Show that if \(R\) is Noetherian then every ideal is finitely generated.
Solution
Let \(I\) be an ideal of a Noetherian ring \(R\). Suppose \(I\) were not finitely generated. Choose \(x_1\in I\). Since \((x_1)\ne I\), choose \[ x_2\in I\setminus(x_1). \] Inductively, having chosen \(x_1,\dots,x_n\) with \[ I_n=(x_1,\dots,x_n)\subsetneq I, \] choose \[ x_{n+1}\in I\setminus I_n. \] Then \[ I_1\subsetneq I_2\subsetneq I_3\subsetneq\cdots \] is an infinite strictly ascending chain of ideals, contradicting the ACC on ideals.
Hence every ideal of \(R\) is finitely generated.