Show that every prime ideal is irreducible.
Hint
If \(\mfp = J\cap K\) with \(a\in J\sm\mfp\) and \(b\in K\sm\mfp\), then \(ab\in J\cap K=\mfp\).
Solution
1. Recall that \(I\) is irreducible when \(I = J \cap K\) for ideals \(J, K\) forces \(I = J\) or \(I = K\).
2. Let \(\mfp\) be prime and suppose \(\mfp = J \cap K\) with \(\mfp \neq J\) and \(\mfp \neq K\).
3. There are \(a \in J \sm \mfp\) and \(b \in K \sm \mfp\).
\(\mfp = J \cap K \subseteq J\), so \(\mfp \neq J\) gives an \(a \in J\) outside \(\mfp\), and likewise for \(K\).
4. \(ab \in \mfp\).
4.1. \(ab \in J\), because \(a \in J\) and \(J\) is an ideal.
4.2. \(ab \in K\), because \(b \in K\) and \(K\) is an ideal.
4.3. So \(ab \in J \cap K = \mfp\).
5. Q.E.D.
3. Hence \(\mfp = J\) or \(\mfp = K\), and \(\mfp\) is irreducible.