Compute the class number of .
Recall that the ring of integers in this field is , and that is an integral basis. We will now give an upper bound on the constant which is proven to exist in Theorem 10.2.
Let be an ideal in , and let . That is, is the largest integer such that . Now consider . Since , there exist some in this set such that . Say , where (clearly) . By the triangle inequality, . By our proof of Theorem 10.3, every class of ideals in contains an ideal whose norm is at most 7. We will now construct all of the ideals having norm where .
There is only one ideal of norm 1, namely .
If is an ideal of norm 2, then divides . Note that . Now , so that . By Corollary 9.15, is prime. Thus is the only ideal of norm 2.
If is an ideal of norm 3, then divides . Note that . Now , so that . By Corollary 9.15, is prime. Thus is the only ideal of norm 3.
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 such ideal is . If is prime, then by Theorem 9.19, divides . But no prime ideals dividing have norm 4. So the only ideal of norm 4 is .
If is an ideal of norm 5, then divides . Note that , since . We claim that both of these factors are proper. Indeed, if , then for some . Comparing coefficients mod 5, we have , a contradiction. Thus is proper. Similarly, is proper. Now , and neither factor is 1. So . Thus and are the only possible ideals of norm 5 in . Note that the discriminant of this field is not divisible by 5, so that these factors are distinct by Theorem 9.6. (Though this last step is not necessary.)
If is an ideal of norm 6, then divides . Thus the prime factors of come from the set . The only such ideal is .
If is an ideal of norm 7, then divides . Note that . Moreover, by Corollary 9.11, , and by Corollary 9.15, these ideals are prime. So the only ideals of norm 7 are and .
Certainly we have .
Note that and that . Thus .
We also have and , since . So , and thus .
We also that since . So , and thus . So .
Finally, we claim that is not principal. If , then divides both 4 and 6, and so is either 1 or 2. But no element of has norm 2, and is proper; so is not principal, and .
Thus has two ideal classes.