Compute the class number of .
Recall that is the ring of integers in , and that is an integral basis for .
We will now compute an upper bound on the constant which is shown to exist in Theorem 10.2. That is, we will find a number such that, for every ideal , there exists an element such that .
To this end, let – that is, is the largest integer such that . Consider the set of integers . Since by our choice of , there must exist two elements in this set which are congruent mod . Say .
Using the triangle inequality, we have . Thus the constant is at most 4.
By our proof of Theorem 10.3, every class of ideals in contains an ideal of norm at most 4. We will now compute all of the ideals of norm for .
There is only one ideal of norm 1; namely .
Suppose is an ideal of norm 2. By Theorem 9.19, is a divisor of . We claim that ; indeed, . Now , so that . By Theorem 9.15, the ideal is prime, and so is the prime factorization of . So is the only ideal of norm 2 in .
Suppose is an ideal of norm 3. By Theorem 9.19, is a divisor of . Note that , and that . Thus is the prime factorization of , and is the only ideal of norm 3 in .
Suppose is an ideal of norm 4. If is not prime, then it is a product of two prime ideals of norm 2. The only prime ideal of norm 2 is , so that . If is prime, then by Theorem 9.19, . Since has norm 4, in fact , which in this case is a contradiction. So is the only ideal of norm 4.
It is clear that . Moreover, note that , so that .
Thus the class number of is 1.