field so we can compare coefficients on the both side, this gives us =1. Hence the prove.

Note that , is not even an integral domain, hence not a field hence is not an  Euclidean domain. Hence we can't divide there, so this process is not applicable there.