Let \[ I=(f_1,f_2,\ldots)\subset\mathcal O(\mathbb C), \qquad f_n(z)=\frac{\sin(z/n)}{z}. \]
Prove that \(I\ne\mathcal O(\mathbb C)\).
Prove that \(I\) is not contained in any proper principal ideal.
Conclude that \(\mathcal O(\mathbb C)\) is not a PID and that it has maximal ideals which are not of the form \((z-a)\).
Solution
Each \(f_n\) extends holomorphically across \(0\), with \(f_n(0)=1/n\), and its nonzero zeros are \[ Z(f_n)=\{\pi nk:k\in\mathbb Z\setminus\{0\}\}. \]
1. Any finite collection \(f_{n_1},\ldots,f_{n_m}\) has a common nonzerozero.
Let \[ L=\operatorname{lcm}(n_1,\ldots,n_m). \] Then \(L/n_j\) is a positive integer for every \(j\), so \[ f_{n_j}(\pi L) =\frac{\sin(\pi L/n_j)}{\pi L} =0. \] Thus \(\pi L\) is a common zero.
2. The ideal \(I\) is proper.
Every element of \(I\) is a finite sum \[ h_1f_{n_1}+\cdots+h_mf_{n_m}, \qquad h_j\in\mathcal O(\mathbb C). \] By 1, this function vanishes at \(\pi L\) for \(L=\operatorname{lcm}(n_1,\ldots,n_m)\). Hence no element of \(I\) can be the constant function \(1\). Therefore \[ I\neq\mathcal O(\mathbb C). \]
3. The family \((f_n)_{n\ge1}\) has no common zero.
The point \(0\) is not a zero of any \(f_n\), since \(f_n(0)=1/n\). If \(z\neq0\) were a common zero, then for every \(n\) one would have \[ z=\pi nk_n \] for some nonzero integer \(k_n\). In particular \(z/\pi\) would be an integer divisible by every positive integer \(n\), which is impossible for a fixed nonzero integer. Thus there is no common zero.
4. The ideal \(I\) is not contained in any proper principal ideal.
Suppose \[ I\subset(g) \] for some entire function \(g\). Then \(g\) divides every \(f_n\). Consequently every zero of \(g\) would be a zero of every \(f_n\). By 3, \(g\) has no zeros. Therefore \(1/g\) is entire, so \(g\) is a unit and \[ (g)=\mathcal O(\mathbb C). \] Thus no proper principal ideal can contain \(I\).
5. The ring \(\mathcal O(\mathbb C)\) is not a PID.
4. Hence thering has a nonprincipal ideal and is not a PID.
6. There is a maximal ideal containing \(I\) which is not of the form\((z-a)\).
3. Hence this maximal ideal is not an evaluationideal \((z-a)\).