Ideal Classes and Matrix Conjugation over F q [X]

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1 Ideal lasses and Matrix onjugation oer F q [X] hristian Berghoff, Georg-August-Uniersität Göttingen Trais Morrison, Penn State Uniersity Yujia Qiu, Ruprecht-Karls-Uniersität Heidelerg Thomas Sicking, Georg-August-Uniersität Göttingen Decemer 10, Introduction In a commutatie ring R two matrices M, N M n (R) are called conjugate, when there is a matrix S GL n (R), such that M = S 1 NS. Howeer, gien two matrices M and N as aoe, it is not always easy to see whether they are conjugate or not. onjugate matrices will hae the same characteristic polynomial, ut if two matrices hae the same characteristic polynomial, they are not necessarily conjugate. So the question is: Gien a polynomial f(x) R[X], how many conjugacy classes of matrices with that characteristic polynomial do we hae and how can we find them? In the case when R = Z it is known from a paper y. Latimer and.. MacDuffee (see [2]) that the conjugacy classes of matrices in M n (Z) with a characteristic polynomial f(x) Z[X] of degree n are in one-to-one correspondence with the Z[α]-ideal classes in Q(α) with α a root of f. In the following paper we will deal with that prolem when the ring R is F q [X] =: A, the polynomial ring oer a finite field with q elements. F will denote the quotient field of A, so F = F q (X), and it will turn out that the theorem of Latimer and MacDuffee also holds if we exchange Z y A and Q y F. This is asically due to the fact that A, like Z, is a unique factorization domain. In the proof of the theorem the map giing the one-to-one correspondence will e explicitly gien, so gien a polynomial we will e ale to compute the conjugacy classes of matrices with that polynomial as their characteristic polynomial, and the other way round, gien a matrix we will find the class of fractional ideals corresponding to its conjugacy class (see section 3). The proof of the main theorem is asically adapted from Keith onrad s paper [1], which can e found on his home page. 2 Main section Theorem 1. Let K e a field and f(t ) K[T ] e monic and irreducile of degree n. Then (a) For a matrix M in M n (K), the characteristic polynomial of M is f(t ) if and only if f(m) = 0. 1

2 () All matrices in M n (K) with characteristic polynomial f(t ) are conjugate in M n (K). We can make K n into a K[T ]-module y defining g(t ) 0 = g(m) 0. Denote such a module y V M. To proe the aoe theorem, we will make use of the following lemma: Lemma 2. For M, N M n (K), V M = VN if and only if M and N are conjugate oer M n (K). Proof (of the lemma). Let M, N M n (K) e conjugate with N = S 1 MS. S induces a linear isomorphism of K n as a ector space oer K; we use this to define a map φ : V N V M y φ() = S. We check that this is indeed an isomorphism of modules. We need only show that for g(t ) K[T ], φ(g(t ) VN ) = g(t ) VM φ(). Using the fact that N = S 1 MS, we hae SN k = M k S for k N, so Sg(N) = g(m)s. Thus φ(g(t ) VN ) = φ(g(n)) = Sg(N) = g(m)s = g(m)φ() = g(t ) VM φ() and we conclude that V M = VN as modules oer K[T ]. Now, let V M and V N e two K[T ]-modules, with φ : V M V N a module isomorphism. Fix a asis B = (x 1,..., x n ) for V M oer K as a ector space, then φ(b) is a asis of V N as a ector space. Let S = [φ] e the matrix representatie of φ as a linear transformation. We claim that S 1 NS = M. Let K n. Then we hae S 1 NS() = S 1 N(φ()) = S 1 (T VN φ()) = φ 1 φ(t VM ) = T VM = M and since was aritrary, we conclude S 1 NS = M and the matrices M and N are conjugate. Proof of Theorem 1. (a) Suppose M has characteristic polynomial f(t ); then y the ayley-hamilton theorem, f(m) = 0. onersely, if f(m) = 0, then f(t ) diides χ M (T ) (the characteristic polynomial of M) since f(t ) must e the minimal polynomial of M in F [T ] as it is monic and irreducile. Because oth polynomials are monic and of the same degree, they must e equal. () The idea to proe this is to show that V M and K[T ]/f(t ) are isomorphic as K[T ]-modules, and therefore y our lemma any other matrix N M n (K) with characteristic polynomial f(t ) is conjugate to M, since we would hae V M = VN. Let M M n (K) hae f(t ) as its characteristic polynomial and let V M e the K[T ]-module defined aoe. Since we hae f(t ) = 0 for any K n, we can instead consider this as a module oer K[T ]/(f(t )), where g(t ) = g(m) for g(t ) K[T ]/(f(t )). This in fact makes K n a K[T ]/(f(t ))-ector space, since K[T ]/(f(t )) is a field ecause f(t ) is irreducile. For an aritrary nonzero 0 K n, the multiples K[T ] 0 = (K[T ]/(f(t )) 0 are suspaces of K n ; the K-dimension is dim K (K[T ]/(f(t ))) = n and therefore (K[T ]/(f(t )) 0 = K n. Since 0 is nonzero and K[T ]/(f(t )) is a field, we also hae K[T ]/(f(t )) = K[T ]/(f(t )) 0 as K[T ]-modules. Thus K n = K[T ]/(f(t )) as K[T ]-modules, as desired. 2

3 From Theorem 1 we can conclude that all matrices in M n (F ) (recall F = F q (X)) sharing the same characteristic polynomial are conjugate oer F, ut unfortunately it is not true for general surings of F, in particular, the ring we want to work on, A = F q [X]. So the following theorem is on our inestigation into the additional criteria when two matrices in M n (A) which hae the same characteristic polynomial, say f(t ) A[T ], are conjugate oer A. And it turns out to e strongly related to the A[α]-ideal classes in F (α) with α a root of the characteristic polynomial f in a certain extension of F. This theorem is an analogon for A and F of what Latimer and MacDuffee proed for Z and Q. Theorem 3. Let f(t ) A[T ] e monic irreducile of degree n. (a) A matrix M M n (A) has characteristic polynomial f(t ) if and only if f(m) = 0. () onjugacy classes of matrices in M n (A) with characteristic polynomial f(t ) are in ijection with the A[α]-ideal classes in F (α), where α denotes a root of f(t ). Proof. To proe theorem 3a, we need the following fact: Let R e an integrally closed domain, K its field of fractions, and f(t ) R[T ] e monic. Then if f is reducile in K[T ], it is reducile in R[T ]. This gies us that if f R[T ] is monic and irreducile oer R[T ], it is irreducile in K[T ]. Once we hae this, the proof of Theorem 1a follows through word for word, since the ayley-hamilton theorem holds in any commutatie ring. To proe this, we need the following: Suppose f(t ) R[T ] is monic and irreducile, L e a field extension oer K such that f has a root α L. Then the minimal polynomial g(t ) oer K for α in fact has coefficients from R. Indeed, if α L is a zero of the monic polynomial f(t ) R[T ], then its minimal polynomial g(t ) diides f(t ) in K[T ] (write f(t ) = g(t )q(t ) + r(t ), with deg(r) < deg(g) or r = 0, then we see r(α) = 0 and thus r(t ) = 0 since g is minimal and monic). Thus the zeroes of g(t ) are integral oer R, since they are also zeroes of f R[T ]. Then we hae g(t ) = (T β 1 )... (T β k ) with β i R. The coefficients of g(t ) are sums and products of the β i and are therefore in R as well, thus g(t ) (R K)[T ] = R[T ], since R is integrally closed. Suppose now that f(t ) R[T ] and f(t ) = k i=1 pei i (T ) with p i(t ) irreducile oer K[T ]. Then we know there is a field extension L such that each p i has a root α i in L. Then α i is also a root of f, and since f R[T ], α i is integral oer R. By our lemma, we know that this implies that in fact, p i (T ) R[T ]. Thus f(t ) is reducile oer R[T ]. () For any A[α]-fractional ideal F (α) multiplication y α gies an A- linear map m α :. hoosing an A-asis of (which is an A-module of rank n), we can represent m α y a matrix [m α ] M n (A). A different A-asis gies a conjugate matrix, so independent of the choice of a asis we can associate to a fractional ideal a conjugacy class in M n (A), namely the conjugacy class of a matrix representation of m α. For any M in this conjugacy class we hae f(m) = f([m α ]) = [m f(α) ] = [m 0 ] = 0. For an equialent fractional ideal c = x with x F the conjugacy class is the same as the one for, ecause the matrix for m α with respect to the A-asis {e 1,..., e n } of is the same as the matrix for m α with respect to the A-asis 3

4 {xe 1,..., xe n } of c, so the map A[α]-ideal classes in F (α) conjugacy classes of M M n (A) such that f(m) = 0 (1) gien y the rule: pick a fractional ideal of an ideal class and write a matrix representation for m α with respect to a asis and map the ideal class of the ideal to the conjugacy class of the matrix is well defined. We will show that this map is a ijection. We show surjectiity first. Let B M n (A) with characteristic polynomial f(t ) and let K = F (α) = F [α]. Make F n into a K-ector space in the following way: For c K write c = g(α) for a g(t ) F [T ]. For F n set c = g(α) := g(b) (2) This is well defined, as for a different polynomial h(t ) F [T ] satisfying c = h(α), we hae g(α) = h(α), so g(t ) h(t ) is diisile y the minimal polynomial of α, namely f(t ). But f(b) = 0, so g(b) = h(b) and g(b) = h(b). If A n then α = B A n, so A n is a finitely generated A[α]-sumodule of F n. The equation α = B and the way F n is defined as a K-ector space tells us that the matrix representation of m α with respect to the standard asis of A n is B. We then hae n = dim F (F n ) = [K : F ] dim K (F n ) = n dim K (F n ) (3) so F n is a 1-dimensional K-ector space. Thus for any 0 F n that is nonzero the K-linear map ϕ 0 : K F n gien y ϕ 0 (c) = c 0 is an isomorphism of 1-dimensional ector spaces. The inerse image := ϕ 1 0 (A n ) = {c K : c 0 A n } is a finitely generated A[α]-sumodule of K, i.e. a fractional ideal, since A n has these properties inside F n. Since B is the matrix representation of m α on A n with respect to the standard asis {e 1,..., e n }, B is also the matrix representation of m α on with respect to the A-asis {ϕ 1 0 (e 1 ),..., ϕ 1 0 (e n )} of. So the matrix B is indeed a representation for m α on a fractional A[α]-ideal, so the map (1) is onto. Now we intend to show that (1) is injectie. To do so, suppose and c are two fractional A[α]-ideals in K, for which the matrices B and representing m α with respect to A-ases of and c are conjugate in M n (A). Injectiity of (1) is equialent to and c eing in the same ideal class, i. e. z K : = zc. Since we know B represents the map m α : with respect to a certain A-asis B of we otain the following commutatie diagram m α [ ] B [ ] B B where the map [ ] B is the coordinate isomorphism identifying B with the standard asis A n. onsidering the asis of c with respect to which m α : c c is represented y the matrix, we derie the commutatie diagram 4

5 m α c c [ ] [ ] in an analog way. Furthermore, the conjugacy of B and in M n (A) implies = UBU 1, U GL n (A), i. e. the diagram A n U B A n U commutes. Juxtaposing these diagrams yields: [ ] B U [ ] 1 c m α B [ ] B U [ ] 1 The whole diagramm commutes ecause each of its squares is commutatie. The top and ottom maps are A-linear isomorphisms. Hence, the composite map c on the top and ottom is an A-linear isomorphism which also commutes with m α y commutatiity of the diagram. This property implies the composite map c is not only A-linear, ut een A[α]-linear which means and c are isomorphic as A[α]-modules. Isomorphic fractional A[α]-ideals are scalar multiples of each other, i. e. z K : = zc. So we hae proed that map (1) is injectie. c m α 3 Examples In this section, we are going to present four examples with different class numers and different structures of the class groups. To simplify the computation, we only choose examples in which the characteristic polynomial has degree 2. The examples 2-4 were detected y making extensie calculations with the computer algera system Magma. Example 4. As the first example, we take A to e the polynomial ring oer F 2 and we look at the following polynomial oer A: f(y) = y 2 + y + x 5 + x 3 + 1, and we denote y α a zero of f. As one of the examples gien in [3], Hayes showed that the class numer of A[α] is 1. Thus, all matrices in M 2 (A) haing f as their characteristic polynomial are conjugate. The same situation applies to the other three cures with class numer 1 gien in [3]. 5

6 Example 5. Let us denote y A the polynomial ring oer F 3 and we look at the following polynomial oer A: f(y) = y 2 x 3 + x 2 + 1, which has α as a zero. We are looking for the conjugacy classes of matrices M M 2 (A) satisfying M 2 + ( x 3 + x 2 + 1)I 2 = 0. The class numer of A[α] is 2 and the representaties of classes of fractional ideals are (1), (x + 1, α), for which we can compute the corresponding representing matrices associated to the map m α. We hae an A-asis of the ideal (1) as {1, α}, and multiplying y α yields: 1 α = α; α α = (x 3 x 2 1) α. So the multiplication y α is represented y the matrix M 1 := ( ) 0 x 3 x The matrix corresponding to the ideal class (x+1, α) has A-asis {x+1, α}, thus we can compute the representing matrix as follows: (x + 1) α = 0 (x + 1) + (x + 1) α; α α = (x 2 + x 1) (x + 1) + 0 α. So the multiplication y α is represented y the matrix M 2 := ( ) 0 x 2 +x 1 x+1 0. Thus, y theorem 3, any M in M 2 (A) haing f as its characteristic polynomial is conjugate to exactly one of those two matrices. Example 6. Now we want to erify the calculations we just made y going the other way round, i. e. conerting our matrices into fractional ideals y using the proof of the surjectiity of map (1). In example 5 we hae computed that the matrices M 1 := ( ) 0 x 3 x and M2 := ( ) 0 x 2 +x 1 x+1 0 satisfy the polynomial M 2 + ( x 3 + x 2 + 1)I 2 = 0. Let s start with M 1. First make F 2 into a F (α)-ector space, where α denotes a root of the polynomial f(y) = y 2 x 3 + x 2 + 1, y defining ( ) ( ) ( ( )) ( ) u u 0 x (a + α) := (a + M 1 ) = a + 3 x 2 1 u 1 0 ( ) ( ) a (x = 3 x 2 1) u a Set 0 = ( 1 0 ), so there is an isomorphism F (α) F 2 as F (α)-ector spaces y c c 0. The fractional ideal we are interested in is the inerse image of A 2 under this isomorphism F (α) F 2. This is {c F (α) : c 0 A 2 }. Writing c = a + α we otain ( 1 c 0 = (a + M 1 ) 0 ) ( a = Therefore, haing c 0 A 2 is equialent to c = a + α for some a, A, so the A(α)-fractional ideal in F (α) corresponding to M 1 is A + αa = (1, α) ). 6

7 and this is exactly the ideal that gae rise to M 1 in example 5. Turning to M 2 we proceed as follows. F 2 is made into a F (α)-ector space y defining ( ) ( ) ( ) ( ) u u a (x (a + α) := (a + M 2 ) = 2 + x 1) u. (x + 1) a As efore, we set 0 = ( 1 0 ) and compute {c F (α) : c 0 A 2 }. Writing c = a + α we otain ( ) ( ) ( ) ( ) 1 a 1 0 a c 0 = (a + M 2 ) = =, 0 (x + 1) 0 x + 1 so c 0 A 2 ( a ) ( x + 1 ) 1 A 2 = ( x+1 ) A 2. This means c 0 A 2 is equialent to c = u+ x+1α, u, A. Thus, the fractional ideal we get is {u + α : u, A} = A + α x + 1 x + 1 A. Another ideal from the same class is (x + 1)A + αa = (x + 1, α), which is the ideal used in example 5 to derie M 2. Example 7. Take A to e F 5 [x] and we look at the following polynomial oer A: f(y) = y 2 x 3 + x + 2, which has α as a zero. We are looking for the conjugacy classes of matrices M M 2 (A) satisfying M 2 + ( x 3 + x + 2)I 2 = 0. The class numer of A[α] is 3 and the representaties of classes of fractional ideals are (1), ((x + 3) 2, x α), ((x 2 + x 1) 2, x 3 x 2 + 3x α), to which we can compute the corresponding representing matrices associated to the map m α. We hae an A-asis of the ideal (1) as {1, α}, and multiplying y α yields: 1 α = α; α α = (x 3 x 2) α. So the multiplication y α is represented y the matrix M 1 := ( ) 0 x 3 x The matrix corresponding to the ideal class ((x + 3) 2, x α) has A-asis {(x + 3) 2, x α}, thus we can compute the representing matrix as follows: (x + 3) 2 α = ( (x + 1)) (x + 3) 2 + (x + 3) 2 (x α); (x α) α = (x + 3) (x + 3) 2 + (x + 1) (x α). 7

8 So the multiplication y α is represented y the matrix M 2 := ( (x+1) x+3 ) (x+3) 2 x+1. In a similar manner, we can also compute the matrices corresponding to the third ideal class, namely M 3 := ( (x 3 x 2 +3x+3) x 4 +2x 3 +x 2 1 ) (x 2 +x 1) 2 x 3 x 2 +3x+3. Thus, y theorem 3, any M in M 2 (A) haing f as its characteristic polynomial is conjugate to exactly one of M 1, M 2 and M 3. Example 8. Take q to e 7 and we look at the following polynomial in A[y]: f(y) = y 2 + y x 4 x + 2, which has α as a zero. We are looking for the conjugacy classes of matrices M M 2 (A) satisfying M 2 + M + ( x 4 x + 2)I 2 = 0. The class numer of A[α] is 4 and the class group is isomorphic to Z/2Z Z/2Z. The representaties of classes of fractional ideals are (1), (x(x + 5), x α), (x(x + 1), 4 + α), (x 2 + 6x + 5, 3x + α), to which we can compute the corresponding representing matrices associated to the map m α. We hae an A-asis of the ideal (1) as (1, α), and multiplying y α yields: 1 α = α; α α = (x 4 + x 2) 1 1 α. So the multiplication y α is represented y the matrix M 1 := ( ) 0 x 4 +x The matrix corresponding to the ideal class (x(x + 5), x α) can e computed as follows: x(x + 5) α = ( (x + 4)) x(x + 5) + x(x + 5) (x α); (x α) α = (x 2 + 2x + 3) x(x + 5) + (x + 3) (x α). So the multiplication y α is represented y the matrix M 2 := ( ) (x+4) x 2 +2x+3 x(x+5) x+3. In a similar manner, we can also compute the matrices corresponding to the other two ideal classes, namely M 3 := ( ) 4 x 2 x+1 x(x+1) 3 and M4 := ( ) 3x (x 2 +x+1) x 2 +6x+5 3x 1. Thus, y theorem 3, any M in M 2 (A) haing f as its characteristic polynomial is conjugate to exactly one of the four aoe matrices. References [1] K. onrad: Ideal classes and matrix conjugation oer Z [2]. Latimer,.. MacDuffee: A correspondence etween classes of ideals and classes of matrices, Ann. of Math. 34 pp , 1933 [3] D. Hayes: Explicit class field theory in gloal function fields, Rota, G.., Studies in Algera and Numer theory, pp , New York London: Academic Press

Math 581 Problem Set 9

Math 581 Problem Set 9 Math 581 Prolem Set 9 1. Let m and n e relatively prime positive integers. (a) Prove that Z/mnZ = Z/mZ Z/nZ as RINGS. (Hint: First Isomorphism Theorem) Proof: Define ϕz Z/mZ Z/nZ y ϕ(x) = ([x] m, [x] n