ALMOST TRIANGULAR MATRICES OVER DEDEKIND DOMAINS

Internat. J. Math. & Math. Sci.

Vol. 22, No. 4 (1999) 761–764 S 0161-17129922761-5

©Electronic Publishing House

ALMOST TRIANGULAR MATRICES OVER DEDEKIND DOMAINS

FRANK DEMEYER and HANIYA KAKAKHAIL (Received 25 November 1997 and in revised form 13 March 1998)

Abstract.Every matrix over a Dedekind domain is equivalent to a direct sum of matrices A=(a_i,j), wherea_i,j=0 wheneverj > i+1.

Keywords and phrases. Matrices, Dedekind domains, equivalence.

1991 Mathematics Subject Classification. 13F05, 15A21.

1. Introduction. Twom×nmatricesAand Bover a ringR are called equivalent if B=PAQ for invertible matrices P and Q over R. From now on, assume that R denotes a Dedekind domain with quotient fieldK. IfI= a,bis a non principal ideal inR, then, in contrast with the situation for Principal Ideal Domains, the 1×2 matrix a,b

is not equivalent overRto a matrix whose off diagonal entries are 0. Using the separated divisor theorem in the form given by Levy in [2], other facts about matrices over Dedekind domains in [2], and elementary properties of ideals in Dedekind domain [1], we show that anym×nmatrix over a Dedekind domain is equivalent to a direct sum of matrices A=(ai,j)with ai,j =0 when j > i+1. If the direct summand A has rankr, then the number of rows, respectively columns, ofAis eitherrorr+1.

The corresponding result for similarity of matrices over principal ideal rings is that everyn×nmatrix over a principal ideal ring is similar to an upper triangular matrix [3, p. 42].

2. Diagonalization of matrices. IfAis anm×nmatrix, thenAcan be viewed as anR-module homomorphismA:Rⁿ→R^m by left multiplication. IfMA denotes the submodule ofR^mgenerated by the columns ofA, thenMAis the image ofAinR^mand the isomorphism class of the cokernelSA=R^m/MAofAdetermines the equivalence class ofA.

Separated divisor theorem[2]. There is a chain of integralR-idealsL1⊆L2⊆

··· ⊆Lr and a fractionalR-idealHsuch that

SA=







⊕^r_i=1L^R_i⊕H⊕R^m−r−1, m < r

⊕^r_i=1L^R_i, m=r , (2.1)

whereH=_r

i=1Liifr=nandHRifr=0 orr=m.

The isomorphism class ofSA, the ideals{Li}^r_i=1(as sets), and the isomorphism class ofHboth determine and are determined by the equivalence class ofA.

762 F. DEMEYER AND H. KAKAKHAIL

We also need the following elementary facts about ideals in Dedekind domains.

Lemma1[1, p. 150, 154]. LetI,Jbe integral ideals inR. Then

(1)There is anαin the quotient fieldKofRsuch thatαIis integral andαI+J=R;

(2)There is anR-module isomorphismγ:IJ⊕R→I⊕J, given byγ(u,v)=(x1v−u, αu−x2v), whereαis as in(1)andx1∈I,x2∈Jare chosen withαx1−x2=1.

Note. TheR-linear homomorphismγis given by the matrix_{−1 x}

α −x12 , whereα∈K.

Theorem2.2. Everym×nmatrix Aover a Dedekind domain is equivalent to a direct sum of matrices(aij)withaij=0wheneverj > i+1.

Proof. Anm×nmatrixA is called indecomposable ifAis not equivalent to a matrix of the form_B

1 0

0 B2 for any matricesB1,B2. That is,A is not equivalent to a direct sum of matricesB1,B2. IfA=0, the result is clear. Assume that A≠0. It is sufficient to verify the result for indecomposable matrices. In this case, if r is the rank ofAover the quotient fieldKofR, then [2, Lem. 2.1] asserts thatm=rorr+1 andn=r orr+1. There are then four possible cases to check.

Case1. Assume thatm=r andn=r. ThenSA= ⊕^r_i=1R/Li, withL1,...,Lrintegral R-ideals withL1⊆L2⊆ ··· ⊆Lr and_r

i=1LiR. Thus, _r

i=1Li= ais a principal ideal generated bya∈R. Letφ0:R^r →_r

i=1Li⊕R^r−1 be given byφ0(r1,...,rr)= (ar1,r2,...,rr) and let φj :L1⊕ ··· ⊕Lj−1⊕_r

i=jLi⊕R⊕R^r−j−1→L1⊕ ··· ⊕Lj⊕ _r

i=j+1Li⊕R^r−j−1 be given byφj=Ij−1⊕γj⊕Ir−j−1, where γj:_r

i=jLi⊕R→Lj⊕ _r

i=j+1Liis the map given in Lemma 1 andIj−1,Ir−j−1are the identity maps of indi- cated rank. Letφ:R^r→L1⊕···⊕Lr⊂R^rbe given byφ=φr−1φr−2···φ1φ0. Then the matrix[φ]ofφ, with respect to the standard bases forR^r, is:[φ]=[φr−1]···[φ] [φ0].

While[φi]may have entries which are not inR, [φ]has all its entries inRsince eachLjis integral. If we write

φj

=









Ij 0 0 0

0 −1 x^j_i 0 0 αj −x^j₂ 0

0 0 0 Ir−j−1







, (2.2)

then a direct calculation shows that

[φ]=

















−a x₁¹ 0 0 0 0 0 0 0

−aα1 −x₂¹ x²₁ 0 0 0 0 0 0

−aα1α2 α2x2¹ x²2 x1³ 0 0 0 0 0

−aα1α2α3 α2α3x²₂ x³₂ x₁⁴ 0 0 0 0 0

... ... ...

−a_r−1

i=1αi ··· αr−2x₂^r−2 x₂^r−1

















. (2.3)

Since[φ]has the same number of rows and columns and the same cokernel asA,[φ]

is equivalent toA.

ALMOST TRIANGULAR MATRICES OVER DEDEKIND DOMAINS 763 Remark. Assume thatLi= aiis principal for eachi,i=1,...,r andai∈R. The isomorphismγj:_r

i=jLi⊕R⊕ →Lj⊕_r

i=j+1Lican be given asγj(u,v)=(αju,βjv), whereαj=1/

i=j+1aiand βj=_r

i=j+1ai. In this case,[φ]=diag{a1,...,ar}with ai|ai+1for 1≤i≤r. This is the only case which occurs ifRis a PID.

Case2. Assume thatm=r andn=r+1. ThenSA= ⊕^r_i=1R/LiwithLi,1≤i≤r integral ideals andL1⊆L2⊆ ··· ⊆Lr. LetLr+1 be integral ideal with_r+1

i=1L= a principal, then⊕^r+1_i=1LiRⁿand there is a chain ofR-homomorphisms

R^{n φ}→L1⊕···⊕Lr⊕Lr+1 π

→L1⊕···⊕Lr⊆R^r, (2.4) whereπis the projection onL1⊕···⊕LralongLr+1. The matrix ofπ◦φis anm×n matrix obtained by deleting the last row of[φ]and, thus, has the same form as in Case 1. Since the cokernel ofπφis the same asAand[πφ]has the same number of rows and columns asA,[πφ]is equivalent toA.

Case 3. Assume that m = r+1 and n = r. Then SA = ⊕^r_i=1R/Li⊕H, where Li, 1≤ i≤ r are integral ideals and H _r

i=1Li. Choose a∈ R with LrH⁻¹a in- tegral. Note that LrH⁻¹a is a submodule of H⁻¹a. From Case 1, we construct an R-isomorphism φr : R^r →L1⊕ ··· ⊕Lr−1⊕LrH⁻¹a ⊂R^r+1 whose matrix has the same form as that of[φ]in Case 1. By Lemma 1, there is a chain of isomorphisms ψ:H⁻¹a⊕H→H⁻¹Ha⊕R→R⊕RcarryingLrH⁻¹aonto a submoduleN ofR⊕R.

By [1, Cor. 18.24],(H⁻¹a⊕H)/LrH⁻¹aR/Lr⊕H. LetΦ=(Ir−1⊕ψ)◦φr:Rⁿ→R^m. The matrix ofΦism×nand the firstr=nrows are the same as[φr]. The last row does not contribute any entries above the main diagonal. So, for eachj > i+1, the i,jth entry of[Φ]is 0. Since the cokernel of[Φ]isSAand[Φ]has the same number of rows and columns asA,[Φ]andAare equivalent.

Case4. Let SA=⊕^r_i=1R/Li⊕H, where L1,...,Lr are integral ideals withL1⊆ ··· ⊆ Lr and by replacingH (if necessary) by an isomorphic copy,H is an integral ideal.

By [1, Thm. 18.20], there is an integral idealHowithHoH principal andHo+H=R.

There is an a∈R such thatJ =(_r

i=1Li·Ho)⁻¹a⊆ H. As in Case 1, there is an isomorphismφr+1:R^r+1→L1⊕···⊕Lr−1⊕LrHo⊕J. ViewLi≤Rfor 1≤i≤r ,LrHo≤ Ho. As in Case 3, there is an isomorphismψ:Ho⊕H→R⊕Rwithψ(LrHo)=N≤R⊕R andR⊕R/NR/Lr⊕H. LetΦ=(Ir−1⊕ψ)◦φr+1. ThenΦ:R^r+1→R^r+1and all the rows, except possibly the last two of[Φ], are the same as that of[φ]in Case 1. So, for eachj > i+1, thei,jth entry of[Φ]is 0. Since the cokernel ofΦisSA,[Φ]andA are equivalent.

Remark. While we could have given explicit formula for the entries in the matri- ces constructed in Cases 2, 3, and 4 as in Case 1, these entries are not canonically determined byAas a result of the many choices made in their construction. In par- ticular, the choices ofαandx1,x2in Lemma 1 are not canonically determined by the idealsI,J.

References

[1] C. W. Curtis and I. Reiner,Representation theory of finite groups and associative algebras, Pure and Applied Mathematics, vol. 11, Interscience Publishers, a division of John Wiley & Sons, New York, London, 1962. MR 26#2519. Zbl 131.25601.

764 F. DEMEYER AND H. KAKAKHAIL

[2] L. S. Levy,Almost diagonal matrices over Dedekind domains, Math.-Z.124(1972), 89–99.

MR 45 3437. Zbl 211.36903.

[3] M. Newman,Integral matrices, Pure and Applied Mathematics, vol. 45, Academic Press, New York, London, 1972. MR 49 5038. Zbl 254.15009.

Demeyer: Department of Mathematics, Colorado State University, Fort Collins CO 80523, USA

E-mail address:[email protected]

Kakakhail: Department of Mathematics, Metropolitan State College, Denver CO 80217, USA

E-mail address:[email protected]