Let and let be the ring of integers in . Let and . Exhibit bases for and , compute and , and compute the discriminant of and .
We claim that is a basis for over . It is clear that , and if , then evidently . So . Now if and , then , so that . Hence is a basis for over . The discriminant of is then .
We now claim that is proper. To see this, suppose to the contrary that ; then for some integers and ; comparing coefficients, we have and . But then , which has no integer solutions; thus is proper.
Now let be arbitrary in and let mod 3, where . Now mod , so that . Since is proper, and mod . If , then we have , a contradiction; so mod . Thus .
We claim now that is a basis for over . Certainly , and if , then . So . Moreover, if , then . So is a basis for over . The discriminant of is then .
We claim that is proper. Indeed, if , then for some integer , a contradiction.
Now let be arbitrary in . Say and mod 2, where . Then mod . Thus . Since is proper, we have and mod . If , then we have for some integer , a contradiction. Thus and mod . Likewise, if , we have a contradiction, so that and mod .
We claim that . To see this, define by . Certainly is surjective. Now , so that . Now if , then . Thus mod 2. By the first isomorphism theorem for rings, .