Let \(R\) be a PID.
- Prove that every nonzero prime ideal is maximal.
- Prove that \(R\) is Noetherian.
- Prove that every nonzero nonunit factors into finitely many irreducibles.
- Prove that every nonzero maximal ideal is generated by an irreducible element.
- Prove that \(\mathbb Z\) is Noetherian but not Artinian.
Solution
1. Every nonzero prime ideal is maximal.
Let \((p)\) be a nonzero prime ideal and suppose \[ (p)\subseteq(a)\subseteq R. \] Then \(p=ra\). Since \(p=ra\in(p)\) and \((p)\) is prime, either \(a\in(p)\) or \(r\in(p)\). In the first case \((a)=(p)\). In the second, \(r=sp\), so \[ p=spa. \] Since \(R\) is a domain and \(p\ne0\), \(sa=1\), hence \((a)=R\). Thus \((p)\) is maximal.
2. Every PID is Noetherian.
Every ideal is principal, hence finitely generated. A ring in which every ideal is finitely generated is Noetherian.
3. Every nonzero nonunit factors into finitely many irreducibles.
Suppose not. Choose a nonzero nonunit \(a_0\) with no factorization into irreducibles. Then \(a_0\) is reducible, say \[ a_0=b_1a_1 \] with both factors nonunits. At least one factor, say \(a_1\), also has no factorization into irreducibles. Then \[ (a_0)\subsetneq(a_1). \] Repeating produces an infinite strictly ascending chain of ideals, contradicting Noetherianity. Hence the process terminates and gives a finite irreducible factorization.
4. Every nonzero maximal ideal is generated by an irreducible.
Write \(\mathfrak m=(m)\ne(0)\). Since \(\mathfrak m\) is proper, \(m\) is not a unit. If \(m=ab\), then \[ (m)\subseteq(a)\subseteq R. \] Maximality gives \((a)=(m)\) or \((a)=R\), so respectively \(b\) or \(a\) is a unit. Thus \(m\) is irreducible.
5. \(\mathbb Z\) is Noetherian but not Artinian.
It is a PID, hence Noetherian. But \[ (2)\supsetneq(4)\supsetneq(8)\supsetneq\cdots \] is an infinite strictly descending chain of ideals, so \(\mathbb Z\) is not Artinian.