The Hermite ring conjecture and special linear groups for valuation rings
aa r X i v : . [ m a t h . A C ] N ov The Hermite ring conjecture and special linear groups forvaluation rings ∗ Jinwang Liu Dongmei Li + Licui ZhengSchool of Mathematics and Computing Sciences, Hunan Universityof Science and Technology, Xiangtan, Hunan, China, 411201
ABSTRACT
In this paper we prove that the Hermite ring conjecture holdsfor valuation rings V , and the special liner group SL n ( V [ x ]) coincides with the groupgenerated by elementary matrices E n ( V [ x ]) for n ≥
3. For any arithmetical ring R and n ≥
3, we show SL n ( R [ x ]) = SL n ( R ) · E n ( R [ x ]). KEYWORDS.
Hermit ring Conjecture; Suslin’s stability theorem; special linear groups, valuationrings; arithmetical rings.
Serre’s conjecture, proposed by J. P. Serre in 1955, asserts that finitely generatedprojective modules of polynomial rings with finitely many variables over a field, areactually free. In 1976, Quillen [1] and Suslin [2] independently gave an affirmative an-swer to this famous conjecture via completely different approaches. Note that Serre’sconjecture is equivalent to saying that any unimodular row ω in K × n [ x , · · · , x m ]can be completed to an invertible square matrix, where K is a field. In [3] Suslinestablished the following K -analogue of Serre’s conjecture (Suslin’s stability The-orem): If R is a commutative Noetherian ring and n ≥ max(3 , dim( R ) + 2) thenSL n ( R [ x , · · · , x m ]) = E n ( R [ x , · · · , x m ]).Recall that a ring R is called a Hermite ring if every finite generated stably freemodule is free. It is fairly easy to see that R is Hermite if any unimodular row ω ∈ R × n can be completed to an invertible square matrix over R . In [4, 5], Lampresented the following Hermite ring conjecture: If R is a Hermite ring, then R [ x ] isHermite as well. This conjecture seems rather untractable since there does not seemto exist much evidence for its truth, and so far it has only been verified for somespecial cases. For instances, Quillen-Suslin Theorem implies that the Hermite ringconjecture is true for K [ x , x , · · · , x n ], where K is a field, and Yengui [6] provedthat it also holds for rings of Krull dimension ≤ K -analogue question has been proposed in [6]. ∗ This research is supported by the National Science Foundation of China(11501192,11471108) and Scientific Research Fund of Hunan province education Department(2017JJ3084).+Corresponding author: Dongmei Li, [email protected] onjecture 1 Suppose that R is a ring of Krull dimension ≤
1, and n ≥ σ ( x ) ∈ SL n ( R [ x ]) is congruent to σ (0) modulo E n ( R [ x ]). Thatis, for every σ ( x ) ∈ SL n ( R [ x ]), there is E ∈ E n ( R [ x ]) such that σ ( x ) · E = σ (0).In fact, by virtue of local-global principle for elementary matrices [5], Conjecture1 is equivalent to the following Conjecture 2. Conjecture 2
Suppose R is a local ring of Krull dimension ≤
1, and M = p q r s
00 0 1 ∈ SL ( R [ x ])Then M ∈ E ( R [ x ]).Liu [7] has proved that Conjectures 1 and 2 are true for valuation rings of Krulldimension ≤
1. Since it is well-known that a valuation ring is Hermite, it is naturalfor us to ask the following three question:(1) Is the Hermite ring conjecture true for valuation rings?(2) Are conjectures 1 and 2 true for any valuation ring V ? Furthermore, do wehave SL n ( V [ x ]) = E n ( V [ x ]) for any n ≥ All rings considered in this paper are unitary and commutative. The undefinedterminology is standard as in [4]. Recall that a commutative ring V is said to be a valuation ring if principal ideals of V are totally ordered by inclusion; that is, anytwo elements of V are comparable under division. Obviously, a valuation ring is alocal ring. A commutative ring R is said to be arithmetical if the localization R m of R at m is a valuation ring for every maximal ideal m of R . For any commutativering R and n ≥ U m n ( R ) denotes the set of length n unimodular rows over R . LetSL n ( R ) denote the special linear group, and E n ( R ) denotes the subgroup of SL n ( R )generated by elementary matrices of the form I n + a e ij , where I n is the identitymatrix, a ∈ R , i = j , and e ij is the matrix whose only nonzero entry is the 1 R atposition ( i, j ). We know that E n ( R ) is a normal subgroup of SL n ( R ). In this section, let R be an unitary and commutative ring, and V be a valuationring. For f ( x ) ∈ V [ x ], we say that f ( x ) is primitive if at least one coefficient in f ( x )is a unit.We first introduce several necessary lemmas.2 emma 3.1. ([4]) Let G = E n ( R ) , and γ = ( c , c , · · · , c n ) ∈ U m n ( R ) , n ≥ , b , b ∈ R , such that c b + c b is a unit modulo c R + · · · + c n R . Then ( c , c , c , · · · , c n ) ∼ G ( b , b , c , · · · , c n ) Lemma 3.2.
Let G = E n ( R ) , and γ = ( c , c , · · · , c n ) ∈ U m n ( R ) , n ≥ . If u ∈ R is a unit modulo c R + · · · + c n R , then ( c , c , · · · , c n ) ∼ G ( uc , uc , c , · · · , c n ) Proof.
Because ( c , c , · · · , c n ) ∈ U m n ( R ), u ∈ R is a unit modulo c R + · · · + c n R ,then there exist x , x , · · · , x n , v, y , y , · · · , y n ∈ R such that c x + c x + c x + · · · + c n x n = 1 ,vu + c y + · · · + c n y n = 1Then v (( uc ) x + ( uc ) x ) = vu ( c x + c x )= (1 − c y − · · · − c n y n ) · (1 − c x − · · · − c n y n ) ≡ c R + · · · + c n R ) (1)From the equation above, we see that c x + c x and ( uc ) x + ( uc ) x are unitsmodulo c R + · · · + c n R . Applying Lemma 3.1 twice, we get γ ∼ G ( x , x , c , · · · , c n ) ∼ G ( uc , uc , c , · · · , c n ) . For any unitary commutative ring R, we see that R [ x ] is also a unitary commu-tative ring, so the two lemmas above are also true for R [ x ]. Lemma 3.3.
Let e ( x ) ∈ R [ x ] such that deg e ( x ) ≥ and e (0) is a unit. Then wehave:1. x is a unit modulo e ( x ) R [ x ] ;2. if ( x s f ( x ) , x s f ( x ) , e ( x )) ∈ U m ( R [ x ]) , then ( x s f ( x ) , x s f ( x ) , e ( x )) ∼ E ( R [ x ]) ( x s − k f ( x ) , x s − k f ( x ) , e ( x )) , where s , s , k are positive integers, and s , s ≥ k ;3. if ( f ( x ) , f ( x ) , e ( x )) ∈ U m ( V [ x ]) , then ( f ( x ) , f ( x ) , e ( x )) ∼ E ( R [ x ]) ( x k f ( x ) , x k f ( x ) , e ( x )) , where k is a positive integer. roof. Let e ( x ) = b n x n + · · · + b x + b such that b n = 0 and b is a unit, and g ( x ) = ( e ( x ) − b ) /x . Then x ( − b − g ( x )) + b − e ( x ) = 1, so x is a unit modulo e ( x ) R [ x ]. Clearly, ( x s − k f ( x ) , x s − k f ( x ) , e ( x )) ∈ U m ( R [ x ]). By lemma 3.2, thedesired result is obtained. Lemma 3.4.
Let a ∈ R , ( af ( x ) , ag ( x ) , h ( x )) ∈ U m ( R [ x ]) . Then ( af ( x ) , ag ( x ) , h ( x )) ∼ E ( R [ x ]) ( f ( x ) , g ( x ) , h ( x )) Proof.
Firstly, there exist f ( x ) , f ( x ) , f ( x ) ∈ R [ x ] such that af ( x ) · f ( x ) + ag ( x ) · f ( x ) + h ( x ) f ( x ) = 1 , that is, a ( f ( x ) · f ( x ) + g ( x ) · f ( x )) + f ( x ) · h ( x ) = 1 . Thus a is a unit modulo h ( x ) R [ x ], and hence ( f ( x ) , g ( x ) , h ( x )) ∈ U m ( R [ x ]). ByLemma 3.2, the desired result is obtained. Lemma 3.5. ([10])
Let g ( x ) ∈ R [ x ] be of degree n > such that g (0) is a unit in R .Then for any f ( x ) ∈ R [ x ] and k ≥ deg f ( x ) − deg g ( x ) + 1 , there exists h k ( x ) ∈ R [ x ] of degree < n such that f ( x ) ≡ x k · h k ( x ) mod g ( x ) R [ x ] . Lemma 3.6. ([4])
Let R be a commutative local ring and f = ( f , f , · · · , f n ) ∈ U m n ( R [ x ]) , where n ≥ , f is unitary. Then f ( x ) ∼ E n ( R [ x ]) f (0) ∼ E n ( R ) (1 , , · · · , . Theorem 3.1.
Let V be a valuation ring and ( f ( x ) , g ( x ) , e ( x )) ∈ U m ( V [ x ]) . Then ( f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) (1 , , Proof.
Note that ( f (0) , h (0) , e (0)) ∈ U m ( V [ x ]). Since V is a valuation ring andhence V is local, we may assume that e (0) = 1.We proceed by induction on deg e ( x ). If deg e ( x ) = 0, the result is obviouslytrue. Now assume that the conclusion holds for e ( x ) with degree ≤ n −
1. In thecase that deg e ( x ) = n ≥
1, write e ( x ) = c n x n + c n − x n − + · · · + c x + 1. By Lemma3.5, for any s ∈ N which is big enough, there are h ( x ) , h ( x ) ∈ V [ x ] such that f ( x ) ≡ x s h ( x ), g ( x ) ≡ x s h ( x ) mod e ( x ) V [ x ], and deg h ( x ) , deg h ( x ) ≤ n − f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) ( x s h ( x ) , x s h ( x ) , e ( x )) . (2)Since V is a valuation ring, We may assume that h ( x ) = af ( x ) , h ( x ) = bg ( x ), where a, b ∈ V , f ( x ) and g ( x ) are primitive polynomials, and deg f ( x ) =deg h ( x ), deg g ( x ) = deg h ( x ). Since V is a valuation ring, one knows that a | b or b | a . Without loss of generality we may assume that a | b . Set b = a · c . Then h ( x ) = a · cg ( x ), for any k ∈ N ( k < s ). By (2) and Lemma 3.3, 3.4, we have that( f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) ( ax s f ( x ) , ax s cg ( x ) , e ( x )) ∼ E ( V [ x ]) ( x s f ( x ) , x s cg ( x ) , e ( x )) ∼ E ( V [ x ]) ( x s − k f ( x ) , cx s − k g ( x ) , e ( x )) . (3)4ince f ( x ) is primitive, we may suppose that f ( x ) = a t x t + · · · + a m x m + · · · + a ,where a t = 0, t ≤ n −
1, and a m is a unit. Pick k such that s − k + t = n , i.e., s − k = n − t ≥ t ≤ n − a t | c n , let c n = d · a t , and e ( x ) = e ( x ) − d · x s − k f ( x ). Since s − k + t = n ,deg e ( x ) ≤ n −
1, and e (0) = e (0) = 1. By (3), we have( f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) ( x s − k f ( x ) , cx s − k g ( x ) , e ( x )) . Since deg e ( x ) ≤ n −
1, by the induction hypothesis, we have( f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) (1 , , a t ∤ c n , then c n | a t . Set a t = d · c n . Then d is not a unit. Let f ( x ) = x n − t f ( x ) − d · e ( x ). we obtain that f ( x ) = ( a t − − dc n − ) x n − + · · · + ( a m − dc n − t + m ) x n − t + m + · · · + ( − d ) . Since d is not a unit, a m − dc n − t + m is a unit, namely that the degree of the termwhere the coefficient is a unit in f ( x ) is greater than that of f ( x ) by one (since n − t ≥ f ( x ) ≤ n −
1. Since n − t = s − k , by (3), we have( f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) ( f ( x ) , cx s − k g ( x ) , e ( x )) . (4)Suppose that deg f ( x ) = t . By Lemma 3.3 and (4), we have( f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) ( x n − t f ( x ) , cx n + s − t − k g ( x ) , e ( x ))Note that f ( x ) is also primitive, and the degree of the term where the coefficientis a unit in f ( x ) is greater than that of f ( x ) by one. If c n can be divided by theleading coefficient of f ( x ), by case (1), the result is true. Otherwise, by repeatingthe preceding procedure, we obtain that f i ( x ) such that the leading coefficient of f i ( x ) ( i ≤ n − m + 1) will be a unit, and( f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) ( f i ( x ) , cx p g ( x ) , e ( x ))Since f i ( x ) is unitary, by Lemma 3.6,( f ( x ) , g ( x ) , e ( x )) ∼ E ( V [ x ]) (1 , , . Thus the conclusion is true.From Theorem 3.1, we obtain the following easy result.
Theorem 3.2.
The Hermite ring conjecture is true for valuation rings.Proof.
Let ω = ( f ( x ) , g ( x ) , e ( x )) ∈ V [ x ] be a unimodular row. It is well-knownthat V is a Hermite ring. By Theorem 3.1, there is P ∈ E ( V [ x ]) such that( f ( x ) , g ( x ) , e ( x )) · P = (1 , , N = (cid:18) (cid:19) · P − , A = (cid:18) f ( x ) g ( x ) e ( x ) N (cid:19) , Then A · P = I . So A is an invertible square matrix, and ω can be completed toan invertible square matrix A . Thus V [ x ] is also a Hermite ring.5 Special linear Groups for valuation rings
Throughout this section R is a unitary commutative ring. We first introduce fivewell-known lemmas. Lemma 4.1. ([4])
Let A = (cid:18) I n − I n (cid:19) , A = (cid:18) − I n I n (cid:19) ∈ E n ( R ) , Then A , A ∈ E n ( R ) , where I n is the n × n identity matrix over R . Lemma 4.2. ([9])
Let a, a ′ , b ∈ R such that aa ′ d − bc = 1 for any c, d ∈ R . We have aa ′ b c d
00 0 1 ≡ a b c d
00 0 1 · a ′ b c d
00 0 1 mod E ( R ) where d = a ′ d, d = ad . In fact, if we transpose aa ′ and b, c, d , the conclusion still holds. For the conve-nience of the reader, we give the following lemma. Lemma 4.3. ([7] , [9]) Let a, a ′ , b ∈ R such that aa ′ d − bc = 1 for any c, d ∈ R . Then d b c aa ′
00 0 1 ≡ d b c a
00 0 1 · d b c a ′
00 0 1 mod E ( R ) where d = a ′ d, d = ad . Lemma 4.4. ([10])
Let b, y ∈ R . If b · y is nilpotent, then ∈ h b, y i ⇔ ∈ h b + y i . Lemma 4.5. ([5])
Let R be a commutative local ring, and let M = f ( x ) g ( x ) 0 p ( x ) q ( x ) 00 0 1 ∈ SL ( R [ x ]) Assume that there is at least one monic polynomial among f ( x ) , g ( x ) , p ( x ) , q ( x ) .Then M ∈ E ( R [ x ]) . Let V be a valuation ring, and M = f ( x ) g ( x ) 0 p ( x ) q ( x ) 00 0 1 ∈ SL ( V [ x ])Observe that f (0) q (0) − g (0) p (0) = 1. since V is local, we may assume that g (0) and p (0) are units (in this case, f (0) or q (0) is not a unit). When deg g ( x ) ≤ deg p ( x ) (ordeg p ( x ) ≤ deg g ( x )), we say that g ( x ) (or p ( x )) is a minimal entry of M , denoted by6 ( m ). When deg M ( m ) = 0, M ( m ) is a unit, and it is obvious that M ∈ E ( V [ x ]).By lemma 4.1, we may assume that g ( x ) is a minimal entry of M . LetSL ( V [ x ]) n , { M ∈ SL ( V [ x ]) | ∃ M , · · · , M t s.t. deg M i ( m ) ≤ ni = 1 , · · · , t, and M ≡ M · · · M t mod E ( V [ x ]) } Obviously, SL ( V [ x ]) n is a multiplicative semigroup, SL ( V [ x ]) = E ( V [ x ]), and E ( V [ x ]) ⊆ SL ( V [ x ]) n for any positive integer n . Lemma 4.6.
Let V be a valuation ring, M ( m ) = g ( x ) , and M = f ( x ) g ( x ) 0 p ( x ) q ( x ) 00 0 1 ∈ SL ( V [ x ]) n , where f ( x ) is primitive and deg f ( x ) < deg g ( x ) . Then M ∈ SL ( V [ x ]) n − .Proof. Let f ( x ) = a l x l + · · · + a m x m + · · · + a g ( x ) = b n x n + b n − x n − + · · · + b where a l = 0, b n = 0, l < n , a m and b are units.(1) Suppose that a l | b n . Let b n = c · a l and g ( x ) = g ( x ) − cx n − l f ( x ). Then g (0) = g (0) is a unit, and deg g ( x ) ≤ n −
1. By Lemma 3.5, for any s ≥ deg q ( x ) − deg g ( x ) + 1, there is q ( x ) such that q ( x ) ≡ x s q ( x ) mod g ( x ) V [ x ], and deg q ( x ) ≤ n − s ≥ n − l . By Lemmas 4.3 and 4.5, M ≡ f ( x ) g ( x ) 0 p ( x ) x s q ( x ) 00 0 1 = f ( x ) g ( x ) 0 p ( x ) x n − l · x s − n + l q ( x ) 00 0 1 ≡ x n − l f ( x ) g ( x ) 0 p ( x ) x s − n + l q ( x ) 00 0 1 · x s − n + l q ( x ) f ( x ) g ( x ) 0 p ( x ) x n − l
00 0 1 ≡ x n − l f ( x ) g ( x ) 0 p ( x ) x s − n + l q ( x ) 00 0 q ≡ x n − l f ( x ) g ( x ) 0 p ( x ) q ( x ) 00 0 1 , M mod ( V [ x ])where q ( x ) = x s − n +1 q ( x ) − cp ( x ). Obviously, p (0) is necessarily a unit anddeg M ( m ) ≤ deg g ( x ) ≤ n −
1, so M ∈ SL ( V [ x ]) n − and M ∈ SL ( V [ x ]) n − .72) If a l ∤ b n , then b n | a l and a l = b n · c , where c is not a unit. Let f ( x ) = x n − l f ( x ) − c · g ( x ). Then M ≡ x n − l f ( x ) g ( x ) 0 p ( x ) x s − n + l q ( x ) 00 0 1 ≡ f ( x ) g ( x ) 0 p ( x ) x s − n + l q ( x ) 00 0 1 mod E ( V [ x ])Where f ( x ) = ( a l − − cb n − ) x n − + · · · + ( a m − cb m + n − l ) x m + n − l + · · · + ( − cb ).Since c is not a unit, a m − cb m + n − l is a unit, namely the degree of the term wherethe coefficient is a unit in f ( x ) is greater than that of f ( x ) by one (since n − l ≥ f ( x ) < n , and f ( x ) is also primitive. If b n can be divided by the leadingcoefficient of f ( x ), then by case (1), M ∈ SL ( V [ x ]) n − . Otherwise, by repeatingthe preceding procedure, we obtain f i ( x ) such that the leading coefficient of f i ( x )( i ≤ n − m ) will be a unit, and M ≡ f ( x ) g ( x ) 0 p ( x ) x s − n + l q ( x ) 00 0 1 ≡ f i ( x ) g ( x ) 0 g p ( x ) g q ( x ) 00 0 1 mod E ( V [ x ])By lemma 4.5, M ∈ E ( R [ x ]), and certainly M ∈ SL ( V [ x ]) n − . Lemma 4.7.
Let V be a valuation ring, M ( m ) = g ( x ) , and M = a g ( x ) 0 p ( x ) q ( x ) 00 0 1 ∈ SL ( V [ x ]) n , where a ∈ V . Then M ∈ SL ( V [ x ]) n − Proof. (1) If a = 0 or a is a unit, the result is obviously true.(2) If a is nilpotent, then by lemma 4.4, 1 ∈ h g ( x ) + a i , so there exists f ( x ) suchthat f ( x ) · ( g ( x ) + a ) = 1, and M = a g ( x ) 0 p ( x ) q ( x ) 00 0 1 ≡ a g ( x ) + a p ( x ) q ( x ) 00 0 1 ≡ a g ( x ) + a p ( x ) + f ( x ) q ( x ) q ( x ) 00 0 1 , M mod E ( V [ x ])Since 1 + a is a unit, M ∈ E ( V [ x ]), and hence M ∈ E ( V [ x ]) ⊆ SL ( V [ x ]) n − .(3) If a is neither a unit nor nilpotent, we set8 ( x ) = b n x n + b n − x n − + · · · + b p ( x ) = a m x m + a m − x m − + · · · + a q ( x ) = c s x s + c s − x s − + · · · + c where b n = 0, a m = 0, c s = 0. Furthermore, by elementary transformation, wemay assume that a ∤ b n , a m . By virtue of valuation rings, there exist d , d ∈ V such that a = a m · d = b n · d and d , d are not units. If a m · b n = 0, then a = a m d b n d = a m b n d d = 0, so a is nilpotent, a contradiction. Therefore, wecan assume that a m · b n = 0. Since aq ( x ) − p ( x ) g ( x ) = 1, s ≥ m + n , and a · c s = a · c s − = · · · = a · c m + n +1 = 0 , a · c m + n = a m b n . Obviously, c s , c s − , · · · , c m + n +1 are not units.For any k , i ( m + n + 1 ≤ i ≤ s ), it is necessary that a k | c i . Otherwise, thereexists p i ∈ V such that a k = c i · p i , and a k +1 = a · a k = a · c i · p i = 0, a contradiction.We show that b n ∤ c m + n . If this is not true, then b n | c m + n , c m + n = b n d , a m b n = ac m + n = a m d b n d = a m b n d d , a m b n (1 − d d ) = 0. Since d is not a unit,then 1 − d d is a unit, and a m b n = 0, a contradiction. Thus b n ∤ c m + n , c m + n | b n ,and we assume that b n = c m + n · d , where d ∈ V is not a unit.Now, we show that there exists u ( x ) ∈ V ( x ) such that h ( x ) = q ( x ) − u ( x ) · g ( x )satisfying that deg h ( x ) = m + n , and the leading coefficient of h ( x ) is c m + n · ε ,where ε ∈ V is a unit. Because b n | a , a k | c i for any positive integer k , and i = m + n + 1 , · · · , s , then b kn | c s , and there exists e s ∈ V such that c s = b n · e s , e s is not a unit. Similarly, for any k , a k | e s since otherwise a k = e s · d , and a k +2 = a · a · a k = a · b n · d · e s · d = a · ( b n · e s ) · d · d = a · c s · d · d = 0so a is nilpotent, a contradiction.Let h ( x ) = q ( x ) − e s x s − n · g ( x ). Then deg h ( x ) ≤ s −
1, and the coefficient c ′ s − i of x s − i in h ( x ) satisfies that a k | c ′ s − i (since c ′ s − i = c s − i − e s · b n − i , and a k | c s − i , e s ),where i = 1 , , · · · , s − m − n − c ′ m + n = c m + n − e s · b m +2 n − s = c m + n (1 − b m +2 n − s · e s /c m + n ) . Since c m + n | b n , b n | a , and a k | e s , then c km + n | e s . So e s /c m + n is not a unit, and(1 − b m +2 n − s · e s /c m + n ) is a unit. Since b n | a, a k | c ′ k − , then b kn | c ′ k − , by repeatingthe process above, we obtain that h ( x ) , q ( x ) − u ( x ) g ( x ) = c m + n · ε · x m + n + · · · + c , where ε is a unit and c ∈ V .Let h ( x ) , h ( x ) − c b − g ( x ), then h ( x ) = x ( c m + n · εx m + n − + · · · + c (1)0 ) . h ( x ) , h ( x ) − c ′ b − · x · g ( x ) = h ( x ) − ( c b − + c (1)0 b − x ) g ( x ) , so h ( x ) = x ( c m + n · εx m + n − + · · · + c (2)0 ) . By repeating the process above m + 1 times, we obtain that h ( x ) = h ( x ) − u ( x ) g ( x ) = x m +1 (( c m + n · ε + b n · q ) x n − + · · · + c ( m +1)0 ) , where q = − c ( m )0 b − , c (1)0 , c (2)0 , · · · , c ( m +1)0 ∈ V . Since c m + n ε + b n q = c m + n ε + c m + n dq = c m + n ( ε + dq ) , where ε is a unit, and d is not a unit. Then ε , ε + d · q is a unit. Let u ( x ) = u ( x ) + u ( x ), q ( x ) = c m + n · εx n − + · · · + c (1)0 ). Then h ( x ) = x m +1 q ( x ), and h ( x ) = h ( x ) − u ( x ) g ( x ) = q ( x ) − ( u ( x ) + u ( x )) g ( x ) = q ( x ) − u ( x ) g ( x ) . Define p ( x ) , p ( x ) − au ( x ) ,g ( x ) , g ( x ) − d · ¯ ε − · x · q ( x ) ,f ( x ) , ax m +1 − d · ¯ ε − · x · p ( x ) . Since b n = c m + n · d = d · ε − · c m + n · ε , we know that deg g ( x ) ≤ n − g (0) = g (0)is a unit. Applying elementary transformations and Lemma 4.3, 4.5, we have M = a g ( x ) 0 p ( x ) q ( x ) 00 0 1 ≡ a g ( x ) 0 p ( x ) h ( x ) 00 0 1 ≡ a g ( x ) 0 p ( x ) x m +1 q ( x ) 00 0 1 ≡ aq ( x ) g ( x ) 0 p ( x ) x m +1
00 0 1 · ax m +1 g ( x ) 0 p ( x ) q ( x ) 00 0 1 ≡ ax m +1 g ( x ) 0 p ( x ) q ( x ) 00 0 1 ≡ f ( x ) g ( x ) 0 p ( x ) q ( x ) 00 0 1 , M mod E ( V [ x ]) . Since f ( x ) = ax m +1 − d · ε − · x · p ( x ), then f (0) = 0, and p (0) must be aunit. Thus deg M ( m ) ≤ deg g ( x ) ≤ n − M ∈ SL ( V [ x ]) n − , and hence M ∈ SL ( V [ x ]) n − . Theorem 4.1.
Let V be a valuation ring, and M = f ( x ) g ( x ) 0 p ( x ) q ( x ) 00 0 1 ∈ SL ( V [ x ]) Then M ∈ E ( V [ x ]) . roof. We assume that M ( m ) = g ( x ) with M ∈ SL ( V [ x ]) n . By Lemma 3.5,for any s > deg f ( x ) − deg g ( x ) + 1, there exists f ( x ) such that f ( x ) = x s f ( x )mod g ( x ) V [ x ], and deg f ( x ) < deg g ( x ) = n . Furthermore, let f ( x ) = af ( x ),where a ∈ V , f ( x ) is primitive, and deg f ( x ) ≤ n −
1. By Lemma 4.2, 4.5, we have M ≡ ax s f ( x ) g ( x ) 0 p ( x ) q ( x ) 00 0 1 ≡ ax s g ( x ) 0 p ( x ) f ( x ) q ( x ) 00 0 1 f ( x ) g ( x ) 0 p ( x ) ax s q ( x ) 00 0 1 ≡ a g ( x ) 0 p ( x ) x s f ( x ) q ( x ) 00 0 1 x s g ( x ) 0 p ( x ) af ( x ) q ( x ) 00 0 1 f ( x ) g ( x ) 0 p ( x ) ax s q ( x ) 00 0 1 ≡ a g ( x ) 0 p ( x ) x s f ( x ) q ( x ) 00 0 1 f ( x ) g ( x ) 0 p ( x ) ax s q ( x ) 00 0 1 mod E ( V [ x ]))Let M = a g ( x ) 0 p ( x ) x s f ( x ) q ( x ) 00 0 1 , M = f ( x ) g ( x ) 0 p ( x ) ax s q ( x ) 00 0 1 Since a ∈ V , f ( x ) is primitive and deg f ( x ) ≤ n −
1, by Lemmas 4.6 and 4.7, M , M ∈ SL ( V [ x ]) n − , then M ∈ SL ( V [ x ]) n − . By repeating the procedureabove, we have that M ∈ SL ( V [ x ]) n − , · · · , SL ( V [ x ]) = E ( V [ x ]). Theorem 4.2.
Let V ba a valuation ring. Then SL ( V [ x ]) = E ( V [ x ]) .Proof. Firstly, it is obvious that E ( V [ x ]) ⊆ SL ( V [ x ]). For any A ( x ) ∈ SL ( V [ x ]),we set A ( x ) = a ( x ) a ( x ) a ( x ) a ( x ) a ( x ) a ( x ) a ( x ) a ( x ) a ( x ) Then α = ( a ( x ) , a ( x ) , a ( x )) is a unimodular row in V × [ x ]. By Theorem 3.1,there exists E ∈ E ( V [ x ]) such that α · E = (0 , , A ( x ) ≡ f ( x ) g ( x ) 0 p ( x ) q ( x ) 00 0 1 mod E ( V [ x ])By Theorem 4.1, A ( x ) ∈ E ( V [ x ]). Thus SL ( V [ x ]) = E ( V [ x ]). Theorem 4.3.
Let V be a valuation ring. Then SL n ( V [ x ]) = E n ( V [ x ]) for n ≥ .Furthermore, every matrix σ ( x ) ∈ SL n ( V [ x ]) is congruent to σ (0) modulo E n ( V [ x ]) .Proof. By Theorem 4.2, we easily obtain that SL n ( V [ x ]) = E n ( V [ x ]) for n ≥
3. Foran arbitrary σ ( x ) ∈ SL n ( V [ x ]), we have that β ( x ) = σ (0) − · σ ( x ) ∈ SL n ( V [ x ]) = E n ( V [ x ]), and σ ( x ) = σ (0) · β ( x ). Thus σ ( x ) is congruent to σ (0) modulo E n ( V [ x ]).11 emma 4.8. ([5]GL n -Patching Theorem ) Let n ≥ and σ ( x ) ∈ SL n ( R [ x ] , ( x )) . If σ m ( x ) ∈ E n ( R m [ x ]) for every maximal ideal m of R , then σ ( x ) ∈ E n ( R [ x ]) . For any σ ( x ) ∈ SL n ( R [ x ]), obviously, β ( x ) = σ (0) − · σ ( x ) ∈ SL n ( R [ x ] , ( x )).Combining Theorem 4.3 and the GL n -Patching Theorem, we now can give an answerto question (3). Theorem 4.4.
Let R be an arithmetical ring and n ≥ . Then SL n ( R [ x ]) = SL n ( R ) · E n ( R [ x ]) .Proof. Let σ ( x ) ∈ SL n ( R [ x ]). Then β ( x ) = σ (0) − · σ ( x ) ∈ SL n ( R [ x ] , ( x )). Since R m is a valuation ring, by Theorem 4.3, β m ( x ) ∈ E n ( R m [ x ]) for every maximalideal m of R , by the virtue of GL n -Patching Theorem, β ( x ) ∈ E n ( R [ x ]). Thus σ ( x ) = σ (0) · β ( x ) ∈ SL n ( R ) · E n ( R [ x ]), and hence SL n ( R [ x ]) = SL n ( R ) · E n ( R [ x ]).From Theorem 4.4, we can obtain the following result directly. Theorem 4.5.
Let R be an arithmetical ring and n ≥ . Then every matrix σ ( x ) ∈ SL n ( R [ x ]) is congruent to σ (0) modulo E n ( R [ x ]) . Valuation rings form a class of important local rings. We conclude that Questions(1), (2) and (3) are completely solved by Theorems 3.2, 4.1,4.3 and 4.5.
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