Let be an ideal in the ring of integers of . Find an algebraic integer such that , where .

We saw in the text that the class number of is 2. Note that , as indeed . Let . By our proof of Theorem 10.6, .

unnecessary lemmas. very sloppy. handwriting needs improvement.

Let be an ideal in the ring of integers of . Find an algebraic integer such that , where .

We saw in the text that the class number of is 2. Note that , as indeed . Let . By our proof of Theorem 10.6, .

Let and be algebraic number fields with , and having integer rings and , respectively. Suppose is an ideal such that is principal. Show that is principal in .

We will let and denote the ideal class groups of and , respectively. For brevity, if is a subset, then we will denote by the ideal generated by in and by the ideal generated by in .

We claim that if , then . To see this, suppose we have and such that . Then . Now if , then each is in , and so has the form . So . Conversely, , and so .

The mapping then induces a well-defined mapping given by . Moreover, since , is a group homomorphism.

In particular, if in , then in , as desired.

Let be the ring of integers in an algebraic number field of class number . Let be an ideal. Show that if is minimal such that is principal, then .

This is precisely the order of in the class group of . The result then follows by Lagrange’s Theorem.

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.

Determine which of the following ideals of are equivalent: , , , and .

First we argue that is principal. Indeed, . Now , so that . Indeed, and , and .

In particular, .

Now we claim that and are both principal. Indeed, . Since and , we have . So . Similarly, , so that . Now , so that .

Finally, we claim that is not principal. If , then and , so that divides both 4 and 6. So . Note, however, that no element of has norm 2 since the equation has no solutions in . If , then we have . But then , so that , which is impossible mod 2. So is not principal.

In summary, we have , , and .

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.

Let be the set of integers in an algebraic number field , and let denote the set of equivalence classes of ideals in under the relation if and only if for some nonzero . Prove that is an abelian group under the usual ideal product of class representatives.

We will denote by the equivalence class containing the ideal . We showed in this previous exercise that is a well-defined binary operator on .

For all , we have , so that our product is associative.

For all , we have , and similarly .

For all , there exists (by Theorem 8.13) an ideal such that is principal. Thus , and so every element of has an inverse.

Hence is a group.

Moreover, we have , so that is abelian.

Let be the ring of integers in an algebraic number field . Recall that we defined a relation on the set of ideals in as follows: if and only if there exist nonzero such that . Show that if and , then .

We have such that and . Then , so that .

Characterize the irreducible elements in .

Suppose is irreducible. By our proof of Theorem 8.5, it follows that for some unique positive rational prime . That is, for some . Now , so that either or . (Recall that is a rational integer and not 1.) Let .

Suppose . Then , so that is a rational prime.

Suppose . Now , so that (by Theorem 7.3) is a unit. (In Theorem 7.9 we saw that the units in have the form where .) Thus is an associate of in . We claim that in this case, there does not exist a solution of the equation . If so, then . But since is irreducible, either or is a unit- so , a contradiction.

In summary, if is irreducible in , then either (1) is a rational prime or (2) is the square of a rational prime and the equations have no solutions in .

Conversely, if is prime, then is irreducible. If where is a rational prime and if has no solution, then no element of has norm , so that is irreducible.

Can there exist an algebraic integer ring with a prime ideal such that ?

Recall that if is prime, then it is maximal, and that is precisely . Since is maximal, is a field. We claim that no field of order 12 exists.

Suppose to the contrary that is a field of order 12. The characteristic of is then either 2 or 3, since it must be a prime dividing 12. Moreover, by Cauchy’s Theorem, contains nonzero elements and of (additive) order 2 and 3, respectively. That is, and , but no smaller multiple of or is 0. If has characteristic 2, then , and if has characteristic 3, then , both contradictions.

Since no field of order 12 exists, no prime ideal can have norm 12.