Show that a Dedekind domain \(R\) is a PID iff \(R\) is a UFD.
Solution
1. If \(R\) is a PID, then \(R\) is a UFD.
In a PID every nonzero nonunit factors into irreducibles, because the ascending chain condition on principal ideals holds, and every irreducible element is prime, because it generates a maximal ideal. Hence factorization is unique up to order and associates.
2. Suppose now that \(R\) is a Dedekind domain and a UFD. Then every nonzero prime ideal of \(R\) is principal.
A Dedekind domain is Noetherian and every nonzero prime ideal has height \(1\). In a Noetherian UFD, every height-\(1\) prime ideal is principal. Thus for every nonzero prime ideal \(\mathfrak p\subset R\) there is a prime element \(\pi_{\mathfrak p}\in R\) with \[ \mathfrak p=(\pi_{\mathfrak p}). \]
3. Every nonzero ideal of \(R\) is principal.
Let \(0\neq I\subseteq R\). Since \(R\) is Dedekind, \(I\) has a factorization \[ I=\mathfrak p_1^{e_1}\cdots \mathfrak p_r^{e_r} \] into nonzero prime ideals. By 2, write \[ \mathfrak p_i=(\pi_i). \] Then \[ I=(\pi_1)^{e_1}\cdots(\pi_r)^{e_r} =\bigl(\pi_1^{e_1}\cdots\pi_r^{e_r}\bigr), \] so \(I\) is principal. The zero ideal is \((0)\), hence is principal as well.
4. Therefore a Dedekind domain is a PID if and only if it is a UFD.
1. For the reverse implication, 3 shows that every ideal of \(R\) is principal, so \(R\) is a PID.