Let F be a finite field say F a1 a2 an Let m x a1x a2x

Let F be a finite field, say F = {a_1, a_2, ..., a_n}. Let m = (x - a_1)(x - a_2)...(x - a_n) F[x] and A = {f F[x] | f(a_i) = 0 for all i = 1, 2,..., n}. Show that A = m, i.e., that A is the principal ideal generated by m.

Solution

Consider the set A

It consists of elements f such that f(ai) =0

Since m = (x-a1)(x-a2)...(x-an)

and f(ai) = 0 for i = 1,2,3....n

we conclude that A is contained in m.

Similarly let us consider mf

If f(ai) =0 then mf(ai) =0 for i = 1,2,3...n

Hence m is contained in A

So it follows that A = <m>