Show that if \(J\in \Id(R)\) (with \(J\contains I\)) is radical/prime/maximal iff \(\bar J \in \Id(R/I)\) is radical/prime/maximal.
Solution
Let \(I\subseteq J\) be ideals of the commutative ring \(R\), and write \[ \overline J=J/I\subseteq R/I. \]
1. The quotient correspondence gives\[ (R/I)/(J/I)\cong R/J. \]
Define \[ \phi:R/I\longrightarrow R/J, \qquad r+I\longmapsto r+J. \] Because \(I\subseteq J\), this map is well-defined and surjective. Its kernel is exactly \(J/I\). The first isomorphism theorem therefore gives the displayed isomorphism.
2. The ideal \(J\) is prime in \(R\) if and only if \(J/I\) is prime in \(R/I\).
An ideal is prime exactly when its quotient ring is an integral domain. By 1, \[ R/J\cong (R/I)/(J/I). \] Hence one quotient is a domain exactly when the other is.
3. The ideal \(J\) is maximal in \(R\) if and only if \(J/I\) is maximal in \(R/I\).
An ideal is maximal exactly when its quotient ring is a field. Again 1 identifies the two quotient rings, so the two maximality conditions are equivalent.
4. The ideal \(J\) is radical in \(R\) if and only if \(J/I\) is radical in \(R/I\).
Suppose first that \(J\) is radical. If \[ (r+I)^n\in J/I, \] then \(r^n+I\in J/I\), which means \(r^n\in J\). Radicality of \(J\) gives \(r\in J\), hence \(r+I\in J/I\). Thus \(J/I\) is radical.
Conversely, suppose \(J/I\) is radical and \(r^n\in J\). Then \[ (r+I)^n=r^n+I\in J/I. \] Hence \(r+I\in J/I\), so \(r\in J\). Therefore \(J\) is radical.
5. Thus radical, prime, and maximal ideals containing \(I\) correspond to radical, prime, and maximal ideals of \(R/I\) under \(J\mapsto J/I\).
Combine <1>2–<1>4 with the ordinary ideal correspondence for the quotient map \(R\to R/I\).