What is a Noetherian ring? If \(I\) is an ideal in a Noetherian ring with a unit, what is the intersection of \(I^n\) over all positive integers \(n\)?
Solution
A commutative ring \(R\) is Noetherian if every ideal is finitely generated; equivalently, every ascending chain of ideals stabilizes.
Let \(I\trianglelefteq R\) and set \[ J=\bigcap_{n\ge1} I^n. \] Since \(R\) is Noetherian, the ideal \(J\) is finitely generated. By Artin–Rees applied to the inclusion \(J\subseteq R\), there is \(k\) such that \[ I^n\cap J=I^{\,n-k}(I^k\cap J)\qquad(n\ge k). \] Because \(J\subseteq I^n\) for every \(n\), taking \(n=k+1\) gives \[ J=IJ. \] The determinant trick applied to the finitely generated module \(J\) now yields some \(x\in I\) such that \[ (1-x)J=0. \] Thus every \(a\in J\) satisfies \((1-x)a=0\) for some \(x\in I\).
Conversely, if \((1-x)a=0\) for some \(x\in I\), then \(a=xa=x^2a=\cdots=x^na\) for every \(n\), so \(a\in I^n\) for every \(n\). Hence \[ \bigcap_{n\ge1}I^n = \{a\in R:(1-x)a=0\text{ for some }x\in I\}. \]
In particular, if \(I\) is contained in the Jacobson radical (for example, if \((R,\mathfrak m)\) is local and \(I\subseteq\mathfrak m\)), then \(1-x\) is a unit for every \(x\in I\), so the intersection is \(0\). If \(R\) is a domain and \(I\ne R\), then \(1-x\ne0\) and \((1-x)a=0\) again forces \(a=0\), so the intersection is also \(0\).