Factoring the Semigroup Determinant of a Finite Commutative Chain Ring Jay A. Wood? Department of Mathematics, Computer Science & Statistics Purdue University Calumet Hammond, Indiana 46323{2094 USA [email protected] http://www.calumet.purdue.edu/public/math/wood/

Abstract. The semigroup determinant of a nite commutative chain ring factors completely into linear factors involving the characters of the group of units of the ring. The factorization proves the equivalence of two approaches to the extension theorem for general weight functions on linear codes over nite commutative chain rings.

1 Introduction In 1896, Frobenius factored the group determinant of a nite group [2] (also see [1] and [3] and the references therein). If G is a nite group and ag , g 2 G, are indeterminates, the group determinant of G is det(agh,1 ), i.e., the determinant of the jGj  jGj matrix whose entry in position (g; h) is the indeterminate agh,1 indexed by the product gh,1 2 G. The matrix A = (agh,1 ) represents the left regular representation of G with respect to the basis of the group algebra C[G] consisting of the elements of G. The factorization of det A re ects the decomposition of C[G] into irreducible representations. In fact, Frobenius' work on factoring det A led directly to his development of the representation theory for nite groups. If G is abelian, the basis of characters of G will diagonalize A with typical diagonal entry X a^() = ag (g) ; g2G

the Fourier coecient of a with respect to . Thus det A = ?

Y

2Gb

a^() ;

Partially supported by NSA grant MDA904-96-1-0067, and by Purdue University Calumet Scholarly Research Awards. This paper is an expanded version of results presented at the International Conference on Coding Theory, Cryptography and Related Areas, Guanajuato, Mexico, April 21, 1998. The paper is in nal form and no version of it will be submitted for publication elsewhere.

in the abelian case. Now suppose that S is a nite semigroup and that as , s 2 S , are indeterminates. As in the group case, form a matrix A = (ast ) of size jS j  jS j, where the (s; t) entry is the indeterminate indexed by the product st 2 S . The main result of this paper is a factorization of det A in the case where the semigroup S is the multiplicative semigroup of a nite commutative chain ring, i.e., of a nite, commutative, local ring R with principal maximal ideal. Unlike in the group case, the matrix A does not represent the regular representation. The factorization is explicit, and it involves the Fourier coecients of a (thought of as a function on S ) restricted to certain orbits of the group U of units of R. The key to the matter is to pick a good basis for the reduced semigroup algebra C0 [S ]. Multiplication by U partitions S into U -orbits. Each orbit can be identi ed with a quotient space of U which, since U is abelian, is actually a quotient group of U . The characters of the quotient group provide a good basis along the orbit, and the union of all the characters of all the orbits provides a good basis for C0 [S ]. Expressed in terms of this good basis, the matrix A takes on an especially simple form. The formula for det A then follows directly. In [5], the author discusses the extension theorem for general weight functions on linear codes over nite commutative chain rings. In that paper there are two approaches to the extension problem, one involving det A and the other involving the regular representation of C0 [S ]. The factorization of det A given here shows that those two approaches are equivalent. Acknowledgments. I thank the referee for several improvements to this paper.

2 Preliminaries Let S be a multiplicative semigroup with 0 element. The complex semigroup algebra C[S ] is a complex vector space with basis es , s 2 S . By de ning a multiplication of basis elements by es et = est (where the latter is indexed by the semigroup product st) and extending linearly, C[S ] becomes a complex algebra. Observe that the ideal (e0 ) = Ce0 generated by e0 is a two-sided ideal. The quotient algebra C0 [S ] = C[S ]=(e0 ) is the reduced semigroup algebra of S ; it has dimension jS j , 1 over C. Let R be a nite commutative chain ring, that is, a nite ring with 1 that is commutative, local, and whose maximal ideal m is principal (say, m = (m) = Rm). It is easy to see ([4, Lemma 13]) that the ideals of R t into the following chain: R  (m)  (m2 ) 


 (ml )  (0) : Here, l is the smallest nonnegative integer such that ml+1 = 0. Let U be the group of units of R. Then U acts on R by multiplication, and the orbits of U are exactly the set-theoretic dierences O0 = U = R n (m), O1 = (m) n (m2 ), O2 = (m2 ) n (m3 ), : : :, Ol = (ml ) n (0), Ol+1 = (0). If x 2 Oi , the stabilizer subgroup of x is Ux = fu 2 U : ux = xg. Because U is abelian, if x; y 2 Oi , then Ux = Uy . We will simply write Ui = Stab(Oi ) for this common

stabilizer subgroup of the orbit Oi . Note that Oi = fumi : u 2 U =Ui g provides a bijection between Oi and U =Ui . Observe that Ui  Ui+1 . The complex characters of U form a group denoted by Ub. Given a subgroup U of U , the annihilator of U is ( Ub : U ) = f 2 Ub : (u) = 1; u 2 U g. It follows that (Ub : U )  = (U =U ) b , so that (Ub : U ) = jUj = jU j. A pair (; Oi ) consisting of a character  of U and a U -orbit Oi is admissible if Ui  ker , i.e., if  2 (Ub : Ui ). Since Ui  Ui+1 , every  2 Ub has a largest non-negative integer i such that (; Oi ) is admissible. We make one exception. Denote by 0 the trivial character of U ; i.e., 0 (u) = 1, for all u 2 U . The admissible pair (0 ; Ol+1 = f0g), is called the trivial admissible pair. We exclude this trivial pair, and we set i0 = l for the trivial character 0 .

3 The factorization

Let R be a nite commutative chain ring with group of units U . For every element r 2 R assign an indeterminate ar . Form a square matrix A of size jRj, the semigroup matrix, as follows. The rows and columns of A are indexed by the elements of R. The entry in row r and column s is the indeterminate ars associated to the element rs 2 R. In a similar fashion, form a square matrix A0 of size jRj , 1, the reduced semigroup matrix. Within A0 , set a0 = 0. The rows and columns of A0 are indexed by the non-zero elements of R. The entry in row r and column s is again ars (which will equal 0 if rs = 0). We wish to factor det A and det A0 . Because A0 is more relevant to coding theory (see Remark 9), we will focus our attention on det A0 , and then describe brie y what happens for det A. Given an admissible pair (; Oi ), set X a^(; i) = (u)aumi : u2U =Ui

Since (; Oi ) is admissible, this expression is well-de ned. In fact, a^(; i) may be viewed as the Fourier coecient of the restriction of a to the orbit Oi (which is bijective to the quotient group U =Ui ;  2 (Ub : Ui )  = (U =Ui ) b ). Theorem 1. For any nite commutative chain ring R with group of units U , det A0 factors into linear factors: Y det A0 = C a^(; i )1+i ; 2Ub

where C is a non-zero integral constant. Theorem 2. For any nite commutative chain ring R with group of units U , det A factors into linear factors: Y det A = C a0 (^a(0 ; i0 ) , d a0 )1+i0 a^(; i )1+i ; 2Ub;6=0

where C is a non-zero integral constant and d = [U : Ul ] = jOl j is the index of

Ul in U .

These theorems are proved at the end of Sect. 5.

4 A good basis In order to calculate det A0 , we actually calculate P ,1 A0 P for a particularly nice choice of matrix P . Factoring det A0 = det(P ,1 A0 P ) is then an easy task. For notational convenience, set Mi = jOi j.

Lemma 3. The number of non-trivial admissible pairs (; Oi ),  2 Ub, equals jRj , 1. Proof. For a xed i, (; Oi ) is admissible if and only if  2 (Ub : Ui ). There are jUj = jUi j such 's. But since Ui is the stabilizer subgroup of the orbit Oi , jUj = jUi j is also equal to Mi = jOi j. Since jRj = Pi Mi , the result follows once we exclude the trivial admissible pair. ut

We now de ne a matrix P . The rows of P are indexed by the non-zero elements of R while the columns of P are indexed by the non-trivial admissible pairs (; Oi ),  2 Ub. By Lemma 3, P is a square matrix. The entry pr;(;Oi) in row r and column (; Oi ),  2 Ub, is

pr;(;Oi) =



0; if r 62 Oi ; i  ( u Mi ); if r = um 2 Oi : 1

Since (; Oi ) is admissible,  2 (Ub : Ui ). This implies that (u) is well-de ned, depending only on r. Let the matrix Q be an \unscaled" version of P :



i ; qr;(;Oi) = (u0);; ifif rr 62= Oum i 2 Oi :

Lemma 4. P , = Qt, the conjugate transpose of Q. 1

Proof. This follows easily from the orthogonality relations on characters, viewing  2 (Ub : Ui ) as a character of U =Ui . ut

The columns of P are a good basis for C0 [S ], where S is the multiplicative semigroup of the nite commutative chain ring R. For each orbit Oi , we use the characters of U =Ui .

5 Calculations We begin with a technical lemma. Lemma 5. Suppose 2 (Ub : Uj ) = (U =Uj ) b . Then

(

X

b (x) = jUi+j j = jUj j ; if 2 (Ub : Ui+j ) ; 0; if 26 (U : Ui+j ) : x2Ui+j =Uj Proof. This is an application of the following general result. If H is a subgroup of G and  2 Gb, then (

X

b : H) ; (h) = jH0j ;; ifif  262 ((G Gb : H ) : h2H

The proof is an exercise for the reader. ut Theorem 6. Let b be the entry of P ,1A0 P in row (; Oi ) and column ( ; Oj ), ; 2 Ub. Then 8 0; if 6=  ; < 0; if =  , but (; Oi+j ) is not admissible ; b=: C a^( ; i + j ); if =  and (; Oi+j ) is admissible ; where C is a non-zero integral constant (that depends only on i, j ). Proof. The entry b is given by the following summation:

b=

X r6=0

qr;(;Oi)

X s6=0

ars ps;( ;Oj ) :

The q-term vanishes unless r 2 Oi , so that r = mi ,  2 U =Ui . Similarly, the p-term vanishes unless s = mj , 2 U =Uj . Thus X X 1 b=  () ( )a i+j : 2U =Ui 2U =Uj

Mj

 m

Now we isolate the coecient of a mi+j , 2 U =Ui+j . This coecient has the form X c = M1  () ( ) ; j

where the sum runs over pairs of  2 U =Ui , 2 U =Uj , satisfying  = in U =Ui+j . Since Ui  Ui+j , we see that Ui+j =Ui is the kernel of the natural projection U =Ui  U =Ui+j . Similarly, Ui+j =Uj is the kernel of U =Uj  U =Ui+j . Fix an  2 U =Ui , with image ~ 2 U =Ui+j . Since U =Ui+j is a group, set ~ = ~,1 2 U =Ui+j . Every in the preimage of ~ contributes a term to the coecient c. If 0 is one

particular preimage of ~ ( 0 depends on ), then every other preimage has the form = 0 ,  2 Ui+j =Uj . Then X X 1 X 1 X c=  () ( ) =  () ( ) ( ) : 2U =Ui 2Ui+j =Uj

Mj

0



Mj

0



By Lemma 5, c = 0 unless 2 (Ub : Ui+j ), in which case ( 0 ) is independent of the particular choice of preimage of ~. Repeating this argument with the roles of ; reversed, we see that c = 0 unless both  and belong to (Ub : Ui+j ). In that case () = (~) and ( ) = ( ~) are well-de ned. Thus X 1 c = jU =U jjU =U j  (~) ( ~) : i+j

But

X ~ ~=

(~) ( ~) =

i

X ~

i+j

j

~ ~=

Mj

 (~) (~,1 ) =

X ~

(~) (~)

!

( ) :

By the orthogonality relations of characters, this last expression equals 0 unless =  , in which case it equals jU =Ui+j j ( ) = jU =Ui+j j  ( ). In conclusion, the entry b equals 0 unless =  belongs to (Ub : Ui+j ). This means that (; Oi+j ) must be admissible. In particular, i + j  i is required. If these conditions are satis ed, then X ( )a i+j b = jU =U j jU =U j jU =U j 1 i+j

i

= jUi+j =Ui j

i+j

X

j

2U =Ui+j

i+j

Mj 2U =Ui+j

m

( )a mi+j ;

which has the form b = C a^( ; i + j ), where C is the non-zero integral constant jUi+j =Ui j. ut

Proof of Theorem 1. Fix a character  2 Ub. The submatrix of P ,1 A0 P with row indices (; Oi ) and column indices (; Oj ), 0  i; j  i , has the form (neglecting the non-zero constants)

0 a^(; 0) a^(; 1) : : : a^(; i , 1) a^(; i ) 1 BB a^(; 1) a^(; 2) : : : a^(; i ) 0  CC BB ... .. C : . C B@ a^(; i , 1) a^(; i ) : : : 0 C A 0   a^(; i )

0

:::

0

0

The determinant of this submatrix is a^( ; i )1+i times a non-zero integral constant. From this, Theorem 1 follows. ut

Proof of Theorem 2. When dealing with A, one uses a matrix P which is square of size jRj, using all elements of R and all admissible pairs, including the trivial pair (0 ; f0g). In calculating P ,1 AP , the only dierence arises in the submatrix involving the trivial character 0 . Neglecting the non-zero constants, that submatrix is 0 a^( ; 0) a^( ; 1) : : : a^( ; i , 1) a^( ; i ) a 1 BB a^(00 ; 1) a^(00 ; 2) : : : a^(00; 0i0 ) 0a0 0 a00 CC BB . .. .. C . .C BB a^( ; i.. , 1) a^( ; i ) : : : C: a0 a0 a0 C B@ a^(0 ; 0i ) 0a 0 : : : C a0 a0 a0 A 0 0 0

a0

a0

:::

a0

a0

a0 In any given row, the (neglected) coecients for a0 all equal d times the (neglected) coecient for a^(0 ; i0 ). Evaluating the determinant is then an easy

exercise involving elementary row and column operations.

ut

6 A relative case We discuss brie y the situation where some symmetry is imposed on the indeterminates ar . To be precise, suppose U is a subgroup of the group of units U of a nite commutative chain ring R. Multiplication de nes an action of U on R, and we denote the space of orbits by R=U . We impose the conditions aur = ar , u 2 U , r 2 R, on the indeterminates. Notice that ar now depends only upon the U -orbit orb(r) of r. Form a square matrix AU of size jR=U j, as follows. The rows and columns of AU are indexed by the U -orbits. The entry in row orb(r) and column orb(s) is the indeterminate ars associated to orb(rs). This is well-de ned. Similarly, form a reduced square matrix AU0 of size jR=U j , 1, by setting a0 = 0 and indexing the rows and columns of AU0 by the non-zero U -orbits. The entry in row orb(r) and column orb(s) is again the indeterminate ars . This too is well-de ned. When U = f1g, the matrices AU and AU0 are just A and A0 . We now wish to factor det AU and det AU0 . Given an admissible pair (; Oi ) with  2 (Ub : U ), set

a(; i) =

X

u2U =U Ui

(u)aumi :

Notice that a^(; i) = Ei a(; i), where Ei equals the size of a U -orbit in Oi . Theorem 7. For any subgroup U of the group of units U of a nite commutative chain ring R, det AU0 factors into linear factors: det AU0 = C

Y

2(Ub:U )

where C is a non-zero integral constant.

a(; i )1+i ;

Theorem 8. For any subgroup U of the group of units U of a nite commutative chain ring R, det AU factors into linear factors: det AU = C a0 (a(0 ; i0 ) , d a0 )1+i0

Y

2(Ub:U );6=0

a(; i )1+i ;

where C is a non-zero integral constant and d = [U : U Ul ] is the index of U Ul in U . Idea of proofs. View everything as living on the quotient group U =U and apply Theorems 1 and 2. ut Remark 9. By applying Theorem 7, one obtains the equivalence of the two approaches discussed in [5] to the extension theorem for general weight functions on linear codes over nite commutative chain rings.

7 Examples Example 10. To illustrate Theorems 1 and 6, consider the nite commutative chain ring R = ZZ=(9). The maximal ideal is m = (3), and m2 = (0), so that l = 1. The group of units is U = f1; 2; 4; 5; 7; 8g, which is cyclic of order 6. Either 2 or 5 generates U . The characters of U are determined by their values at the generator 2 2 U . Let  = exp(2i=6), a primitive 6th root of unity in C. Of course,  3 = ,1. Then the formulas j (2) =  j , j = 0; 1; : : :; 5, determine the 6 characters of U . Observe that

ker 0 = f1; 2; 4; 5; 7; 8g ; ker 1 = ker 5 = f1g ; ker 2 = ker 4 = f1; 8g ; ker 3 = f1; 4; 7g : The U -orbits Oi and their stabilizer subgroups Ui are as follows.

O = f1; 2; 4; 5; 7; 8g U = f1g O = f3; 6g U = f1; 4; 7g O = f0g U = f1; 2; 4; 5; 7; 8g 0

0

1

1

2

2

Comparing Ui to ker j , we see that i0 = 1 (remember that (0 ; O2 ) is trivial and thus excluded), i1 = i2 = i4 = i5 = 0, and i3 = 1.

In writing down the matrix A0 , we will use the order 1; 2; 4; 8; 7; 5; 3; 6 on the non-zero elements of R. Remember that a0 = 0. Then

0a BB a BB a Ba A =B BB a BB a @a

1

a2 a4 a8 a7 a5 a3 a6 C 2 a4 a8 a7 a5 a1 a6 a3 C C a a a a a a a 4 8 7 5 1 2 3 6C C: a a a a a a a 8 7 5 1 2 4 6 3C C a a a a a a a 7 5 1 2 4 8 3 6C C 5 a1 a2 a4 a8 a7 a6 a3 C A 3 a6 a3 a6 a3 a6 0 0 a6 a3 a6 a3 a6 a3 0 0

0

1

To write down the matrix P , we need to order the non-trivial admissible pairs (j ; Oi ). We choose to write the real-valued characters rst, followed by the others, grouped in conjugate pairs. Thus the order used is (0 ; O0 ), (0 ; O1 ), (3 ; O0 ), (3 ; O1 ), (1 ; O0 ), (5 ; O0 ), (2 ; O0 ), (4 ; O0 ). Then

01 0 1 B 1 0 ,1 B B 10 1 B B 1 1 0 ,1 P = 6B B 10 1 B B 1 0 ,1 B @0 3 0

1

0 1 1 1 1 0  , 2  2 , C C 0  2 , ,  2 C C 0 ,1 ,1 1 1 C C: 0 ,  2  2 , C C 0 , 2  ,  2 C C 3 0 0 0 0A 0 3 0 ,3 0 0 0 0

The reader will verify that P ,1 A0 P has the block diagonal form

0T 0 0 B0 T 0 P, A P = B @0 0 T 0

1

3

0

1

1

0 0C C 0A ;

0 0 0 T2

where

 a^( ; 0) 3^a( ; 1)  T = a^( ; 1) 0 ; T  0 a^( ; 0)  0

0

0

3

0

T1 = a^( ; 0) 1

5

0

;

 a^( ; 0) 3^a( ; 1)  = a^( ; 1) 0 ;  0 a^( ; 0)  3

3

3

T2 = a^( ; 0) 2

4

0

:

(1)

Thus det A0 = 9^a(0 ; 1)2 a^(1 ; 0)^a(2 ; 0)^a(3 ; 1)2a^(4 ; 0)^a(5 ; 0). Example 11. To illustrate Theorem 2, we continue to examine R = ZZ=(9). In writing down the matrix A, we will use the order 1; 2; 4; 8; 7; 5; 3; 6; 0 on the elements of R. In the matrix A, a0 need no longer be zero. The order on admissible pairs used to write down the matrix P is (0 ; O0 ), (0 ; O1 ), (0 ; O2 ), (3 ; O0 ),

(3 ; O1 ), (1 ; O0 ), (5 ; O0 ), (2 ; O0 ), (4 ; O0 ). The matrices A and P are then 0a a a a a a a a a 1 01 0 0 1 0 1 1 1 11 BB a12 a24 a48 a87 a75 a51 a36 a63 a00 CC BB 1 0 0 ,1 0  ,2 2 , CC BB a4 a8 a7 a5 a1 a2 a3 a6 a0 CC BB 1 0 0 1 0 2 , , 2 CC BB a8 a7 a5 a1 a2 a4 a6 a3 a0 CC 1 BB 1 0 0 ,1 0 ,1 ,1 1 1 CC BB a7 a5 a1 a2 a4 a8 a3 a6 a0 CC ; BB 1 0 0 1 0 , 2 2 , CC : BB a5 a1 a2 a4 a8 a7 a6 a3 a0 CC 6 BB 1 0 0 ,1 0 ,2  , 2 CC BB a3 a6 a3 a6 a3 a6 a0 a0 a0 CC BB 0 3 0 0 3 0 0 0 0 CC @ a6 a3 a6 a3 a6 a3 a0 a0 a0 A @ 0 3 0 0 ,3 0 0 0 0 A a0 a0 a0 a0 a0 a0 a0 a0 a0 006 0 0 0 0 0 0 Then the reader will verify that P ,1 AP has the block diagonal form 0T 0 0 0 1 0 B C; 0 , 1 B P AP = @ 0 T03 T0 00 C A 1 0 0 0 T2 where 0 a^( ; 0) 3^a( ; 1) 6a 1 0 0 0 T0 = @ a^(0 ; 1) 2a0 2a0 A ;

a0

a0

a0

and the other Ti are as in (1). Thus det A = 9a0 (^a(0 ; 1) , 2a0 )2 a^(1 ; 0)^a(2 ; 0)^a(3 ; 1)2 a^(4 ; 0)^a(5 ; 0) : Example 12. To illustrate Theorem 7, we consider the subgroup U = f1; 8g  U inside R = ZZ=(9). Only the characters 0 , 2 , 4 belong to (U : U ). The U -orbits are f1; 8g, f2; 7g, f4; 5g, f3; 6g, f0g. The matrix AU0 is

0a a a a Ba a a a AU = B @a a a a 0

1

2

4

3

2

4

1

3

4

1

2

3

a3 a3 a3 0

1 CC : A

To write down P , we use only those non-trivial admissible orbits with  2 (U : U ). They are (0 ; O0 ), (0 ; O1 ), (2 ; O0 ), and (4 ; O0 ). We write ! =  2 , a primitive 3rd root of unity in C. Then 01 0 1 1 1 B 1 0 ! !2 CC : P = 13 B @ 1 0 !2 ! A 03 0 0 The reader will verify that 0 a( ; 0) 3a( ; 1) 0 1 0 0 0 B a(0 ; 1) 0 0 0 C C P ,1 AU0 P = B @ 0 0 0 a(4 ; 0) A : 0 0 a(2 ; 0) 0

Thus, det AU0 = 3a(0 ; 1)2 a(2 ; 0)a(4 ; 0). Example 13. To illustrate Theorem 8, we indicate how to change the appropriate matrices from the previous example.

0a BB a U A =B B@ aa

1

0

1

a2 a4 a3 a0 100 1 1 C B 1 0 0 ! !2 C 2 a4 a1 a3 a0 C B C: 1 C B P = 3 B 1 0 0 !2 ! C 4 a1 a2 a3 a0 C ; A @ 0 3 0 0 0 CA 3 a3 a3 a0 a0 a0 a0 a0 a0 a0 003 0 0 1

Then

0 a( ; 0) 3a( ; 1) 3a BB a( ; 1) a a a a P , AU P = B B@ a0 0 0 1

0

0

0

0

0

0

0

0

0

0

0

0 0 0 0

0 a(2 ; 0)

1 CC CC : a( ; 0) A 0 0 0 4

0

Thus, det AU = 3a0 (a(0 ; 1) , a0 )2 a(2 ; 0)a(4 ; 0).

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