ALGEBRA ISRN Algebra 2090-6293 International Scholarly Research Network 520148 10.5402/2012/520148 520148 Research Article Cogredient Standard Forms of Symmetric Matrices over Galois Rings of Odd Characteristic Cao Yonglin Kelarev A. V. Kressner D. Rodrigues W. A. School of Sciences Shandong University of Technology Shandong, Zibo 255091 China sdut.edu.cn 2012 19 7 2012 2012 20 03 2012 13 05 2012 2012 Copyright © 2012 Yonglin Cao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let R=GR(ps,psm) be a Galois ring of characteristic ps and cardinality psm, where s and m are positive integers and p is an odd prime number. Two kinds of cogredient standard forms of symmetric matrices over R are given, and an explicit formula to count the number of all distinct cogredient classes of symmetric matrices over R is obtained.

1. Introduction and Preliminaries

Let p be a prime number, s and m be positive integers, and R=GR(ps,psm) a Galois ring of characteristic ps and cardinality psm. Then GR(ps,psm) is isomorphic to the ring ps[x]/(h(x))   for any basic irreducible polynomial h(x) of degree m over ps. It is clear that R=𝔽pm, that is, a finite field of pm elements, if s=1, and R=ps, that is the ring of residue classes of modulo its ideal ps, if m=1.

We denote by R* the group of units of R. R is a local ring with the maximal ideal (p)=pR, and all ideals of R are given by (0)=(ps)(ps-1)(p)(p0)=R. By [1, Theorem 14.8], there exists an element ξR* of multiplicative order pm-1, which is a root of a basic primitive polynomial h(x) of degree m over ps and dividing xpm-1-1 in ps[x], and every element aR can be written uniquely as (1.1)a=a0+a1p++an-1pn-1,a0,a1,,an-1T, where 𝒯={0,1,ξ,,ξpm-2}. Moreover, a is a unit if and only if a00, and a is a zero divisor or 0 if and only if a0=0. Define the p-exponent of a by τ(0)=s and τ(a)=i if a=aipi++an-1pn-1 with ai0. By [1, Corollary 14.9], R*ξ×[1+(p)], where ξ is the cyclic group of order pm-1, and 1+(p)={1+xx(p)} is the one group of Galois ring R, so |R*|=(pm-1)p(s-1)m.

For a fixed positive integer n, let Mn(R) and GLn(R) be the set of all n×n matrices and the multiplicative group of all n×n invertible matrices over R, and denote by I(n) and 0(n) the n×n identity matrix and zero matrix, respectively. In this paper, for l×n matrix A and q×r matrix B over R, we adopt the notation AB:=(A00B) which is a (l+q)×(n+r) matrix over R.

For any matrix AMn(R), A is said to be symmetric if AT=A, where AT is the transposed matrix of A. We denote the set of all n×n symmetric matrices over R by 𝒮(n,R). Then (𝒮(n,R),+) is a group under the addition of matrices. For any matrices S1,S2Mn(R), if there exists matrix PGLn(R) such that PS1PT=S2, we say that S1 is cogredient to S2 over R. It is clear that S1𝒮(n,R) if and only if S2𝒮(n,R). So cogredience of matrices over R is an equivalent relation on 𝒮(n,R). If S1𝒮(n,R), we call {PS1PTPGLn(R)} the cogredient classes of 𝒮(n,R) containing S1 over R. Let 𝒮0={0}, 𝒮1,,𝒮d be all distinct cogredient classes of 𝒮(n,R). As in  we define relations on 𝒮(n,R) by (1.2)Γi:={(A,B)A,BS(n,R),A-BSi},i=0,1,,d. Then the system (𝒮(n,R),{Γi}0id) is an association scheme of class d on the set 𝒮(n,R) and denoted by Sym(n,R).

Let p stand for an odd prime number in the following. When s=1, we know that the class number of Sym(n,𝔽pm) is given by d=2n and the association scheme Sym(n,𝔽pm) has been investigated in . When m=1, two kinds of cogredient standard forms of symmetric matrices over ps are given in [3, 4]. If n2, s>1 and p1 (mod 4), a complex formula to count the number of all distinct cogredient classes of 𝒮(n,ps) is given in , which shows that, for example,

if m is odd and s is odd, then (1.3)d+1=(m-12+1)+s10,orsi,i(m-12-s1-s2+s3+s4+s5+ε2+1)×[(s-11)+(s-12)(s1-11)++(s-1s1)]×(s-12s2)(s+12s3)(s-12s4)(s+12s5), where the meanings of m,s1,s2,s3,s4,s5,ε and formulas for other cases are referred to .

Then two problems arise. (1) Is there a simple and explicit formula to count the number of all distinct cogredient classes of 𝒮(n,ps)? (2) For arbitrary Galois ring R, in order to determine precisely the class number d of the association scheme Sym(n,R), we have to count the number of all distinct cogredient classes of 𝒮(n,R).

In Section 2 we give two kinds of cogredient standard forms for every symmetric matrix over arbitrary Galois ring R of odd characteristic. In Section 3 we obtain an explicit formula to count the number of all distinct cogredient classes of 𝒮(n,R), which is simpler than that of  for the special case R=ps.

Now, we list some properties for the Galois ring R which will be needed in the following sections. For general theory of Galois rings, one can refer to .

Lemma 1.1 (see [<xref ref-type="bibr" rid="B6">1</xref>, Theorem 14.11]).

R * = G 1 × G 2 where G1 is a cyclic group of order pm-1, and G2=1+p is a group of order p(s-1)m.

Proposition 1.2.

( i ) R * 2 is a subgroup of R* with index [R*:R*2]=2.

( ii ) For any zR*R*2, R*R*2=zR*2, and |R*2|=|zR*2|=(1/2)|R*|.

( iii ) For any uR* and ap, there exists cR* such that c2(u+a)=u.

Proof .

In the notation of Lemma 1.1. Let ξ be a generator of the cyclic group G1. Then ξ is of order pm-1. Since p is odd and pm-1 is even, ξ2 is of order (1/2)(pm-1) and G12=ξ2. Since p(s-1)m is odd and G2 is a commutative group of order p(s-1)m by Lemma 1.1, for every aG2, there exists a unique bG2 such that a=b2, so G22=G2. Moreover, by Lemma 1.1 each uR* can be uniquely expressed as u=gh where gG1 and hG2.

For every u=ghR* where gG1 and hG2, uR*2 if and only if there exist g1G1 and h1G2 such that gh=(g1h1)2=g12h12, which is then equivalent to that g=g12 and h=h12. So uR*2 if and only if uG12×G2 by Lemma 1.1. Then R*2=G12×G2 and so |R*2|=|G12|·|G2|=(1/2)(pm-1)·p(s-1)m=(1/2)|R*|. Hence, [R*:R*2]=2 by group theory.

Since [R*:R*2]=2, for any zR*R*2, we have R*=R*2zR*2 and R*2zR*2= by group theory. So |zR*2|=|R*|-|R*2|=(1/2)|R*| by the proof of (i).

Let uR* and ap. Then u-1(u+a)=1+u-1a1+p=G2. From this and by Lemma 1.1, there exists a unique element bG2R* such that u-1(u+a)=b2. Now, let c=b-1. Then cR* satisfying c2(u+a)=u.

Proposition 1.3.

Let -1R*2. Then for any zR*R*2, there exist x,yR* such that z=(1+x2)y2.

Proof.

Let uR*. Suppose that 1+u2R*. Then there exists aR such that 1+u2=ap. So u2=-(1-ap). Since p is odd and ps=0 in R, there exists bR such that (ups)2=-(1-ap)ps=-(1-ppsb)=-1. From upsR* we deduce -1R*2, which is a contradiction. Hence 1+u2R*. Therefore, σ:w1+w (for  allwR*2) is a mapping from R*2 to R*. Suppose that σ(R*2)R*2. Then for 1R*2, there exists w0R*2 such that σ(w0)=1+w0=1, which implies that w0=0, and we get a contradiction. So there exists xR* such that 1+x2R*2, that is, 1+x2R*R*2=zR*2 by Proposition 1.2. Then there exists cR* such that 1+x2=zc2, so (1+x2)y2=z, where y=c-1R*.

2. Cogredient Standard Forms of Symmetric Matrices

In this section, we give two kinds of cogredient standard forms of symmetric matrices over R corresponding to that of cogredient standard forms of symmetric matrices over finite fields (see , or , Theorems 1.22 and 1.25).

Notation 1.

For any nonnegative integer ν and zR*R*2, define (2.1)H2ν=(0I(ν)I(ν)0),H2ν+2,Δ=H2νΔ,whereΔ=(100-z),H2ν+1,(1)=H2ν(1),H2ν+1,(z)=H2ν(z).

Lemma 2.1.

For any ν+ and zR*R*2, zI(2ν) is cogredient to I(2ν).

Proof.

Let -1R*2. Then there exists uR* such that u2=-1, that is, 1+u2=0. Since p is an odd prime number, we have gcd(2,ps)=1 and so 2R*. Let P=2-1((1+z)u-1(1-z)u(1-z)(1+z)). Since R is a commutative ring, we have detP=(2-1)2[(1+z)(1+z)-u-1(1-z)u(1-z)]=(2-1)2·2·2z=zR*. Hence, PGL2(R). Then by (u-1)2=(u2)-1=-1 and u(1-z2)+u-1(1-z2)=u-1(u2+1)(1-z2)=0, we get (2.2)PI(2)PT=(2-1)2((1+z)u-1(1-z)u(1-z)(1+z))((1+z)u(1-z)u-1(1-z)(1+z))=(2-1)2(22z0022z)=zI(2), so zI(2) is cogredient to I(2).

Let -1R*2. Then by Proposition 1.3 there exist x,yR* such that (1+x2)y2=z. Let Q=(xyy-yxy). Then detQ=(1+x2)y2=zR* and so QGL2(R). By (1+x2)y2=z, a matrix computation shows that QI(2)QT=QQT=zI(2). Hence, zI(2) is cogredient to I(2) as well.

Then zI(2ν)=zI(2)zI(2)νs is cogredient to I(2ν)=I(2)I(2)νs.

Lemma 2.2.

Let zR*R*2 and ν+.

If -1R*2, then I(2ν) is cogredient to H2ν.

If -1R*2, then I(ν)zI(ν) is cogredient to H2ν.

Proof.

We select P1=2-1(I(ν)-I(ν)I(ν)I(ν)) and denote that M=2(I(ν)00-I(ν)). From P1(I(ν)I(ν)0I(ν))=(2-1I(ν)02-1I(ν)I(ν)) we deduce detP1=det(2-1I(ν))=(2-1)νR*. Hence P1GL2ν(R). Then by P1MP1T=2-1(I(ν)-I(ν)I(ν)I(ν))(I(ν)I(ν)I(ν)-I(ν))=H2ν, we see that M is cogredient to H2ν.

By -1R*2 there exists uR* such that -1=u2. Then M is cogredient to 2I(2ν). If 2R*2, 2I(2ν) is cogredient to I(2ν) by Lemma 2.1. If 2R*2, there exists wR* such that 2=w2, so 2I(2ν) is cogredient to I(2ν) as well. Therefore, I(2ν) is cogredient to H2ν in this case.

Let -1R*2. Then by Proposition 1.2 there exists cR* such that -1=zc2. Hence I(ν)zI(ν) is cogredient to (I(ν)00-I(ν)). If 2R*2, there exists wR* such that 2=w2, so (I(ν)00-I(ν)) is cogredient to M. If 2R*2, then -2=(-1)2R*2, and hence there exists aR* such that -2=a2, so (aI(2ν))H2ν(I(ν)00-I(ν))H2νT(aI(2ν))=M. Hence, (I(ν)00-I(ν)) is cogredient to M as well. Therefore, I(ν)zI(ν) is cogredient to H2ν.

Lemma 2.3.

Let zR*R*2 and D=diag(u1,,ur), where uiR*, i=1,,r and r+. Then, One has the following.

D is necessarily cogredient to either I(r) or I(r-1)(z). Moreover, these two matrices are not cogredient over R.

If r=2ν+1 is odd, then D is necessarily cogredient to either H2ν+1,(1) or H2ν+1,(z). Moreover, these two matrices are not cogredient. If r=2ν is even, then D is necessarily cogredient to either H2ν or H2(ν-1)+2,Δ. Moreover, these two matrices are not cogredient.

Proof.

(i) We may assume that u1,,utR*2 and ut+1,,urzR*2, where 0tr. Then D is cogredient to I(t)zI(r-t). If r-t is even, by Lemma 2.1  zI(r-t) is cogredient to I(r-t) and hence D is cogredient to I(t)I(r-t)=I(r). Now, let r-t be odd. If r-t=1, D is obviously cogredient to I(1)(z). If r-t3, by Lemma 2.1  zI(r-t-1) is cogredient to I(r-t-1), and hence D is cogredient to I(t)I(r-t-1)(z)=I(r-1)(z).

Suppose that I(r) is cogredient to I(r-1)(z) over R. Then there exists QGLr(R) such that QI(r)QT=I(r-1)(z). From this and by detQR*, we obtain that z=(detQ)2R*2, which is a contradiction. So I(r) and I(r-1)(z) are not cogredient over R.

(ii) We have one of the following two cases.

Let r=2ν+1 be an odd number. Then r-1=2ν is even and we have one of the following two cases.

Let -1R*2. Then I(2ν) is cogredient to H2ν by Lemma 2.2(i). From this and by (i) we deduce that D is cogredient to H2ν+1,(1) when D is cogredient to I(r), or D is cogredient to H2ν+1,(z) when D is cogredient to I(r-1)(z).

Let -1zR*2. Then we have one of the following two cases.

Let (1/2)(r-1)=ν be even. Then I(ν) is cogredient to zI(ν) by Lemma 2.1, so I(2ν) is cogredient to I(ν)zI(ν). Since I(ν)zI(ν) is cogredient to H2ν by Lemma 2.2(ii), by (i) we see that: D is cogredient to H2ν+1,(1) when D is cogredient to I(r), or D is cogredient to H2ν+1,(z) when D is cogredient to I(r-1)(z).

Let (1/2)(r-1)=ν be odd. Then ν=2ω+1 for some nonnegative integer ω and so r-1=4ω+2. By Lemma 2.1 we see that I(2ω) is cogredient to zI(2ω), and I(2) is cogredient to zI(2). Hence I(r)=I(2ω)I(2ω)I(2)(1) is cogredient to I(2ω)zI(2ω)zI(2)(1), which is then cogredient to I(2ω+1)zI(2ω+1)(z). Since I(2ω+1)zI(2ω+1) is cogredient to H2(2ω+1)=H2ν by Lemma 2.2(ii), I(r) is cogredient to H2ν+1,(z). Moreover, I(r-1)(z)=I(2ω)I(2ω)I(2)(z) is cogredient to I(2ω)zI(2ω)I(2)(z), which is then cogredient to I(2ω+1)zI(2ω+1)(1). Since I(ν)zI(ν) is cogredient to H2ν by Lemma 2.2(ii), I(r-1)(z) is cogredient to H2ν+1,(1). Therefore, D is necessarily cogredient to either H2ν+1,(1) or H2ν+1,(z) by (i).

Let r=2ν be an even number. Then r-2=2(ν-1) is also even and we have one of the following two cases.

Let -1R*2. Then -1=u2 for some uR* and so (100z) is cogredient to (100-z)=Δ. By Lemma 2.2(i) D is cogredient to H2ν when D is cogredient to I(r), or D is cogredient to H2(ν-1)+2,Δ when D is cogedient to I(r-1)(z)=I(2(ν-1))(100z).

Let -1zR*2. Then -1=zc2 for some cR*. By 1=(-z)c2, we see that I(2) is cogredient to Δ. Now, we have one of the following two cases.

Let ν be even. Then I(ν) is cogredient to zI(ν) by Lemma 2.1 and so I(r)=I(ν)I(ν) is cogredient to I(ν)zI(ν). From this and by Lemma 2.2(ii), we see that I(r) is cogredient to H2ν. Let ν=2. Since I(2) is cogredient to Δ and I(1)(z) is cogredient to H2 by Lemma 2.2(ii), I(3)(z)=I(2)I(1)(z) is cogredient to H2Δ=H2·1+2,Δ. Now, let ν4. Since ν-2 is even, I(ν-2) is cogredient to zI(ν-2) by Lemma 2.1, so I(ν-2)I(ν-2) is cogredient to I(ν-2)zI(ν-2). Hence, I(r-1)(z)=I(ν-2)I(ν-2)I(3)(z) is cogredient to I(ν-2)zI(ν-2)I(3)(z), which is then cogredient to I(ν-1)zI(ν-1)I(2). Since I(2) is cogredient to Δ, we see that I(r-1)(z) is cogredient to H2(ν-1)+2,Δ by Lemma 2.2(ii). Therefore, D is necessarily cogredient to either H2ν or H2(ν-1)+2,Δ by (i).

Let ν be odd. Then there exists nonnegative integer ω such that ν=2ω+1 and so r=4ω+2. Since I(2ω) is cogredient to zI(2ω) by Lemma 2.1, I(r)=I(2ω)I(2ω)I(2) is cogredient to I(2ω)zI(2ω)Δ, that is then cogredient to H2(2ω)+2,Δ=H2(ν-1)+2,Δ by Lemma 2.2(ii). Now, I(r-1)(z)=I(2ω)I(2ω)(1)(z) is cogredient to I(2ω)zI(2ω)(1)(z) by Lemma 2.1, which is then cogredient to I(2ω+1)zI(2ω+1). Hence I(r-1)(z) is cogredient to H2(2ω+1)=H2ν by Lemma 2.2(ii). Therefore, D is necessarily cogredient to either H2ν or H2(ν-1)+2,Δ by (i).

Theorem 2.4.

Let zR*R*2. Then every n×n symmetric matrix A over R is necessarily cogredient to one of the following matrices: (2.3)D(n,k,t;k1,,kt;r1,,rt):=diag(pr1D1,pr2D2,,prtDt,0(n-k)), where 0tkn, Di=I(ki) or I(ki-1)(z) for all i=1,,t, 0r1<r2<<rts-1, and ki+ satisfy Σi=1tki=k.

Proof.

The statement holds obviously if A=0 (corresponding to the case k=0) or n=1. Now, let n2 and A=(aij)n×n0. Then, there exist 1i0,j0n such that ai0j00 and τ(ai0j0)=min{τ(aij)aij0,  1i,jn}. Let s1=ν(ai0j0). Then 0s1s-1, and there exists P1GLn(R) such that P1AP1T=diag(u1ps1,B) where u1R* and B=(bij) is a (n-1)×(n-1) symmetric matrix over R satisfying B=0 or τ(bij)s1 for all bij0, 1i,jn-1. By induction there exists XGLn-1(R) such that XBXT=diag(u2ps2,,ukpsk,0(n-k)), where u2,,ukR* and s2sks-1. Then P=diag(1,X)P1GLn(R) satisfies PAPT=diag(u1ps1,,ukpsk,0(n-k)).

Now, there must exist t,ki+, i=1,,t and 0r1<<rts-1 such that s1==sk1=r1<sk1+1==sk1+k2=r2<<sk1+k2++kt-1+1==sk1+k2++kt-1+kt=rt. Then Σi=1tki=k and A is cogredient to M=diag(pr1M1,pr2M2,,prtMt,0(n-k)), where Mi=diag(uk1++ki-1+1,,uk1++ki-1+ki) is a ki×ki matrix over R for all i=1,,t. Since Mi is cogredient to Di for every 1it by Lemma 2.3(i), we deduce that A is cogredient to diag(pr1D1,pr2D2,,prtDt,0(n-k)).

Theorem 2.5.

Let zR*R*2. Then every n×n symmetric matrix A over R is necessarily cogredient to one of the following matrices: (2.4)H(n,k,t;k1,,kt;r1,,rt):=diag(pr1H1,pr2H2,,prtHt,0(n-k)), where Hi is a ki×ki matrix over R such that Hi is equal to either H2νi+1,(1) or H2νi+1,(z) when ki=2νi+1 is odd, and Hi is equal to either H2νi or H2(νi-1)+2,Δ when ki=2νi is even, for all i=1,,t; 0tkn, 0r1<r2<<rts-1, and ki+ satisfy Σi=1tki=k.

Proof.

It follows from Theorem 2.4 and the proof of Lemma 2.3(ii).

For any n×n symmetric matrix A, we call D(n,k,t;k1,,kt;r1,,rt) the cogredient standard form of kind (I) of A if A is cogredient to D(n,k,t;k1,,kt;r1,,rt), and call H(n,k,t;k1,,kt;r1,,rt) the cogredient standard form of kind (II) of A if A is cogredient to H(n,k,t;k1,,kt;r1,,rt).

3. The Number of Cogredient Classes of Symmetric Matrices

In order to count the number of all distinct cogredient classes of n×n symmetric matrices over R, we show that every n×n symmetric matrix over R has only one cogredient standard form of kind (I) first, then the number of all distinct cogredient classes of n×n symmetric matrices over R is equal to the number of all cogredient standard forms of kind (I) by Theorem 2.4.

Theorem 3.1.

The number 𝒞s,n of all distinct cogredient classes of n×n symmetric matrices over R is given by the following:

If ns, then 𝒞s,n=1+j=0n-1i=jn-1(ij)(sj+1)2j+1;

If ns+1, then 𝒞s,n=1+j=0s-1i=jn-1(ij)(sj+1)2j+1.

Proof.

Let D^:=diag(pr^1D^1,pr^2D^2,,pr^t^D^t^,0(n-k^)), where D^i=I(k^i) or I(k^i-1)(z) for all i=1,,t^, 0t^k^n, 0r^1<r^2<<r^t^s-1, and k^i+ satisfy Σi=1t^k^i=k^. In the notation of Theorem 2.4, by [7, Theorem D], it follows that D=D^ if D:=D(n,k,t;k1,,kt;r1,,rt) is cogredient to D^ over R. Hence, every n×n symmetric matrix over R has only one cogredient standard form of kind (I).

For any 1tkn, denote that S1={(k1,,kt)ki+,Σi=1t=k} and S2={(r1,,rt)ri,0r1<r2<<rts-1}. Then |S1|=(k-1t-1), |S2|=(st) if ts and, |S2|=0 if ts. From this and by Theorem 2.4 it follows that 𝒞s,n=1+k=1n(t=1k|S1|·|S2|·2t). Therefore, 𝒞s,n=1+j=0n-1i=jn-1(ij)(sj+1)2j+1 if ns and, 𝒞s,n=1+j=0s-1i=jn-1(ij)(sj+1)2j+1 if ns+1.

In the notations of Section 1, we see that the class number d of the association scheme Sym(n,R) is determined by d+1=𝒞s,n. Then by Theorem 3.1, we have the following corollary.

Corollary 3.2.

The class number of the association scheme Sym(n,R) is given by the following.

If ns, then d=j=0n-1i=jn-1(ij)(sj+1)2j+1;

If ns+1, then d=j=0s-1i=jn-1(ij)(sj+1)2j+1.

Example 3.3.

Let p be an odd prime number and s=2. Then by Theorem 3.1 the number 𝒞2,2 of all cogredient classes of 2×2 symmetric matrices over Galois ring GR(p2,p2m) is given by 𝒞2,2=1+j=01i=j1(ij)(2j+1)2j+1=13. In fact, for a fixed element zR*R*2, all cogredient standard forms of kind (I) of 2×2 symmetric matrices over GR(p2,p2m) are given by the following: (3.1)(0000),(1000),(z000),(p000),(zp000),(1001),(100z),(p00p),(p00zp),(100p),(z00p),(100zp),(z00zp). The number 𝒞2,3 of all cogredient classes of 3×3 symmetric matrices over GR(p2,p2m) is given by 𝒞2,3=1+j=01i=j2(ij)(2j+1)2j+1=25. In fact, all cogredient standard forms of kind (I) of 3×3 symmetric matrices over GR(p2,p2m) are given by the following: (J000) where J is one of matrices in (3.1), and (3.2)(100010001),(10001000z),(p000p000p),(p000p000zp),(10001000p),(1000z000p),(10001000zp),(1000z000zp),(1000p000p),(z000p000p),(1000p000zp),(z000p000zp).

Example 3.4.

Let p be an odd prime number and s=5. Then by Theorem 3.1 the number 𝒞5,4 of all cogredient classes of 4×4 symmetric matrices over Galois ring GR(p5,p5m) is given by 𝒞5,4=1+j=03i=j3(ij)(5j+1)2j+1=681; the number 𝒞5,7 of all cogredient classes of 7×7 symmetric matrices over GR(p5,p5m) is given by 𝒞5,7=1+j=04i=j6(ij)(5j+1)2j+1=6943.

Acknowledgment

This reaserach is supported in part by the National Science Foundation of China (No. 10971160) and Natural Science Foundation of Shandong provence (Grant No. ZR2011AQ004).

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