# Norms and Traces

Before discussing norms and traces we introduce some notation for field extensions. If are number fields, we let denote the dimension of  viewed as a -vector space. If  is a number field and , let be the number field generated by , which is the smallest number field that contains . If then  has a minimal polynomial , and the of  are the roots of . For example the element has minimal polynomial and the Galois conjugates of are .

Suppose is an inclusion of number fields and let . Then left multiplication by  defines a -linear transformation . (The transformation is -linear because is commutative.)

Definition 5.2.1 (Norm and Trace)   The and of  from  to  are

and

It is standard from linear algebra that determinants are multiplicative and traces are additive, so for we have

and

Note that if is the characteristic polynomial of , then the constant term of is , and the coefficient of is .

Proposition 5.2.2   Let and let , where , be the distinct field embeddings that fix every element of . Then

and

Proof. We prove the proposition by computing the characteristic polynomial  of . Let be the minimal polynomial of  over , and note that  has distinct roots (since it is the polynomial in of least degree that is satisfied by ). Since  is irreducible, , and  satisfies a polynomial if and only if  does, the characteristic polynomial of acting on is . Let be a basis for over and note that is a basis for , where . Then is a basis for over , and left multiplication by acts the same way on the span of as on the span of , for any pair . Thus the matrix of on is a block direct sum of copies of the matrix of acting on , so the characteristic polynomial of on  is . The proposition follows because the roots of are exactly the images , with multiplicity (since each embedding of into extends in exactly ways to by Exercise 9).

The following corollary asserts that the norm and trace behave well in towers.

Corollary 5.2.3   Suppose is a tower of number fields, and let . Then

and

Proof. For the first equation, both sides are the product of , where runs through the embeddings of  into . To see this, suppose fixes . If is an extension of to , and are the embeddings of into that fix , then are exactly the extensions of to . For the second statement, both sides are the sum of the .

The norm and trace down to  of an algebraic integer  is an element of  , because the minimal polynomial of  has integer coefficients, and the characteristic polynomial of  is a power of the minimal polynomial, as we saw in the proof of Proposition 5.2.2.

Proposition 5.2.4   Let  be a number field. The ring of integers is a lattice in , i.e., and is an abelian group of rank .

Proof. We saw in Lemma 5.1.8 that . Thus there exists a basis for , where each  is in . Suppose that as varies over all elements of the denominators of the coefficients are arbitrarily large. Then subtracting off integer multiples of the , we see that as varies over elements of with between 0 and , the denominators of the are also arbitrarily large. This implies that there are infinitely many elements of in the bounded subset

Thus for any , there are elements such that the coefficients of are all less than (otherwise the elements of would all be a distance'' of least from each other, so only finitely many of them would fit in ).

As mentioned above, the norms of elements of are integers. Since the norm of an element is the determinant of left multiplication by that element, the norm is a homogenous polynomial of degree in the indeterminate coefficients . If the get arbitrarily small for elements of , then the values of the norm polynomial get arbitrarily small, which would imply that there are elements of with positive norm too small to be in , a contradiction. So the set contains only finitely many elements of . Thus the denominators of the are bounded, so for some , we have that has finite index in . Since is isomorphic to , it follows from the structure theorem for finitely generated abelian groups that is isomorphic as a -module to , as claimed.

Corollary 5.2.5   The ring of integers of a number field is Noetherian.

Proof. By Proposition 5.2.4, the ring is finitely generated as a module over , so it is certainly finitely generated as a ring over . By the Hilbert Basis Theorem, is Noetherian.

William Stein 2004-05-06