Let r be a pid and let p be a prime ideal of r prove that

Question

1. (a) Let R be a PID and let P be a prime ideal of R. Prove that the quotient ring R/P is again a PID.

(b) Give a counter example to disprove the statement: If R is an integral domain and P is a prime ideal of R such that R/P is a PID, then R is a PID.

2. Let d be a square free positive integer (i.e., d is not divisible by the square of any prime number). Let

Z[√-d] = {a + b√-d : a, b ∈ Z}.

It is given that Z[√-d] is an integral domain. Let N : Z[√-d] → Z⁺ U {0} be the function (called norm) defined by

N(a + b√-d) = (a + b√-d)(a - b√-d) = a² + db².

(a) Show that N(αβ) = N(α)N(β) for all α, β ∈ Z[√-d].

(b) Show that if α, β, ∈ Z[√-d] are nonzero and α | β, then 1 ≤ N(α) ≤ N(β).

(c) Prove that a ∈ Z[√-d] is a unit if and only if N(α) = 1.

(d) Show that 2 and 1 + √-d are irreducibles in Z[√-d].

(e) Prove that every nonzero element in Z[√-d] that is not a unit can be written as a product of irreducibles. [Hint: Adapt the proof of Chapter 4 Theorem 3, using properties of the norm N in parts (a) to (c) above.]

3. Consider the integral domain

Z[√-5 ] = {a + b√-5 : a, b ∈ Z},

i.e., set d = 5 in the ring in Question 2 above. Let I be the ideal of Z[√-5] generated by 2 and 1 + √-5, i.e., I = (2, 1 + √-5).

(a) Prove that I is not a principal ideal. [Hint: Prove by contradiction. Assume that I = (a + b√-5). Since 2, 1 + √-5 ∈ I, we have 2 = [a + b√-5]α and 1+ √-5 = [a+b√-5]β for some α, β ∈ Z[-5]. Taking norms on both sides of the first equation, and also using the second equation, obtain a contradiction]

(b) Prove that I² = II = (2), the principal ideal generated by 2.

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