next up previous
Next: 6.4 Lattices and ideals Up: 6 Algebraic Number Fields Previous: 6.2 Algebraic Number Fields

6.3 Orders and Maximal orders

We now look a the subring R of K consisting of matrices with integer entries; R is called an order in K. Now, for any invertible n×n matrix g it is clear that gKg-1 is isomorphic to K; but g or g-1 may have entries with denominators. So the ring Rg consisting of matrices in gKg-1 with integer entries need not be that same as R. Thus, one can look for a maximal order. We will see below that one such exists and is unique. It is usually called the ring of integers in K and is denoted by $ \mathcal {O}$K.

There is a natural symmetric pairing on n×n matrices given by

$\displaystyle \langle$A, B$\displaystyle \rangle$ = Trace(A . B)

We study the restriction of this to K. This pairing is non-degenerate; i. e. for any non-zero A there is a B so that < A, B > $ \neq$ 0. For example, if $ \alpha$ in K, then $ \langle$$ \alpha$,$ \alpha^{-1}_{}$$ \rangle$ = Trace(1) = n which is non-zero! (Clearly, a different argument is required when the base field is not $ \mathbb {Q}$ but a finite field). From the non-degeneracy it also follows that for any additive map from K to the rationals $ \mathbb {Q}$ there is an $ \alpha$ in K so that the map is precisely $ \beta$ $ \mapsto$ $ \langle$$ \alpha$,$ \beta$$ \rangle$.

Now, R is a subgroup of the finitely-generated free abelian group of n×n matrices with integer coefficients; thus R is a finitely-generated free abelian group as well. If $ \alpha$ is any element of K we can clear denominators to find an integer d so that d$ \alpha$ is a matrix with integer entries. It follows that R contains a basis of K as a vector space over $ \mathbb {Q}$. Thus R is of the form $ \mathbb {Z}$ . w1 + ... + $ \mathbb {Z}$ . wn; moreover, K = $ \mathbb {Q}$ . w1 + ... + $ \mathbb {Q}$ . wn. Let $ \check{R}$ denote the collection of all elements $ \alpha$ in K so that < $ \alpha$,$ \beta$ > is an integer for all $ \beta$ in R. Finding such an $ \alpha$ is clearly equivalent to solving the system of equations

r1 . $\displaystyle \langle$w1, w1$\displaystyle \rangle$ + ... + rn . $\displaystyle \langle$wn, w1$\displaystyle \rangle$ = p1
$\displaystyle \vdots$       $\displaystyle \vdots$   $\displaystyle \vdots$
r1 . $\displaystyle \langle$w1, wn$\displaystyle \rangle$ + ... + rn . $\displaystyle \langle$wn, wn$\displaystyle \rangle$ = pn

By Cramer's rule, this requires the inversion of the matrix ($ \langle$wi, wj$ \rangle$)i = 1, j = 1n, n. The determinant of this matrix is called the discriminant of the order R and is denoted by DR. Note that if v1, ..., vn is another basis for R and A is the matrix that gives the ``change of co-ordinates'', then the determinant of ($ \langle$vi, vj$ \rangle$)i = 1, j = 1n, n differs from the earlier determinant by det(A)2. Since A and A-1 have integer entries, det(A) = ±1. Hence DR is independent of the choice of basis. Clearly, R $ \subset$ $ \check{R}$ and $ \check{R}$/R is a finite group with | DR| elements.

Now suppose R $ \subset$ S, where S is another order (i. e. an Rg for some g). We clearly have the sequence of inclusions R $ \subset$ S $ \subset$ $ \check{S}$ $ \subset$ $ \check{R}$. It follows that DS divides DR; by decreasing induction we see that there is a maximal order. We also note that by duality, S/R and $ \check{R}$/$ \check{S}$ have the same order, so that DR is the multiple of DS by the square of an integer. Let $ \mathcal {O}$K be the collection of all elements of K whose characteristic polynomials have integer coefficients; one can show that this is closed under addition and multiplication. It is clear that $ \mathcal {O}$K contains R since very matrix with integer entries has a characteristic polynomial with integer coefficients. By the above, we see that $ \mathcal {O}$K is contained in $ \check{R}$, hence it is finitely generated; let $ \mathcal {O}$k = $ \mathbb {Z}$ . u1 + ... + $ \mathbb {Z}$ . un. Let v be any non-zero column vector and consider the basis ui . v of the space of column vectors. With this change of basis, each each element of $ \mathcal {O}$K is represented by a matrix with integer entries. Thus $ \mathcal {O}$K is an order and the unique maximal order.

An extension of the example we looked at for fields is to associate an order with an irreducible polynomial P(T) = Tn + a1Tn - 1 + ... + an where the ai are all integers. We continue the notation of the previous subsection. It follows that $ \alpha_{P}^{}$ is a matrix with integer coefficients; with a little effort one can also show that the natural order RP in $ \mathbb {Q}$($ \alpha_{P}^{}$) is precisely the collection of all integer linear combinations of the powers 1, $ \alpha_{P}^{}$, ..., $ \alpha_{P}^{n-1}$. The discriminant of this order is also the discriminant of the polynomial P(T) and is denoted as DP. Unlike the case of fields, however, it is not true that every order has the form RP for some polynomial P(T).

next up previous
Next: 6.4 Lattices and ideals Up: 6 Algebraic Number Fields Previous: 6.2 Algebraic Number Fields
Kapil Hari Paranjape 2002-10-20