An algorithm for unimodular completion over Noetherian rings Abdessalem Mnif (1), Ihsen Yengui (2) April 9, 2010

Abstract We give an algorithm for the well-known result asserting that if R is a polynomial ring in a finite number of variables over a Noetherian ring A of Krull dimension d < ∞, then for n ≥ max(3, d+2), SLn (R) acts transitively on Umn (R). For technical reasons we demand that the Noetherian ring A has a theory of Gr¨obner bases and contains an infinite set E = {y1 , y2 , ...} such that yi − yj is invertible when i $= j. The most important guiding examples are affine rings K[x1 , . . . , xm ]/I over an infinite field K. Moreover, we give an algorithmic proof of Suslin’s stabilty theorem over these rings.

MSC 2000 : 13C10, 19A13, 14Q20, 03F65. Key words : Quillen-Suslin’s Theorem, Suslin’s Stability Theorem, Constructive Mathematics, Computer Algebra.

Introduction It has been known since 1958 that projective modules over R = K[X1 , . . . , Xk ], K a field, are stably free. In 1976, Quillen and Suslin proved independently that such modules are free, giving a positive answer to Serre’s conjecture [18] which became subsequently known as the Quillen-Suslin theorem ([10] is an excellent exposition). This is in fact equivalent to the fact that GLn (R) acts transitively on Umn (R) = { t (x1 , . . . , xn ) ∈ Rn such that &x1 , . . . , xn ' = R} the set of unimodular vectors with n entries in R. Our principal motivation is to obtain an algorithmic proof of the following result (see [9, 10, 19]) of which the Quillen-Suslin theorem is a particular case: If R is a polynomial ring in a finite number of variables over a Noetherian ring A of Krull dimension d < ∞, then for n ≥ max(3, d + 2), SLn (R) acts transitively on Umn (R).

For the purpose of carrying out all the necessary computations algorithmically we will suppose that we have a theory of Gr¨obner bases over A. More precisely, following [1] (page 204), we will suppose that A is a Noetherian ring such that: (i) Given a, a1 , . . . , am ∈ A, there is an algorithm to determine whether a ∈ &a1 , . . . , am ' and if it is, to compute b1 , . . . , bm ∈ A such that a = a1 b1 + · · · + am bm ; (ii) Given a, a1 , . . . , am ∈ A, there is an algorithm that computes a set of generators for the A-module Syz(a1 , . . . , am ) = {(b1 , . . . , bm ) ∈ Am such that a1 b1 + · · · + am bm = 0}.

Such a ring A will be called a computable Noetherian ring. For technical reasons, we will in addition suppose that A contains an infinite set E = {y1 , y2 , ...} such that yi − yj is invertible when i $= j. This property was studied in [3]. In case A has only finitely many maximal ideals M1 , . . . , Ms , it is equivalent (see Theorem 2.2 of [3]) to the fact that all the residue fields A/Mi are infinite. 1 2

Departement of Mathematics, Faculty of Sciences of Sfax, 3038 Sfax, TUNISIA. Departement of Mathematics, Faculty of Sciences of Sfax, 3038 Sfax, TUNISIA, email: [email protected].

1

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Unimodular completion over Noetherian rings

For us, the most important guiding examples are affine rings over infinite fields, that is quotients of polynomial rings K[x1 , . . . , xm ]/I, where K is an infinite field and I is an ideal of K[x1 , . . . , xm ] (see the paragraph “ Algorithmic computations in K[x1 , . . . , xn ]/I ” of [4]). Some of the obtained results are the missing keys for obtaining an algorithm for unimodular completion using elementary operations like what Park and Woodburn [16] did in case A is a field: If R is a polynomial ring in a finite number of variables over a Noetherian ring A of Krull dimension d < ∞, then for n ≥ max(3, d+2), En (R) (the subgroup of SLn (R) generated by elementary matrices) acts transitively on Umn (R). This result together with some established lemmas enable us to give an algorithmic proof of Suslin’s stability theorem for computable Noetherien rings of Krull dimension 1 and containing an infinite set E = {y1 , y2 , ...} such that yi − yj is invertible when i $= j: Suslin’s stability theorem : If R is a polynomial ring in a finite number of variables over a Noetherien rings A of Krull dimension 1, then every matrix in SLn (R), where n ≥ 3, is a product of elementary matrices.

It is worth pointing out that the results obtained by Gago-Vargas [7] about Suslin’s stability theorem for A[X1 , . . . , Xn ], A a euclidean domain, clearly does not cover quotients of polynomial rings. We think that the main features of the obtained algorithm for unimodular completion are: new, simple, explicit, does not use prime nor maximal ideals. Another issue raised by this paper is a constructive deciphering of a lemma of Suslin which played a central role in the resolution of Serre’s conjecture. This lemma says that for a commutative ring A if &v1 (X), . . . , vn (X)' = A[X] where v1 is monic and n ≥ 3, then there exist γ1 , . . . , γ! ∈ En−1 (A[X]) such that &Res(v1 , e1 .γ1 t (v2 , . . . , vn )), . . . , Res(v1 , e1 .γ! t (v2 , . . . , vn ))' = A. The constructive proof we give may be a model for miming constructively abstract proofs in which one works modulo each maximal ideal to prove that a given ideal contains 1. The undefined terminology is standard as in [9, 10].

1

Suslin’s lemma

It is worth pointing out that the first two subsections of this section are extracted from a non published note of the second author [20].

1.1

A constructive deciphering of a lemma of Suslin

One principal motivation is to obtain a dynamical constructive rereading of a lemma of Suslin [19] (Lemma 2.3) which played a central role in Suslin’s solution of Serre’s conjecture. This lemma says that for a commutative ring A, if &v1 (X), . . . , vn (X)' = A[X] where v1 is monic and n ≥ 3, then there exist finitely many γi ∈ En−1 (A[X]), the subgroup of SLn−1 (A[X]) generated by elementary matrices, such that &Res(v1 , ei .γ1 t (v2 , . . . , vn )), 1 ≤ i ≤ "' = A. Contrary to the papers [13, 14], more close to the Quillen’s techniques [17], in which the authors use dynamical rereading of “localizing at all primes”, the goal here is to mime an abstract proof in which one works modulo each maximal ideal in order to find the hidden algorithms. In other words the classical proof to decipher, instead of using all possible local rings that are localizations of some given ring, uses all possible residue fields (that is, quotient by maximal ideals) of this ring. In fact, the lemma cited above is the principal non constructive step in Suslin’s proof of Serre’s conjecture. In the literature, in order to surmount the obstacle of this lemma which is true for any ring A, constructive mathematicians interested in Suslin’s techniques for the Suslin’s stability and QuillenSuslin theorems are restricted to few rings satisfying additional conditions and in which one knows effectively the form of all the maximal ideals. For instance, in [7, 11, 16], the authors utilize the facts

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Mnif and Yengui

that for a discrete field K, the ring K[X1 , . . . , Xk ] is Noetherian and has an effective Nullstellensatz (see the proof of Theorem 4.3 of [16]). That is why we think that a constructive proof of Suslin’s lemma without any restriction on the ring A will enable the extension of the algorithms for the Suslin’s stability and Quillen-Suslin theorems for a wider class of rings. Another feature of the method used in our proof is that it may be a model for miming constructively abstract proofs passing to the residue fields (that is, quotients by maximal ideals). Note that, in the literature, only the localization at all maximal ideals, which is one of the two main aspects of utilization of maximal ideals, has been treated constructively (see the concrete local-global principles developed in [13]). We are convinced that when one rewrites constructively an abstract proof relying on maximal ideals, he can always avoid the use of maximal ideals and thus obtains a “fully” constructive proof. Lemma 1 (Suslin’s Lemma, [19] Lemma 2.3) Let A be a commutative ring. If &v1 (X), . . . , vn (X)' = A[X] where v1 is monic and n ≥ 2, then there exist γ1 , . . . , γ! ∈ En−1 (A[X]) such that:

&Res(v1 , e1 .γ1 (v2 , . . . , vn )t , . . . , Res(v1 , e1 .γ! (v2 , . . . , vn )t ' = A. Here e1 .x, where x is a column vector, stands for the first coordinate of x. Proof For n = 2, see Lemma 5.

Suppose n ≥ 3. We can without loss of generality suppose that all the vi for i ≥ 2 have degrees < d = deg v1 . For the sake of simplicity, we write vi instead of vi . Let u1 (X), . . . , un (X) ∈ A[X] such that v1 u1 + · · · + vn un = 1. Set w = v3 u3 + · · · + vn un and V = t (v2 , . . . , vn ). We suppose that v1 has degree d and for 2 ≤ i ≤ n, the formal degree of vi is di < d. This means that vi has no coefficient of degree > di but one does not guarantee that deg vi = di (it is not necessary to have a zero test inside A). We proceed by induction on min2≤i≤n {di }. To simplify, we always suppose that d2 = min2≤i≤n {di }.

For d2 = −1, v2 = 0 and by one elementary operation, we put w in the second coordinate. We have Res(v1 , w) = Res(v1 , v1 u1 + w) = Res(v1 , 1) = 1 and we are done. Now, suppose that we can find the desired elementary matrices for d2 = m − 1 and let show that we can do the job for d2 = m. Let a be the coefficient of degree m of v2 and consider the ring B = A/&a'. In B, all the induction hypotheses are satisfied without changing the vi nor the ui . Thus, we can obtain Γ1 , . . . , Γk ∈ En−1 (B[X]) such that &Res(v1 , e1 .Γ1 V ), . . . , Res(v1 , e1 .Γk V )' = B. It follows that, denoting by Υ1 , . . . , Υk the matrices in En−1 (A[X]) lifting respectively Γ1 , . . . , Γk , we have Let b ∈ A such that

&Res(v1 , e1 .Υ1 V ), . . . , Res(v1 , e1 .Υk V ), a' = A.

ab ≡ 1 mod &Res(v1 , e1 .Υ1 V ), . . . , Res(v1 , e1 .Υk V )' = J and consider the ring C = A/J. Note that in C, we have ab = 1. By an elementary operation, we replace v3 by its remainder modulo v2 , say v3# , and then we exchange v2 and −v3# . The new column V # obtained has as first coordinate a polynomial with formal degree m − 1. The induction hypothesis applies and we obtain ∆1 , . . . , ∆r ∈ En−1 (C[X]) such that

&Res(v1 , e1 .∆1 V # ), . . . , Res(v1 , e1 .∆r V # )' = C. Since V is the image of V by a matrix in En−1 (C[X]) (this matrix is in fact the product of two elementary matrices in E2 (C[X]) transforming t (v2 , v3 ) into t (−v3# , v2 )), we obtain matrices Λ1 , . . . , Λr ∈ En−1 (C[X]) such that #

&Res(v1 , e1 .Λ1 V ), . . . , Res(v1 , e1 .Λr V )' = C. The matrices Λj lift in En−1 (A[X]) as, say Ψ1 , . . . , Ψr . Finally, we obtain

&Res(v1 , e1 .Ψ1 V ), . . . , Res(v1 , e1 .Ψr V )' + J = A,

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Unimodular completion over Noetherian rings

the desired conclusion. ! Remark 2 It is easy to see that in Lemma 1, with the hypothesis deg vi ≤ d for 1 ≤ i ≤ n, the number " of matrices γj in the group En−1 (A[X]) is bounded by 2d . Moreover, each γj is the product of at most 2d elementary matrices.

1.2

A more general strategy (by “Back-tracking”)

As already mentioned above, contrary to the local-global principles explained in [13], we do not reread a proof in which one localizes at each prime ideal but a proof in which one passes separately to all the residue fields. Our purpose in this subsection is to describe a general strategy applicable in situations analogous to Suslin’s lemma. Suppose that we want to mime constructively a classical proof which demonstrates that for a class of commutative rings stable by quotient by finite-type ideals (for example, the class of all commutative rings like in Suslin’s lemma, or the class of Noetherian rings), an ideal J = &bj , j ∈ E' contains 1, where the bj are concrete objects we are able to construct. We suppose that the classical proof passes to all the residue fields and uses essentially a finite number of tests of type “xi = 0 or xi is invertible ”, such tests will be called “maximal tests ”. Suppose that n is the maximal number of consecutive maximal tests we need to obtain the desired result (n is computed in a formal way). Of course, this number decreases if instead of the initial ring A, we consider its quotient by a finite-type ideal. Let a1 be the element of A where the first maximal test is indispensable. We force it to be zero by passage to the ring A/&a1 ' and we do so with all the elements we meet consecutively in the proof and where the maximal test is indispensable. At the end, we have elements a1 , . . . , an in A such that in the ring A/&a1 , . . . , an ' the proof works. Thus, in A/&a1 , . . . , an ', we have b1,n , . . . , bin ,n ∈ J such 1'. that &b1,n , . . . , bin ,n ' = &¯ Set Jn = &b1,n , . . . , bin ,n '. In the ring (A/Jn )/&a1 , . . . , an−1 ' = A/&a1 , . . . , an−1 , Jn ', only one maximal test suffices to obtain the desired result. Thus, the fact that an is invertible enables us to find b1,n−1 , . . . , bin−1 ,n−1 ∈ J such that, setting Jn−1 = &b1,n−1 , . . . , bin−1 ,n−1 ', an−1 is invertible modulo &a1 , . . . , an−2 , Jn , Jn−1 ', and so on, by “back-tracking”, until finding the desired result. Remark 3 In the constructive proof above, the dynamical assessment of the classical proof does not produce an assessment tree as in [13]. Indeed, in [13], at each step one opens two branches, the first of type “xi = 0” and the second of type “xi is invertible”. Of course, the inverse one introduces is not in the initial ring but in its current localization. At the end, one goes back from the leaves to the tree’s root by technical dynamical gluing lemmas. On the other hand, in the method described above, one uses systematically and consecutively “xi = 0” and then, by “back-tracking”, comes up to the initial ring using systematically and consecutively “xi is invertible ”. The inverses used are not artificial since they are computed in concrete quotients of the initial ring, that is, computed in the initial ring modulo a finite-type ideal.

1.3

A particular case of Suslin’s lemma

Let us recall the following lemma [15] (Theorem 1) stating that under the hypotheses of the lemma of Suslin cited above, if A contains a set E = {y1 , . . . , y! } of cardinal " = deg v1 + 1, such that y − y # is invertible for each y $= y # in E (for instance if A is a K-algebra over an infinite field K), then the identity &Res(v1 , e1 .γ1 (v2 , . . . , vn )t , . . . , Res(v1 , e1 .γ! (v2 , . . . , vn )'t = A can simply be achieved by the elementary operations L1 → L1 + yi where u1 v1 + . . . + un vn = 1.

n−1 ! j=2

uj+1 Lj , 1 ≤ i ≤ ",

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Mnif and Yengui

Note the crucial role played by these type of elementary operations all along this paper (Lemma 4, Lemma 8, Theorem 9, Corollary 10,..). These operations seem to be very efficient and bring useful simplifications to the problem of unimodular completion. Lemma 4 (Suslin’s Lemma, particular case, [15] Theorem 1) Let A be a commutative ring, V, v, U, u, w ∈ A[X] such that V v + U u + w = 1 and v is monic. Denote " = deg v + 1 and suppose that A contains a set E = {y1 , . . . , y! } such that yi − yj is invertible for each i $= j. For each 1 ≤ i ≤ ", denoting ri = Res(v, u + yi w), then &r1 , . . . , r! ' = A, that is, there exist α1 , . . . , α! ∈ A such that α1 r1 + · · · + α! r! = 1. We also recall the following lemma which was used in the proof of Lemma 1.

Lemma 5 ([15], Lemma 2) Let v, a, b, w ∈ A[X] with v monic and av + bw = 1, then Res(v, w) is invertible in A. Corollary 6 Let A be a commutative ring, v1 , . . . , vn , u1 , . . . , un ∈ A[X] such that u1 v1 + . . . + un vn = 1, v1 is monic and n ≥ 3. Denote " = deg v1 + 1 and suppose that A contains a set E = {y1 , . . . , y! } such "n that yi − yj is invertible for each i $= j. For each 1 ≤ i ≤ ", denoting ri = Res(v1 , v2 + yi j=3 uj vj ), then &r1 , . . . , r! ' = A.

Example 7 If A is a Q-algebra, and (v1 , . . . , vn )t , (u1 , . . . , un )t ∈ Umn (A[X]) such that u1 v1 + · · · + un vn = 1, then the γi of Lemma 1 considered as elements of En (A[X]) can be chosen from the following list of elementary operations: L2 → L2 L2 → L2 + u3 L3 + · · · un Ln L2 → L2 + 2(u3 L3 + · · · un Ln ) .. . L2 → L2 + (deg v1 )(u3 L3 + · · · un Ln ).

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Unimodular completion over Noetherian rings

Let us recall some equivalent definitions of Krull dimension [5, 6, 12]. First, in a ring A, a sequence (x1 , . . . , x! ) is said to be pseudo-singular if there exist a1 , . . . , a! ∈ A and m1 , . . . , m! ∈ N such that m! m2 1 xm 1 (x2 · · · (x! (1 + a! x! ) + · · · + a2 x2 ) + a1 x1 ) = 0.

(x1 , . . . , x! ) is said to be pseudo-regular if it is not pseudo-singular. Recall also the definition of regular sequence, we will make a slight modification on it. We will say that a sequence (x1 , . . . , x! ) of elements in A is regular if &x1 , . . . , x! ' = A or x1 is not a zero divisor, x2 is not a zero divisor modulo &x1 ', x3 is not a zero divisor modulo &x1 , x2 ', and so on. Note that if &x1 , . . . , x! ' = $ A and (x1 , . . . , x! ) is regular then it is pseudo-regular [12]. Recall also that if a ∈ A, the boundary of a is the ideal Na of A generated by a and the elements b such that ab is nilpotent [6].

A ring A is said to have a Krull dimension ≤ −1 (in short, dim A ≤ −1) if it is trivial, that is 1 = 0 in A. For " ∈ N, the following assertions are equivalent and characterize the fact that dim A ≤ " [5, 6, 12]: (a) For each chain p0 ⊂ p1 ⊂ · · · ⊂ ps of prime ideals of A, we have s ≤ ". (b) Each sequence (x1 , . . . , x!+1 ) of elements in A is pseudo-singular. (c) For all a ∈ A, dim A/Na ≤ " − 1.

Assertions (b) and (c) can be seen as constructive inductive definitions of Krull dimension [5, 6, 12]. Now, we go back to our unimodular vectors. The following key lemma will play an important role in this paper.

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Unimodular completion over Noetherian rings

Lemma 8 Suppose that uv + w = 1 in a Noetherian ring A containing an infinite set E = {y1 , y2 , ...} such that yi − yj is invertible when i $= j. Then there exists i such that v + yi w is not a zero divisor. Proof We will prove by induction on n that for n ≥ 2, if all v + y1 w, v + y2 w, . . . , v + yn w are zero divisors with di (v + yi w) = 0 and di ∈ A \ {0}, then we obtain a strictly increasing chain &d1 ' ⊂ &d1 , d2 ' ⊂ · · · ⊂ &d1 , ..., dn ' of ideals in A. For n = 2, suppose that both v + y1 w and v + y2 w are zero divisors. Write d1 (v + y1 w) = 0 and d2 (v + y2 w) = 0 with d1 , d2 $= 0. If d2 = d1 δ1 for some δ1 ∈ A then d2 (v + y1 w) = 0. Together with d2 (v + y2 w) = 0 this would imply that (y1 − y2 )d2 w = 0, d2 w = 0 and d2 v = 0. Since 1 = uv + w then we would have d2 = 0 which is not true. Thus, &d1 ' ⊂ &d1 , d2 '. For n ≥ 3, suppose that all v+y1 w, v+y2 w, . . . , v+yn w are zero divisors, that is there exist di ∈ A\{0}, 1 ≤ i ≤ n, such that di (v + yi w) = 0. We can suppose that for any σ ∈ Sn (the permutation group of {1, . . . , n}), dσ(n) $∈ &dσ(1) , . . . , dσ(n−1) '. Passing to the ring A/&d1 ' and using the induction hypothesis, we infer that d¯n $∈ &d1 , . . . , dn−1 ' and thus dn ∈ / &d1 , . . . , dn−1 '. Together with the fact that we have a strictly increasing chain &d1 ' ⊂ &d1 , d2 ' ⊂ · · · ⊂ &d1 , ..., dn−1 ' (by induction hypothesis), we deduce that &d1 ' ⊂ &d1 , d2 ' ⊂ · · · ⊂ &d1 , ..., dn '. Now, the fact that A is Noetherian guaranties that we finish by finding yi ∈ E such that v + yi w is not a zero divisor. ! Theorem 9 Suppose that u1 v1 + u2 v2 + · · · + un vn = 1 in a Noetherian ring A containing an infinite set E = {y1 , y2 , ...} such that yi − yj is invertible when i $= j. Then there exist y1 , . . . , yn−1 ∈ E such that the sequence (v1 + y1 ξ0 (u2 v2 + · · · + un vn ), v2 + y2 ξ1 (u3 v3 + · · · + un vn ), . . . , vn−1 + yn−1 ξn−2 un vn ) is regular, where ξ0 = 1 and ξk+1 = ξk (1 − yk+1 uk+1 ξk ). Proof Denote wi = ui vi + · · · + un vn . Applying Lemma 8, we can find y1 ∈ E such that v1 + y1 w2 is not a zero divisor. Note that in A we have (v1 + y1 w2 )u1 + (1 − y1 u1 )w2 = 1. In the ring A/&v1 + y1 w2 ', we have: (1 − y1 u1 )u2 v2 + (1 − y1 u1 )u3 v3 + · · · + (1 − y1 u1 )un vn = 1. Applying Lemma 8 again, we can find y2 ∈ E such that v2 + y2 (1 − y1 u1 )w3 is not a zero divisor in A/&v1 + y1 w2 ' and so on we construct the desired regular sequence. Suppose that in the ring A/&v1 + y1 ξ0 w2 , v2 + y2 ξ1 w3 , . . . , vk + yk ξk−1 wk+1 ', we have: ξk wk+1 = 1, and then (vk+1 + yk+1 ξk wk+2 )ξk uk+1 + ξk (1 − yk+1 ξk uk+1 )wk+2 = 1. Thus, in the ring A/&v1 + y1 ξ0 w2 , v2 + y2 ξ1 w3 , . . . , vk + yk ξk−1 wk+1 , vk+1 + yk+1 ξk wk+2 ', we have: ξk (1 − yk+1 ξk uk+1 )wk+2 = 1. It follows that the sequence (ξk ) satisfies the relation: #

ξ0 = 1 ξk+1 = ξk (1 − yk+1 uk+1 ξk ). !

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Mnif and Yengui

Recall that for any ring A and n ≥ 1, Umn (A) denotes the set of unimodular rows in A, that is Umn (A) = {(x1 , . . . , xn ) ∈ An such that &x1 , . . . , xn ' = A}. En (A) denotes the subgroup of SLn (A) generated by elementary matrices. For i $= j, Ei,j (a) is the matrix corresponding to the elementary operation Li → Li + aLj . Corollary 10 Let A be a Noetherian ring with Krulldim A = d < ∞ and suppose that A contains an infinite set E = {y1 , y2 , ...} such that yi − yj is invertible when i $= j. If u1 v1 + u2 v2 + · · · + un vn = 1 in A with n ≥ d + 2, then denoting wk = uk vk + · · · + un vn , there exist y1 , . . . , yd+1 ∈ E such that 1 ∈ &v1 + y1 ξ0 w2 , v2 + y2 ξ1 w3 , . . . , vd+1 + yd+1 ξd wd+2 ',

where ξ0 = 1 and ξk+1 = ξk (1 − yk+1 uk+1 ξk ). In particular, there exists M ∈ En (A) such that M t (v1 , v2 , . . . , vn ) = t (1, 0, . . . , 0).

More explicitly, if q1 , . . . , qd+1 ∈ A are such that then denoting

q1 (v1 + y1 ξ0 w2 ) + · · · + qd+1 (vd+1 + yd+1 ξd wd+2 ) = 1,

M1 :=

$1

k=d+1 Ek,k+1 (yk ξk−1 uk+1 ) · · · Ek,n (yk ξk−1 un ),

M2 := En,1 ((vn − 1)q1 ) · · · En,d+1 ((vn − 1)qd+1 ), M3 := E1,n (−v1 − y1 ξ0 w2 ) · · · Ed+1,n (−vd+1 − yd+1 ξd wd+2 ) Ed+2,n (−vd+2 ) · · · En−1,n (−vn−1 ),

we have

En,1 (−1) E1,n (1) M3 M2 M1 t (v1 , v2 , . . . , vn ) = t (1, 0, . . . , 0). Proof Using Theorem 9, there exist y1 , . . . , yd+1 ∈ E such that the sequence (v1 + y1 ξ0 w2 , v2 + y2 ξ1 w3 , . . . , vd+1 + yd+1 ξd wd+2 ) is regular. The result follows since Krulldim A = d. For the computation of M , just observe that: M1 t (v1 , v2 , . . . , vn ) = t (v1 + y1 ξ0 w2 , . . . , vd+1 + yd+1 ξd wd+2 , vd+2 , . . . , vn ) := V1 , M2 V1 = t (v1 + y1 ξ0 w2 , . . . , vd+1 + yd+1 ξd wd+2 , vd+2 , . . . , vn−1 , 1) := V2 , M3 V2 = t (0, . . . , 0, 1).

!

If A is a ring, the ring A&X' denotes the localization of the polynomial ring A[X] at the multiplicative subset of monic polynomials. By induction, we define A&X1 , . . . , Xk ' := A&X1 , . . . , Xk−1 '&Xk '.

It is in fact the localization of the polynomial ring A[X1 , . . . , Xk ] at the multiplicative subset Uk = {f ∈ A[X1 , . . . , Xk ] such that LC(f ) = 1},

where LC(f ) denotes the leading coefficient of f related to the lexicographic monomial order with Xk > Xk−1 > · · · > X1 .

It is well-known that if A is Noetherian then Krulldim A&X1 , . . . , Xk ' = Krulldim A [8], but up to now we did not manage to obtain a constructive proof of this result using one of the equivalent constructive definitions of the Krull dimension quoted above. So, the following result is only algorithmic and not fully constructive. All the necessary computations are guaranteed by the fact that the basic ring is a computable Noetherian ring and so Gr¨obner bases techniques can be utilized [1].

8

Unimodular completion over Noetherian rings

Corollary 11 (Producing a monic coordinate) Let A be a computable Noetherian ring with Krulldim A = d < ∞ and suppose that A contains an infinite set E = {y1 , y2 , ...} such that yi − yj is invertible when i $= j. If u1 v1 + u2 v2 + · · · + un vn = 1 in A[X1 , . . . , Xk ] with n ≥ d + 2, then denoting wk = uk vk + · · · + un vn , there exist y1 , . . . , yd+1 ∈ E such that &v1 + y1 ξ0 w2 , v2 + y2 ξ1 w3 , . . . , vd+1 + yd+1 ξd wd+2 ' ∩ Uk $= ∅, where ξ0 = 1 and ξk+1 = ξk (1 − yk+1 uk+1 ξk ). In particular, via a change of variables and by elementary operations we can transform t (v1 , v2 , . . . , vn ) into a unimodular row whose first coordinate is monic at Xk . Proof Just use Corollary 10 and the fact that Krulldim A&X1 , . . . , Xk ' = Krulldim A. Moreover, suppose that we have already computed an F ∈ &v1 + y1 ξ0 w2 , v2 + y2 ξ1 w3 , . . . , vd+1 + yd+1 ξd wd+2 ' ∩ Uk . It is well-known (see [19] for example) that by a change of variables “`a la Nagata”, that is of type 2 (X1 , . . . , Xk−1 , Xk ) → (Y1 , . . . , Yk−1 , Xk ) with Xk−1 = Yk−1 + Xkm , Xk−2 = Yk−2 + Xkm , . . . , X1 = k−1 Y1 + Xkm for sufficiently large m, F becomes monic at Xk . In fact, in order to avoid the explosion of the degrees of the considered polynomials, one has to make a change of variables of type (X1 , . . . , Xk−1 , Xk ) → (Y1 , . . . , Yk−1 , Xk ) with Xk−1 = Yk−1 + Xkn1 , Xk−2 = Yk−2 + Xkn2 , . . . , X1 = n Y1 + Xk k−1 and (n1 , . . . , nk−1 ) ∈ Nk−1 as small as possible. Of course, if possible, it should be better to use a linear change of variables so that the polynomials considered keep the same total degree. Now, suppose that F is monic at Xk , degXk F = δ, degXk vn = δ # , δ” = max(δ, δ # ), and let q1 , . . . , qd+1 ∈ A such that q1 (v1 + y1 ξ0 w2 ) + · · · + qd+1 (vd+1 + yd+1 ξd wd+2 ) = F. Set M1 :=

so that

$1

k=d+1 Ek,k+1 (yk ξk−1 uk+1 ) · · · Ek,n (yk ξk−1 un ),

M2 := En,1 (Xkδ”−δ+1 q1 ) · · · En,d+1 (Xkδ”−δ+1 qd+1 ), M1 t (v1 , v2 , . . . , vn ) = t (v1 + y1 ξ0 w2 , . . . , vd+1 + yd+1 ξd wd+2 , vd+2 , . . . , vn ),

and E1,n (1) En,1 (−1)M2 M1 t (v1 , v2 , . . . , vn ) has as first coordinate a monic polynomial at Xk with degree δ” + 1. ! Now we give our main algorithm for unimodular completion. It is based on [13, 15, 19], Lemma 4, Corollary 10, and Corollary 11. We use the notation X = (X1 , . . . , Xk ). An algorithm for unimodular completion Input: Two columns V = V(X) = t (v1 (X), . . . , vn (X)), U = U(X) = t (u1 (X), . . . , un (X)) ∈ A[X]n such that V t U = 1. We assume that A is a computable Noetherian ring, A contains infinitely many yi such that yi − yj is invertible for i $= j, Krulldim A = d < ∞ and n ≥ max(3, d + 2). Output: A matrix M in SLn (A[X]) such that M V = t (1, 0, . . . , 0). Step 1: Make a change of variables and elementary operations on V so that v1 becomes monic at Xk (follow the algorithm given in Corollary 11).

9

Mnif and Yengui

Step 2: For 1 ≤ i ≤ " = degXk v1 + 1, set wi := v2 + yi (u3 v3 + · · · + un vn ), compute ri := ResXk (v1 , wi ) and find α1 , . . . , α! ∈ A[X1 , . . . , Xk−1 ] such that α1 r1 + · · · + α! r! = 1 (here we use the Gr¨obner bases techniques [1] or the constructive proof of Lemma 4). Note that one should consider only the ri which are not nilpotent (see Remark 13). In order to lighten the notations, we will suppose that all the ri are non nilpotent. For 1 ≤ i ≤ ", compute fi , gi ∈ A[X] such that fi v1 + gi wi = ri . Step 3: For 1 ≤ i ≤ ", set

Hi := En,1 (−1)E1,n (1)En−1,n (−vn−1 ) · · · E3,n (−v3 )E2,n (−wi )E1,n (−v1 )En,2 (−ri−1 (vn − 1)gi ) En,1 (−ri−1 (vn − 1)fi )E2,n (yi un ) · · · E2,3 (yi u3 ).

(Comment: Of course, we consider only the ri which are nonzero and we have Hi V = t (1, 0, . . . , 0)). ˜i := Hi (X1 , . . . , Xk−1 , 0)−1 Hi (X1 , . . . , Xk−1 , Xk ) so that H ˜i V(X1 , . . . , Xk−1 , Xk ) = Set H V(X1 , . . . , Xk−1 , 0). ˜i are in A[X] 12 . Note that the coefficients of Hi are in the module A[X] r1i and those of H r i

Step 4: For 1 ≤ i ≤ ", find ki ∈ N and Gi ∈ SLn (A[X, Y ]) such that Gi V(X1 , . . . , Xk−1 , Xk + riki Y ) = V(X). In more details, following Step 3, we deduce that ˜i (X) V(X1 , . . . , Xk−1 , Xk ) = V(X1 , . . . , Xk−1 , 0) H ˜i (X1 , . . . , Xk−1 , Xk + ri2n Y ) V(X1 , . . . , Xk−1 , Xk + ri2n Y ), =H and so

˜i (X)−1 H ˜i (X1 , . . . , Xk−1 , Xk + r2n Y ) V(X1 , . . . , Xk−1 , Xk + r2n Y ) = V(X). H i i

˜i (X)−1 H ˜i (X1 , . . . , Xk−1 , Xk + r2n Y ). Thus, we take ki = 2n and Gi = H i In fact, we have simply:

Gi = Hi (X)−1 Hi (X1 , . . . , Xk−1 , Xk + ri2n Y ), with Hi−1 := E2,3 (−yi u3 ) · · · E2,n (−yi un )En,1 (ri−1 (vn − 1)fi )En,2 (ri−1 (vn − 1)gi )E1,n (v1 )) E2,n (wi )E3,n (v3 )) · · · En−1,n (vn−1 ))E1,n (−1)En,1 (1).

˜i are in A[X, Y ] then Gi ∈ SLn (A[X, Y ]) (see Lemma 12). Moreover, since all the coefficients of ri2 H In practice, as mentioned in Remark 13, one can compute Gi as follows: Consider a new variable t, set (X1 , . . . , Xk−1 , Xk + t2n Y, t) = X (t) , and compute i (X) G#i (X, t) := E2,3 (−yi u3 (X)) · · · E2,n (−yi un (X))En,1 ( (vn (X)−1)f ) t (vn (X)−1)gi (X) En,2 ( )E1,n (v1 (X)))E2,n (wi (X))E3,n (v3 (X))) · · · En−1,n (vn−1 (X))) t

(t) (t) (t) (t) (t) G#−1 i (X ) := En−1,n (−vn−1 (X )) · · · E3,n (−v3 (X ))E2,n (−wi (X ))E1,n (−v1 (X ))

En,2 ( (1−vn (X

(t) ))g (X (t) ) i

t

)En,1 ( (1−vn (X

(t) ))f

t

i (X

(t) )

)E2,n (yi un (X (t) ) · · · E2,3 (yi u3 (X (t) ),

(t) so that by Lemma 12, Gi ”(X, Y, t) := G#i (X, t)G#−1 i (X ) ∈ SLn (A[X, Y, t]).

Now take Gi := Gi ”(X, Y, ri ). To sum up, the main properties of Gi are #

Gi ∈ SLn (A[X, Y ]) Gi V(X1 , . . . , Xk−1 , Xk + ri2n Y ) = V(X).

10

Unimodular completion over Noetherian rings

˜ = G(X ˜ 1 , . . . , Xk−1 , Xk , Y ) ∈ SLn (A[X, Y ]) such that GV(X ˜ Step 5: Find G 1 , . . . , Xk−1 , Xk + Y ) = V(X). More precisely, let β1 , . . . , β! ∈ A[X1 , . . . , Xk−1 ] such that β1 r12n + · · · + β! r!2n = 1 (β1 , . . . , β! can be deduced from the identity α1 r1 + · · · + α! r! = 1, or directly using Gr¨obner basis techniques [1]). Set ! % 2n ˜= G Gi (X1 , . . . , Xk−1 , Xk + (β1 r12n + · · · + βi−1 ri−1 )Y, βi Y ). i=2

Remark that

2n )Y, β! Y ) V(X1 , . . . , Xk−1 , Xk + Y ) G! (X1 , . . . , Xk−1 , Xk + (β1 r12n + · · · + β!−1 r!−1

2n 2n )Y, β! Y ) V(X1 , . . . , Xk−1 , Xk +(β1 r12n +· · ·+β!−1 r!−1 )Y +β! r!2n Y ) = G! (X1 , . . . , Xk−1 , Xk +(β1 r12n +· · ·+β!−1 r!−1

and so on until getting

2n = V(X1 , . . . , Xk−1 , Xk + (β1 r12n + · · · + β!−1 r!−1 )Y ),

˜ GV(X 1 , . . . , Xk−1 , Xk + Y ) = V(X).

˜ 1 , . . . , Xk−1 , 0, Xk ). Step 6: G := G(X ˜ (Comment: Since GV(X 1 , . . . , Xk−1 , Xk + Y ) V(X1 , . . . , Xk−1 , 0)).

=

V(X),

then GV(X1 , . . . , Xk−1 , Xk )

=

Step 7: Repeat for V(X1 , . . . , Xk−1 , 0). Note that V(X1 , . . . , Xk−1 , 0) t U(X1 , . . . , Xk−1 , 0) = 1.

Step 8: (basic step) Follow the algorithm given in Corollary 10 to transform V(0, . . . , 0) into t (1, 0, . . . , 0) using elementary operations. For sake of completeness, we add the following lemma which is a more precise formulation of a lemma originally given in [13] and was used in Step 4 of the above algorithm. Lemma 12 (Lemma 15 of [13]) Let X, Y, t be three variables over a ring A, and H(X) ∈ SLn (A[ 1t ][X]) such that tm H(X) ∈ Mn (A[X, t]) for some m ∈ N. Then H(X + tmn Y )H(X)−1 ∈ SLn (A[X, Y, t]). Proof Let L(X, t) = tm H(X) ∈ Mn (A[X, t]) and denote M (X, t) the cotranspose of L(X, t). Then H(X + tmn Y )H(X)−1 = L(X + tmn Y )L(X)−1 =

1 1 L(X + tmn Y )M (X, t) := mn BX,t (tmn Y ), tmn t

with BX,t (0) = tmn In . Thus, we can write BX,t (tmn Y ) = tmn In + tmn Y CX,t (Y ), for some matrix CX,t (Y ) ∈ Mn (A[X, Y, t]. It follows that H(X + tmn Y )H(X)−1 ∈ SLn (A[X, Y, t]).

!

Remark 13 In this formal proof, the condition that A is integral is not necessary. In our concern, when we use this lemma in the algorithm for unimodular completion above, we consider only the ri which are not nilpotent since in any ring: nilpotent + unit = unit. Moreover one has a nilpotency test in any computable Noetherian ring A, since taking a new variable y on A: a is nilpotent in A ⇔ 1 ∈ &ay − 1' in A[y].

11

Mnif and Yengui

3

Suslin’s stability theorem

Before stating & ' our next result, recall that for any ring T , by identifying N ∈ SL2 (T ) with N 0 ∈ SLn (T ), we can regard SL2 (T ) as a subgroup of SLn (T ). 0 In−2 Proposition 14 Let A be a ring. Suppose (v1 (X), . . . , vn (X))t ∈ Umn (A[X]), v1 is monic, and n ≥ 3. Then there exist B1 ∈ SL2 (A[X]) and B2 ∈ En (A[X]) such that 

   v1 (X) v1 (0)    ..  .. B1 B2   =  .  ∈ Umn (A). . vn (X) vn (0)

Proof Let γ1 , . . . , γl ∈ En−1 (A[X]) and α1 , . . . , αl ∈ A as in Lemma 1 or Lemma 4 if the ring A contains an infinite set E = {y1 , y2 , . . .} such that yi − yj is invertible for each i $= j. Set ci = res(v1 , e1 .γi (v2 , . . . , vn )t ). " " We have 1 = li=1 αi ci , and thus X = li=1 αi ci X. Set: bl = 0, bl−1 = αl Xcl , bl−2 = bl−1 + αl−1 Xcl−1 , .. . b0 = b1 + α1 Xc1 = X. Since En (A[X]).SL2 (A[X]) ⊆ SL2 (A[X]).En (A[X]) (see [16], Corollary 2.8) whose proof is valid for any ring A), it suffices to prove that  there  exists Bi,1 ∈ SL2 (A[X]) and Bi,2 ∈ En (A[X]) such that v1   v(bi ) = Bi,1 Bi,2 v(bi−1 ), where v =  ... . vn    w2 v2     Set γi  ...  =  ... , and denote wn vn wj (bi−1 ) − wj (bi ) = ci fi (fi ∈ A[X]) = σj v1 (bi−1 ) − τj w2 (bi−1 ), σj , τj ∈ A[X]. Denote by Γi ∈ En (A[X]) the matrix corresponding to the & elementary operation: ' 1 0 . Lj → Lj + σj L1 + τj L2 , 3 ≤ j ≤ n; and denote Bi,2 = Γi 0 γi (bi−1 )   v1 (bi−1 )  w2 (bi−1 )      We have Bi,2 v(bi−1 ) =  w3 (bi ) .   ..   . 

wn (bi ) By&Lemma 2.1 'of [19], we can & 'explicitly find Ci ∈ SL2 (A[X]) such that v1 (bi ) v1 (bi−1 ) = . Ci w2 (bi−1 ) w2 (bi ) & '& ' 1 0 Ci 0 Take Bi,1 = . 0 γi (bi )−1 0 In−2

!

Using Corollaries 11 and 10, we obtain an algorithmic proof of the following theorem for unimodular completion by elementary matrices.

12

Unimodular completion over Noetherian rings

Theorem 15 Let X1 , . . . , Xk be k variables over a computable Noetherian ring A of Krull dimension d < ∞ containing an infinite set E = {y1 , y2 , . . .} such that yi − yj is invertible for each i $= j, and set R = A[X1 , . . . , Xk ]. Then, for n ≥ max(3, d + 2), the group En (R) acts transitively on Umn (R). Proof We proceed as in Theorem 2.6 of [19]. Since the proof holds for k = 0 (by Corollary 10), we may assume by induction that the statement   v1  ..  holds for k − 1. Let T = A[X1 , . . . , Xk−1 ], X = Xk , and V =  .  ∈ Umn (T[X]).

vn We may assume that v1 is monic by multiplying V by an elementary matrix and changing variables (here we use the algorithmic proof of Corollary 11). Now by Proposition 14, we can find B1 ∈ SL2 (T[X]) and B2 ∈ En (T[X]) such that B1 B2 V(X) = V(0) ∈ T. By the inductive hypothesis, we can find B ∈ En (T) such that BB1 B2 V(X) = t (1, 0, . . . , 0). Using the normality of En (T[X]) in SLn (T[X]), we can write B1−1 B −1 = CB1−1 , for some C ∈ En (T[X]). Thus, V = B2−1 B1−1 B −1 t (1, 0, . . . , 0) = (B2−1 C)B1−1 t (1, 0, . . . , 0) = B2−1 C t (1, 0, . . . , 0), where B2−1 C ∈ En (T[X]). ! We can now give an algorithmic proof of Suslin’s stability theorem for computable Noetherian rings of Krull dimension 1 containing an infinite set E = {y1 , y2 , . . .} such that yi − yj is invertible for each i $= j. Examples of such rings are one-dimensional finitely generated algebras over an infinite field. Theorem 16 (Suslin’s stability theorem) Let X1 , . . . , Xk be k variables over a computable Noetherian ring A of Krull dimension 1 containing an infinite set E = {y1 , y2 , . . .} such that yi − yj is invertible for each i $= j, and set R = A[X1 , . . . , Xk ]. Then, for n ≥ 3, any n × n matrix in SLn (R) can be written as a product of elementary matrices in En (R). Proof As in 4. Reduction toSL3 of [16], using Theorem 15, by induction we see that it suffices to  p q 0 express a matrix  r s 0  ∈ SL3 (T[X]), with p monic, as a product of elementary matrices with 0 0 1 T a commutative ring. This has been explained in [7] (proof of Theorem 3.3) and explicitly obtained in [2] (Theorem 12). !

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[10] T.Y. Lam, Serre’s conjecture, Lecture Notes in Mathematics 635, Springer-Verlag, Berlin-New York, 1978. [11] Logar A., Sturmfels B. Algorithms for the Quillen-Suslin theorem. J. Algebra 145 231–239.

no. 1, (1992),

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