(February 1982) Prove that if an ideal \(I\) of the ring \(\mathbb{Z}\) contains two relatively prime integers, then \(I=\mathbb{Z}\).
Solution
Proof. By hypothesis, there exist a,b∈I with gcd(a,b ) = 1. Thus, there are integers m,n∈ Z such thatma +nb = 1. But ma,nb∈I by the sticky property, so 1 = ma +nb∈I sinceI is closed under +. For any x∈ Z, then, we have x =x· 1∈I by the sticky property. So I = Z QED