What are the prime ideals and maximal ideals of \(\ZZ[x]\)?
Solution
The prime ideals of \(\ZZ[x]\) are exactly:
- \((0)\);
- \((p)\) for a rational prime \(p\);
- \((f)\) for a primitive irreducible nonconstant polynomial \(f\in\ZZ[x]\);
- \((p,f)\) where \(p\) is prime and the reduction \(\bar f\in\FF_p[x]\) is irreducible.
Among these, the maximal ideals are exactly the ideals \((p,f)\) in the last family.
1. Let \(P\subseteq\ZZ[x]\) be prime. Then\[ P\cap\ZZ=(0) \quad\text{or}\quad P\cap\ZZ=(p) \]for some prime integer \(p\).
The contraction of a prime ideal along the inclusion \(\ZZ\hookrightarrow\ZZ[x]\) is prime in \(\ZZ\). The prime ideals of \(\ZZ\) are \((0)\) and \((p)\).
2. If \(P\cap\ZZ=(p)\), then either \(P=(p)\) or\[ P=(p,f) \]with \(\bar f\) irreducible in \(\FF_p[x]\).
Prime ideals of \(\ZZ[x]\) containing \((p)\) correspond to prime ideals of \[ \ZZ[x]/(p)\cong\FF_p[x]. \] Since \(\FF_p[x]\) is a PID, its prime ideals are \((0)\) and \((\bar f)\) for irreducible \(\bar f\). Pulling them back gives the stated ideals.
3. If \(P\cap\ZZ=(0)\) and \(P\ne(0)\), then\[ P=(f) \]for a primitive irreducible nonconstant \(f\in\ZZ[x]\).
Localize at \(S=\ZZ\setminus\{0\}\). Since \(P\cap S=\varnothing\), the localization \(S^{-1}P\) is a nonzero prime ideal of \[ S^{-1}\ZZ[x]=\QQ[x]. \] Hence \[ S^{-1}P=(g) \] for an irreducible \(g\in\QQ[x]\). Multiplying by a rational scalar, choose a primitive \(f\in\ZZ[x]\) associated to \(g\). By Gauss’s lemma, \(f\) is irreducible in \(\ZZ[x]\), and the contraction of \((g)\) back to \(\ZZ[x]\) is \((f)\). Since prime ideals disjoint from \(S\) are recovered by contracting their localizations, \(P=(f)\).
4. Every listed ideal is prime.
The quotient by \((0)\) is the domain \(\ZZ[x]\). The quotient by \((p)\) is the domain \(\FF_p[x]\). Since \(\ZZ[x]\) is a UFD, a primitive irreducible \(f\) is prime, so \((f)\) is prime. Finally, \[ \ZZ[x]/(p,f)\cong\FF_p[x]/(\bar f), \] which is a field when \(\bar f\) is irreducible.
5. The maximal ideals are exactly the \((p,f)\) with \(\bar f\) irreducible modulo \(p\).
The quotient in 4 shows each \((p,f)\) is maximal.
The ideal \((0)\) is not maximal. The ideal \((p)\) is not maximal because its quotient \(\FF_p[x]\) is not a field. Finally, let \((f)\) be of the third type. Choose a prime \(\ell\) for which the reduction of \(f\) modulo \(\ell\) is still nonconstant; such a prime exists because some nonconstant coefficient of \(f\) is nonzero. If \(\ell\) were invertible in \(\ZZ[x]/(f)\), there would be \(a,b\in\ZZ[x]\) with \[ \ell a+fb=1. \] Reducing modulo \(\ell\) would make the nonconstant polynomial \(\bar f\) a unit in \(\FF_\ell[x]\), impossible. Hence \(\ZZ[x]/(f)\) is not a field, so \((f)\) is not maximal.