La conjecture des anneaux de Hermite en dimension 1 Ihsen Yengui D´ epartement de Math´ ematiques, Facult´ e des Sciences de Sfax, Tunisie Email: ihsen.yengui@fss.rnu.tn JNCF 08, Luminy, Octobre 2008 1
The Hermite ring conjecture 1972: If R is an Hermite ring, then R [ X ] is also Hermite. � If R is a ring and v = ( v 0 ( X ) , . . . , v n ( X )) is a unimodular row over R [ X ] such that v (0) = (1 , 0 , . . . , 0) , then v can be completed to a ma- trix in GL n +1 ( R [ X ]) . Recall that a ring A is said to be Hermite if any finitely generated stably free A -module is free. Examples of Hermite rings are local rings, rings of Krull dimension ≤ 1, polynomial rings over Bezout domains. 2
I will prove constructively that for any finite- dimensional ring R and n ≥ dim R +2, the group E n ( R [ X ]) acts transitively on Um n ( R [ X ]). In particular, I obtain, without any Noetherian hypothesis, that for any finite-dimensional ring R , all finitely generated stably free modules over R [ X ] of rank > dim R are free. 3
More particularly, I obtain that for any ring R with Krull dimension ≤ 1, all finitely generated stably free modules over R [ X ] are free. This settles the Hermite ring conjecture for rings of Krull dimension ≤ 1. The proof relies heavily on the very nice paper of M. Roitman “ On stably extended projec- tive modules over polynomial rings , Proc. Amer. Math. Soc. 97 (1986) 585-589”. 4
Lemma 0: Let R be a ring, and f, g ∈ R [ X ] with f a monic polynomial. Then Res( f, g ) ∈ R × . � f, g � = R [ X ] ⇐ ⇒ Proof: “ ⇐ ” This is an immediate consequence of the fact that Res( f, g ) ∈ � f, g � ∩ R . “ ⇒ ” Let u, v ∈ R [ X ] such that uf + vg = 1. Since f is a monic polynomial, we have Res( f, vg ) = Res( f, v ) Res( f, g ) = Res( f, vg + uf ) = Res( f, 1) = 1 . 5
Lemma 1: Let R be a ring, and I an ideal in R [ X ] that contains a monic polynomial. Let J be an ideal in R such that I + J [ X ] = R [ X ] . Then ( I ∩ R ) + J = R . Classical Proof: Use the “going-up” property of integral extensions. Constructive Proof: Let us denote by f a monic polynomial in I . Since I + J [ X ] = R [ X ], there exist g ∈ I and h ∈ J [ X ] such that It follows that � ¯ g + h = 1. f, ¯ g � = ( R /J )[ X ] where the classes are taken modulo J [ X ]. By virtue of Lemma 0, we obtain that Res( ¯ g ) ∈ f, ¯ ( R /J ) × . As f is a monic polynomial, Res( ¯ f, ¯ g ) = Res( f, g ), and thus � Res( f, g ) � + J = R . The desired conclusion follows from the fact that Res( f, g ) ∈ I ∩ R . 6
Lemma 2 (Roitman): Let R be a ring, f ( X ) ∈ R [ X ] n > 0 , and of degree such R × . ∈ ∈ f (0) g ( X ) that Then for any R [ X ] and k ≥ deg g ( X ) − deg f ( X ) + 1 , ∃ h k ( X ) ∈ R [ X ] of degree < n such that g ( X ) ≡ X k h k ( X ) mod � f ( X ) � . Let f ( X ) = a 0 + · · · + a n X n , g ( X ) = Proof: Let g ( X ) − c 0 a − 1 c 0 + · · · + c m X m . 0 f ( X ) = Xh 1 ( X ). Then g ( X ) ≡ Xh 1 ( X ) mod � f ( X ) � and deg h 1 ( X ) < max( m, n ). Similarly we obtain h 2 ( X ) such that h 1 ( X ) ≡ Xh 2 ( X ) mod � f ( X ) � , g ( X ) ≡ X 2 h 2 ( X ) � f ( X ) � , deg h 2 ( X ) < mod max( m − 1 , n ), and so on. 7
Lemma 3 (Vaserstein): Let R be a ring, ( x 0 , . . . , x r ) ∈ Um r +1 ( R ) , ≥ 2 , and r and t let be an element of R which is invert- mod � x 0 , . . . , x r − 2 � . ible Then there ex- ists E ∈ E r +1 ( R ) such that E ( x 0 , . . . , x r ) = ( x 0 , . . . , x r − 1 , t 2 x r ) . Proof: This is also Proposition III.6.1.(b) of T. Y. Lam ’s book “ Serre’s Problem on Projective Modules. Springer Monographs in Mathematics, 2006”. The proofs given by Lam and Roitman are constructive and free of any Noetherian hy- pothesis. 8
Lemma 4 (Bass): Let k ∈ N , R a ring, f 1 , . . . , f r ∈ R [ X ] with degrees ≤ k − 1 , and f r +1 ∈ R [ X ] monic with degree k . If the co- efficients of f 1 , . . . , f r generate the ideal R of R , then � f 1 , . . . , f r , f r +1 � contains a monic with degree k − 1 . Proof: Let us denote by a = � f 1 , . . . , f r , f r +1 � and b the ideal formed by the coefficients of X k − 1 of the elements of a having degree ≤ k − 1. It suffices to prove that b = R . In fact we will prove that b contains all the coefficients of f 1 , . . . , f r . For 1 ≤ i ≤ r , denoting by f i = b 0 + b 1 X + · · · + b k − 1 X k − 1 and f r +1 = a 0 + · · · + a k − 1 X k − 1 + X k , we have b k − 1 ∈ b and k − 1 X k − 1 ∈ a f ′ i = Xf i − b k − 1 f = b ′ 0 + b ′ 1 X + · · · + b ′ with b ′ Thus, b ′ j ≡ b j − 1 mod � b k − 1 � . k − 1 = b k − 2 − a k − 1 b k − 1 ∈ b , b k − 2 ∈ b , and so on until getting that all the b i ’s are in b . 9
Lemma 5 (Suslin): Let A be a commutative If � v 1 ( X ) , . . . , v n ( X ) � = A [ X ] where v 1 is ring. monic and n ≥ 2 , then there exist γ 1 , . . . , γ ℓ ∈ E n − 1 ( A [ X ]) such that, denoting by w i the first coordinate of γ i t ( v 2 , . . . , v n ) , we have � Res( v 1 , w 1 ) , . . . , Res( v 1 , w ℓ ) � = A . Proof: A constructive proof is given in I. Y “ Making the use of maximal ideals constructive . Theoretical Computer Science 392 (2008) 174- 178”. 10
Stable range Theorem: Let R be a ring of di- mension ≤ d , n ≥ d +1 , and let v = ( v 0 , . . . , v n ) ∈ Um n +1 ( R ) . Then there exists E ∈ E n +1 ( R ) such that E v = (1 , 0 , . . . , 0) . Stable range Theorem, bis: For any ring R with Krull dimension ≤ d , all finitely generated stably free R -modules of rank > d are free. Proof: A constructive proof is given by T. Coquand, H Lombardi and C. Quitt´ e in “ Generating non-noetherian modules construc- Manuscripta mathematica 115 (2004) tively. 513–520”. 11
Recall that the boundary ideal of an element a of a ring R is the ideal I ( a ) of R generated by a and all the y ∈ R such that ay is nilpotent. Moreover, dim R ≤ d ⇔ dim ( R / I ( a )) ≤ d − 1 ∀ a ∈ R . This defines the Krull dimension recursively ini- tializing with “dim R ≤ − 1 ⇔ R being trivial”. See the paper by T. Coquand T, H. Lom- bardi, and M.-F. Roy “ An elementary chara- terization of Krull dimension , From sets and types to analysis and topology: towards practi- cable foundations for constructive mathematics (L. Corsilla, P. Schuster, eds), Oxford University Press, 2005”. 12
Main Theorem: R Let be a ring of di- mension ≤ d , n ≥ d + 1 , and let v ( X ) = ( v 0 ( X ) , . . . , v n ( X )) ∈ Um n +1 ( R [ X ]) . Then there exists E ∈ E n +1 ( R [ X ]) such that E v ( X ) = (1 , 0 , . . . , 0) . Proof: By virtue of the Stable range Theo- rem, it suffices to prove that there exists E ∈ E n +1 ( R [ X ]) such that E v ( X ) = v (0). For this, by the local-global principle for elementary ma- trices, we can suppose that R is local. More- over, it is clear that we can suppose that R is reduced. 13
We prove the claim by double induction on the number of nonzero coefficients of N v 0 ( X ) , . . . , v n ( X ) and d , starting with N = 1 (in that case the result is immediate) and d = 0 (in that case the result is well-known). We will first prove a first claim: v ( X ) can be transformed by elementary operations into a vector with one constant entry. 14
Let N > 1 and d > 0. We may as- sume that v 0 (0) ∈ R × . Let us denote by the leading coefficient of and := a v 0 m 0 ∈ R × deg v 0 . If a then the result follows from Suslin’s lemma. So we may assume a ∈ Rad( R ). By the induction hypothesis applied to the ring R / � a � , we can assume that v ( X ) ≡ (1 , 0 , . . . , 0) mod ( a R [ X ]) n +1 . By Roitman’s Lemma, we assume now v i = X 2 k w i , where deg w i < m 0 for 1 ≤ i ≤ n . By Vaserstein’s Lemma, we assume deg v i < m 0 . If m 0 ≤ 1, our first claim is established. Assume now that m 0 ≥ 2. 15
Let ( c 1 , . . . , c m 0 ( n − 1) ) be the coefficients 1 , X, . . . , X m 0 − 1 of in the polynomials v 2 ( X ) , . . . , v n ( X ). By Lemma 1, the ideal generated in R a by R a ∩ ( v 0 R a [ X ] + v 1 R a [ X ]) and the c i ’s is R a . As m 0 ( n − 1) ≥ 2 d > dim R a , by the Stable range Theorem, ∃ ( c ′ 1 , . . . , c ′ m 0 ( n − 1) ) ≡ ( c 1 , . . . , c m 0 ( n − 1) ) mod ( v 0 R [ X ] + v 1 R [ X ]) ∩ R such that c ′ 1 R a + · · · + c ′ m 0 ( n − 1) R a = R a . 16
Assume that we have already c 1 R a + · · · + c m 0 ( n − 1) R a = R a . By Bass’ Lemma, the ideal � v 0 , v 2 , . . . , v n � of R [ X ] contains a polynomial w ( X ) of degree m 0 − 1 which is unitary in R a . Let us denote the leading coefficient of w by ua k where u ∈ R × and that of v 1 by b . Using Vaser- stein’s Lemma, we can by elementary operations make the following transformations ( v 0 , a 2 k v 1 , . . . , v n ) ( v 0 , v 1 , . . . , v n ) → → ( v 0 , a 2 k v 1 + (1 − a k u − 1 b ) w, v 2 , . . . , v n ) . Now, a 2 k v 1 + (1 − a k u − 1 b ) w is unitary in R a . So wa can assume that v 1 unitary in R a , and deg( v 1 ) := m 1 < m 0 . 17
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