Section 4.2 Prime and maximal ideals in \(\Z\)
Proof.
Suppose that \(\mfp\) is a nonzero prime ideal. By Example 1.6.6, there exists \(n\in\N\) such that \(\mfp=n\Z\text{.}\) Let \(k\in\N\) be such that \(k\mid n\text{,}\) i.e., there exists \(\ell\in\Z\) such that \(n=k\ell\text{.}\) Since \(\mfp=n\Z\) is a prime ideal, and \(n=k\ell\in\mfp\text{,}\) either \(k\in\mfp\) or \(\ell\in\mfp\) (see Lemma 4.1.1). Thus either \(n\mid k\) or \(n\mid\ell\text{.}\) If \(n\mid k\) then we get \(k=n\) because, by our assumption, \(k\mid n\text{.}\) If \(n\mid\ell\text{,}\) say \(\ell=ns\) for some \(s\in\Z\text{,}\) then \(n=kns\text{,}\) i.e., \(n(1-ks)=0\in\Z\text{.}\) As \(\Z\) is an integral domain, \(ks=1\) and hence \(k=1\text{.}\) Thus a divisor of \(n\) is either \(n\) or \(1\text{,}\) i.e., \(n\) is a prime number.
Conversely let \(p\) be a prime number and consider the ideal generated by \(p\text{,}\) \(p\Z\text{.}\) We show that \(p\Z\) is a prime ideal. Let \(a,b\in\Z\) be such that \(ab\in p\Z\text{,}\) i.e., \(p\mid ab\) (see Lemma 4.1.1). As \(p\) is a prime number either \(p\mid a\) or \(p\mid b\text{.}\) By Lemma 4.1.1 either \(a\in p\Z\) or \(b\in p\Z\text{.}\)
Remark 4.2.2.
In view of the above result it is clear that there are infinitely many prime ideals in \(\Z\text{.}\)As a consequence of Theorem 4.2.1 we obtain the following result.
Theorem 4.2.3. (A nonzero prime ideal if and only if maximal).
A nonzero ideal \(\mfp\) in \(\Z\) is a prime ideal if and only if \(\mfp\) is a maximal ideal.Proof.
Suppose that \(\mfp\) is a nonzero prime ideal. By Theorem 4.2.1, \(\mfp\) is generated by a prime number. By Proposition 4.1.2, \(\Z/\mfp\) is a field. If \(\mfa\) is an ideal containing \(\mfp\) then \(\mfa+\mfp\) is an ideal of \(\Z/\mfp\) (see Theorem 3.2.2). By Example 1.6.7 only ideals of fields are the trivial ideal and the whole field we must have either \(\mfa=\mfp\) or \(\mfa=\Z\text{.}\) Hence \(\mfp\) is a maximal ideal.
Suppose that \(\mfm\) is a maximal ideal, say generated by \(a\text{,}\) i.e., \(\mfm=a\Z\text{.}\) Let \(b\in\N\) be a divisor of \(a\text{,}\) i.e., \(b\mid a\text{.}\) By Lemma 4.1.1, \(\mfm=a\Z\subseteq b\Z\text{.}\) As \(\mfm\) is maximal either \(\mfm=b\Z\) or \(b\Z=\Z\text{,}\) i.e., either \(a=b\) or \(b=1\) (see Lemma 4.1.1). Hence \(a\) must be a prime number. By Theorem 4.2.1, \(\mfm\) is a prime ideal.
