Use Theorem 27.24 to prove the equivalence of these two theorems: 

Fundamental Theorem of Algebra: Every nonconstant polynomial in C[x] has a zero in C.

Nullstellensatz for C[x]: Let f₁(x), ···, f_r(x) ∈ C[x] and suppose that every α ∈ C that is a zero of all r of these polynomials is also a zero of a polynomial g(x) in C[x]. Then some power of g(x) is in the smallest ideal of C[x] that contains the r polynomials f₁(x ), ··· , f_r(x).

Data from Theorem 27.24

If F is a field, every ideal in F[x] is principal. 

Proof Let N be an ideal of F[x]. If N = {0}, then N = (0). Suppose that N ≠ {0}, and let g(x) be a nonzero element of N of minimal degree. If the degree of g(x) is 0, then g(x) ∈ F and is a unit, so N = F[x] = (1) by Theorem 27.5, so N is principal. If the degree of g(x) is ≥ 1, let f(x) be any element of N. Then by Theorem 23.1, f(x) = g(x)q(x) + r(x), where r(x) = 0 or (degree r(x)) < (degree g(x)). Now f(x) ∈ N and g(x) ∈ N imply that f (x) - g(x)q(x) = r(x) is in N by definition of an ideal. Since g(x) is a nonzero element of minimal degree in N, we must have r(x) = 0. Thus f(x) = g(x)q(x) and N = (g(x)}.