effective normality criteria for algebras of linear type

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Journal of Algebra 273 (2004) 640–656

www.elsevier.com/locate/jalgebr

Effective normality criteriafor algebras of linear type

Joseph Brennana and Wolmer V. Vasconcelosb,∗,1

a Department of Mathematics, North Dakota State University, Fargo, ND 58105-5075, USAb Department of Mathematics, Rutgers University, 110 Frelinghuysen Road, Piscataway, NJ 08854-801

Received 13 January 2003

Communicated by Craig Huneke

Dedicated to Jürgen Herzog on the occasion of his sixtieth-second birthday

Abstract

The algebras studied here are subalgebras of rings of polynomials generated by 1-formcalled Rees algebras), with coefficients in a Noetherian ring. Given a normal domainR and a

torsionfree moduleE with a free resolution,· · · → F2ψ→ F1

ϕ→ F0 →E → 0, we study the roleof the matrices of syzygies in the normality of the Rees algebra ofE. When the Rees algebraR(E)

and the symmetric algebraS(E) coincide, the main results characterize normality in terms ofideal Ic(ψ)S(E) and of the completeness of the firsts symmetric powers ofE, wherec = rankψ ,and s = rankF0 − rankE. It requires thatR be a regular domain. Special results, under broaconditions onR, are still more effective. 2004 Published by Elsevier Inc.

Keywords:Algebra of linear type; Complete module; Normal algebra; Rees algebra; Symmetric algebra

1. Introduction

LetR be a commutative, Noetherian ring and letA be a (commutative) finitely generategradedR-algebra generated by its component of degree one,

* Corresponding author.E-mail addresses:[email protected] (J. Brennan), [email protected]

(W.V. Vasconcelos).1 Partially supported by the NSF.

0021-8693/$ – see front matter 2004 Published by Elsevier Inc.doi:10.1016/j.jalgebra.2003.10.012

J. Brennan, W.V. Vasconcelos / Journal of Algebra 273 (2004) 640–656 641

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s of

A=R +A1 +A2 + · · · =R[A1].

There is a natural surjectionτ :SR(A1) → A, from the symmetric algebra ofA1 ontoA,andA is said to be oflinear typeif τ is an isomorphism. Such algebras are naturally defiin terms of generators and relations of its degree 1 component. SettingE =A1, if

Rm ϕ−→ Rn →E → 0

is a presentation ofE as anR-module,

S(E)= SR(E)R[T1, . . . , Tn]/(f1, . . . , fm),

where thefi ’s are linear forms in theTi -variables, obtained from the matrix multiplicatio[T1, . . . , Tn] ·ϕ. The equalitiesSt (E)At , t � 1, impose several known restrictions on tmatrix ϕ (see [16, Chapter 1]). We are going to assume thatR is a (normal) domain antheAt are torsionfreeR-modules, when these algebras can also be described as foLet K be the field of fractions ofR, and letE be a finitely generated torsionfreerank r, that isK ⊗R E ∼= Kr . This provides an embeddingE ↪→ Rr which allows usto define theRees algebra ofE as the subalgebraR(E) of S(Rr) generated by the formin E (S(·) denotes the symmetric algebra functor):R(E) is a standard graded algebwhose component of degreen, R(E)n ⊂ Sn(R

r), is a module ofn-forms of the ring ofpolynomialsS(Rr ) = R[T1, . . . , Tr ]. It is a natural extension of the blowup algebra ofidealI , R(I)= ⊕

n�0 In. When we have the isomorphismS(E)R(E) (which with the

fixed embedding we view simply as an equality), it is standard terminology to referE

as anR-module oflinear type. Looking atR(E) as a symmetric algebra or as a subringa ring of polynomials provides us with two windows through which the properties oalgebra can be examined.

In this paper we develop effective criteria of normality for Rees algebras, especiamodules of linear type. In particular, given a normal domainR and a torsionfree moduleEwith a free resolution,

· · · → F2ψ−→ F1

ϕ−→ F0 →E → 0,

we study the role of the matrices of syzygies in the normality of the Rees algebraE.This goal being, in general, too open ended, we will restrict ourselves to those algwhich are of linear type, that is,R(E)= S(E).

Let us describe the contents of this paper. In Section 2 we introduce thedeterminant of a moduleE, and discuss some of its properties in relation to the inteclosure ofE. Since it is needed later, we establish them-fullness property of completmodules. (At this point, we warn the reader for using the qualifiercompletein very distinctconcepts: complete modules, complete intersection modules and complete interalgebras.)

In Section 3, we embark on our analysis of the normality of the following familiealgebras:

642 J. Brennan, W.V. Vasconcelos / Journal of Algebra 273 (2004) 640–656

mality.provide

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ation:

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• Algebras of complete intersection modules.• Complete intersection algebras.• Almost complete intersection algebras.• General algebras of linear type.

The first three classes are described by data very amenable to verification of norFurthermore, should the algebras turn out to be not normal, the processes involvedproper integral extensions.

The main results of this paper (Theorems 6.1 and 6.2) characterize normality ofR(E),over a regular domainR, in terms of the idealIc(ψ)S(E) and of the completeness of thfirst s symmetric powers ofE, wherec = rankψ , ands = rankF0 − rankE. It requiresthatR be a regular domain. The other results (Theorems 3.1, 4.1, and 4.2), are varof this result for special classes of modules but under less restrictive hypothesesbase rings. Moreover, they are conveniently framed for computation. For this claalgebras, certification of normality of this method outperforms the direct approaJacobian criteria. In addition, the methods are mostly characteristic free.

2. Complete modules

This section has a preliminary character in which we extend some propertcomplete ideals to modules. Most of the facts we are going to review quickly catraceable to [12]. For basic facts and terminology we shall use [3] and [5]. As for notlet (R,m) be a Noetherian local ring and letM be a finitely generatedR-module; weset ν(M) for the minimal number of generators ofM, λ(M) for its length if M has afinite composition series. IfM is a torsionfreeR-module with Rees algebraR(M), itsanalytic spread�(M) is the dimension of the special fiber of the Rees algebra ofM,R(M) ⊗R (R/m). If M is a module of linear type,�(M) = dimSR(M)/mSR(M) =SR/m(M/mM)= ν(M).

ThroughoutR will be a Noetherian domain with a field of fractionsK. Let E be asubmodule ofRr , of rankr, and letR(E) be its Rees algebra. In analogy with the caseideals, one has the following key notion.

Definition 2.1. Let U ⊂ E be a submodule. We say thatU is a reduction of E or,equivalently,E is integral overU , if R(E) is integral over theR-subalgebra generatebyU .

Alternatively, the integrality condition is expressed by the equationsR(E)s+1 = U ·R(E)s , s � 0. The least integers � 0 for which this equality holds is called thereductionnumberof E with respect toU and denoted byrU (E). For any reductionU of E themoduleE/U is torsion, henceU has the same rank asE. This follows from the fact that amodule of linear type, such as a free module, admits no proper reductions.

LetE be a submodule ofRr . Theintegral closureof E in Rr is the largest submoduE ⊂ Rr havingE as a reduction.E is integrally closedor completeif E = E. If R(E) isthe Rees algebra ofE,E is said to benormalif all componentsR(E)n are complete. Sinc


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ce the

R(E) is a standard gradedR-algebra, andR(E)1 = E, without risk of confusion we seR(E)n =En. WhenE is of linear type,En = Sn(E).

We introduce an ideal measuring the embeddingE ↪→ Rr , and discuss its role.

Definition 2.2. Let E be a finitely generatedR-module of rankr. Theorder determinantof the embeddingE ϕ−→ Rr is the idealI defined by the image of the mapping∧rϕ,

image(∧rϕ

) = I · ∧r(Rr

).

When the embedding is clear we will writeI = det0(E).

Proposition 2.3 [12]. Let F ⊂ E be torsionfreeR-modules of rankr andE ↪→ Rr anembedding. Denote bydet0(F ) anddet0(E) the corresponding order determinants.F is areduction ofE if and only ifdet0(F ) is a reduction ofdet0(E).

Proof. LetS =R(Rr )=R[T1, . . . , Tr ], andR(F )⊂R(E) the Rees algebras ofF andE.Since the three algebras have the same fields of fractions, they have the same voverrings. For a valuation ringV of S, letV = V ∩K. We note thatVR(F ),VR(E),V S

are rings of polynomials over the free modulesVF ⊂ VE ⊂ V r , and therefore arintegrally closed. It follows thatF is a reduction ofE if and only if VF = VE for each ofthe valuations such asV .

We check that the order determinants detect this situation. First, note that for any dT that is an overring ofR, one has that det0(E)T = det0(T E). This follows becauseT Eis the image ofT ⊗R E in T ⊗R R

r = T r and exterior powers commute with base chanFinally we observe that in the embedding of free modules

VF α−→ VEβ−→ V r,

the two idealsV det0(E) andV det0(F ) are generated by det(β) and det(β · α), respecti-vely. Thus our conditions are equivalent to det(α) being a unit ofV . ✷

Making use of the notion of Rees valuations (see a discussion in [6]) we redunumber of valuation rings in the description of the integral closure of a module.

Corollary 2.4. LetR be an integral domain and letE be a submodule of rankr of the freemoduleRr . The integral closure ofE in Rr is given by

E =Rr ∩⋂V

VE,

whereV runs over the Rees valuations ofdet0(E).

We have the following application to the natural extension ofm-full ideals to modules.

Definition 2.5. Let (R,m) be a Noetherian local ring and letE be a submodule ofRr . E isanm-full moduleif there is an elementx ∈ m such thatmE :Rr x =E.


kestegral

r

r

.

Proposition 2.6. Suppose(R,m) is a Noetherian local domain andE ↪→ Rr is a moduleof rank r. If the residue field ofR is infinite, complete modules arem-full, that is thereexistsx ∈ m such thatmE :Rr x =E.

Proof. SetL= det0(E). The proof is similar to the ideal case, according to [6], but tainto account the role of the order determinantal ideal in the description of the inclosure observed above.

Let (V1,p1), . . . , (Vn,pn) be the Rees valuations associated toL. Observe that since, foeachi, mVi �= 0, we have thatm containsLi = pimVi ∩R properly. SinceR/m is infinite,we have the comparison of vector spaces

m/m2 �=n∑i=1

Li + m2/m2.

Thus we can choosex ∈ m that is not contained in⋃n

i=1 pimVi . This means that foeachi, mVi = xVi . We claim this element will do. Lety ∈ mE :Rr x. For eachVi wehaveyx ∈ mViE = xViE. Thusy ∈ ⋂n

i=1ViE ∩Rr , and thereforey ∈E =E. ✷As a consequence one has:

Corollary 2.7. Let E be anm-full submodule ofRr of rank r. SupposeG is a module,E ⊂G⊂Rr that satisfiesλ(G/E) <∞. Thenν(G)� ν(E).

Proof. Using the notation above, like in the ideal case, consider the exact sequence

0 →E/mE →G/mE ·x−→G/mE →G/(xG,mE)→ 0.

We have

ν(E)= ν(E/mE)= ν(G/(xG+ mE)

)� ν(G),

as desired. ✷

3. Complete intersection modules

Let R be a Cohen–Macaulay ring and letE be acomplete intersection module. Thismeans that there exists a mapping of rankr

ϕ :Rm → Rr,

with E = imageϕ, with m = r + c − 1, for c � 2, and the idealI = det0(E) hascodimensionc.

Our purpose is to describe, entirely in terms of the idealI , whenE is integrally closedIt turns out that more is achieved.


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Theorem 3.1. Let R be a Cohen–Macaulay integrally closed domain and letE be acomplete intersection module. The following conditions are equivalent:

(a) E is integrally closed.(b) E is normal.(c) det0(E) is an integrally closed generic complete intersection.

If moreoverdet0(E) is a prime ideal the conditions above hold.

The proof of the theorem for modules will be inspired by the arguments in [6],some technicalities stripped away due to the assumption thatR is a Cohen–Macauladomain.

Proof. LetR=R(E) be the Rees algebra of the moduleE. According to [9, Lemma 3.2or [14, Theorem 5.6],E is a module of linear type andR is a Cohen–Macaulay ring.

To check the normality ofR, we need to understand the height 1 prime idealsP of R.Setp = P ∩ R and localize atRp. If p = 0,R(E)p is a ring of polynomials over the fielof fractions ofR, and we have nothing to be concerned about. We may assume that(R,m)

is a local ring and thatP = mR(E). If dimR = 1,R is a discrete valuation ring andE is afreeR-module.

We may thus assume dimR � 2. If E ⊂ Rr contains a free summand ofRr , E =E0 ⊕Rs , R(E) would be a Rees algebra of a complete intersection module of rankr − s,over the ring of polynomialsR[T1, . . . , Ts]. Thus all conditions would be preserved inE0.We may finally assume thatE ⊂ mRr .

We claim that dimR = d = c, that ism is a minimal prime ofI , in particular the moduleRr/E has finite length. This follows simply from the equality

heightmR= dimR− �(E)= (d + r)− (c+ r − 1)= (d − c)+ 1,

sinceE is of linear type and therefore its analytic spread�(E) is equal to its minimanumber of generators.

Let us assume that condition (a) holds, that isE is integrally closed. We denote byd0the embedding dimension ofR. Among other things, we must show thatd = d0, whichmeans thatR is a regular local ring. According to Proposition 2.7,E is m-full. Sinceλ(mRr/E) <∞, by Corollary 2.7, we must have

ν(E)� ν(mRr

).

Because the modulemRr is minimally generated byrd0 elements, whileE itself isgenerated byr + d − 1 elements, we have

r + d − 1� rd0,

which means that

(r − 1)(d − 1)� r(d − d0)� 0.


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r

Since we assumedd � 2, this means thatd = d0 and thatr = 1. In other words,E isisomorphic to a system of parameters of a regular ring. The assertion thus follows froideal case [6].

To prove the other implications, we observe that given an embeddingE ↪→ Rr , to verifythatE = E, it suffices to consider localizationsRp for the associated primes ofRr/E.Thus to show (c)⇒ (a), or (b), we first recall the equality of associated primes[8, Theorem 3.3], [2, Proposition 3.3])

∀n Ass(Sn

(Rr

)/En

) = Ass(R/det0(E)

).

Localizing at one of the associated primes of det0(E), we may assume that(R,m) is a localring and det0(E) is a m-primary parameter ideal. Since we may assume that dimR � 2,by the criterion of [6], det0(E) �⊂ m2. This gives a decompositionE Rr−1 ⊕ det0(E),and thereforeR(E) R(det0(E))[T1, . . . , Tr−1], which is a normal ring as det0(E) is anormal ideal, again by [6].

Finally, the ideal p = det0(E) has finite projective dimension (Eagon–Northccomplex) and therefore ifp is prime,Rp is a regular local ring and (c) will hold.✷Example 3.1. Let p be a perfect prime ideal of codimension 2 generated by the maxminors of then× (n + 1) matrix ϕ. According to the assertion above, the columns oϕgenerate a normal submodule ofRn. In contrast,p is rarely a normal ideal. This also showthat the Rees algebra generated by the column vectors of a generic matrix is always

Complete intersection ideals can be tested for completeness (or even normalityefficiently:

Theorem 3.2. LetR be a Cohen–Macaulay integral domain and letI be a height unmixedgeneric complete intersection ideal of codimensionc. I is integrally closed if and only ithe following conditions hold:

(i) height ann∧c+1√I � c+ 1;

(ii) height ann∧2(√I/I)� c+ 1.

Proof. The meaning of these numerical conditions is simply the following: (i) for eminimal primep of I , Rp is a regular local ring. The condition (ii) means that at each sprimep, Ip has at leastc− 1 of the elements in a minimal generating set ofpRp. Togetherthey imply that the primary components ofI are integrally closed according to [6].✷

4. Complete intersection algebras

LetR be a Noetherian ring andA a finitely generatedR-algebra,A=R[T1, . . . , Tn]/J .We will say thatA is a complete intersectionR-algebra ifJ is generated by a regulasequence. LetR be an integrally closed Cohen–Macaulay domain and letE be a finitely


e

n

lo

ms of

n

generated torsionfreeR-module. Locally, the condition that the Rees algebraR(E) be acomplete intersection can be characterized by the projective presentation ofE: if

Rm ϕ−→Rn →E → 0, (1)

is a minimal presentation, them 1-forms(f1, . . . , fm) = [T1, . . . , Tn] · ϕ are necessarilyminimal generators in any (algebra) presentation ofR(E). Since dimR(E) = dimR +(n − m), R(E) is a complete intersection if and only if thefj form a regular sequencgenerating a prime ideal. Furthermore,E is of linear type (see [1]).

Conversely, according to [16, Theorem 3.1.6], forS(E) to be a complete intersectiowe must have

gradeIt (ϕ)� rank(ϕ)− t + 1=m− t + 1, 1� t �m.

Moreover, sinceS(E) = R(E) is an integral domain, this requirement will hold moduany nonzero element ofR, so it will be strengthened to

gradeIt (ϕ)� rank(ϕ)− t + 2=m− t + 2, 1� t �m.

It can be rephrased in terms of the local number of generators as

ν(Ep)� n−m+ heightp − 1, p �= 0.

For these modules, the graded components of the Koszul complex of the forA=R[T1, . . . , Tn] (see [1]),

[f1, . . . , fm] = [T1, . . . , Tn] · ϕ,giveR-projective resolutions of the symmetric powers ofE:

0 → ∧sRm → ∧s−1Rm ⊗Rn → ·· · → Rm ⊗ Ss−1(Rn

) → Ss(Rn

) → Ss(E)→ 0.

In particular we have:

Proposition 4.1. LetR be a Cohen–Macaulay normal domain and letE be a finitely gen-erated torsionfreeR-module such that the Rees algebraR(E) is a complete intersectiodefined bym equations. The non-normalR-locus ofR(E) has codimension at mostm+1.

Proof. It suffices to consider one observation. For any torsionfreeR-moduleG, containedin a free moduleF , the embedding

G/G ↪→ F/G

shows that ifG has projective dimension� r, the associated primes ofG/G havecodimension at mostr + 1. In the case of the symmetric powersSs(E) of E, the projectivedimensions are bounded bym, from the comments above.✷


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Our next aim is to discuss the normality of the algebraR(E) versus the completeneof the moduleE and of a few other symmetric powers.

We will begin our analysis with the special case of a hypersurface. Assume thmoduleE is torsionfree with one defining relation

0→ Ra → Rn →E → 0, a = (a1, . . . , an).

The ideal generated by theai has grade at least 2, and

S(E)=R[T1, . . . , Tn]/(f ), f = a1T1 + · · · + anTn.

Proposition 4.2. In the setting above,S(E) is normal if and only if for every prime idealp

of codimension two that contains(a1, . . . , an), the following conditions hold:

(i) Rp is a regular local ring.(ii) (a1, . . . , an) �⊂ p(2).

Proof. Since S(E) is Cohen–Macaulay, to test for normality if suffices to checklocalizationsS(E)P , whereP is a prime ideal ofA = R[T1, . . . , Tn] of height 2 thatcontainsf . Let p = P ∩R. If heightp � 1,Ep is a freeRp-module, andS(Ep) is a ring ofpolynomials over a DVR. OtherwiseP = pA, with heightp = 2.

The listed conditions just express the fact that(pA/(f ))P is a cyclic module. ✷Corollary 4.3. Suppose further thatR = k[x1, . . . , xd ] is a ring of polynomials over a fielof characteristic zero.S(E) is normal if and only if

height

(a1, . . . , an,

∂ai

∂xj, i = 1, . . . , n, j = 1, . . . , d

)� 3.

While the normality is straightforward, the completeness ofE requires a different kindof analysis; see also [10].

Theorem 4.1. Let R be an integrally closed Cohen–Macaulay domain that is regulacodimension at most two. IfE is a module defined by a single relation as above, thenE iscomplete if and only ifE is normal.

Proof. According to the previous discussion,E is not normal precisely when there exisa prime ideal of codimension twop ⊃ (a1, . . . , an), such that(pA/(f ))pA is not principal.We will argue that in such case there exists an elementh in the integral closure ofS(E), ofdegree 1 but not lying inE. In other words,E is not complete.

We replaceR byRp, and denote byx, y a set of generators of its maximal ideal. In tone-dimensional local ringB = S(E)pS(E), P = pB, one have that HomB(P,P ) = P−1,sinceP is not principal. It will suffice to find elements of degree 1 in(pS(E))−1 that donot belong toE = S1.


rix is

n

at all

l(i)

f

Let z ∈ p, so thatx, f is a regular sequence inpA. Writing

z= ax + by,

f = cx + dy,

wherec andd are elements ofE, we obtain

(z, f ) : pA= (z, f, ad − bc).

The fractionh= (ad − bc)z−1 /∈E and has the desired property.✷WhenS(E) is defined by more than one hypersurface, the role of the defining mat

more delicate.

Theorem 4.2. Let R be an integrally closed Cohen–Macaulay domain and letE be atorsionfreeR-module of linear type with a projective resolution

0 → Rm ϕ−→Rn →E → 0.

R(E) is normal if and only if the following conditions hold:

(i) For every prime idealp ⊃ It (ϕ) of heightm− t + 2, Rp is a regular local ring.(ii) The modulesSs(E) are complete, fors = 1, . . . ,m.

Proof. Let us assume thatR(E)(= S(E)) is normal, and establish (i). Letp be a primeideal as in (i). Since heightIt−1(ϕ)�m− t + 3, localizing atp, we obtain a presentatioof the module in the form (we set stillR =Rp)

0 →Rm−t+1 ϕ′−→ Rn−t+1 →E → 0,

where the entries ofϕ′ lie in p. Changing notation, this means that we can assume thentries ofϕ lie in the maximal idealp. SettingA=R[T1, . . . , Tn], andP = pA,

S(E)P = (A/(f1, . . . , fm)

)P,

and therefore ifSP is a DVR,AP must be a regular local ring, and thereforeR will also bea regular local ring.

For the converse, letP be a prime ideal of the ring of polynomialsA, of heightm+ 1,containing the formsfi ’s, and setp = P ∩ R. The normality ofS(E) means that for alsuchP , S(E)P is a DVR. The claim is that failure of this to hold is controlled by eitheror (ii).

We may localize atp. Suppose that heightp = m + 1. In this case, there existsz ∈ p

so thatz, f1, . . . , fm is a regular sequence inpA. If x1, . . . , xm+1 is a regular system oparameters of the local ringR, from a representation

[z, f1, . . . , fm] = [x1, . . . , xm+1] ·ψ,


ure

s

s,e

of finite

give a

nstruct

as in Theorem 4.1, we obtain the socle equality, according to [11]

(z, f1, . . . , fm) : pA= (z, f1, . . . , fm,detψ).

The imageu of z−1 detψ , provides us with a nonzero form of degreem, in the field offractions ofS(E). If S(E)P is not a DVR,

HomS(E)

(PS(E)P ,PS(E)P

) = (PS(E)P

)−1,

and, given thatu ∈ (PS(E)P )−1, we obtained a fresh element in the integral clos

of Sm(E).On the other hand, if heightp �m, we may assume that for some 1< t �m, It (ϕ)⊂ p,

but It−1(ϕ) �⊂ p. This implies that heightp =m− t + 2. We can localize atp, and argue ain the previous case.✷

5. Almost complete intersection algebras

We shall now consider modules of linear type whose Rees algebras arealmost completeintersections. Let (R,m) be a Cohen–Macaulay integral domain and letE be a torsionfreeR-module of finite projective dimension. Let

Rp ψ−→Rm ϕ−→ Rn →E → 0

be a minimal presentation ofE. If E is of linear type and the ideal of relation(f1, . . . , fm) = [T1, . . . , Tn] · ϕ, has codimensionm − 1, the linear relations among thfj has rank 1, and therefore has a single generator being a second syzygy moduleprojective dimension. This means thatp = 1 andψ is injective.

For the moduleE to be of linear type the roles of the determinantal idealsIt (ϕ) and ofI = I1(ψ) are less well behaved than in the case of complete intersections. Let ussummary of some of the known results, according to [16, Section 3.4]:

Theorem 5.1. Let R be a Cohen–Macaulay integral domain and letE be a torsionfreeR-module whose second Betti number is1 (in particular of projective dimension2). Thefollowing hold:

(a) If S(E)=R(E) thenheightI1(ψ) is odd.(b) If I is a strongly Cohen–Macaulay ideal of codimension3, satisfying the conditionF1,

andE∗ is a third syzygy module, thenS(E) is a Cohen–Macaulay integral domain.

This shows the kind of requirement that must be present when one wants to cointegrally closed Rees algebras in this class.

Example 5.1. LetR = k[x1, . . . , xd ] be a ring of polynomials. In [18], for eachd � 4, it isdescribed an indecomposable vector bundle on the punctured spectrum ofR, of rankd−2.


madeer withn.

e

his,

t

n-

e. If

n–texact

e

Its moduleE of global sections has a resolution

0 → Rψ−→Rd ϕ−→ R2d−3 →E → 0,

with ϕ having linear forms as entries, andψ(1) = [x1, . . . , xd]. If d is odd, according to[13, Corollary 3.10],E is of linear type and normal.

The analysis of the normality of the two previous classes of Rees algebras wassimpler because they were naturally Cohen–Macaulay. This is not the case any longalmost complete intersections, requiring that theS2 condition be imposed in some fashioWe pick one closely related to normality.

Theorem 5.2. LetR be a regular integral domain and letE be a torsionfree module whossecond Betti number is1. SupposeE is of linear type. If the idealI1(ψ)R(E) is principalat all localizations ofR(E) of depth1 thenR(E) satisfies the conditionS2 of Serre.

Proof. The condition onL= I1(ψ)R(E) means that for any localizationSP , S = R(E),with depthSP = 1, the idealLP is principal. There are several global ways to recast tsuch as(L ·L−1)−1 = S.

We may assume thatR is a local ring, and that the resolution ofE is minimal. Pickin A = R[T1, . . . , Tn] a prime idealP for which depthSP = 1. We must show thadimSP = 1. As in the other cases, we may assume thatP ∩R is the maximal ideal ofR.

We derive now a presentation of the idealJ = J (ϕ)= (f1, . . . , fm)A, moduloJ :

0 →K → (A/J )m = Sm → J/J 2 → 0,

and analyze the element ofS induced byv = ψ(1). This is a nonzero ‘vector’ whose etries inSP generateLP , which by assumption is a principal ideal. This means thatv = αv0,for some nonzeroα ∈ SP , wherev0 is an unimodular elementSP . Sincev ∈K, this meansthat the imageu of v0 in (J/J 2)P , is a torsion element of the module. Two cases arisu = 0, (J/J 2)P is a freeSP since it has also rankm− 1, and thereforeJP is a completeintersection in the regular local ringAP , according to [15]. This implies that the CoheMacaulay local ringSP has dimension 1. On the other hand, ifu �= 0,u is a torsion elemenof (J/J 2)P which is also a minimal generator of the module. We thus have that in thesequence

0 → torsion(J/J 2)

P→ (

J/J 2)P

→ C → 0,

C is a torsionfreeSP -module, of rankm− 1, generated bym− 1 elements. We thus havthat the idealJP of the regular local ringAP has the property

JP /J2P (AP /JP )

m−1 ⊕ (torsion).

According to [15] again,JP is a complete intersection since it has codimensionm − 1.This shows thatSP must have dimension 1.✷


erom

ctiondules.ase of

se.

h

In studying the normality in an algebraS = S(E) (E torsionfree) of linear type, thadvantage ofS2 holding is extremely useful. To recall briefly some technical facts f[16, p. 138]. For a prime idealp ⊂R there is an associated prime ideal inS(E) defined by

T (p)= ker(SR(E)→ SR/p(E/pE)p

).

Since R is a domain,T (p) is the contractionpS(E)p ∩ S(E), so heightT (p) =heightpS(E).

The prime ideals we are interested in are those of height 1. IfR is equidimensional, wesee that

heightT (p)= 1 if and only if ν(Ep)= heightp + rank(E)− 1.

Let us quote [16, Proposition 5.6.2]:

Proposition 5.2. LetR be a universally catenarian Noetherian ring and letE be a finitelygenerated module such thatS(E) is a domain. Then the set

{T (p) | heightp � 2 and heightT (p)= 1

}

is finite. More precisely, for any presentationRm ϕ−→ Rn →E → 0, this set is in bijectionwith

{p ⊂R | Ep not free,p ∈ Min

(R/It (ϕ)

)and heightp = rank(ϕ)− t + 2

},

where1 � t � rank(ϕ).

6. Effective criteria

Having already given effective criteria of normality in the cases of complete intersemodules and modules of projective dimension 1, we now consider more general moWe have found convenient to break down the proofs into two cases, the special cmodules of second Betti number 1 serving as guide in tracking the more general ca

Remark 6.1. There is also a technical reason for grouping these results. IfE is a torsionfreeR-module of linear type, a prominent role (checking the conditionR1 of Serre) is playedby the prime ideals ofS(E) of codimension one, which we denotedT (p): heightT (p)= 1if and only if dimRp + rank(E)= ν(Ep)+1, a condition equivalent to (whenEp has finiteprojective dimension�= 1,0) the second Betti number ofEp is 1.

Theorem 6.1. LetR be a regular integral domain and letE be a torsionfree module wita free resolution

0 → Rψ−→ Rm ϕ−→Rn →E → 0.


it

yat

r

e11].

f

f

SupposeE is of linear type.E is normal if and only if the following conditions hold:

(i) The idealI1(ψ)S(E) is principal at all localizations ofS(E) of depth1.(ii) The modulesSs(E) are complete, fors = 1, . . . ,m− 1.

Proof. We setA = R[T1, . . . , Tn] andJ = J (ϕ) = (f1, . . . , fm) = [T1, . . . , Tn] · ϕ. Weonly have to show that (i) and (ii) imply thatE is normal. In view of Theorem 5.2,suffices to verify the conditionR1 of Serre.

Let P ⊂ A be a prime ideal such that its image inS = A/J has height 1. We maassume thatP ∩ R is the maximal ideal ofR and thatE has projective dimension 2, this I1(ψ)⊂ m as otherwise we could apply Theorem 4.2.

From Proposition 5.2, and the paragraph preceding it, dimR = rank(ϕ)− t + 2 on theone hand and dimR = n− r + 1 on the other. Thus,t = 1 andm= dimR. Pick 0 �= a ∈ m

and consider the idealI = (a, f1, . . . , fm) ⊂ A. With P = mA, setL = I : P . For a setx1, . . . , xm of minimal generators ofm, we have

[a,f1, . . . , fm] = [x1, . . . , xm] ·B(Φ), (2)

whereB(Φ) is am × (m+ 1) matrix whose first column has entries inR, and the othecolumns are linear forms in theTi ’s. We denote byL0 the ideal ofA generated by theminors of orderm that fix the first column ofB(Φ). These are all forms of degreem− 1,andL0 ⊂ L. When we localize atP however,L0AP = LAP , since by condition (i) and thproof of Theorem 5.2,JP is a complete intersection and the assertion follows from [This means that the imageC of a−1L0 in the field of fractions ofS is not contained inSand has the property thatC · PSP ⊂ SP , giving rise to two possible outcomes:

C · PSP ={PSP ,

SP .(3)

In the first case,C would consist of elements in the integral closure ofSP but it notcontained inSP . This cannot occur since by condition (ii) all the symmetric powers oE,up to orderm− 1, are complete. This means that the second possibility occurs, andSP isa DVR. ✷Theorem 6.2. Let R be a regular integral domain and letE be a torsionfree module orank r, with a free presentation

Rp ψ−→ Rm ϕ−→Rn →E → 0.

SupposeE is of linear type.R(E) is normal if and only if the following conditions hold:

(i) The idealIc(ψ)S(E), c = m + r − n, is principal at all localizations ofS(E) ofdepth1.

(ii) The modulesSs(E) are complete, fors = 1, . . . , n− r.


omplete

ral

e the

ude

6.1.f

droof of

re areich is

Proof. As it was emphasized in Remark 6.1, as far as the conditionR1 is concerned, theRees algebras we are considering behave as if they are either complete or almost cintersections. We focus therefore on the conditionS2.

We note thatn− r is the height of the defining idealJ (ϕ) of S, while c =m+ r − n isthe rank of the second syzygy module.

We consider the complex induced by tensoring the tail of the presentation byS,

Spψ−→ Sm → J/J 2 → 0.

Note thatψ has rankc, whileJ/J 2 has rankn−r. This means that the kernel of the natusurjectionC = coker(ψ)→ J/J 2 is a torsionS-module.

Lemma 6.2. LetE be a module as above, setA= R[T1, . . . , Tn] andJ = (f1, . . . , fm)=[T1, . . . , Tn] · ϕ the defining ideal ofS(E) = A/J . Let P ⊃ J be a prime ideal ofA. IfIc(ψ)SP is a principal ideal thenJP is a complete intersection ideal.

Proof. LetC be the module defined above. According to [17, Proposition 2.4.5], sincFitting idealIc(ψ)P is principal, the moduleCP decomposes as

CP Sn−rP ⊕ (torsion).

This means that we have a surjection

Sn−rP → (JP /J

2P

)/(torsion)

of torsionfreeSP -modules of the same rank. Therefore

JP /J2P Sn−rP ⊕ (torsion).

At this point, sinceJP has finite projective dimension, we invoke [15] again to conclthatJP is a regular sequence.✷

The rest of the proof of Theorem 6.2 would proceed as in the proof of TheoremChoosingP = mA so that SP has dimension 1,JP is a complete intersection ocodimensionn − r = dimR − 1 = d − 1. As in setting up Eq. (2), we pick 0�= a ∈ m,pick a minimal set{x1, . . . , xd} of generators form, and define the matrixB(Φ)

[a,f1, . . . , fm] = [x1, . . . , xd ] ·B(Φ). (4)

Note that when localizing atP the ideals(a, f1, . . . , fm)P and(x1, . . . , xd)P are generateby regular sequences and we can use the same argument employed in the pTheorem 6.1. ✷

A significant difference between Theorems 6.1 and 6.2 lies in the fact that thenatural constructions of algebras arising from modules of second Betti number 1, whlacking in the general case.


artlyes ofproof

n

t

ams 6.1

f

rdical

requireto

Remark 6.3. It is rather natural to assume regularity in our discussion. It arises, pfrom a result in [4] (rediscovered in [7]), that the localizations of the associated primcomplete ideals are regular local rings. Jooyoun Hong has pointed out to us how thein [7] extends readily to modules. We are grateful to her for this observation.

Remark 6.4. It may be worthwhile to rephrase this criterion algorithmically: letR be aregular integral domain and letE be a torsionfree module of rankr, with a free presentatio

Rp ψ−→ Rm ϕ−→Rn →E → 0.

SetJ = J (ϕ)= [T1, . . . , Tn] ·ϕ ⊂A=R[T1, . . . , Tn]. SupposeE is of linear type, a pretesfor which could be formulated as follows:

If 0 �= z ∈ In−r (ϕ) thenJ (ϕ) : z= J (ϕ).

ThenE is normal if and only if it passes the following tests:

(i) Let c =m+ r − n, pick 0 �= b ∈ I = Ic(ψ), setL= (b, J ) : I . Then

grade(b−1L(J, I)R(E)

)� 2.

This condition is equivalent to saying thatS has theS2 condition of Serre and it iscomplete intersection in codimension one. It also sets up, as the proofs of Theoreand 6.2 make explicit, procedures providing elements of the integral closure ofS(E)

should the test fail.(ii) Pick 0 �= a ∈ Ir (ϕ). For each integer 1� t � r, let Lt be the component o

codimensionr − t + 2 of the radical ofIt (ϕ). Then

height(a−1((a, J ) : Lt

)Lt, J

)� r + 2.

If one could be assured of the completeness of theSs(E), for s � n− r, by some othemeans, this cumbersome step, requiring a great deal of work in identifying racomponents (fortunately of ideals ofR) could be avoided.

This should be contrasted (in characteristic zero) with the Jacobian criterion: IfK =Jac(R(E)), then

gradeKR(E)� 2.

Its computation could be rather cumbersome. Another point, Theorem 6.2 does notcharacteristic zero or thatR be a ring of polynomials, just that it be regular upcodimensionr + 1.


erous

a 341

c. 74

995.

ls, Proc.

l ring,

(1995)

Math.

0.

) 610–

195,

ringer,

n, Arch.

Acknowledgment

The authors are very grateful to the referee for the detailed reading and numsuggestions that lead to a more transparent manuscript.

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effective normality criteria for algebras of linear type

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