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On the ideal case of a conjecture of Huneke and Wiegand Naoki - - PowerPoint PPT Presentation

1 Introduction 2 Weakly m -full ideals 3 Proof of Theorem 1.8 4 Integrally closed ideals References On the ideal case of a conjecture of Huneke and Wiegand Naoki Taniguchi Waseda University Joint work with O. Celikbas, S. Goto and


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§1 Introduction §2 Weakly m-full ideals §3 Proof of Theorem 1.8 §4 Integrally closed ideals References

On the ideal case of a conjecture of Huneke and Wiegand

Naoki Taniguchi

Waseda University

Joint work with O. Celikbas, S. Goto and R. Takahashi The 10th Japan-Vietnam jonint seminar on Commutative Algebra September 13, 2018

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§1 Introduction For a moment R an integral domain M, N finitely generated torsion-free R-modules Recall that M is called torsion-free, if the natural map 0 → M → M ⊗R Q(R) is injective. Question When is the tensor product M ⊗R N torsion-free?

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§1 Introduction For a moment R an integral domain M, N finitely generated torsion-free R-modules Recall that M is called torsion-free, if the natural map 0 → M → M ⊗R Q(R) is injective. Question When is the tensor product M ⊗R N torsion-free?

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Let (−)∗ = HomR(−, R) be the algebraic dual. Conjecture 1.1 (Huneke-Wiegand, 1994) Let (R, m) be a Gorenstein local domain with dim R = 1, M a finitely generated torsion-free R-module. If M ⊗R M∗ is torsion-free, then M is free.

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Theorem 1.2 (Auslander, 1961) Let R be a Noetherian normal domain, M a finitely generated R-module. Then M is projective if and only if M ⊗R M∗ is reflexive. Theorem 1.3 (Huneke-Wiegand, 1994) Let R be a hypersurface domain, M, N finitely generated R-modules. If M ⊗R N is torsion-free, then either M or N is free. Theorem 1.4 (Celikbas, 2011) Let R be a complete intersection domain, M a finitely generated torsion-free R-module with bounded Betti numbers. If M ⊗R M∗ is torsion-free, then M is free.

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For a commutative Noetherian local ring R, we define (HWC) For every finitely generated torsion-free R-module M, if M ⊗R M∗ is reflexive, then M is free. (ARC) For every finitely generated R-module M, if Ext>0

R (M, M ⊕ R) = (0), then M is free.

Theorem 1.5 (Celikbas-Dao, Cekilbas-Takahashi, Huneke-Wiegand) Consider the following conditions. (1) (HWC) holds for all Gorenstein local domains. (2) (HWC) holds for all one-dimensional Gorenstein local domains. (3) (ARC) holds for all Gorenstein local domains. Then the implications (1) ⇐ ⇒ (2) = ⇒ (3) hold.

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Conjecture 1.6 Let (R, m) be a Gorenstein local domain with dim R = 1, I an ideal of R. If I ⊗R I ∗ is torsion-free, then I is principal.

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Fact 1.7 Conjecture 1.6 holds for the following cases. (1) e(R) ≤ 6 (Goto-Takahashi-T-Truong, 2015) (2) R = k[[H]], I monomial, e(R) ≤ 7 (Goto-Takahashi-T-Truong, 2015) (3) e(R) ≤ 8 (Huneke-Iyengar-Wiegand, 2018) (4) I ∼ = trace ideal (Lindo, 2017)

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The main result of this talk is stated as follows. Theorem 1.8 (Celikbas-Goto-Takahashi-T, 2018) Let (R, m) be a Cohen-Macaulay local ring with dim R = 1, I an m-primary ideal of R. Suppose that I is weakly m-full, that is, mI : m = I. If I ⊗R I ∗ is torsion-free, then I is principal, and hence R is a DVR.

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Contents (1) Introduction (2) Weakly m-full ideals (3) Proof of Theorem 1.8 (4) Integrally closed ideals

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§2 Weakly m-full ideals Throughout, let (R, m) be a Noetherian local ring I an ideal of R Definition 2.1 (1) I is called m-full, if mI : x = I for ∃ x ∈ m (Rees) (2) I is called weakly m-full, if mI : m = I (Celikbas-Iima-Sadeghi-Takahashi) Fact 2.2 (Goto, Rees) Suppose |R/m| = ∞. If I = I, then I = √ (0) or I is m-full.

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Remark 2.3 The implication I is m-full = ⇒ I = I does not hold. Example 2.4 Let R = k[[X, Y ]] be the formal power series ring over a field k. Then I = (X 3, X 2Y 3, XY 4, Y 5) is m-full, but I ̸= I.

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Example 2.5 Let J be an ideal of R and set I = J : m. Then I is a weakly m-full ideal. Remark 2.6 The implication I is weakly m-full = ⇒ I is m-full does not hold.

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Proposition 2.7 Let (R, m) be a Cohen-Macaulay local ring, Q a parameter ideal of R. We set I = Q : m. If µR(m) > dim R + r(R), then I is not m-full. Proof. Note that R is not regular. Then mI = mQ, so that µR(I) = dim R + r(R) because ℓR(I/Q) = r(R). If I is m-full, then µR(I) ≥ µR(m) which makes a contradiction.

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Example 2.8 Let k be a field and set R = k[[t5, t6, t7, t9]]. Then I = (t5, t9, t13) = (t5) : m is weakly m-full, but not m-full.

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Theorem 2.9 (Goto-Hayasaka, 2002) Suppose that I is m-full and depth R/I = 0. If idR I < ∞, then R is a RLR. Theorem 2.10 (Celikbas-Iima-Sadeghi-Takahashi, 2018) Suppose that I is weakly m-full and depth R/I = 0. If idR I < ∞, then R is a RLR.

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§3 Proof of Theorem 1.8 Setting 3.1 (R, m) a Noetherian local ring I an m-primary ideal of R M a finitely generated R-module Proposition 3.2 (cf. Corso-Huneke-Katz-Vasconcelos, 2006) Suppose that I is weakly m-full. If TorR

t (M, R/I) = 0 for ∃ t ≥ 0

then pdR M < t.

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Proof of Proposition 3.2 If t = 0, then M = 0 and pdR M = −∞. Thus we may assume t > 0. Consider a minimal free resolution of M · · · − → Ft+1 − → Ft

∂t

− → Ft−1 → · · · → F0 → 0. Applying (−) = (−) ⊗R R/I, we obtain · · · − → F t+1 − → F t

∂t

− → F t−1 → · · · → F 0 → 0. Suppose ∂t = 0. Then F t = mF t, whence F t = (0) and Ft = (0). Hence pdR M < t. We now assume Im ∂t ̸= 0 and seek a contradiction. Since (Im ∂t)I ⊆ IFt−1, (Im ∂t)ms ⊆ IFt−1 for ∃ s > 0. Let us choose the integer s as small as possible.

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Proof of Proposition 3.2 Since (Im ∂t)ms−1 ⊈ IFt−1, we choose u ∈ (Im ∂t)ms−1 s.t. u / ∈ IFt−1. Then mu ⊆ mIFt−1 and hence u ∈ (mI : m)Ft−1 = IFt−1 which is a contradiction. Therefore Im ∂t = 0, as desired.

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Corollary 3.3 Suppose that I is weakly m-full and depth R > 0. Then I ⊗R M is torsion-free ⇐ ⇒ M is free. Proof. (⇒) Applying (−) ⊗R M to 0 → I → R → R/I → 0, we get 0 − → TorR

1 (R/I, M) −

→ I ⊗R M. Since I contains a NZD on R, TorR

1 (R/I, M) is torsion. Hence

TorR

1 (R/I, M) = (0)

which implies pdR M ≤ 0.

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We are now ready to prove Theorem 1.8. Theorem 1.8 (Celikbas-Goto-Takahashi-T, 2018) Let (R, m) be a Cohen-Macaulay local ring with dim R = 1, I an m-primary ideal of R. Suppose that I is weakly m-full. If I ⊗R I ∗ is torsion-free, then I is principal, and R is a DVR.

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Proof of Theorem 1.8 Since I ⊗R I ∗ is torsion-free, by Corollary 3.3, I ∗ is free. Claim 3.4 Let (R, m) be a Cohen-Macaulay local ring with dim R = 1, M a finitely generated torsion-free R-module. Then M∗ is free ⇐ ⇒ M is free

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Proof of Theorem 1.8 Proof of Claim 3.4. We may assume M is indecomposable. Since M is torsion-free, if M∗ = (0), then M = (0). We may assume M∗ ̸= (0). Let F1 − → F0 − → M − → 0 be the minimal presentation of M. Then 0 − → M∗ − → F ∗

0 −

→ F ∗

1 −

→ TrM − → 0. Since pdR(TrM) < ∞, we have pdR(TrM) ≤ 1. Thus Ext2

R(TrM, R) = (0)

which implies M − → M∗∗ − → 0 and hence M ∼ = M∗∗ as claimed.

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Proof of Theorem 1.8 Consequently, I is principal. Therefore, I = (f ) for ∃ R-NZD f ∈ m. Since I ∼ = R, we have pdR(R/I) ≤ 1, so that TorR

2 (R/m, R/I) = (0).

Hence, by Proposition 3.2, pdR(R/m) ≤ 1, which yields that R is a DVR. Remark 3.5 (1) Let (R, m) be a Noetherian local ring with depth R > 0, I a weakly m-full m-primary ideal. If µR(I) = 1, then R is a DVR. (2) Assertion (1) holds for the case where I = I.

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§4 Integrally closed ideals Theorem 4.1 Let R be a Noetherian ring, I an ideal of R with htR I > 0. Assume R satisfies Serre’s condition (S2). Then TFAE. (1) I = I and [I] ∈ Pic R. (2) Rp is a DVR for ∀p ∈ AssR R/I and [I] ∈ Pic R. (3) I = I and I ⊗R I ∗ is reflexive. When this is the case, I = ∩

p∈AssR R/I

p(n(p)) where n(p) ≥ 1 for ∀p ∈ AssR R/I and p(n(p)) is a symbolic power of p.

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Recall that a Noetherian ring R satisfies (Sn) if depth Rp ≥ min{n, htR p} for ∀p ∈ Spec R. We set

Pic R = {finitely generated projective module M s.t. Mp ∼ = Rp for ∀p ∈ Spec R}/ ∼ =

and call it the Picard group of R.

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Lemma 4.2 Let R be a Noetherian ring satisfying (S1), I an ideal of R. Assume IRp ∼ = Rp for ∃ p ∈ AssR R/I. Then Rp is a Cohen-Macaulay local ring with dim Rp = 1. If, furthermore, I = I, then Rp is a DVR. Proof. Note that depthRp(Rp/IRp) = 0. Since IRp is principal, we have depth Rp = 1. As R satisfies (S1), we conclude that Rp is a Cohen-Macaulay local ring with dim Rp = 1. If I = I, then Remark 3.5 (2) shows Rp is a DVR.

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Theorem 4.1 Let R be a Noetherian ring, satisfying (S2), I an ideal of R with htR I > 0. Then TFAE. (1) I = I and [I] ∈ Pic R. (2) Rp is a DVR for ∀p ∈ AssR R/I and [I] ∈ Pic R. (3) I = I and I ⊗R I ∗ is reflexive. When this is the case, I = ∩

p∈AssR R/I

p(n(p)) where n(p) ≥ 1 for ∀p ∈ AssR R/I.

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Proof of Theorem 4.1 (1) ⇒ (2): It follows from Lemma 4.2. (2) ⇒ (3): Suppose that I/I ̸= (0) and choose p ∈ AssR I/I. Then p ∈ AssR R/I so that Rp is a DVR by assumption. Hence IRp = IRp = IRp. This is a contradiction, because (I/I)p ̸= (0). Thus I = I. Since [I] ∈ Pic R, I is projective. Note that R-duals and tensor products of projective modules are projective projective modules are reflexive so we conclude that I ⊗R I ∗ is reflexive. (2) ⇒ (1): It follows from the fact that (2) implies I = I.

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Proof of Theorem 4.1 (3) ⇒ (2): Suppose that I = I and I ⊗R I ∗ is reflexive. Claim 4.3 Let p ∈ Spec R with htR p ≤ 1. Then IRp ∼ = Rp and SuppR I = Spec R. We now proceed to show [I] ∈ Pic R, by using Theorem 4.4. Theorem 4.4 (Auslander, Huneke-Wiegand) Let (R, m) be a Noetherian local ring satisfying (S2), M a finitely generated torsion-free R-module. Suppose that MP is a free RP-module for P ∈ Spec R with htR P ≤ 1. If M ⊗R M∗ is reflexive, then M is free.

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Proof of Theorem 4.1 For q ∈ Spec R, let P ∈ Spec Rq with htRq P ≤ 1. Then P = pRq for ∃ p ∈ Spec R with dim Rp ≤ 1. It follows that (IRq)P ∼ = IRp ∼ = Rp ∼ = (Rq)P. Moreover IRq ⊗Rq (IRq)∗ is reflexive. Theorem 4.4 implies IRq is Rq-free. Since SuppR I = Spec R, we see that IRq ∼ = Rq. This shows I is projective, i.e., [I] ∈ Pic R.

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Proof of Theorem 4.1 Let p ∈ AssR R/I. Since [I] ∈ Pic R, we have IRp ∼ = Rp. By Lemma 4.2, Rp is a Cohen-Macaulay ring with dim Rp = 1. As IRp is free, we have TorRp

2 (Rp/pRp, Rp/IRp) = 0.

By Proposition 3.2, Rp is a DVR. This completes the proof of (3) ⇒ (2). Let us make sure of the last assertion. Let p ∈ AssR R/I. Then, since Rp is a DVR, we have IRp = pn(p)Rp for ∃ n(p) ≥ 1. Thus IRp ∩ R = p(n(p)), and hence the result follows.

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Thank you so much for your attention.

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References

[1]

  • M. Auslander, Modules over unramified regular local rings, Illinois J. Math., 5 (1961),

631–647. [2]

  • O. Celikbas, K.-i. Iima, A. Sadeghi, and R. Takahashi, On the ideal case of a conjecture of

Auslander and Reiten, Bull. Sci. Math., 142 (2018), 94–107. [3]

  • O. Celikbas and R. Takahashi, Auslander-Reiten conjecture and Auslander-Reiten duality,
  • J. Algebra, 382 (2013), 100–114.

[4]

  • O. Celikbas, S. Goto, R. Takahashi, and N. Taniguchi, On the ideal case of a conjecture of

Huneke and Wiegand, Proc. Edinb. Math. Soc., (to appear). [5]

  • A. Corso, C. Huneke, D. Katz, and W. V. Vasconcelos, Integral closure of ideals and

annihilators of homology, Commutative algebra, Lect. Notes Pure Appl. Math., 244, Chapman & Hall/CRC, Boca Raton, FL, 2006, 33–48. [6]

  • S. Goto, R. Takahashi, N. Taniguchi, and H. L. Truong, Huneke-Wiegand conjecture and

change of rings, J. Algebra, 422 (2015), 33–52. [7]

  • C. Huneke and R. Wiegand, Tensor products of modules and the rigidity of Tor, Math.

Ann., 299 (1994), no.3, 449–476.

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