An Algorithmic Proof of Suslin s Stability Theorem for Polynomial Rings

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1 An Algorithmic Proof of Suslin s Stability Theorem for Polynomial Rings Hyungju Park Cynthia Woodburn Abstract Let k be a field Then Gaussian elimination over k and the Euclidean division algorithm for the univariate polynomial ring k[x] allow us to write any matrix in SL n (k) or SL n (k[x]), n 2, as a product of elementary matrices Suslin s stability theorem states that the same is true for SL n (k[x 1,, x m ]) with n 3 and m 1 In this paper, we present an algorithmic proof of Suslin s stability theorem, thus providing a method for finding an explicit factorization of a given polynomial matrix into elementary matrices Gröbner basis techniques may be used in the implementation of the algorithm 1 Introduction Immediately after proving the famous Serre s Conjecture (which is now known as the Quillen-Suslin theorem) in 1976 [Sus76], AA Suslin went on to prove the following K 1 -analogue of Serre s Conjecture [Sus77] Suslin s Stability Theorem Let R be a commutative Noetherian ring and n max(3, dim(r) + 2) Then, any n n matrix A = (f ij ) of determinant 1, with f ij elements of the polynomial ring R[x 1,, x m ], can be written as a product of elementary matrices over R[x 1,, x m ] Dept of Mathematics, University of California, Berkeley, CA 94720; park@mathberkeleyedu Dept of Mathematics, Pittsburg State University, Pittsburg, KS 66762; cwoodbur@mailpittstateedu 1

2 Recall that for any ring R, an n n elementary matrix E ij (a) over R is a matrix of the form I + a e ij where i j, a R and e ij is the n n matrix whose (i, j) entry is 1 and all other entries are zero Let SL n (R) be the group of all the n n matrices of determinant 1 whose entries are elements of R, and let E n (R) be the subgroup of SL n (R) generated by the elementary matrices Then Suslin s stability theorem can be expressed as SL n (R[x 1,, x m ]) = E n (R[x 1,, x m ]) for all n max(3, dim(r) + 2) In this paper, we develop an algorithmic proof of the above assertion over a field R = k For a given A SL n (k[x 1,, x m ]) with n 3, the algorithm produces elementary matrices E 1,, E t E n (k[x 1,, x m ]) such that A = E 1 E t Implementation of this algorithm involves use of the method of Gröbner bases, an important tool in computational algebra and computational algebraic geometry See [CLO92] and [Mis93] for more information on Gröbner bases and applications Definition 11 A square matrix A over a ring R is called realizable, if A can be written as a product of elementary matrices over R The contents of this paper are as follows: In Section 2, an algorithmic proof of the normality of E n (k[x 1,, x m ]) in SL n (k[x 1,, x m ]) for n 3 is given In Section 3, we give an algorithm for the Quillen Induction Process, a standard way of reducing a given problem over a ring to an easier problem over a local ring Using this Quillen Induction Algorithm, we reduce our realization problem over the polynomial ring R[x m ] to one over R M [x m ] s, where R = k[x 1,, x m 1 ] and M ranges over a finite set of a maximal ideals of R In Section 4, an algorithmic proof of the Elementary Column Property, a stronger version of the Unimodular Column Property, is given, and we note that this algorithm gives another constructive proof of the Quillen-Suslin theorem Using the Elementary Column Property, we 2

3 show that a realization algorithm for SL n (k[x 1,, x m ]) is obtained from a realization algorithm for matrices of the special form p q 0 r s 0 SL 3 (k[x 1,, x m ]), where p is monic in the last variable x m In Section 5, in view of the results in the preceding two sections, we note that a realization algorithm over k[x 1,, x m ] can be obtained from a p q 0 realization algorithm for the matrices of the special form r s 0 over R[X], where R is now a local ring and p is monic in X A realization algorithm for this case was already found by MP Murthy in [GM80] We reproduce Murthy s algorithm in this section In Section 6, we suggest using the Steinberg relations from algebraic K-theory to lower the number of elementary matrix factors in a factorization produced by our algorithm We also mention an ongoing effort of using our algorithm in Signal Processing 2 E n (k[x 1,, x m ]) SL n (k[x 1,, x m ]), for n 3 Lemma 21 The Cohn matrix A = ( ) A 0 SL (k[x, y]) is ( 1 + xy x 2 y 2 1 xy ) is not realizable, but Proof: The nonrealizability of A was first proved by P M Cohn in [Coh66], and a complete algorithmic criterion for the realizability of matrices in SL 2 (k[x 1,, x m ]) was recently developed in [THK94] Now consider ( ) A 0 = xy x 2 0 y 2 1 xy 0 3

4 Noting that 1 + xy x 2 0 y 2 1 xy 0 = I + x y 0 (y, x, 0), we see that the realizability of this matrix is a special case of Lemma 23 below Definition 22 Let n 2 A Cohn-type matrix is a matrix of the form I + av (v j e i v i e j ) v 1 where v = (k[x 1,, x m ]) n, i < j {1,, n}, a k[x 1,, x m ], v n and e i = (0,, 0, 1, 0,, 0) with 1 occurring only at the i-th position Lemma 23 Any Cohn-type matrix for n 3 is realizable Proof: First, consider the case i = 1, j = 2 In this case, B = I + a = = v 1 v n (v 2, v 1, 0,, 0) 1 + av 1 v 2 av av av 1 v av 3 v 2 av 3 v 1 I n 2 av n v 2 av n v av 1 v 2 av av av 1 v So, it is enough to show that I n A = 1 + av 1 v 2 av av av 1 v n E l1 (av l v 2 )E l2 ( av l v 1 ) l=3

5 is realizable for any a, v 1, v 2 k[x 1,, x m ] Let indicate that we are applying elementary operations, and consider the following: A = 1 + av 1 v 2 av1 2 0 av2 2 1 av 1 v av1 2 v av 1 v 2 v 2 av v 2 0 av av 1 v av 1 v 2 av1 2 v 1 av2 2 1 av 1 v 2 v v v 2 av 2 av v v v 2 0 av av 1 v 2 Keeping track of all the elementary operations involved in 21, we get (21) A = E 13 ( v 1 )E 23 ( v 2 )E 31 ( av 2 )E 32 (av 1 )E 13 (v 1 )E 23 (v 2 )E 31 (av 2 )E 32 ( av 1 ) In general (ie, for arbitrary i < j), v 1 B = I + a (0,, 0, v j, 0,, 0, v i, 0,, 0) v n with v j occuring at the i-th position and v i occuring at the j-th position Therefore, we have B = 1 av 1 v j av 1 v i av i v j avi 2 avj 2 1 av i v j v n v j v n v i 1 5

6 = av i v j avi 2 1 av i v j av 2 j 1 l n l i,j E li (av l v j )E lj ( av l v i ) = E it ( v i )E jt ( v j )E ti ( av j )E tj (av i )E it (v i )E jt (v j )E ti (av j )E tj ( av i ) E li (av l v j )E lj ( av l v i ), 1 l n l i,j where t {1,, n} can be chosen to be any number other than i or j Since a Cohn-type matrix is realizable, any product of Cohn-type matrices is also realizable This observation motivates the following generalization of the above lemma Definition 24 Let R be a ring and v = (v 1,, v n ) t R n for some n IN Then v is called a unimodular column vector if its components generate R, ie if there exist g 1,, g n R such that v 1 g v n g n = 1 Remark 25 When R = k[x 1,, x m ] is a polynomial ring and v R n is a unimodular vector, we can explicitly find these g i s by either using the effective Nullstellensatz (see [FG90]) or Gröbner bases (see [CLO92]) Lemma 26 Suppose that A SL n (k[x 1,, x m ]) with n 3 can be written in the form A = I +v w for a unimodular column vector v and a row vector w over k[x 1,, x m ] such that w v = 0 Then A is realizable Proof: Since v = (v 1,, v n ) t is unimodular, we can find g 1,, g n k[x 1,, x m ] such that v 1 g v n g n = 1 This, combined with w v = w 1 v w n v n = 0, yields a new expression for w: w = i<j a ij (v j e i v i e j ) 6

7 where a ij = w i g j w j g i Now, A = I + v a ij (v j e i v i e j ) i<j = I + v a ij (v j e i v i e j ) i<j = ( I + v aij (v j e i v i e j ) ) i<j Each factor on the right hand side of this equation is a Cohn-type matrix and thus realizable, so A is also realizable Corollary 27 BE ij (a)b 1 is realizable for any B GL n (k[x 1,, x m ]) with n 3 and a k[x 1,, x m ] Proof: Note that i j, and BE ij (a)b 1 = I + (i-th column vector of B) a (j-th row vector of B 1 ) Let v be the i-th column vector of B and w be a times the j-th row vector of B 1 Then (i-th row vector of B 1 ) v = 1 implies v is unimodular, and w v is clearly zero since i j Therefore, BE ij (a)b 1 = I + v w satisfies the condition of the above lemma, and is thus realizable Corollary 28 For n 3, E n (k[x 1,, x m ]) is a normal subgroup of SL n (k[x 1,, x m ]) Proof: Let A SL n (k[x 1,, x m ]) and E E n (k[x 1,, x m ]) Then the above corollary gives us an algorithm for finding elementary matrices E 1,, E t such that A 1 EA = E 1 E t 3 Gluing of Local Realizability Let R = k[x 1,, x m 1 ], X = x m and M Max(R) ={maximal ideals of R} For A SL n (R[X]), we let A M SL n (R M [X]) be its image under the canonical mapping SL n (R[X]) SL n (R M [X]) We will occasionally write A = A(X) to emphasize that we are viewing the entries of the matrix A as polynomials in one variable Now consider the following analogue of Quillen s patching theorem for elementary matrices: 7

8 Suppose n 3 and A SL n (R[X]) Then A is realizable over R[X] if and only if A M SL n (R M [X]) is realizable over R M [X] for every M Max(R) While a non-constructive proof of this assertion is given in [Sus77] and a more general functorial treatment of this Quillen Induction Process can be found in [Knu91], we will give a constructive proof for it here, thus providing a patching algorithm with input certain local factorizations of a given matrix A and output a global factorization of A into elementary matrices Since the necessity of the condition is clear, we have to prove the following theorem Theorem 31 (Quillen Induction Algorithm) Let A SL n (R[X]) If A M E n (R M [X]) for every M Max(R), then A E n (R[X]) Proof: Let a 1 = (0,, 0) k m 1, and let M 1 = {g k[x 1,, x m 1 ] g(a 1 ) = 0} be the corresponding maximal ideal Then by assumption, A M1 is realizable over R M1 [X] Hence, we can write A M1 = j ( ) cj E sj t j d j where c j, d j R, d j M 1 Letting r 1 = j d j / M 1, we can rewrite 22 as A M1 = j ( ) cj k j d k E sj t j E n (R r1 ) E n (R M1 ) r 1 (22) Denote an algebraic closure of k by k Inductively, let a j k m 1 be a common zero of r 1,, r j 1 and M j = {g k[x 1,, x m 1 ] g(a j ) = 0} be the corresponding maximal ideal of R for each j 2 (See [CLO92] for details and references for using Gröbner bases to find a common zero of a finite set of polynomials) Define r j / M j in the same way as r 1 above, so that A Mj E n (R rj [X]) Since a j is a common zero of r 1,, r j 1 in this construction, we immediately see that r 1,, r j 1 M j = {g R g(a j ) = 0} But noting r j / M j, we conclude that r j / r 1 R+ +r j 1 R Now, since R is Noetherian, we will get to some l after a finite number of steps such that r 1 R + + r l R = R (We 8

9 can use Gröbner bases to determine when 1 R is in the ideal r 1 R + + r l R, eg see [CLO92]) Let d be a natural number Then since r d 1R + + r d l R = R, we can find g 1,, g l R such that r d 1 g r d l g l = 1 Now, we express A(X) SL n (R[X]) in the following way: A(X) = A(X Xr d 1 g 1) [A 1 (X Xr d 1 g 1)A(X)] = A(X Xr d 1g 1 Xr d 2g 2 ) [A 1 (X Xr d 1g 1 Xr d 2g 2 )A(X Xr d 1g 1 )] [A 1 (X Xr d 1 g 1)A(X)] = = l l A(X Xri d g i ) [A 1 (X Xr d l 1 i g i )A(X Xri d g i )] i=1 i=1 i=1 [A 1 (X Xr1 d g 1)A(X)] Note here that the first matrix A(X l i=1 Xr d i g i ) = A(0) on the right hand side is in E n (R) by the induction hypothesis We will now show that for a sufficiently large d, each bracketed expression in the above equation is actually in E n (R[X]), so that A itself is in E n (R[X]) To this end, we let A Mi = A i and identify A SL n (R[X]) with A i SL n (R Mi [X]) Then each bracketed expression is of the form A 1 i (cx)a i ((c + ri d g)x) Claim: For any c, g R, we can find a sufficiently large d such that A 1 i (cx)a i ((c + ri dg)x) E n(r[x]) for i = 1,, l Let D i (X, Y, Z) = A 1 i (Y X)A i ((Y + Z) X) E n (R ri [X, Y, Z]) and write D i in the form h D i = E sj t j (b j + Zf j ) j=1 where b j R ri [X, Y ] and f j R ri [X, Y, Z] From now on, the elementary matrix E sj t j (a) will be simply denoted as E j (a) for notational convenience 9

10 For p = 1,, h, define C p by p C p = E j (b j ) E n (R ri [X, Y ]) j=1 Then the C p s satisfy the following recursive relations: E 1 (b 1 ) = C 1, E p (b p ) = Cp 1 1 p C h = I (2 p h), Hence, using the fact that E ij (a + b) = E ij (a)e ij (b), we have D i = = h E j (b j + Zf j ) j=1 h E j (b j )E j (Zf j ) j=1 = [E 1 (b 1 )E 1 (Zf 1 )][E 2 (b 2 )E 2 (Zf 2 )] [E h (b h )E h (Zf h )] = [C 1 E 1 (Zf 1 )][C1 1 C 2 E 2 (Zf 2 )] [C 1 h 1 C he h (Zf h )] h = C j E j (Zf j )Cj 1 j=1 Now in the same way as in the proof of Lemma 26 and Corollary 27, we can write C j E j (Zf j )Cj 1 as a product of Cohn-type matrices, ie for any given j {1,, h}, let v = v 1 v n C j E sj t j (Zf j )C 1 j = be the s j-th column vector of C j Then 1 γ<δ n [I + v Zf j a γδ (v γ e δ v δ e γ )] for some a γδ R ri [X, Y ] Also, we can find a natural number d such that v γ = v γ r d i, a γδ = a γδ, f ri d j = f j ri d 10

11 for some v γ, a γδ R[X, Y ], f j R[X, Y, Z] Now, replacing Z by r4d i g, we see that all the Cohn-type matrices in the above expression for C j E j (Zf j )Cj 1 have denominator-free entries Therefore, C j E j (ri 4d gf j )Cj 1 E n (R[X, Y ]) Since this is true for each j, we conclude that for a sufficiently large d, D i (X, Y, r d i g) = h j=1 C j E j (r d i gf j)c 1 j E n (R[X, Y ]) Now, setting Y = c proves the claim, and completes the proof of the theorem 4 Reduction to SL 3 (k[x 1,, x m ]) Let A SL n (k[x 1,, x m ]) with n 3, and let v be its last column vector Then v is unimodular (Recall that the cofactor expansion along the last column gives a required relation) Now, if we can reduce v to e n = (0, 0,, 0, 1) t by applying elementary operations, ie if we can find B E n (k[x 1,, x m ]) such that Bv = e n, then 0 BA = à 0 p 1 p n 1 1 for some à SL n 1(k[x 1,, x m ]) and p i k[x 1,, x m ] for i = 1,, n 1 Hence, BAE n1 ( p 1 ) E n(n 1) ( p n 1 ) = ( ) à Therefore, our problem of expressing A SL n (k[x 1,, x m ]) as a product of elementary matrices is now reduced to the same problem for à SL n 1 (k[x 1,, x m ]) By repeating this process, we get to the problem of 11

12 expressing A = p q 0 r s 0 SL 3 (k[x 1,, x m ]) as a product of elementary matrices, which is the subject of the next section In this section, we will develop an algorithm for finding elementary operations that reduce a given unimodular column vector v (k[x 1,, x m ]) n to e n Also, as a corollary to this Elementary Column Property, we give an algorithmic proof of the Unimodular Column Property which states that for any given unimodular column vector v (k[x 1,, x m ]) n, there exists a unimodular matrix B, ie a matrix with constant nonzero determinant, over k[x 1,, x m ] such that Bv = e n Recently, A Logar, B Sturmfels in [LS92] and N Fitchas, A Galligo in [FG90], [Fit93] have given different algorithmic proofs of this Unimodular Column Property, thereby giving algorithmic proofs of the Quillen- Suslin theorem Therefore, our algorithm gives another constructive proof of the Quillen-Suslin theorem The second author has given a different algorithmic proof of the Elementary Column Property based on a localization and patching process in [Woo94] Definition 41 For a ring R, Um n (R) = {n-dimensional unimodular column vectors over R} As in Section 3, let R = k[x 1,, x m 1 ] and X ( = x m Then ) k[x 1,, x m ] = A 0 R[X] By identifying A SL 2 (R[X]) with SL 0 I n (R[X]), we n 2 can regard SL 2 (R[X]) as a subgroup of SL n (R[X]) As before, we use the notation A = A(X) and v = v(x) to emphasize that we are viewing the entries of a matrix A or a vector v as polynomials in one variable Now consider the following lemma and theorem, which will be used to prove the Elementary Column Property Lemma 42 Let f 1, f 2, b, d R[X] and let r be the resultant of f 1 and f 2 Then there exists B SL 2 (R[X]) such that B ( ) f1 (b) f 2 (b) = ( ) f1 (b + rd) f 2 (b + rd) Proof: By a property of the resultant of two polynomials, we can find g 1, g 2 R[X] such that f 1 g 1 + f 2 g 2 = r (See [CLO92] or [GM80] for details) Let 12

13 s 1, s 2, t 1, t 2 R[X, Y, Z] be the polynomials defined by Now, define f 1 (X + Y Z) = f 1 (X) + Y s 1 (X, Y, Z), f 2 (X + Y Z) = f 2 (X) + Y s 2 (X, Y, Z), g 1 (X + Y Z) = g 1 (X) + Y t 1 (X, Y, Z), g 2 (X + Y Z) = g 2 (X) + Y t 2 (X, Y, Z) B 11 = 1 + s 1 (b, r, d) g 1 (b) + t 2 (b, r, d) f 2 (b), B 12 = s 1 (b, r, d) g 2 (b) t 2 (b, r, d) f 1 (b), B 21 = s 2 (b, r, d) g 1 (b) t 1 (b, r, d) f 2 (b), B 22 = 1 + s 2 (b, r, d) g 2 (b) + t 1 (b, r, d) f 1 (b) ( ) B11 B Then one checks easily that B = 12 satisfies the desired property B 21 B 22 and that B SL 2 (R[X]) v 1 (X) Theorem 43 Suppose v(x) = Um n(r[x]), and v 1 (X) is v n (X) monic in X Then there exists B 1 SL 2 (R[X]) and B 2 E n (R[X]) such that B 1 B 2 v(x) = v(0) Proof: Let a 1 = (0,, 0) k m 1 Define M 1 = {g k[x 1,, x m 1 ] g(a 1 ) = 0} and k 1 = R/M 1 as the corresponding maximal ideal and residue field, respectively By hypothesis v (R[X]) n is a unimodular column vector, so its image v in (k 1 [X]) n = ((R/M 1 )[X]) n is also unimodular Since k 1 [X] is a principal ideal ring, the ideal < v 2,, v n > is generated by a single element, G 1 = gcd( v 2,, v n ) Then v 1 and G 1 generate the unit ideal in k 1 [X] since v 1, v 2,, v n generate the unit ideal Using the Euclidean division algorithm for k 1 [X], we can find E 1 E n 1 (k 1 [X]) such that E 1 v 2 v n = 13 G 1 0 0

14 By identifying k 1 with a subring of R, we may regard E 1 as an element of E n (R[X]) and G 1 as an element of R[X] Then, v 1 ( ) G q 12 v = q 13 0 E 1 for some q 12,, q 1n M 1 [X] Now, define r 1 R by r 1 = Res(v 1, G 1 + q 12 ), the resultant of v 1 and G 1 + q 12, and find f 1, h 1 R[X] such that q 1n f 1 v 1 + h 1 (G 1 + q 12 ) = r 1 Since v 1 is monic, and v 1 and G 1 k 1 [X] generate the unit ideal, we have r 1 = Res(v 1, G 1 + q 12 ) = Res( v 1, G 1 ) 0 Therefore, r 1 / M 1 Denote an algebraic closure of k by k Inductively, let a j k m 1 be a common zero of r 1,, r j 1 and M j be the corresponding maximal ideal of R for each j 2 Define r j / M j in the same way as above Define also, E j E n 1 (k j [X]), G j k j [X], f j, h j R[X], and q j2,, q jn M j [X] in an analogous way As we saw in the proof of Theorem 31, there is a finite l such that r 1 R+ +r l R = R Find g i s in R such that r 1 g 1 + +r l g l = 1 Now, define b 0, b 1,, b l R[X] as b 0 = 0 b 1 = r 1 g 1 X b 2 = r 1 g 1 X + r 2 g 2 X b l = r 1 g 1 X + r 2 g 2 X + + r l g l X = X Then these b i s satisfy the recursive relations: b 0 = 0 b i = b i 1 + r i g i X for i = 1,, l 14

15 Claim: For each i {1,, l}, there exists B i SL 2 (R[X]) and B i E n (R[X]) such that v(b i ) = B i B i v(b i 1) Using this and the fact that E n (R[X]) SL 2 (R[X]) SL 2 (R[X]) E n (R[X]) (Corollary 27), we inductively get v(x) = v(b l ) = B l B l v(b l 1) = BB v(b 0 ) = BB v(0) for some B SL 2 (R[X]) and B E n (R[X]) Therefore, it is enough to prove the above claim Proof of claim: Let G i = G i + q i2 Then For 3 j n, we have ( ) 1 0 v(x) = 0 E(X) v 1 (X) G i (X) q i3 (X) q in (X) q ij (b i ) q ij (b i 1 ) (b i b i 1 ) R[X] = r i g i X R[X] Since r i R doesn t depend on X, we have r i = f i (X)v 1 (X) + h i (X) G i (X) = f i (b i 1 )v 1 (b i 1 ) + h i (b i 1 ) G i (b i 1 ) = a linear combination of v 1 (b i 1 ) and G i (b i 1 ) over R[X] Therefore, we see that for 3 j n, q ij (b i ) = q ij (b i 1 ) + a linear combination of v 1 (b i 1 ) and G i (b i 1 ) overr[x] 15

16 Hence we can find C E n (R[X]) such that C ( ) 1 0 v(b 0 E(b i 1 ) i 1 ) = C v 1 (b i 1 ) G i (b i 1 ) q i3 (b i 1 ) q in (b i 1 ) = v 1 (b i 1 ) G i (b i 1 ) q i3 (b i ) q in (b i ) Now, by Lemma 42, we can find B SL 2 (R[X]) such that B ( ) ( ) v1 (b i 1 ) v1 (b i ) = G i (b i 1 ) G i (b i ) Finally, define B SL n (R[X]) as follows: B = Then this B satisfies ( ) ( ) ( ) 1 0 B E(b i ) 1 C 0 I n 2 0 E(b i ) Bv(b i 1 ) = v(b i ), and, by using the normality of E n (R[X]) again, we see that B SL 2 (R[X])E n (R[X]), which proves the claim and completes the proof of the theorem Remark 44 Note that the groups GL n (k[x 1,, x m ]) and E n (k[x 1,, x m ]) act on the set Um n (k[x 1,, x m ]) by matrix multiplication If the group action of E n (k[x 1,, x m ]) on Um n (k[x 1,, x m ]) is transitive, then we get a desired algorithm in the following way: For any n-dimensional unimodular column vectors v, v over k[x 1,, x m ], we can find B E n (k[x 1,, x m ]) such that Bv = v Now, let v = e n Theorem 45 (Elementary Column Property) For n 3, the group E n (k[x 1,, x m ]) acts transitively on the set Um n (k[x 1,, x m ]) 16

17 Corollary 46 (Unimodular Column Property) For n 2, the group GL n (k[x 1,, x m ]) acts transitively on the set Um n (k[x 1,, x m ]) Proof: For n 3, the Elementary Column Property clearly implies the Unimodular Column Property since a product of elementary matrices is always unimodular, ie has a constant nonzero determinant If n = 2, for any v = (v 1, v 2 ) t Um 2 (k[x 1,, x m ]), by using Buchberger s algorithm for computing a Gröbner basis, we can find g 1, g 2 k[x ( 1,, x) m ] such that v 1 g 1 + v 2 g 2 = 1 Then the unimodular matrix U v = v2 v 1 satisfies U g 1 g v v = e 2 Therefore we see that, for any v, w 2 Um 2 (k[x 1,, x m ]), Uw 1U v v = w where Uw 1U v GL 2 (k[x 1,, x m ]) Proof of Theorem 45: Since the Euclidean division algorithm for k[x 1 ] proves the theorem for m = 1, we may assume by induction the statement of the theorem for R = k[x 1,, x m 1 ] Let X = x m and v = v 1 v n Um n (R[X]) We may also assume that v 1 is monic by applying a change of variables (as in the proof of the Noether Normalization Lemma) Now by Theorem 43, we can find B 1 SL 2 (R[X]) and B 2 E n (R[X]) such that B 1 B 2 v(x) = v(0) R Then by the inductive hypothesis, we can find B E n (R) such that Therefore, we get B v(0) = e n v = B 1 2 B 1 1 B 1 e n By the normality of E n (R[X]) in SL n (R[X]) (Corollary 27), we can write B1 1 B 1 = B B1 1 for some B E n (R[X]) Since p q 0 0 r s 0 0 B1 1 = 0 0 I n

18 for some p, q, r, s R[X], we have v = B2 1 1 B 1 e n = (B 1 = (B 1 2 B )B1 1 e n 2 B ) = (B 1 2 B )e n p q 0 0 r s I n where B2 1 B E n (R[X]) Since we have this relationship for any v Um n (R[X]), we get the desired transitivity 5 Realization Algorithm for SL 3 (R[X]) Now, we give a realization algorithm for matrices of the form p q 0 r s 0 SL 3 (k[x 1,, x m ]) Again, by applying a change of variables, we may assume that p k[x 1,, x m ] is a monic polynomial in the last variable x m In view of the Quillen Induction Algorithm developed in Section 3, we see that it is enough to develop a realization algorithm for matrices of the form p q 0 r s 0 SL 3 (R[X]), in the case where R is a commutative local ring and p R[X] is a monic polynomial A realization algorithm for this case was obtained by MP Murthy [GM80] We present below a slightly modified version Lemma 51 Let L be a commutative ring, and a, a, b L Then, the following are true 1 (a, b) and (a, b) are unimodular over L if and only if (aa, b) is unimodular over L 18

19 2 For any c, d L such that aa d bc = 1, there exist c 1, c 2, d 1, d 2 L such that ad 1 bc 1 = 1, a d 2 bc 2 = 1, and aa b 0 c d 0 a b 0 c 1 d 1 0 a b 0 c 2 d 2 0 (mod E 3 (L)) Proof: (1) If (aa, b) is unimodular over L, there exist h 1, h 2 L such that h 1 (aa ) + h 2 b = 1 Now (h 1 a ) a + h 2 b = 1 implies (a, b) is unimodular, and (h 1 a) a + h 2 b = 1 implies (a, b) is unimodular Suppose, now, that (a, b) and (a, b) are unimodular over L Then, we can find h 1, h 2, h 1, h 2 L such that h 1 a + h 2 b = 1, h 1a + h 2b = 1 Now, let g 1 = h 1 h 1, g 2 = h 2 + a h 2 h 1, and consider g 1 aa + g 2 b = h 1 h 1aa + (h 2 + a h 2 h 1)b = h 1 a (h 1 a + h 2 b) + h 2 b = h 1 a + h 2 b = 1 So we have a desired unimodular relation (2) If c, d L satisfy aa d bc = 1, then (aa, b) is unimodular, which in turn implies that (a, b) and (a, b) are unimodular Therefore, we can find c 1, d 1, d 1, d 2 L such that ad 1 bc 1 = 1 and a d 2 bc 2 = 1 For example, we can let Now, consider aa b 0 c d 0 c 1 = c 2 = c, d 1 = a d, d 2 = ad = E 21 (cd 1 d 2 d(c 2 + a c 1 d 2 )) aa b 0 c 2 + a c 1 d 2 d 1 d 2 0 = E 21 (cd 1 d 2 d(c 2 + a c 1 d 2 ))E 23 (d 2 1)E 32 (1)E 23 ( 1) a b 0 a b 0 c 1 d 1 0 E 23 (1)E 32 ( 1)E 23 (1) c 2 d 2 0 E 23 ( 1)E 32 (1)E 23 (a 1)E 31 ( a c 1 )E 32 ( d 1 ) 19

20 This explicit expression shows that aa b 0 a b 0 c d 0 c 1 d 1 0 a b 0 c 2 d 2 0 (mod E 3 (L)) Theorem 52 Suppose (R, M) is a commutative local ring, and A = SL 3 (R[X]) where p is monic Then A is realizable over R[X] p q 0 r s 0 Proof: We induct on deg(p) If deg(p) = 0, then p = 1, and A is clearly realizable Now, suppose deg(p) = d > 0 and deg(q) = l Since p R[X] is monic, we can find f, g R[X] such that q = fp + g, deg(g) < d Then, AE 12 ( f) = p q fp 0 r s fr 0 = p g 0 r s fr 0 Hence we may assume deg(q) < d Now, we note that either p(0) or q(0) is a unit in R, otherwise, we would have p(0)s(0) q(0)r(0) M, which contradicts the fact that ps qr = p(0)s(0) q(0)r(0) = 1 Consider these two cases separately Case 1: Suppose that q(0) is a unit We have AE 21 ( q(0) 1 p(0)) = p q(0) 1 p(0)q q 0 r q(0) 1 p(0)s s 0 So, we may assume p(0) = 0 Now, write p = Xp Then, by Lemma 51, we can find c 1, d 1, c 2, d 2 R[X] such that Xd 1 qc 1 = 1, p d 2 qc 2 = 1 and p q 0 X q 0 p q 0 r s 0 c 1 d 1 0 c 2 d 2 0 (mod E 3 (R[X])) 20

21 Since p is monic and deg(p ) < d, the second matrix on the right hand side is realizable by the induction hypothesis As for the first matrix, we may assume that q is a unit of R since we can assume deg(q) < deg(x) = 1 and q(0) is a unit Then invertibility of q leads easily to an explicit factorization of X q 0 c 1 d 1 0 into elementary matrices Case 2: Suppose that q(0) is not a unit First we claim that there exist p, q R[X] such that deg(p ) < l, deg(q ) < d and p p q q = 1 To prove this claim, we let r R be the resultant of p and q Then there exist f, g R[X] with deg(f) < l, deg(g) < d such that fp + gq = r Since p is monic and p, q R[X] generate the unit ideal, we see that r / M, hence r is a unit or R Now, setting p = f/r, q = g/r proves the claim Also note that since p (0)p(0) q (0)q(0) = 1 and q(0) M, then p (0) / M This means that q(0) + p (0) is a unit Now, we have that p q 0 r s 0 = E 21 (rp sq ) p q 0 q p 0 = E 21 (rp sq )E 12 ( 1) p + q q + p 0 q p 0 Noting that the last matrix on the right hand side is realizable by Case 1, p q 0 since q(0) + p (0) is a unit and deg(p + q ) = d, we see that r s 0 is also realizable 6 Eliminating Redundancies When applied to a specific matrix, the algorithm presented in this paper will produce a factorization into elementary matrices, but this factorization may not have minimal length The Steinberg relations [Mil71] from algebraic K theory provide a method for improving a given factorization by eliminating 21

22 some of the unnecessary factors The Steinberg relations which elementary matrices satisfy are 1 E ij (0) = I; 2 E ij (a)e ij (b) = E ij (a + b); 3 For i l, [E ij (a), E jl (b)] = E ij (a)e jl (b)e ij ( a)e jl ( b) = E il (ab); 4 For j l, [E ij (a), E li (b)] = E ij (a)e li (b)e ij ( a)e li ( b) = E lj ( ab); 5 For i p, j l, [E ij (a), E lp (b)] = E ij (a)e lp (b)e ij ( a)e lp ( b) = I The realization algorithm of this paper can be implemented together with a Redundancy Elimination Algorithm based on the above set of relations, using existing computer algebra systems As suggested in [THK94], an algorithm of this kind has application in Signal Processing since it gives a way of expressing a given multidimensional filter bank as a cascade of simpler filter banks called elementary ladder steps Acknowledgements The authors wish to thank A Kalker, TY Lam, R Laubenbacher, B Sturmfels and M Vetterli for all the valuable support, insightful discussions and encouragement References [Coh66] PM Cohn On the structure of the GL 2 of a ring Inst Hautes Études Sci Publ Math No 30, , 1966 [CLO92] D Cox, J Little and D O Shea Ideals, Varieties and Algorithms Springer, 1992 [FG90] [Fit93] N Fitchas and A Galligo Nullstellensatz effectif et conjecture de Serre (théorème de Quillen-Suslin) pour le calcule formel Math Nachr 149, , 1990 N Fitchas Algorithmic aspects of Suslin s proof of Serre s conjecture Comput Complexity 3, 31 55,

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