Note that if we let denote the group of fractional ideals, then we have an exact sequence

The following two examples illustrate how to apply Theorem 7.1.2 to compute in simple cases.

sage: K = QQ[sqrt(10)]; K Number Field in sqrt10 with defining polynomial x^2 - 10 sage: B = K.minkowski_bound(); B sqrt(10) sage: B.n() 3.16227766016838Theorem 7.1.2 implies that every ideal class has a representative that is an integral ideal of norm , , or . The ideal is ramified in , so

would have an integer solution. But the squares mod are , so (7.1.1) has no solutions. Thus defines a nontrivial element of the class group, and it has order since its square is the principal ideal . Thus .

To find the integral ideals of norm , we factor modulo , and see that

In summary, we now know that every element of is equivalent to one of

or

Thus the class group is a group of order at most that contains an
element of order . Thus it must have order . We verify this in
sage: K.<sqrt10> = QQ[sqrt(10)]; K Number Field in sqrt10 with defining polynomial x^2 - 10 sage: G = K.class_group(); G Class group of order 2 with structure C2 of Number Field ... sage: G.0 Fractional ideal class (3, sqrt10 + 1) sage: G.0^2 Trivial principal fractional ideal class sage: G.0 == G( (3, 2 + sqrt10) ) True

Before proving Theorem 7.1.2, we prove a few lemmas. The strategy of the proof is to start with any nonzero ideal , and prove that there is some nonzero , with very small norm, such that is an integral ideal. Then will be small, since is small. The trick is to determine precisely how small an we can choose subject to the condition that is an integral ideal, i.e., that .

Let be a subset of
. Then is *convex* if
whenever then the line connecting and lies entirely
in . We say that is *symmetric about the origin* if
whenever then also. If is a lattice in the
real vector space
, then the *volume* of is the
volume of the compact real manifold , which is the same thing as
the absolute value of the determinant of any matrix whose rows form a
basis for .

Next assume that . Then for all there is , since . If then the are all in , which is finite since is bounded and is discrete. Hence there exists nonzero for arbitrarily small . Since is closed, .

Fix a number field with ring of integers .

Let be the real embeddings of and be half the complex embeddings of , with one representative of each pair of complex conjugate embeddings. Let be the embedding

Re Re Im Im |

Note that this is

Re Re Im Im

and whose determinant has absolute value equal to the volume
of . By doing the following three column operations,
we obtain a matrix whose rows are exactly the images of
the under - Add times each column with entries Im to the column with entries Re.
- Multiply all columns with entries Im by , thus changing the determinant by .
- Add each columns that now has entries ReIm to the the column with entries Im to obtain columns ReIm.

Let be a fixed choice of closed, bounded, convex, subset with positive volume that is symmetric with respect to the origin and has positive volume. Since is closed and bounded,

Suppose is any fractional ideal of . Our goal is to prove that there is an integral ideal with small norm. We will do this by finding an appropriate . By Lemma 7.1.8,

Notice that the right hand side is independent of . It depends only on , , , and our choice of . This completes the proof of the theorem, except for the assertion that can be chosen to give the claim at the end of the theorem, which we leave as an exercise.

A prime ramifies in if and only if , so the corollary implies that every nontrivial extension of is ramified at some prime.

William Stein 2012-09-24