# Preliminary Remarks

Let  be a number field of degree . Then there are  embeddings

Let be the product map . Let be the -span of inside .

Proposition 10.1.1   The -vector space  spanned by the image has dimension .

Proof. We prove this by showing that the image is discrete. If were not discrete it would contain elements all of whose coordinates are simultaneously arbitrarily small. The norm of an element is the product of the entries of , so the norms of nonzero elements of would go to 0. This is a contradiction, since the norms of elements of are integers.

The fact that is discrete in implies that has dimension equal to the rank  of , as claimed. This last assertion is not obvious, and requires observing that if  if a free abelian group that is discrete in a real vector space  and , then the rank of equals the dimension of . Here's why this is true. If are a basis for , then has finite index in , since otherwise there would be infinitely many elements of in a fundamental domain for , which would contradict discreteness of . Thus the rank of  is , as claimed.

Since is a lattice in , the volume of is finite. Suppose is a basis for . Then if  is the matrix whose th row is , then is the volume of . (Take this determinant as the definition of the volume--we won't be using volume'' here except in a formal motivating way.)

Example 10.1.2   Let be the ring of integers of . Then , is a basis for . The map is given by

The image is spanned by and . The volume determinant is

Let be the ring of integers of . The map is

and

which has determinant , so the volume of the ring of integers is .

As the above example illustrates, the volume of the ring of integers is not a great invariant of . For example, it need not be an integer. If we consider instead, we obtain a number that is a well-defined integer which can be either positive or negative. In the next section we will do just this.

William Stein 2004-05-06