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Existence of unimodular elements in a projective module
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Existence of unimodular elements in a projective module Manoj K. Keshari a , Md. Ali Zinna b,∗ a b
Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 400076, India Stat-Math Unit, Indian Statistical Institute Kolkata, 203 B. T. Road, Kolkata 700108, India
a r t i c l e
i n f o
Article history: Received 15 September 2016 Received in revised form 19 December 2016 Available online xxxx Communicated by V. Suresh MSC: 13C10
a b s t r a c t (1) Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose I is an ideal of A of height n such that there are two surjections α : P → → I and φ : L ⊕ An−1 → → I. Assume that either (a) k = 1 and n ≥ 3 or (b) k is arbitrary but n ≥ 4 is even. Then P has a unimodular element (see 4.1, 4.3). (2) Let R be a ring containing Q of even dimension n with height of the Jacobson radical of R ≥ 2. Let P be a projective R[T, T −1 ]-module of rank n with trivial determinant. Assume that there exists a surjection α : P → → I, where I ⊂ R[T, T −1 ] is an ideal of height n such that I is generated by n elements. Then P has a unimodular element (see 3.4). © 2017 Elsevier B.V. All rights reserved.
1. Introduction Let R be a commutative Noetherian ring of dimension n. A classical result of Serre [22] asserts that if P is a projective R-module of rank > n, then P has a unimodular element. However, as is shown by the example of projective module corresponding to the tangent bundle of an even dimensional real sphere, this result is best possible in general. Therefore, it is natural to ask under what conditions P has a unimodular element when rank(P ) = n. In [20], Raja Sridharan asked the following question. Question 1.1. Let R be a ring of dimension n and P be a projective R-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I ⊂ R is an ideal of height n such that I is generated by n elements. Does P have a unimodular element? Raja Sridharan proved that the answer to Question 1.1 is affirmative in certain cases (see [20, Theorems 3, 5]) and “negative” in general. * Corresponding author. E-mail addresses: [email protected] (M.K. Keshari), [email protected] (Md. Ali Zinna). http://dx.doi.org/10.1016/j.jpaa.2017.01.009 0022-4049/© 2017 Elsevier B.V. All rights reserved.
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Plumstead [17] generalized Serre’s result and proved that if P is a projective R[T ]-module of rank > n, then P has a unimodular element. Bhatwadekar and Roy [5] extended Plumstead’s result and proved that projective R[T1 , . . . , Tr ]-modules of rank > n have a unimodular element. Mandal [16] proved analogue of Plumstead that projective R[T, T −1 ]-modules of rank > n have a unimodular element. In another direction, Bhatwadekar and Roy [4] proved that projective modules over D = R[T1 , T2 ]/(T1 T2 ) of rank > n have a unimodular element. Later Wiemers [26] extended this result and proved that if D = R[T1 , . . . , Tr ]/I is a discrete Hodge algebra over R (here I is a monomial ideal), then projective D-modules of rank > n have a unimodular element. In view of results mentioned above, it is natural to ask the following question. Let A be either a polynomial ring over R or a Laurent polynomial ring over R or a discrete Hodge algebra over R. Let P be a projective A-module of rank n. Under what conditions P has a unimodular element? We will mention two such results. Bhatwadekar and Raja Sridharan [7, Theorem 3.4] proved: Let R be a ring of dimension n containing an infinite field. Let P be a projective R[T ]-module of rank n. Assume Pf has a unimodular element for some monic polynomial f ∈ R[T ]. Then P has a unimodular element. Das and Raja Sridharan [11, Theorem 3.4] proved: Let R be a ring of even dimension n containing Q. Let P be a projective R[T ]-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I is an ideal of R[T ] of height n such that I is generated by n elements. Assume further that P/T P has a unimodular element. Then P has a unimodular element. Note that when n is odd, the above result is not known. Further, the requirement in the hypothesis that P/T P has a unimodular element is indeed necessary, in view of negative answer of Question 1.1. Motivated by Bhatwadekar–Sridharan and Das–Sridharan, we prove the following results. Theorem 1.2. (See 3.1) Let R be a ring of dimension n containing an infinite field. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n. Assume Pf (Tk ) has a unimodular element for some monic polynomial f (Tk ) ∈ A. Then P has a unimodular element. Theorem 1.3. (See 3.3) Let R be a ring of even dimension n containing Q. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose there is a surjection α : P →→ I, where I is an ideal of A of height n such that I is a surjective image of L ⊕ An−1 . Further assume that P/(T1 , · · · , Tk )P has a unimodular element. Then P has a unimodular element. Theorem 1.4. (See 3.4) Let R be a ring of even dimension n containing Q. Assume that height of the Jacobson radical of R is ≥ 2. Let P be a projective R[T, T −1 ]-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I is an ideal of R[T, T −1 ] of height n such that I is generated by n elements. Then P has a unimodular element. Theorem 1.5. (See 3.6) Let R be a ring of even dimension n containing Q and P be a projective D = R[T1 , T2 ]/(T1 T2 )-module of rank n with determinant L. Suppose there is a surjection α : P →→ I, where I is an ideal of D of height n such that I is a surjective image of L ⊕ Dn−1 . Further, assume that P/(T1 , T2 )P has a unimodular element. Then P has a unimodular element. In view of above results, we end this section with the following question. Question 1.6. Let R be a ring of even dimension n containing Q. (1) Let P be a projective A = R[T, T −1 ]-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I ⊂ A is an ideal of height n such that I is generated by n elements. Assume further that P/(T − 1)P has a unimodular element. Does P has a unimodular element?
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(2) Let D = R[T1 , . . . , Tk ]/I be a discrete Hodge algebra over R and P be a projective D-module of rank n with determinant L. Suppose there is a surjection α : P →→ I, where I is an ideal of D of height n such that I is a surjective image of L ⊕ Dn−1 . Further, assume that P/(T1 , . . . , Tk )P has a unimodular element. Does P has a unimodular element? (3) Generalize question (1) replacing A by R[X1 , . . . , Xr , Y1±1 , . . . , Ys±1 ] and P/(T −1)P by P/(X1 , . . . , Xr , Y1 − 1, . . . , Ys − 1)P . 2. Preliminaries Assumptions. Throughout this paper, rings are assumed to be commutative Noetherian and projective modules are finitely generated and of constant rank. For a ring R, dim(R) and J (R) will denote the Krull dimension of R and the Jacobson radical of R respectively. In this section we state some results for later use. Definition 2.1. Let R be a ring and P be a projective R-module. An element p ∈ P is called unimodular if there is a surjective R-linear map φ : P →→ R such that φ(p) = 1. In particular, a row (a1 , · · · , an ) ∈ Rn is unimodular (of length n) if there exist b1 , · · · , bn in R such that a1 b1 + · · · + an bn = 1. We write U m(P ) for the set of unimodular elements of P . Theorem 2.2. [14, Corollary 3.1] Let R be a ring of dimension n, A = R[X1 , · · · , Xm ] a polynomial ring over R and P be a projective A[T ]-module of rank n. Assume that P/T P and Pf both contain a unimodular element for some monic polynomial f (T ) ∈ A[T ]. Then P has a unimodular element. The following result is a consequence of a result of Ravi Rao [19, Corollary 2.5] and Quillen’s local–global principle [18, Theorem 1]. Proposition 2.3. Let R be a ring of dimension n. Suppose n! is invertible in R. Then all stably free R[T ]-modules of rank n are extended from R. Lemma 2.4. [11, Lemma 3.3] Let R be a ring and I = (a1 , · · · , an ) be an ideal of R, where n is even. Let u, v ∈ R be such that uv = 1 modulo I. Assume further that the unimodular row (v, a1 , · · · , an ) is completable. Then there exists σ ∈ Mn (R) with det(σ) = u modulo I such that if (a1 , · · · , an )σ = (b1 , · · · , bn ), then b1 , · · · , bn generate I. Proposition 2.5. [11, Proposition 4.2] Let R be a ring containing Q of dimension n with ht(J (R)) ≥ 1. Then any stably free R(T )-module of rank n is free, where R(T ) is obtained from R[T ] by inverting all monic polynomials in T . Lemma 2.6. [3, Lemma 2.2] Let R be a ring and I ⊂ R be an ideal of height r. Let P and Q be two projective R/I-modules of rank r and let α : P →→ I/I 2 and β : Q →→ I/I 2 be surjections. Let ψ : P −→ Q be a homomorphism such that β ◦ ψ = α. Then ψ is an isomorphism. Definition 2.7. Let R be a ring and A = R[T, T −1 ]. We say f (T ) ∈ R[T ] is special monic if f (T ) is a monic polynomial with f (0) = 1. Write A for the ring obtained from A by inverting all special monic polynomials of R[T ]. Then it is easy to see that dim(A) = dim(R). Definition 2.8. (The Euler class group E(R[T, T −1 ])). Let R be a ring containing Q of dimension n ≥ 2 and A = R[T, T −1 ]. Assume that height of the Jacobson radical J (R) of R is ≥ 2. Let I ⊂ A be an ideal of height n such that I/I 2 is generated by n elements. Let α and β be two surjections from (A/I)n to I/I 2 .
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We say that α and β are related if there exists σ ∈ SLn (A/I) such that ασ = β. It is easy to see that this is an equivalence relation on the set of surjections from (A/I)n to I/I 2 . Let [α] denote the equivalence class of α. We call such an equivalence class [α] a local orientation of I. If a surjection α from (A/I)n to I/I 2 can be lifted to a surjection θ : An →→ I, then so can any β equivalent to α (by [13, Corollary 3.9]). Therefore, from now on, we shall identify a surjection α with the equivalence class [α] to which it belongs. We call a local orientation [α] of I a global orientation of I, if the surjection α : (A/I)n →→ I/I 2 can be lifted to a surjection θ : An →→ I. Let G be the free abelian group on the set of pairs (I, ωI ), where I ⊂ A is an ideal of height n having the property that Spec(A/I) is connected, I/I 2 is generated by n elements and ωI : (A/I)n →→ I/I 2 is a local orientation of I. Let I ⊂ A be an ideal of height n such that I/I 2 is generated by n elements. Then I can be decomposed as I = I1 ∩ · · · ∩ Ir , where Ik ’s are pairwise comaximal ideals of A of height n and Spec(A/Ik ) is connected. From [9, Lemma 4.4], it follows that such a decomposition is unique. We say that Ik ’s are the connected components of I. Let ωI : (A/I)n →→ I/I 2 be a surjection. Then ωI induces surjections ωIk : (A/Ik )n →→ Ik /Ik2 . By (I, ωI ), we denote the element (Ik , ωIk ) of G. Let H be the subgroup of G generated by the set of pairs (I, ωI ), where I ⊂ A is an ideal of height n and ωI is a global orientation of I. We define the n-th Euler class group of A, denoted by E n (A), to be G/H. By abuse of notation, we will write E(A) for E n (A). Definition 2.9. (The Euler class of a projective R[T, T −1 ]-module). Let P be a projective A(= R[T, T −1 ])∼ module of rank n having trivial determinant. Let χ : A → ∧n (P ) be an isomorphism. To the pair (P, χ), we associate an element e(P, χ) of E(A) as follows: Let λ : P →→ I be a surjection, where I ⊂ A is an ideal of ¯ : P/IP →→ I/I 2 . Since height n. Let “bar” denote reduction modulo I. We obtain an induced surjection λ P has trivial determinant and dim(A/I) ≤ 1, by Serre’s splitting lemma, P/IP is a free A/I-module of ∼ rank n. We choose an isomorphism γ¯ : (A/I)n → P/IP such that ∧n (¯ γ ) = χ. ¯ Let ωI be the surjection ¯ γ : (A/I)n →→ I/I 2 . Let e(P, χ) be the image of (I, ωI ) in E(A). We say that (I, ωI ) is obtained from the λ¯ pair (λ, χ). It has been proved in [13, Lemma 4.2] that the assignment sending the pair (P, χ) to the element e(P, χ), as described above, is well defined. We define the Euler class of (P, χ) to be e(P, χ). The following results were proved in [13]. Theorem 2.10. [13, 4.4, 4.6, 4.8, 4.9] Let R be a ring containing Q of dimension n ≥ 2 and A = R[T, T −1 ]. Assume that ht(J (R)) ≥ 2. Let I ⊂ A be an ideal of height n such that I/I 2 is generated by n elements and ∼ ωI : (A/I)n →→ I/I 2 be a local orientation of I. Let P be a projective A-module of rank n and χ : A → ∧n (P ) an isomorphism. Then following holds. (1) Suppose that the image of (I, ωI ) is zero in E(A). Then ωI is a global orientation of I, i.e., ωI can be lifted to a surjection from An to I. (2) Suppose that e(P, χ) = (I, ωI ) in E(A). Then there exists a surjection α : P →→ I such that ωI is induced from (α, χ). (3) P has a unimodular element if and only if e(P, χ) = 0 in E(A). (4) The canonical map E(A) −→ E(A) is injective, where A is obtained from A by inverting all special monic polynomials in R[T ]. Remark 2.11. Let R be a ring containing Q of dimension n with ht(J (R)) ≥ 2. Let I ⊂ A = R[T, T −1 ] be an ideal of height n ≥ 3 and let ωI : (A/I)n →→ I/I 2 be a local orientation of I. Let θ ∈ GLn (A/I) be such that det(θ) = f . Then ωI ◦ θ is another local orientation of I, which we denote by fωI . On the other hand, if ωI and ω I are two local orientations of I, then by (2.6), it is easy to see that ω I = f ωI for some unit f ∈ A/I.
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The following result is due to Bhatwadekar and Raja Sridharan [7, Theorem 4.5] in domain case. Same proof works in general (See [10, Proposition 5.3]). Theorem 2.12. Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n. Let P be a projective R[T ]-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I is an ideal of R[T ] of height n such that I is generated by n elements. Then P has a unimodular element. One can obtain the following result from [24] and [21, Lemma 2.4], Proposition 2.13. Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n ≥ 3. Let L be a projective R-module of rank one. Then the canonical map E(R, L) −→ E0 (R, L) (from the Euler class group of R with respect to L to the weak Euler class group of R with respect to L) is an isomorphism. Remark 2.14. Let R be a ring containing Q of dimension n ≥ 3. Let P be a projective R[T ]-module of ∼ rank n with determinant L[T ] and χ : L[T ] → ∧n (P ) be an isomorphism. Then the n-th Euler class group E(R[T ], L[T ]) of R[T ] with respect to L[T ] and the Euler class e(P, χ) ∈ E(R[T ], L[T ]) of (P, χ) are defined in [12, Section 4]. The following result is a consequence of local–global principle for Euler class groups [12, Theorem 4.17]. Proposition 2.15. Let R be a ring containing Q with dim(R) = n ≥ 3. Let P be a projective R[T ]-module of rank n with determinant L[T ]. Suppose that P/T P as well as P ⊗ Rm [T ] have a unimodular element for every maximal ideal m of R. Then P has a unimodular element. ∼
Proof. Let χ : L[T ] → ∧n (P ) be an isomorphism and α : P →→ I be a surjection, where I ⊂ R[T ] is an ideal of height n. Let (P, χ)e(P, χ) = (I, ωI ) in E(R[T ], L[T ]), where (I, ωI ) is obtained from the pair (α, χ). As P ⊗ Rm [T ] has a unimodular element for every maximal ideal m of R, the image of e(P, χ) in m E(Rm [T ], Lm [T ]) is trivial. By [12, Theorem 4.17] the following sequence of groups 0 −→ E(R, L) −→ E(R[T ], L[T ]) −→
E(Rm [T ], Lm [T ])
m
is exact. Therefore, there exists (J, ωJ ) ∈ E(R, L) such that (JR[T ], ωJ ⊗ R[T ]) = e(P, χ) in E(R[T ], L[T ]). Then we have e(P/T P, χ ⊗ R[T ]/(T )) = (J, ωJ ) in E(R, L). Since P/T P has a unimodular element, by [6, Corollary 4.4], (J, ωJ ) = 0 in E(R, L). Consequently, e(P, χ) = 0 in E(R[T ], L[T ]). Hence, by [12, Corollary 4.15], P has a unimodular element. 2 The following result is due to Swan [25, Lemma 1.3]. We give a proof due to Murthy. Lemma 2.16. Let R be a ring and P be a projective R[T, T −1 ]-module. Let f (T ) ∈ R[T ] be monic such that Pf (T ) is extended from R. Then P is extended from R[T −1 ]. Proof. If degree of f (T ) is m, then f (T )T −m = 1 +T −1 g1 (T −1 ) := g(T −1 ). Consider the following Cartesian diagram.
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R[T −1 ]
/ R[T −1 ]T −1 (= R[T, T −1 ])
R[T −1 ]g(T −1 )
/ R[T, T −1 ]g(T −1 ) = R[T, T −1 ]f (T )
Since Pf (T ) is extended from R, Pg(T −1 ) is extended from R[T −1 ]g(T −1 ) . Using a standard patching argument, there exists a projective R[T −1 ]-module Q such that Q ⊗ R[T −1 ] R[T, T −1 ] P . This completes the proof. 2 3. Main results In this section, we shall prove our main results stated in the introduction. Theorem 3.1. Let R be a ring of dimension n containing an infinite field. Let P be a projective R[T1 , · · · , Tk ]-module of rank n. Assume Pf (Tk ) has a unimodular element for some monic polynomial f (Tk ) in the variable Tk . Then P has a unimodular element. Proof. When k = 1, we are done by [7, Theorem 3.4]. Assume k ≥ 2 and use induction on k. Since (P/T1 P )f has a unimodular element, by induction on k, P/T1 P has a unimodular element. Also (P ⊗ R(T1 )[T2 , . . . , Tk ])f has a unimodular element and dim (R(T1 )) = n. So again by induction on k, P ⊗ R(T1 )[T2 , . . . , Tk ] has a unimodular element. Since P is finitely generated, there exists a monic polynomial g ∈ R[T1 ] such that Pg has a unimodular element. Applying (2.2), we get that P has a unimodular element. 2 The next result is due to Das–Sridharan [11, Theorem 3.4] when L = R. Proposition 3.2. Let R be a ring of even dimension n containing Q. Let P be a projective R[T ]-module of rank n with determinant L. Suppose there are surjections α : P →→ I and φ : L ⊕ R[T ]n−1 →→ I, where I is an ideal of R[T ] of height n. Assume further that P/T P has a unimodular element. Then P has a unimodular element. Proof. Let Rred = R/n(R), where n(R) is the nil radical of R. It is easy to derive that P has a unimodular element if and only if P ⊗ Rred has a unimodular element. Therefore, without loss of generality we may assume that R is reduced. We divide the proof into two steps. Step 1: Assume that L is extended from R. Since P/T P has a unimodular element, in view of (2.15), it is enough to prove that if m is a maximal ideal of R, then P ⊗ Rm [T ] has a unimodular element. Note that P ⊗ Rm [T ] has trivial determinant and α ⊗ Rm [T ] : P ⊗ Rm [T ] →→ IRm [T ] is a surjection. Using surjection φ ⊗ Rm [T ], we get that IRm [T ] is generated by n elements. Applying [11, Theorem 3.4] to the ring Rm [T ], it follows that P ⊗ Rm [T ] has a unimodular element. Hence we are done. Step 2: Assume that L is not necessarily extended from R. Since R is reduced, we can find a ring R → S → Q(R) such that (1) (2) (3) (4)
The projective S[T ]-module L ⊗ R[T ] S[T ] is extended from S, S is a finite R-module, the canonical map Spec(S) → Spec(R) is bijective, and for every P ∈ Spec(S), the inclusion R/(P ∩ R) −→ S/P is birational.
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Since L ⊗ R[T ] S[T ] is extended from S, by Step 1, P ⊗ R[T ] S[T ] has a unimodular element. Applying [2, Lemma 3.2], we conclude that P has a unimodular element. 2 The following result extends (3.2) to polynomial ring R[T1 , · · · , Tk ]. Theorem 3.3. Let R be a ring of even dimension n containing Q. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose there are surjection α : P →→ I and φ : L ⊕ An−1 →→ I, where I ⊂ A is an ideal of height n. Assume further that P/(T1 , · · · , Tk )P has a unimodular element. Then P has a unimodular element. Proof. When k = 1, we are done by (3.2). Assume k ≥ 2 and use induction on k. We use “bar” when we move modulo (Tk ). Since R contains Q, replacing Xk by Xk − λ for some λ ∈ Q, we may assume that ht (I) ≥ n. Consider the surjection α : P →→ I induced from α. If ht(I) > n, then I = R[T1 , · · · , Tk−1 ] and hence P contains a unimodular element. Assume ht(I) = n. Since P /(T1 , · · · , Tk−1 )P = P/(T1 , · · · , Tk )P has a unimodular element and φ induces a surjection φ : L ⊕ An−1 →→ I, by induction hypothesis, P has a unimodular element. By similar arguments, we get that P/Tk P has a unimodular element. Write A = R(Tk−1 )[T1 , · · · , Tk−2 , Tk ] and P ⊗ A = P. We have surjections α ⊗ A : P →→ IA and φ ⊗ A : (L ⊗ A) ⊕ An−1 →→ IA. If IA = A, then P has a unimodular element. Assume ht(IA) = n. Since P/Tk P = P/Tk P ⊗ R(Tk−1 )[T1 , · · · , Tk−2 ], we get that P/Tk P and hence P/(T1 , · · · , Tk−2 , Tk )P has a unimodular element. Again by induction hypothesis, P has a unimodular element. So, there exists a monic polynomial f ∈ R[Tk ] such that Pf has a unimodular element. Applying (3.1), we get that P has a unimodular element. 2 Now we prove (1.4) mentioned in the introduction. Theorem 3.4. Let R be a ring containing Q of even dimension n with ht(J (R)) ≥ 2. Let P be a projective R[T, T −1 ]-module of rank n with trivial determinant. Assume there exists a surjection α : P →→ I, where I ⊂ R[T, T −1 ] is an ideal of height n such that I is generated by n elements. Then P has a unimodular element. ∼
Proof. Let χ : R[T, T −1 ] → ∧n (P ) be an isomorphism. Let e(P, χ) = (I, ωI ) ∈ E(R[T, T −1 ]) be obtained from the pair (α, χ). By (2.10 (3)), it is enough to prove that e(P, χ) = (I, wI ) = 0 in E(R[T, T −1 ]). Suppose ωI is given by I = (g1 , · · · , gn ) + I 2 . Also I is generated by n elements, say, I = (f1 , · · · , fn ). By (2.11), there exists τ ∈ GLn (R[T, T −1 ]/I) such that (f 1 , · · · , f n ) = (g 1 , · · · , g n )τ , where “bar” denotes reduction modulo I. Let U, V ∈ R[T, T −1 ] be such that det(τ ) = U and U V = 1 modulo I. Claim: The unimodular row (U, f1 , · · · , fn ) over R[T, T −1 ] is completable. First we show that the theorem follows from the claim. Since n is even, by (2.4), there exists η ∈ Mn (R[T, T −1 ]) with det(η) = V modulo I such that if (f1 , · · · , fn )η = (h1 , · · · , hn ), then I = (h1 , · · · , hn ). Further, (g 1 , · · · , g n )τ ◦ η = (h1 , · · · , hn ). Note τ ◦ η ∈ SLn (R[T, T −1 ]/I). Let A be the ring obtained from R[T, T −1 ] by inverting all special monic polynomials. Since dim(A/IA) = 0, we have SLn (A/IA) = En (A/IA). Let Θ ∈ SLn (A) be a lift of (τ ◦ η) ⊗ A. Then IA = (h1 , · · · , hn )Θ−1 and (h1 , · · · , hn )Θ−1 = (g 1 , · · · , g n ). In other words, (IA, ωI ⊗ A) = 0 in E(A). Since ht(J (R)) ≥ 2, by (2.10 (4)), the canonical map E(R[T, T −1 ]) −→ E(A) is injective. Therefore, (I, ωI ) = 0 in E(R[T, T −1 ]). This completes the proof. So we just need to prove the claim. Proof of the claim: Let Q be the stably free R[T, T −1 ]-module associated to the unimodular row (U, f1 , · · · , fn ). If S is the set of all monic polynomials in R[T ], then S −1 R[T, T −1 ] = R(T ). Applying (2.5), we have QS is free and hence there exists a monic polynomial f ∈ R[T ] such that Qf is free. In
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particular, Qf is extended from R. Hence by (2.16), there exists a projective R[T −1 ]-module Q1 such that Q Q1 ⊗ R[T −1 ] R[T, T −1 ]. Since (Q1 ⊕ R[T −1 ])T −1 Q ⊕ R[T, T −1 ] R[T, T −1 ]n+1 , by Quillen and Suslin [18,23], we get Q1 ⊕ R[T −1 ] R[T −1 ]n+1 . By (2.3), Q1 is extended from R, say, Q1 Q2 ⊗ R R[T −1 ]. Since ht(J (R)) ≥ 1, by Bass [1], Q2 is free. Hence Q is free, i.e. (V, f1 , · · · , fn ) is completable. This proves the claim. 2 Next result is the converse of (3.4). Theorem 3.5. Let R be a ring containing Q of dimension n with ht(J (R)) ≥ 2. Let P be a projective R[T, T −1 ]-module of rank n with trivial determinant. Let I ⊂ R[T, T −1 ] be an ideal of height n which is generated by n elements. Suppose that P has a unimodular element. Then there exists a surjection α : P →→ I. ∼
Proof. Let χ : R[T, T −1 ] → ∧n (P ) be an isomorphism and e(P, χ) ∈ E(R[T, T −1 ]) be obtained from the pair (P, χ). Since P has a unimodular element, by (2.10 (3)), e(P, χ) = 0 in E(R[T, T −1 ]). Let I = (f1 , · · · , fn ) and ωI be the orientation of I induced by f1 , · · · , fn . Then (I, ωI ) = 0 = e(P, χ) in E(R[T, T −1 ]). By (2.10 (2)), there exists a surjection α : P →→ I such that (I, ωI ) is induced from (α, χ). 2 Theorem 3.6. Let R be a ring of even dimension n containing Q and D = R[X, Y ]/(XY ). Let P be a projective D-module of rank n with determinant L. Suppose there are surjections α : P →→ I and φ : L ⊕ Dn−1 →→ I, where I is an ideal of D of height n. Assume further that P/(X, Y )P has a unimodular element. Then P has a unimodular element. Proof. Without loss of generality, we may assume that R is reduced. We continue to denote the images of X and Y in D by X and Y . We give proof in two steps. Step 1: Assume L is extended from R. Let “bar” and “tilde” denote reductions modulo (Y ) and (X) : P →→ I induced from α. Note that ht(I) ≥ n and respectively. We get surjections α : P →→ I and α ≥ n. If ht(I) > n, then I = R[X] and P has a unimodular element. Assume ht(I) = n. Since ht(I) D = R[X] and P /XP = P/(X, Y )P has a unimodular element, by (3.2), P has a unimodular element. Similarly, P has a unimodular element. ). Since det(P) = L is extended from R, it follows from [8, Let p1 ∈ U m(P ). Then p1 ∈ U m(P ) is surjective. Therefore there exists Theorem 5.2, Remark 5.3] that the natural map U m(P) U m(P p2 ∈ U m(P) such that p2 = p1 . Since the following square of rings D
/ R[X]
R[Y ]
/ R
is Cartesian with vertical maps surjective, p1 ∈ U m(P ) and p2 ∈ U m(P) will patch up to give a unimodular element of P . Step 2: Assume L is not necessarily extended from R. Since R is reduced, by [15, Lemma 3.7], there exists a ring S such that (1) (2) (3) (4)
R → S → Q(R), S is a finite R-module, R → S is subintegral and L ⊗ R S is extended from S.
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Note that R → S is actually a finite subintegral extension. Hence, by [15, Lemma 2.11], R[X, Y ]/(XY ) → S[X, Y ]/(XY ) is also subintegral. Since L ⊗ R S is extended from S, by Step 1, P ⊗ S[X, Y ]/(XY ) has a unimodular element. Now following the proof of [2, Lemma 3.2], it is easy to see that P has a unimodular element. 2 Using [8, Theorem 5.2, Remark 5.3] and following the proof of (3.6), we can prove the following result. Theorem 3.7. Let R be a ring of even dimension n containing Q and D = R[Y1 , . . . , Ym ]/Y1 (Y2 , . . . , Ym ). Let P be a projective D-module of rank n with determinant L. Suppose there are surjections α : P →→ I and φ : L ⊕ Dn−1 →→ I, where I is an ideal of D of height n. Assume further that P/(Y1 , . . . , Ym )P has a unimodular element. Then P has a unimodular element. 4. Applications In this section, we give some applications of results proved earlier. When L = R, the following result is proved in [7, Theorem 4.5]. Theorem 4.1. Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n ≥ 3. Let P be a projective R[T ]-module of rank n with determinant L. Suppose I is an ideal of R[T ] of height n such that there are two surjections α : P →→ I and φ : L ⊕ R[T ]n−1 →→ I. Then P has a unimodular element. Proof. Let “bar” denote reduction modulo (T ). Then α induces a surjection α : P →→ I(0). We can assume ht(I(0)) ≥ n. If ht(I(0)) > n, then I(0) = R and P has a unimodular element. Assume ht(I(0)) = n and ∼ fix an isomorphism χ : L → ∧n (P ). Let e(P , χ) = (I(0), ωI(0) ) in E(R, L) be induced from (α, χ). Since φ : L ⊕ Rn−1 →→ I(0) is a surjection, by (2.13), e(P , χ) = 0 in E(R, L). By [6, Corollary 4.4], P/T P has a unimodular element. Rest of the proof is exactly as in (3.2) using (2.12). 2 Corollary 4.2. Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n ≥ 3 and D = R[X, Y1 . . . , Ym ]/X(Y1 , . . . , Ym ). Let P be a projective D-module of rank n with determinant L. Suppose I is an ideal of D of height n such that there are two surjections α : P →→ I and φ : L ⊕ Dn−1 →→ I. Then P has a unimodular element. Proof. Follow the proof of (3.6) and use (4.1). 2 Theorem 4.3. Let R be an affine algebra over an algebraically closed field of characteristic 0 such that dim(R) = n ≥ 4 is even. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose I is an ideal of A of height n such that there are two surjections α : P →→ I and φ : L ⊕ An−1 →→ I. Then P has a unimodular element. Proof. By (2.13) and [6, Corollary 4.4], we get P/(T1 , · · · , Tk )P has a unimodular element. By (3.3), we are done. 2 In the following special case, we can remove the restriction on dimension in (3.3). This result improves [7, Theorem 4.4]. Proposition 4.4. Let R be a ring of dimension n ≥ 2 containing Q and P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose M ⊂ A is a maximal ideal of height n such that there are two surjections α : P →→ M and φ : L ⊕ An−1 →→ M. Then P has a unimodular element.
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Proof. Assume k = 1. If M contains a monic polynomial f ∈ R[T1 ], then Pf has a unimodular element. Using [7, Theorem 3.4], P has a unimodular element. Assume M does not contain a monic polynomial. Since T1 ∈ / M, ideal M + (T1 ) = R[T1 ]. By surjection α, we get P/T1 P has a unimodular element. Follow the proof of (3.2) and use [7, Theorem 4.4] to conclude that P has a unimodular element. We are done when k = 1. Assume k ≥ 2 and use induction on k. For rest of the proof, follow the proof of (3.3). 2 References [1] H. Bass, K-theory and stable algebra, Publ. Math. Inst. Hautes Études Sci. 22 (1964) 5–60. [2] S.M. Bhatwadekar, Inversion of monic polynomials and existence of unimodular elements (II), Math. Z. 200 (1989) 233–238. [3] S.M. Bhatwadekar, Cancellation theorems for projective modules over a two dimensional ring and its polynomial extensions, Compos. Math. 128 (2001) 339–359. [4] S.M. Bhatwadekar, Amit Roy, Stability theorems for overrings of polynomial rings, Invent. Math. 68 (1982) 117–127. [5] S.M. Bhatwadekar, A. Roy, Some theorems about projective modules over polynomial rings, J. Algebra 86 (1984) 150–158. [6] S.M. Bhatwadekar, Raja Sridharan, Euler class group of a Noetherian ring, Compos. Math. 122 (2000) 183–222. [7] S.M. Bhatwadekar, Raja Sridharan, On a question of Roitman, J. Ramanujan Math. Soc. 16 (2001) 45–61. [8] S.M. Bhatwadekar, H. Lindel, Ravi A. Rao, The Bass–Murthy question: Serre dimension of Laurent polynomial extensions, Invent. Math. 81 (1) (1985) 189–203. [9] M.K. Das, The Euler class group of a polynomial algebra, J. Algebra 264 (2003) 582–612. [10] M.K. Das, The Euler class group of a polynomial algebra II, J. Algebra 299 (2006) 94–114. [11] M.K. Das, Raja Sridharan, The Euler class groups of polynomial rings and unimodular elements in projective modules, J. Pure Appl. Algebra 185 (2003) 73–86. [12] M.K. Das, Md. Ali Zinna, The Euler class group of a polynomial algebra with coefficients in a line bundle, Math. Z. 276 (2014) 757–783. [13] M.K. Keshari, Euler class group of a Laurent polynomial ring: local case, J. Algebra 308 (2007) 666–685. [14] M.K. Keshari, Md. Ali Zinna, Unimodular elements in projective modules and an analogue of a result of Mandal, J. Commut. Algebra (2017), in press. [15] M.K. Keshari, Md. Ali Zinna, Efficient generation of ideals in a discrete Hodge algebra, J. Pure Appl. Algebra 221 (2017) 960–970. [16] Satya Mandal, Basic elements and cancellation over Laurent polynomial rings, J. Algebra 79 (1982) 251–257. [17] B. Plumstead, The conjecture of Eisenbud and Evans, Am. J. Math. 105 (1983) 1417–1433. [18] D. Quillen, Projective modules over polynomial rings, Invent. Math. 36 (1976) 167–171. [19] Ravi A. Rao, The Bass–Quillen conjecture in dimension 3 but characteristic = 2, 3 via a question of A. Suslin, Invent. Math. 93 (1988) 609–618. [20] Raja Sridharan, Non-vanishing sections of algebraic vector bundles, J. Algebra 176 (1995) 947–958. [21] Raja Sridharan, Projective modules and complete intersections, K-Theory 13 (1998) 269–278. [22] J.P. Serre, Sur les modules projectifs, in: Semin. Dubreil-Pisot, vol. 14, 1960/1961. [23] A.A. Suslin, Projective modules over polynomial ring are free, Sov. Math. Dokl. 17 (1976) 1160–1164. [24] A.A. Suslin, A cancellation theorem for projective modules over affine algebras, Sov. Math. Dokl. 18 (1977) 1281–1284. [25] R.G. Swan, Projective modules over Laurent polynomial rings, Trans. Am. Math. Soc. 237 (1978) 111–120. [26] A. Wiemers, Some properties of projective modules over discrete Hodge algebras, J. Algebra 150 (1992) 402–426.
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Journal of Pure and Applied Algebra www.elsevier.com/locate/jpaa
Existence of unimodular elements in a projective module Manoj K. Keshari a , Md. Ali Zinna b,∗ a b
Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 400076, India Stat-Math Unit, Indian Statistical Institute Kolkata, 203 B. T. Road, Kolkata 700108, India
a r t i c l e
i n f o
Article history: Received 15 September 2016 Received in revised form 19 December 2016 Available online xxxx Communicated by V. Suresh MSC: 13C10
a b s t r a c t (1) Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose I is an ideal of A of height n such that there are two surjections α : P → → I and φ : L ⊕ An−1 → → I. Assume that either (a) k = 1 and n ≥ 3 or (b) k is arbitrary but n ≥ 4 is even. Then P has a unimodular element (see 4.1, 4.3). (2) Let R be a ring containing Q of even dimension n with height of the Jacobson radical of R ≥ 2. Let P be a projective R[T, T −1 ]-module of rank n with trivial determinant. Assume that there exists a surjection α : P → → I, where I ⊂ R[T, T −1 ] is an ideal of height n such that I is generated by n elements. Then P has a unimodular element (see 3.4). © 2017 Elsevier B.V. All rights reserved.
1. Introduction Let R be a commutative Noetherian ring of dimension n. A classical result of Serre [22] asserts that if P is a projective R-module of rank > n, then P has a unimodular element. However, as is shown by the example of projective module corresponding to the tangent bundle of an even dimensional real sphere, this result is best possible in general. Therefore, it is natural to ask under what conditions P has a unimodular element when rank(P ) = n. In [20], Raja Sridharan asked the following question. Question 1.1. Let R be a ring of dimension n and P be a projective R-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I ⊂ R is an ideal of height n such that I is generated by n elements. Does P have a unimodular element? Raja Sridharan proved that the answer to Question 1.1 is affirmative in certain cases (see [20, Theorems 3, 5]) and “negative” in general. * Corresponding author. E-mail addresses: [email protected] (M.K. Keshari), [email protected] (Md. Ali Zinna). http://dx.doi.org/10.1016/j.jpaa.2017.01.009 0022-4049/© 2017 Elsevier B.V. All rights reserved.
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Plumstead [17] generalized Serre’s result and proved that if P is a projective R[T ]-module of rank > n, then P has a unimodular element. Bhatwadekar and Roy [5] extended Plumstead’s result and proved that projective R[T1 , . . . , Tr ]-modules of rank > n have a unimodular element. Mandal [16] proved analogue of Plumstead that projective R[T, T −1 ]-modules of rank > n have a unimodular element. In another direction, Bhatwadekar and Roy [4] proved that projective modules over D = R[T1 , T2 ]/(T1 T2 ) of rank > n have a unimodular element. Later Wiemers [26] extended this result and proved that if D = R[T1 , . . . , Tr ]/I is a discrete Hodge algebra over R (here I is a monomial ideal), then projective D-modules of rank > n have a unimodular element. In view of results mentioned above, it is natural to ask the following question. Let A be either a polynomial ring over R or a Laurent polynomial ring over R or a discrete Hodge algebra over R. Let P be a projective A-module of rank n. Under what conditions P has a unimodular element? We will mention two such results. Bhatwadekar and Raja Sridharan [7, Theorem 3.4] proved: Let R be a ring of dimension n containing an infinite field. Let P be a projective R[T ]-module of rank n. Assume Pf has a unimodular element for some monic polynomial f ∈ R[T ]. Then P has a unimodular element. Das and Raja Sridharan [11, Theorem 3.4] proved: Let R be a ring of even dimension n containing Q. Let P be a projective R[T ]-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I is an ideal of R[T ] of height n such that I is generated by n elements. Assume further that P/T P has a unimodular element. Then P has a unimodular element. Note that when n is odd, the above result is not known. Further, the requirement in the hypothesis that P/T P has a unimodular element is indeed necessary, in view of negative answer of Question 1.1. Motivated by Bhatwadekar–Sridharan and Das–Sridharan, we prove the following results. Theorem 1.2. (See 3.1) Let R be a ring of dimension n containing an infinite field. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n. Assume Pf (Tk ) has a unimodular element for some monic polynomial f (Tk ) ∈ A. Then P has a unimodular element. Theorem 1.3. (See 3.3) Let R be a ring of even dimension n containing Q. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose there is a surjection α : P →→ I, where I is an ideal of A of height n such that I is a surjective image of L ⊕ An−1 . Further assume that P/(T1 , · · · , Tk )P has a unimodular element. Then P has a unimodular element. Theorem 1.4. (See 3.4) Let R be a ring of even dimension n containing Q. Assume that height of the Jacobson radical of R is ≥ 2. Let P be a projective R[T, T −1 ]-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I is an ideal of R[T, T −1 ] of height n such that I is generated by n elements. Then P has a unimodular element. Theorem 1.5. (See 3.6) Let R be a ring of even dimension n containing Q and P be a projective D = R[T1 , T2 ]/(T1 T2 )-module of rank n with determinant L. Suppose there is a surjection α : P →→ I, where I is an ideal of D of height n such that I is a surjective image of L ⊕ Dn−1 . Further, assume that P/(T1 , T2 )P has a unimodular element. Then P has a unimodular element. In view of above results, we end this section with the following question. Question 1.6. Let R be a ring of even dimension n containing Q. (1) Let P be a projective A = R[T, T −1 ]-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I ⊂ A is an ideal of height n such that I is generated by n elements. Assume further that P/(T − 1)P has a unimodular element. Does P has a unimodular element?
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(2) Let D = R[T1 , . . . , Tk ]/I be a discrete Hodge algebra over R and P be a projective D-module of rank n with determinant L. Suppose there is a surjection α : P →→ I, where I is an ideal of D of height n such that I is a surjective image of L ⊕ Dn−1 . Further, assume that P/(T1 , . . . , Tk )P has a unimodular element. Does P has a unimodular element? (3) Generalize question (1) replacing A by R[X1 , . . . , Xr , Y1±1 , . . . , Ys±1 ] and P/(T −1)P by P/(X1 , . . . , Xr , Y1 − 1, . . . , Ys − 1)P . 2. Preliminaries Assumptions. Throughout this paper, rings are assumed to be commutative Noetherian and projective modules are finitely generated and of constant rank. For a ring R, dim(R) and J (R) will denote the Krull dimension of R and the Jacobson radical of R respectively. In this section we state some results for later use. Definition 2.1. Let R be a ring and P be a projective R-module. An element p ∈ P is called unimodular if there is a surjective R-linear map φ : P →→ R such that φ(p) = 1. In particular, a row (a1 , · · · , an ) ∈ Rn is unimodular (of length n) if there exist b1 , · · · , bn in R such that a1 b1 + · · · + an bn = 1. We write U m(P ) for the set of unimodular elements of P . Theorem 2.2. [14, Corollary 3.1] Let R be a ring of dimension n, A = R[X1 , · · · , Xm ] a polynomial ring over R and P be a projective A[T ]-module of rank n. Assume that P/T P and Pf both contain a unimodular element for some monic polynomial f (T ) ∈ A[T ]. Then P has a unimodular element. The following result is a consequence of a result of Ravi Rao [19, Corollary 2.5] and Quillen’s local–global principle [18, Theorem 1]. Proposition 2.3. Let R be a ring of dimension n. Suppose n! is invertible in R. Then all stably free R[T ]-modules of rank n are extended from R. Lemma 2.4. [11, Lemma 3.3] Let R be a ring and I = (a1 , · · · , an ) be an ideal of R, where n is even. Let u, v ∈ R be such that uv = 1 modulo I. Assume further that the unimodular row (v, a1 , · · · , an ) is completable. Then there exists σ ∈ Mn (R) with det(σ) = u modulo I such that if (a1 , · · · , an )σ = (b1 , · · · , bn ), then b1 , · · · , bn generate I. Proposition 2.5. [11, Proposition 4.2] Let R be a ring containing Q of dimension n with ht(J (R)) ≥ 1. Then any stably free R(T )-module of rank n is free, where R(T ) is obtained from R[T ] by inverting all monic polynomials in T . Lemma 2.6. [3, Lemma 2.2] Let R be a ring and I ⊂ R be an ideal of height r. Let P and Q be two projective R/I-modules of rank r and let α : P →→ I/I 2 and β : Q →→ I/I 2 be surjections. Let ψ : P −→ Q be a homomorphism such that β ◦ ψ = α. Then ψ is an isomorphism. Definition 2.7. Let R be a ring and A = R[T, T −1 ]. We say f (T ) ∈ R[T ] is special monic if f (T ) is a monic polynomial with f (0) = 1. Write A for the ring obtained from A by inverting all special monic polynomials of R[T ]. Then it is easy to see that dim(A) = dim(R). Definition 2.8. (The Euler class group E(R[T, T −1 ])). Let R be a ring containing Q of dimension n ≥ 2 and A = R[T, T −1 ]. Assume that height of the Jacobson radical J (R) of R is ≥ 2. Let I ⊂ A be an ideal of height n such that I/I 2 is generated by n elements. Let α and β be two surjections from (A/I)n to I/I 2 .
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We say that α and β are related if there exists σ ∈ SLn (A/I) such that ασ = β. It is easy to see that this is an equivalence relation on the set of surjections from (A/I)n to I/I 2 . Let [α] denote the equivalence class of α. We call such an equivalence class [α] a local orientation of I. If a surjection α from (A/I)n to I/I 2 can be lifted to a surjection θ : An →→ I, then so can any β equivalent to α (by [13, Corollary 3.9]). Therefore, from now on, we shall identify a surjection α with the equivalence class [α] to which it belongs. We call a local orientation [α] of I a global orientation of I, if the surjection α : (A/I)n →→ I/I 2 can be lifted to a surjection θ : An →→ I. Let G be the free abelian group on the set of pairs (I, ωI ), where I ⊂ A is an ideal of height n having the property that Spec(A/I) is connected, I/I 2 is generated by n elements and ωI : (A/I)n →→ I/I 2 is a local orientation of I. Let I ⊂ A be an ideal of height n such that I/I 2 is generated by n elements. Then I can be decomposed as I = I1 ∩ · · · ∩ Ir , where Ik ’s are pairwise comaximal ideals of A of height n and Spec(A/Ik ) is connected. From [9, Lemma 4.4], it follows that such a decomposition is unique. We say that Ik ’s are the connected components of I. Let ωI : (A/I)n →→ I/I 2 be a surjection. Then ωI induces surjections ωIk : (A/Ik )n →→ Ik /Ik2 . By (I, ωI ), we denote the element (Ik , ωIk ) of G. Let H be the subgroup of G generated by the set of pairs (I, ωI ), where I ⊂ A is an ideal of height n and ωI is a global orientation of I. We define the n-th Euler class group of A, denoted by E n (A), to be G/H. By abuse of notation, we will write E(A) for E n (A). Definition 2.9. (The Euler class of a projective R[T, T −1 ]-module). Let P be a projective A(= R[T, T −1 ])∼ module of rank n having trivial determinant. Let χ : A → ∧n (P ) be an isomorphism. To the pair (P, χ), we associate an element e(P, χ) of E(A) as follows: Let λ : P →→ I be a surjection, where I ⊂ A is an ideal of ¯ : P/IP →→ I/I 2 . Since height n. Let “bar” denote reduction modulo I. We obtain an induced surjection λ P has trivial determinant and dim(A/I) ≤ 1, by Serre’s splitting lemma, P/IP is a free A/I-module of ∼ rank n. We choose an isomorphism γ¯ : (A/I)n → P/IP such that ∧n (¯ γ ) = χ. ¯ Let ωI be the surjection ¯ γ : (A/I)n →→ I/I 2 . Let e(P, χ) be the image of (I, ωI ) in E(A). We say that (I, ωI ) is obtained from the λ¯ pair (λ, χ). It has been proved in [13, Lemma 4.2] that the assignment sending the pair (P, χ) to the element e(P, χ), as described above, is well defined. We define the Euler class of (P, χ) to be e(P, χ). The following results were proved in [13]. Theorem 2.10. [13, 4.4, 4.6, 4.8, 4.9] Let R be a ring containing Q of dimension n ≥ 2 and A = R[T, T −1 ]. Assume that ht(J (R)) ≥ 2. Let I ⊂ A be an ideal of height n such that I/I 2 is generated by n elements and ∼ ωI : (A/I)n →→ I/I 2 be a local orientation of I. Let P be a projective A-module of rank n and χ : A → ∧n (P ) an isomorphism. Then following holds. (1) Suppose that the image of (I, ωI ) is zero in E(A). Then ωI is a global orientation of I, i.e., ωI can be lifted to a surjection from An to I. (2) Suppose that e(P, χ) = (I, ωI ) in E(A). Then there exists a surjection α : P →→ I such that ωI is induced from (α, χ). (3) P has a unimodular element if and only if e(P, χ) = 0 in E(A). (4) The canonical map E(A) −→ E(A) is injective, where A is obtained from A by inverting all special monic polynomials in R[T ]. Remark 2.11. Let R be a ring containing Q of dimension n with ht(J (R)) ≥ 2. Let I ⊂ A = R[T, T −1 ] be an ideal of height n ≥ 3 and let ωI : (A/I)n →→ I/I 2 be a local orientation of I. Let θ ∈ GLn (A/I) be such that det(θ) = f . Then ωI ◦ θ is another local orientation of I, which we denote by fωI . On the other hand, if ωI and ω I are two local orientations of I, then by (2.6), it is easy to see that ω I = f ωI for some unit f ∈ A/I.
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The following result is due to Bhatwadekar and Raja Sridharan [7, Theorem 4.5] in domain case. Same proof works in general (See [10, Proposition 5.3]). Theorem 2.12. Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n. Let P be a projective R[T ]-module of rank n with trivial determinant. Suppose there is a surjection α : P →→ I, where I is an ideal of R[T ] of height n such that I is generated by n elements. Then P has a unimodular element. One can obtain the following result from [24] and [21, Lemma 2.4], Proposition 2.13. Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n ≥ 3. Let L be a projective R-module of rank one. Then the canonical map E(R, L) −→ E0 (R, L) (from the Euler class group of R with respect to L to the weak Euler class group of R with respect to L) is an isomorphism. Remark 2.14. Let R be a ring containing Q of dimension n ≥ 3. Let P be a projective R[T ]-module of ∼ rank n with determinant L[T ] and χ : L[T ] → ∧n (P ) be an isomorphism. Then the n-th Euler class group E(R[T ], L[T ]) of R[T ] with respect to L[T ] and the Euler class e(P, χ) ∈ E(R[T ], L[T ]) of (P, χ) are defined in [12, Section 4]. The following result is a consequence of local–global principle for Euler class groups [12, Theorem 4.17]. Proposition 2.15. Let R be a ring containing Q with dim(R) = n ≥ 3. Let P be a projective R[T ]-module of rank n with determinant L[T ]. Suppose that P/T P as well as P ⊗ Rm [T ] have a unimodular element for every maximal ideal m of R. Then P has a unimodular element. ∼
Proof. Let χ : L[T ] → ∧n (P ) be an isomorphism and α : P →→ I be a surjection, where I ⊂ R[T ] is an ideal of height n. Let (P, χ)e(P, χ) = (I, ωI ) in E(R[T ], L[T ]), where (I, ωI ) is obtained from the pair (α, χ). As P ⊗ Rm [T ] has a unimodular element for every maximal ideal m of R, the image of e(P, χ) in m E(Rm [T ], Lm [T ]) is trivial. By [12, Theorem 4.17] the following sequence of groups 0 −→ E(R, L) −→ E(R[T ], L[T ]) −→
E(Rm [T ], Lm [T ])
m
is exact. Therefore, there exists (J, ωJ ) ∈ E(R, L) such that (JR[T ], ωJ ⊗ R[T ]) = e(P, χ) in E(R[T ], L[T ]). Then we have e(P/T P, χ ⊗ R[T ]/(T )) = (J, ωJ ) in E(R, L). Since P/T P has a unimodular element, by [6, Corollary 4.4], (J, ωJ ) = 0 in E(R, L). Consequently, e(P, χ) = 0 in E(R[T ], L[T ]). Hence, by [12, Corollary 4.15], P has a unimodular element. 2 The following result is due to Swan [25, Lemma 1.3]. We give a proof due to Murthy. Lemma 2.16. Let R be a ring and P be a projective R[T, T −1 ]-module. Let f (T ) ∈ R[T ] be monic such that Pf (T ) is extended from R. Then P is extended from R[T −1 ]. Proof. If degree of f (T ) is m, then f (T )T −m = 1 +T −1 g1 (T −1 ) := g(T −1 ). Consider the following Cartesian diagram.
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R[T −1 ]
/ R[T −1 ]T −1 (= R[T, T −1 ])
R[T −1 ]g(T −1 )
/ R[T, T −1 ]g(T −1 ) = R[T, T −1 ]f (T )
Since Pf (T ) is extended from R, Pg(T −1 ) is extended from R[T −1 ]g(T −1 ) . Using a standard patching argument, there exists a projective R[T −1 ]-module Q such that Q ⊗ R[T −1 ] R[T, T −1 ] P . This completes the proof. 2 3. Main results In this section, we shall prove our main results stated in the introduction. Theorem 3.1. Let R be a ring of dimension n containing an infinite field. Let P be a projective R[T1 , · · · , Tk ]-module of rank n. Assume Pf (Tk ) has a unimodular element for some monic polynomial f (Tk ) in the variable Tk . Then P has a unimodular element. Proof. When k = 1, we are done by [7, Theorem 3.4]. Assume k ≥ 2 and use induction on k. Since (P/T1 P )f has a unimodular element, by induction on k, P/T1 P has a unimodular element. Also (P ⊗ R(T1 )[T2 , . . . , Tk ])f has a unimodular element and dim (R(T1 )) = n. So again by induction on k, P ⊗ R(T1 )[T2 , . . . , Tk ] has a unimodular element. Since P is finitely generated, there exists a monic polynomial g ∈ R[T1 ] such that Pg has a unimodular element. Applying (2.2), we get that P has a unimodular element. 2 The next result is due to Das–Sridharan [11, Theorem 3.4] when L = R. Proposition 3.2. Let R be a ring of even dimension n containing Q. Let P be a projective R[T ]-module of rank n with determinant L. Suppose there are surjections α : P →→ I and φ : L ⊕ R[T ]n−1 →→ I, where I is an ideal of R[T ] of height n. Assume further that P/T P has a unimodular element. Then P has a unimodular element. Proof. Let Rred = R/n(R), where n(R) is the nil radical of R. It is easy to derive that P has a unimodular element if and only if P ⊗ Rred has a unimodular element. Therefore, without loss of generality we may assume that R is reduced. We divide the proof into two steps. Step 1: Assume that L is extended from R. Since P/T P has a unimodular element, in view of (2.15), it is enough to prove that if m is a maximal ideal of R, then P ⊗ Rm [T ] has a unimodular element. Note that P ⊗ Rm [T ] has trivial determinant and α ⊗ Rm [T ] : P ⊗ Rm [T ] →→ IRm [T ] is a surjection. Using surjection φ ⊗ Rm [T ], we get that IRm [T ] is generated by n elements. Applying [11, Theorem 3.4] to the ring Rm [T ], it follows that P ⊗ Rm [T ] has a unimodular element. Hence we are done. Step 2: Assume that L is not necessarily extended from R. Since R is reduced, we can find a ring R → S → Q(R) such that (1) (2) (3) (4)
The projective S[T ]-module L ⊗ R[T ] S[T ] is extended from S, S is a finite R-module, the canonical map Spec(S) → Spec(R) is bijective, and for every P ∈ Spec(S), the inclusion R/(P ∩ R) −→ S/P is birational.
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Since L ⊗ R[T ] S[T ] is extended from S, by Step 1, P ⊗ R[T ] S[T ] has a unimodular element. Applying [2, Lemma 3.2], we conclude that P has a unimodular element. 2 The following result extends (3.2) to polynomial ring R[T1 , · · · , Tk ]. Theorem 3.3. Let R be a ring of even dimension n containing Q. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose there are surjection α : P →→ I and φ : L ⊕ An−1 →→ I, where I ⊂ A is an ideal of height n. Assume further that P/(T1 , · · · , Tk )P has a unimodular element. Then P has a unimodular element. Proof. When k = 1, we are done by (3.2). Assume k ≥ 2 and use induction on k. We use “bar” when we move modulo (Tk ). Since R contains Q, replacing Xk by Xk − λ for some λ ∈ Q, we may assume that ht (I) ≥ n. Consider the surjection α : P →→ I induced from α. If ht(I) > n, then I = R[T1 , · · · , Tk−1 ] and hence P contains a unimodular element. Assume ht(I) = n. Since P /(T1 , · · · , Tk−1 )P = P/(T1 , · · · , Tk )P has a unimodular element and φ induces a surjection φ : L ⊕ An−1 →→ I, by induction hypothesis, P has a unimodular element. By similar arguments, we get that P/Tk P has a unimodular element. Write A = R(Tk−1 )[T1 , · · · , Tk−2 , Tk ] and P ⊗ A = P. We have surjections α ⊗ A : P →→ IA and φ ⊗ A : (L ⊗ A) ⊕ An−1 →→ IA. If IA = A, then P has a unimodular element. Assume ht(IA) = n. Since P/Tk P = P/Tk P ⊗ R(Tk−1 )[T1 , · · · , Tk−2 ], we get that P/Tk P and hence P/(T1 , · · · , Tk−2 , Tk )P has a unimodular element. Again by induction hypothesis, P has a unimodular element. So, there exists a monic polynomial f ∈ R[Tk ] such that Pf has a unimodular element. Applying (3.1), we get that P has a unimodular element. 2 Now we prove (1.4) mentioned in the introduction. Theorem 3.4. Let R be a ring containing Q of even dimension n with ht(J (R)) ≥ 2. Let P be a projective R[T, T −1 ]-module of rank n with trivial determinant. Assume there exists a surjection α : P →→ I, where I ⊂ R[T, T −1 ] is an ideal of height n such that I is generated by n elements. Then P has a unimodular element. ∼
Proof. Let χ : R[T, T −1 ] → ∧n (P ) be an isomorphism. Let e(P, χ) = (I, ωI ) ∈ E(R[T, T −1 ]) be obtained from the pair (α, χ). By (2.10 (3)), it is enough to prove that e(P, χ) = (I, wI ) = 0 in E(R[T, T −1 ]). Suppose ωI is given by I = (g1 , · · · , gn ) + I 2 . Also I is generated by n elements, say, I = (f1 , · · · , fn ). By (2.11), there exists τ ∈ GLn (R[T, T −1 ]/I) such that (f 1 , · · · , f n ) = (g 1 , · · · , g n )τ , where “bar” denotes reduction modulo I. Let U, V ∈ R[T, T −1 ] be such that det(τ ) = U and U V = 1 modulo I. Claim: The unimodular row (U, f1 , · · · , fn ) over R[T, T −1 ] is completable. First we show that the theorem follows from the claim. Since n is even, by (2.4), there exists η ∈ Mn (R[T, T −1 ]) with det(η) = V modulo I such that if (f1 , · · · , fn )η = (h1 , · · · , hn ), then I = (h1 , · · · , hn ). Further, (g 1 , · · · , g n )τ ◦ η = (h1 , · · · , hn ). Note τ ◦ η ∈ SLn (R[T, T −1 ]/I). Let A be the ring obtained from R[T, T −1 ] by inverting all special monic polynomials. Since dim(A/IA) = 0, we have SLn (A/IA) = En (A/IA). Let Θ ∈ SLn (A) be a lift of (τ ◦ η) ⊗ A. Then IA = (h1 , · · · , hn )Θ−1 and (h1 , · · · , hn )Θ−1 = (g 1 , · · · , g n ). In other words, (IA, ωI ⊗ A) = 0 in E(A). Since ht(J (R)) ≥ 2, by (2.10 (4)), the canonical map E(R[T, T −1 ]) −→ E(A) is injective. Therefore, (I, ωI ) = 0 in E(R[T, T −1 ]). This completes the proof. So we just need to prove the claim. Proof of the claim: Let Q be the stably free R[T, T −1 ]-module associated to the unimodular row (U, f1 , · · · , fn ). If S is the set of all monic polynomials in R[T ], then S −1 R[T, T −1 ] = R(T ). Applying (2.5), we have QS is free and hence there exists a monic polynomial f ∈ R[T ] such that Qf is free. In
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particular, Qf is extended from R. Hence by (2.16), there exists a projective R[T −1 ]-module Q1 such that Q Q1 ⊗ R[T −1 ] R[T, T −1 ]. Since (Q1 ⊕ R[T −1 ])T −1 Q ⊕ R[T, T −1 ] R[T, T −1 ]n+1 , by Quillen and Suslin [18,23], we get Q1 ⊕ R[T −1 ] R[T −1 ]n+1 . By (2.3), Q1 is extended from R, say, Q1 Q2 ⊗ R R[T −1 ]. Since ht(J (R)) ≥ 1, by Bass [1], Q2 is free. Hence Q is free, i.e. (V, f1 , · · · , fn ) is completable. This proves the claim. 2 Next result is the converse of (3.4). Theorem 3.5. Let R be a ring containing Q of dimension n with ht(J (R)) ≥ 2. Let P be a projective R[T, T −1 ]-module of rank n with trivial determinant. Let I ⊂ R[T, T −1 ] be an ideal of height n which is generated by n elements. Suppose that P has a unimodular element. Then there exists a surjection α : P →→ I. ∼
Proof. Let χ : R[T, T −1 ] → ∧n (P ) be an isomorphism and e(P, χ) ∈ E(R[T, T −1 ]) be obtained from the pair (P, χ). Since P has a unimodular element, by (2.10 (3)), e(P, χ) = 0 in E(R[T, T −1 ]). Let I = (f1 , · · · , fn ) and ωI be the orientation of I induced by f1 , · · · , fn . Then (I, ωI ) = 0 = e(P, χ) in E(R[T, T −1 ]). By (2.10 (2)), there exists a surjection α : P →→ I such that (I, ωI ) is induced from (α, χ). 2 Theorem 3.6. Let R be a ring of even dimension n containing Q and D = R[X, Y ]/(XY ). Let P be a projective D-module of rank n with determinant L. Suppose there are surjections α : P →→ I and φ : L ⊕ Dn−1 →→ I, where I is an ideal of D of height n. Assume further that P/(X, Y )P has a unimodular element. Then P has a unimodular element. Proof. Without loss of generality, we may assume that R is reduced. We continue to denote the images of X and Y in D by X and Y . We give proof in two steps. Step 1: Assume L is extended from R. Let “bar” and “tilde” denote reductions modulo (Y ) and (X) : P →→ I induced from α. Note that ht(I) ≥ n and respectively. We get surjections α : P →→ I and α ≥ n. If ht(I) > n, then I = R[X] and P has a unimodular element. Assume ht(I) = n. Since ht(I) D = R[X] and P /XP = P/(X, Y )P has a unimodular element, by (3.2), P has a unimodular element. Similarly, P has a unimodular element. ). Since det(P) = L is extended from R, it follows from [8, Let p1 ∈ U m(P ). Then p1 ∈ U m(P ) is surjective. Therefore there exists Theorem 5.2, Remark 5.3] that the natural map U m(P) U m(P p2 ∈ U m(P) such that p2 = p1 . Since the following square of rings D
/ R[X]
R[Y ]
/ R
is Cartesian with vertical maps surjective, p1 ∈ U m(P ) and p2 ∈ U m(P) will patch up to give a unimodular element of P . Step 2: Assume L is not necessarily extended from R. Since R is reduced, by [15, Lemma 3.7], there exists a ring S such that (1) (2) (3) (4)
R → S → Q(R), S is a finite R-module, R → S is subintegral and L ⊗ R S is extended from S.
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Note that R → S is actually a finite subintegral extension. Hence, by [15, Lemma 2.11], R[X, Y ]/(XY ) → S[X, Y ]/(XY ) is also subintegral. Since L ⊗ R S is extended from S, by Step 1, P ⊗ S[X, Y ]/(XY ) has a unimodular element. Now following the proof of [2, Lemma 3.2], it is easy to see that P has a unimodular element. 2 Using [8, Theorem 5.2, Remark 5.3] and following the proof of (3.6), we can prove the following result. Theorem 3.7. Let R be a ring of even dimension n containing Q and D = R[Y1 , . . . , Ym ]/Y1 (Y2 , . . . , Ym ). Let P be a projective D-module of rank n with determinant L. Suppose there are surjections α : P →→ I and φ : L ⊕ Dn−1 →→ I, where I is an ideal of D of height n. Assume further that P/(Y1 , . . . , Ym )P has a unimodular element. Then P has a unimodular element. 4. Applications In this section, we give some applications of results proved earlier. When L = R, the following result is proved in [7, Theorem 4.5]. Theorem 4.1. Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n ≥ 3. Let P be a projective R[T ]-module of rank n with determinant L. Suppose I is an ideal of R[T ] of height n such that there are two surjections α : P →→ I and φ : L ⊕ R[T ]n−1 →→ I. Then P has a unimodular element. Proof. Let “bar” denote reduction modulo (T ). Then α induces a surjection α : P →→ I(0). We can assume ht(I(0)) ≥ n. If ht(I(0)) > n, then I(0) = R and P has a unimodular element. Assume ht(I(0)) = n and ∼ fix an isomorphism χ : L → ∧n (P ). Let e(P , χ) = (I(0), ωI(0) ) in E(R, L) be induced from (α, χ). Since φ : L ⊕ Rn−1 →→ I(0) is a surjection, by (2.13), e(P , χ) = 0 in E(R, L). By [6, Corollary 4.4], P/T P has a unimodular element. Rest of the proof is exactly as in (3.2) using (2.12). 2 Corollary 4.2. Let R be an affine algebra over an algebraically closed field of characteristic 0 with dim(R) = n ≥ 3 and D = R[X, Y1 . . . , Ym ]/X(Y1 , . . . , Ym ). Let P be a projective D-module of rank n with determinant L. Suppose I is an ideal of D of height n such that there are two surjections α : P →→ I and φ : L ⊕ Dn−1 →→ I. Then P has a unimodular element. Proof. Follow the proof of (3.6) and use (4.1). 2 Theorem 4.3. Let R be an affine algebra over an algebraically closed field of characteristic 0 such that dim(R) = n ≥ 4 is even. Let P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose I is an ideal of A of height n such that there are two surjections α : P →→ I and φ : L ⊕ An−1 →→ I. Then P has a unimodular element. Proof. By (2.13) and [6, Corollary 4.4], we get P/(T1 , · · · , Tk )P has a unimodular element. By (3.3), we are done. 2 In the following special case, we can remove the restriction on dimension in (3.3). This result improves [7, Theorem 4.4]. Proposition 4.4. Let R be a ring of dimension n ≥ 2 containing Q and P be a projective A = R[T1 , · · · , Tk ]-module of rank n with determinant L. Suppose M ⊂ A is a maximal ideal of height n such that there are two surjections α : P →→ M and φ : L ⊕ An−1 →→ M. Then P has a unimodular element.
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Proof. Assume k = 1. If M contains a monic polynomial f ∈ R[T1 ], then Pf has a unimodular element. Using [7, Theorem 3.4], P has a unimodular element. Assume M does not contain a monic polynomial. Since T1 ∈ / M, ideal M + (T1 ) = R[T1 ]. By surjection α, we get P/T1 P has a unimodular element. Follow the proof of (3.2) and use [7, Theorem 4.4] to conclude that P has a unimodular element. We are done when k = 1. Assume k ≥ 2 and use induction on k. For rest of the proof, follow the proof of (3.3). 2 References [1] H. Bass, K-theory and stable algebra, Publ. Math. Inst. Hautes Études Sci. 22 (1964) 5–60. [2] S.M. Bhatwadekar, Inversion of monic polynomials and existence of unimodular elements (II), Math. Z. 200 (1989) 233–238. [3] S.M. Bhatwadekar, Cancellation theorems for projective modules over a two dimensional ring and its polynomial extensions, Compos. Math. 128 (2001) 339–359. [4] S.M. Bhatwadekar, Amit Roy, Stability theorems for overrings of polynomial rings, Invent. Math. 68 (1982) 117–127. [5] S.M. Bhatwadekar, A. Roy, Some theorems about projective modules over polynomial rings, J. Algebra 86 (1984) 150–158. [6] S.M. Bhatwadekar, Raja Sridharan, Euler class group of a Noetherian ring, Compos. Math. 122 (2000) 183–222. [7] S.M. Bhatwadekar, Raja Sridharan, On a question of Roitman, J. Ramanujan Math. Soc. 16 (2001) 45–61. [8] S.M. Bhatwadekar, H. Lindel, Ravi A. Rao, The Bass–Murthy question: Serre dimension of Laurent polynomial extensions, Invent. Math. 81 (1) (1985) 189–203. [9] M.K. Das, The Euler class group of a polynomial algebra, J. Algebra 264 (2003) 582–612. [10] M.K. Das, The Euler class group of a polynomial algebra II, J. Algebra 299 (2006) 94–114. [11] M.K. Das, Raja Sridharan, The Euler class groups of polynomial rings and unimodular elements in projective modules, J. Pure Appl. Algebra 185 (2003) 73–86. [12] M.K. Das, Md. Ali Zinna, The Euler class group of a polynomial algebra with coefficients in a line bundle, Math. Z. 276 (2014) 757–783. [13] M.K. Keshari, Euler class group of a Laurent polynomial ring: local case, J. Algebra 308 (2007) 666–685. [14] M.K. Keshari, Md. Ali Zinna, Unimodular elements in projective modules and an analogue of a result of Mandal, J. Commut. Algebra (2017), in press. [15] M.K. Keshari, Md. Ali Zinna, Efficient generation of ideals in a discrete Hodge algebra, J. Pure Appl. Algebra 221 (2017) 960–970. [16] Satya Mandal, Basic elements and cancellation over Laurent polynomial rings, J. Algebra 79 (1982) 251–257. [17] B. Plumstead, The conjecture of Eisenbud and Evans, Am. J. Math. 105 (1983) 1417–1433. [18] D. Quillen, Projective modules over polynomial rings, Invent. Math. 36 (1976) 167–171. [19] Ravi A. Rao, The Bass–Quillen conjecture in dimension 3 but characteristic = 2, 3 via a question of A. Suslin, Invent. Math. 93 (1988) 609–618. [20] Raja Sridharan, Non-vanishing sections of algebraic vector bundles, J. Algebra 176 (1995) 947–958. [21] Raja Sridharan, Projective modules and complete intersections, K-Theory 13 (1998) 269–278. [22] J.P. Serre, Sur les modules projectifs, in: Semin. Dubreil-Pisot, vol. 14, 1960/1961. [23] A.A. Suslin, Projective modules over polynomial ring are free, Sov. Math. Dokl. 17 (1976) 1160–1164. [24] A.A. Suslin, A cancellation theorem for projective modules over affine algebras, Sov. Math. Dokl. 18 (1977) 1281–1284. [25] R.G. Swan, Projective modules over Laurent polynomial rings, Trans. Am. Math. Soc. 237 (1978) 111–120. [26] A. Wiemers, Some properties of projective modules over discrete Hodge algebras, J. Algebra 150 (1992) 402–426.