In the final step of the proof, why does the equation `sum_(j = 0)^m b_j(alpha)beta^j = 0` guarantee that `beta` is algebraic over `F(alpha)`:
A) The transcendence of `alpha` over F ensures at least one `b_j(alpha) ne 0`, creating a non-trivial polynomial in `F(alpha)[x]` with `beta` as a root.
B) Since `beta` is transcendental over F, it automatically satisfies a polynomial equation over `F(alpha)` ?