Let R Zx I x 1 Prove that I is prime and not maximalSolut

Let R = Z[x], I = (x - 1). Prove that I is prime, and not maximal.

Solution

Proof: Let a C and consider the ideal hx ai C[x]. One has that C[x]/hx ai = C, thus hx ai is a maximal ideal by problem 4. Thus, we get a map from C to the set of maximal ideals of C[x] by sending a to hx ai. It is clear that this map is an injective map, i.e., if hx ai = hx bi then a = b. (Otherwise we would have x a and x b both in an ideal, but for a 6= b these are relatively prime, and hence the ideal would contain \"> 1 and be the entire ring.) To see this map is surjective, let m be a maximal ideal in C[x]. We know that C[x] is a PID (we saw this with F[x], so just apply that result with F = C.) Let m = hf(x)i. Consider the quotient ring C[x]/hf(x)i. If deg f(x) 2, then f(x) factors into linear polynomials over C and hence is reducible in C[x]. Thus, C[x]/hf(x)i is a field if and only if deg f(x) = 1. Thus maximal ideals are generated by linear polynomials, i.e., they are determined by a complex number, the root of the linear polynomial.
Define a map : Z[x] Z by (f(x)) = f(1). This is a surjective map and is a homomorphism (it is an evaluation map). The kernel of this map is hx 1i, so Z[x]/hx 1i = Z. Since Z is an integral domain, hx 1i is
a prime ideal on the midterm. It is not a maximal ideal because Z is not a field, so hx1i cannot be a maximal ideal.