Give an example of a Principal Ideal Domain (PID) with a unique non-zero prime ideal. Prove that it is a PID and classify all of its ideals.
Solution
Fix a prime integer \(p\) and let \[ R=\mathbb Z_{(p)}=\left\{\frac ab\in\mathbb Q:p\nmid b\right\}. \] Every nonzero \(x\in R\) has a unique form \[ x=p^n u, \] with \(n\ge0\) and \(u\in R^\times\). Indeed, factor the numerator by its \(p\)-adic valuation; every fraction whose numerator and denominator are both prime to \(p\) is a unit.
Let \(0\ne I\trianglelefteq R\). The set \[ \{v_p(x):0\ne x\in I\}\subseteq\mathbb N \] has a minimum \(n\). Choose \(x=p^n u\in I\) with \(u\) a unit. Then \(p^n=xu^{-1}\in I\), so \((p^n)\subseteq I\). By minimality of \(n\), every nonzero \(y\in I\) has valuation at least \(n\), hence \(y\in(p^n)\). Therefore \[ I=(p^n). \] Together with the zero ideal, this gives the complete ideal list \[ (0),\quad R=(1),\quad (p),\quad(p^2),\quad(p^3),\ldots. \] Thus \(R\) is a PID.
Because \(R\) is a domain, \((0)\) is prime. The ideal \((p)\) is maximal since \[ R/(p)\cong\mathbb F_p. \] For \(n\ge2\), \((p^n)\) is not prime because \[ p\cdot p^{n-1}\in(p^n), \qquad p,p^{n-1}\notin(p^n). \] Hence \((p)\) is the unique nonzero prime ideal of \(R\).