The roots of , other than the trivial root, x= 0, form a cyclic set. That is, there exist a root such that consists of all of the non-zero roots of that equation. If all of those roots are not in K, then K is not closed under multiplication.
THEOREM: Let be a field of elements contained in an algebraic closure of . The elements of are precisely the zeros in of the polynomial in .
PROOF: The set of non-zero elements of forms a multiplicative group of order under the field multiplication. For , the order of in this group divides . Thus for we have so . Therefore every element of is a zero of . Since can have atmost zeros, we see that contains precisely the zeros of in .
QUERY: Where have we used the algebraic closureness of . Wouldn't the proof still work if instead of we had any field which is an extension field of the field containing the field .
May be i am missing the obvious but can't figure it out.
The roots of , other than the trivial root, x= 0, form a cyclic set. That is, there exist a root such that consists of all of the non-zero roots of that equation. If all of those roots are not in K, then K is not closed under multiplication.
the point of using the algebraic closure of Zp is to guarantee that we have such a splitting field for EVERY value of n. every "E" can be shown to be the roots of only given some "K" containing that "E", but if we want to consider every possible n, we need a lot of "different" K's.
why, yes, it can! in fact, that is sort of the point, to show that there does in fact exist some finite field of order for every prime p and every positive integer n. one can even go further, because any two such fields are isomorphic, so one often sees the terminology "the field of order ".