Geometrically, one pictures as a discretization of , with evenly spaced integers on a number line punctuating the dense rationals. More generally, forms a *lattice* inside , obtained by taking -linear combinations of the standard basis, rather than -linear combinations. We should like to view similarly in the context of . The following theorem validates this intuition.

**Theorem 2.1:** * is a finitely generated abelian group under addition (i.e., a -module), with rank .*

The approach one takes to prove this result is to embed in some larger object, and then to exploit the properties of this larger object. This is done abstractly, by embedding in a larger -module and invoking the structure theorem for finitely generated abelian groups.

**1. Field norm and trace**

We will need to develop some properties of the field norm and field trace, which are useful linear algebraic invariants that one may associate with any field extension . For any element , there is a natural linear transformation given by . With respect to the -basis of , this transformation has a matrix whose entries are all in , yielding the maps

By definition, these maps inherit the multiplicative and additive properties of determinant and trace, respectively.

As it turns out, given , the elements , may be expressed in terms of the roots of the *minimal polynomial* of over . Recall this minimal polynomial exists since the extension is finite, hence algebraic. We denote the minimal polynomial of as .

To proceed, define the *characteristic polynomial* of as . A simple manipulation shows that for any , one has . Since the map is injective, the Cayley-Hamilton theorem then implies is a root of its characteristic polynomial.

Hence, divides . In general, they need not be equal, as the degree of is whereas the degree of is , which do not have to be same. However, if , then it follows that . This fact may be generalized.

**Proposition 2.2:** *Let . Then .*

*Proof:* A basis for is . Let be a basis for , where . Then an ordered basis for is .

Since has degree , the characteristic polynomial . With respect to the chosen basis for , the matrix is a diagonal block matrix comprised of copies of the matrix . Thus,

Let , , , be the roots of . These are called the *algebraic conjugates* of , and the proposition says that the eigenvalues of are exactly the algebraic conjugates of , each counted with multiplicity . Therefore,

Returning to the case of a number field, suppose . Then is monic with integer coefficients, so the norm and trace of are integers.

**Corollary 2.3:** *The restrictions and have range .*

**2. A nondegenerate symmetric bilinear form**

To embed in a larger -module is to embed in for some , since has no nontrivial elements of finite order. The goal, then, is to produce an injective homomorphism . The trace function restricted to is an additive homomorphism into , so one might naturally think of exploiting its properties. Traditionally, this is done via the *trace pairing*.

We must review a little linear algebra before describing the trace pairing. Recall that a *bilinear form* on a vector space over field is a map such that and are both linear, for any choice of . The bilinear form is *symmetric* if .

If one fixes bases and of , each bilinear form has an associated matrix . The entries of this matrix are , and if is symmetric, then is a symmetric matrix. Assuming is finite dimensional, of dimension , then one may identify via the isomorphisms , which give the coordinates of a vector in bases and respectively. Then one can concretely write

The associated matrix therefore uniquely determines .

The form is called *nondegenerate* if is invertible. As with most nice things in linear algebra, nondegeneracy does not depend on the choice of bases and (exercise!). A coordinate-free formulation of the condition is that there is no nonzero such that is the zero map. In particular, is a nondegenerate form iff

is injective.

Now, we can look at the trace pairing, which is the symmetric bilinear form given by

Observe that if is nonzero, then

Hence, the trace pairing is nondegenerate.

We can now give a proof that is a finitely generated -module.

*Proof 1 (Injecting into a module): *In the previous blog post, we found that for every , there is an integer such that . Accordingly, if is a basis for as a -vector space, there are integers such that is also a basis for , but . Setting , one may define the homomorphism

which is injective, thanks to nondegeneracy of the trace pairing. Restricting this homomorphism to will inject into .

By the structure theorem on finitely generated abelian groups, is a free abelian group of rank at most . But contains , a linearly independent set of size , so the rank of is at least . Hence, we conclude is a free abelian group of rank .

This proves the existence of some linearly independent set whose span in -linear combinations is . One calls such a set an *integral basis* for .

**3. Discriminant of a number field**

Let us define the *discriminant* on . This is the map given by

In particular, if is a basis for , the discriminant of is the determinant of the trace pairing matrix . This determinant is nonzero since the trace pairing form is nondegenerate. On the other hand, one can show that if is linearly dependent, then (exercise!).

Suppose that and are bases for , with change of basis matrix taking coordinates in to coordinates in . Then, by considering the bilinear form , we have the matrix equality

which, upon taking determinants, gives the identity

If and are both actually integral bases of , then the change of basis matrix is an invertible integer matrix, hence it has determinant with absolute value . In this case, . The common value is called the *discriminant of *.

Using properties of the discriminant, we give another proof that has an integral basis.

*Proof 2 (Minimal discriminant): *As in the previous proof, let us begin with a basis that spans . Since the restriction to of trace has range , the discriminant is an integer. Hence, we may choose a basis of such that is minimal.

For some , write . If , then it can be decomposed where is the greatest integer less than , and is the fractional part.

Then, setting , and for , we have another basis for . One easily computes the determinant of the change of basis matrix to be , so that the above identity on discriminants gives that

contradicting the choice of . Thus, each , and the set forms an integral basis for .

This argument yields a useful corollary. To simplify its statement, let us introduce the following definition: if is a basis for as a -vector space, we call an *almost-integral basis*.

**Corollary 2.4:** *Let be an almost-integral basis. Then, the following are equivalent:*

- for all almost-integral bases
- is an integral basis.