Question

Abstract algebra problem... Prove the root theorem for polynomials: For any polynomial f(x) in F[x] and any a in F, f(a)=0 iff x-a is a factor of f(x)

Answer 1

the <= direction is trivial. If (x-a) is a factor of f(x) then we can write f(x) = (x-a)q(x) for some polynominal q(x) and thus ....

Now the => direction. Suppose x=a is a root of f(x) and x-a is not a factor of f(x).

Then we can write f(x) = (x-a)q(x) + r1(x) for some polynominal q(x) and the remainder r(x) where r(x) is not the zero polynominal and is a lower order polynominal than f.

Hence
0 = f(a) = (a-a)q(a) + r1(a) = r(a).
I.e., r1(x) has a root x=a.

We can boot strap this now and repeat the process, with r1(x) replacing p(x). I.e.,

r_(k+1) = (x-a)q_k(x) + r_k(x)

driving down the order of the remainder r_k(x) each time until in a finite number of steps, j, either r_j(x) is of order 1 or 0. But it can't be of order 0 because then
r_j(x) = 0 for all x in F

But it can't be of order 1 either because then r_j(x) = x - a we could have written
r_(j-1)(x) = (x-a) ( q_k(x) + 1 )
which means that r_(j-1)(x) was divisible by x-a, and so on back up our j steps until p(x) was divisible by (x-a). Contradiction.

Therefore, it must be the case that if a is a root of f(x) then in fact ...