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 R_g 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

$$\langle$$A, B$$\rangle$$ = 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 + ... + $\mathbb {Z}$ ^. w_n; moreover, K = $\mathbb {Q}$ ^. w₁ + ... + $\mathbb {Q}$ ^. w_n. 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 . $$\langle$$w1, w1$$\rangle$$	+	...	+	rn . $$\langle$$wn, w1$$\rangle$$	=	p1
$$\vdots$$				$$\vdots$$		$$\vdots$$
r1 . $$\langle$$w1, wn$$\rangle$$	+	...	+	rn . $$\langle$$wn, wn$$\rangle$$	=	pn

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

Now suppose R $\subset$ S, where S is another order (i. e. an R_g for some g). We clearly have the sequence of inclusions R $\subset$ S $\subset$ $\check{S}$ $\subset$ $\check{R}$. It follows that D_S divides D_R; 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 D_R is the multiple of D_S 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 + ... + $\mathbb {Z}$ ^. u_n. Let v be any non-zero column vector and consider the basis u_i ^. 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 + a₁T^{n - 1} + ... + a_n where the a_i 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 R_P 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 D_P. Unlike the case of fields, however, it is not true that every order has the form R_P for some polynomial P(T).