CANONICAL MODULES AND CLASS GROUPS OF REES-LIKE … · 2020. 8. 25. · CANONICAL MODULES AND CLASS...

CANONICAL MODULES AND CLASS GROUPS OF REES-LIKEALGEBRAS

PAOLO MANTERO, JASON MCCULLOUGH, AND LANCE EDWARD MILLER

ABSTRACT. Rees-like algebras have played a major role in settling the Eisenbud-Goto conjecture. This article concerns the structure of the canonical module ofthe Rees-like algebra and its class groups. Via an explicit computation based onlinkage, we provide an explicit and surprisingly well-structured resolution of thecanonical module in terms of a type of double-Koszul complex. Additionally,we give descriptions of both the divisor class group and the Picard group of aRees-like algebra.

1. INTRODUCTION

Rees-like algebras were introduced by I. Peeva and the second author [9]. Givena homogeneous ideal I in a polynomial ring S = K[x1, . . . , xn] over a fieldK, theRees-like algebra is RL(I) := S[It, t2] ⊆ S[t]. Rees-like algebras provide a ma-chine taking as input an arbitrary homogeneous ideal I in a standard graded poly-nomial ring S and producing a homogeneous prime ideal in a non-standard gradedpolynomial ring. A particularly nice advantage of the construction is that its defin-ing equations are explicit, unlike for Rees algebras. Among their applications arethe construction of graded prime ideals with larger than expected regularity, whichmay then be homogenized to produce a negative answer to the Eisenbud-Goto con-jecture [4]. As useful as these algebras are, there remain many questions as to thegeometry of the varieties they define. Towards this end, the authors completed astudy of the singularities of the Rees-like algebras, where again explicit methodswere used to describe the Jacobian and establish various normality properties [7].

A fundamental tool to study the properties of finitely generated algebras over afield is the canonical module. In this paper, we give a complete description of thecanonical module of the Rees-like algebra of an ideal of height at least 2 when thecharacteristic of the base field is not 2. In particular, we give an explicit presenta-tion of ωRL(I) via linkage theory by fully describing the minimal free resolutionof ωRL(I), including explicit differential maps. We show that the resolution hasa surprising self-dual structure. Moreover, we show that, even though the Rees-like algebra is not Cohen-Macaulay when I is not principal, its canonical module,defined as an appropriate Ext module, is Cohen-Macaulay; see Section 3.

Theorem A. (Theorem 3.12) Suppose k is a field with char(k) 6= 2 and S isthe polynomial ring k[x1, . . . , xn]. Let I = (f1, . . . , fm) be an ideal of S with

2010 Mathematics Subject Classification. 13D02,14B05.1

2 P. MANTERO, J. MCCULLOUGH, AND L. E. MILLER

ht(I) ≥ 2. The canonical module ωRL(I) of the Rees-like algebra RL(I) isCohen-Macaulay.

In particular, setting M to be the matrix

M =

[f1t f2t · · · fmt f1 f2 · · · fmf1t

2 f2t2 · · · fmt2 f1t f2t · · · fmt

],

the canonical module of the Rees-like algebraRL(I) isωRL(I) ∼= coker(M),

and thus type(RL(I)) = 2.The fact that ωRL(I) is Cohen-Macaulay is not overly surprising given that the

integral closure S[t] of RL(I) is Cohen-Macaulay; RL(I) is canonically Cohen-Macaulay in the language of Schenzel [12]. Nonetheless, we find the ‘double-Koszul complex’ structure of its resolution over the presenting polynomial ringrather interesting.

We next turn our attention to divisor class groups. This is a somewhat delicatetopic as the literature on class groups primarily limits itself to normal rings, whileRees-like algebras are never normal. Nevertheless, Rees-like algebras are Noether-ian domains, so the codimension-1 Chow group or divisor class group (see e.g. [3,Section 11.5]) is well-defined. First we prove the following general result aboutclass groups for which we could find no reference in the literature.

Theorem B. (Theorem 4.1) Let A be a Noetherian, universally catenary, integraldomain satisfying Serre’s condition (R1). Let A denote the integral closure of A.Then

Cl(A) ∼= Cl(A).Because Rees-like algebras of ideals of height at least two satisfy the (R1) condi-tion [7, Theorem 6], it follows that the class groups of these Rees-like algebras aretrivial; see Corollary 4.6.

Finally, we consider the Picard group ofRL(I). The fundamental approach is toconsider the conductor square, which realizes the Rees-like algebra as a pullback.This is also called a Milnor square, and exploiting a fundamental exact sequencerelating Picard groups and groups of units defined using this square, we show thePicard group of a Rees-like algebra vanishes precisely when I is radical.

Theorem C. (Theorem 4.7) For k a field, S = k[x1, . . . , xn] an S-ideal I isradical if and only if Pic(RL(I)) = 0.

Considering [7, Sec. 5, Thm. 8], where it is shown that I is radical if and onlyif RL(I) is seminormal, Theorem C supports a theme suggesting that Rees-likealgebras are best behaved for radical ideals.

The rest of the paper is structured as follows. In Section 2, we recall somepreliminary results and definition on Rees-like algebras. In Section 3, we computea presentation and free resolution of the canonical module of a Rees-like algebra.Finally, in Section 4 we study the divisor class group and Picard group of a Rees-like algebra.

CANONICAL MODULES AND CLASS GROUPS OF REES-LIKE ALGEBRAS 3

2. PRELIMINARIES

We reserve the following notation. Throughout, unless otherwise stated, k isa field and S = k[x1, . . . , xn] is a standard graded polynomial ring. We alsoreserve bold letters F•,D•, . . . for chain complexes of modules with differentialsdF• , d

D• , . . .

For a homogeneous S-ideal I with generators I = (f1, . . . , fm), recall the Rees-like algebra of I is S[It, t2] ⊆ S[t], where t is a new variable. We denote this byRL(I) := S[It, t2], and we denote by RLP(I) the prime ideal arising as thekernel of the map T → RL(I), where T = S[y1, . . . , ym, z] is a non-standardgraded polynomial ring over S, and the map is determined by sending yi 7→ fitand z 7→ t2. In particular, RL(I) ∼= T/RLP(I), where T has grading defined bydeg(yi) = deg(fi) + 1 and deg(z) = 2. (Later we distinguish between differentpresentations of RL(I), depending on the choice of generators of I .) We quicklyrecall the relevant structure theorem for Rees-like algebras.

Theorem 2.1 (McCullough and Peeva [9, Theorem 1.6, Proposition 2.9]). TheidealRLP(I) is the sumRLP(I)syz +RLP(I)gen with generators

RLP(I)syz =

{rj :=

m∑i=1

cijyi |m∑i=1

cijfi = 0

}and

RLP(I)gen = {yiyj − zfifj | 1 ≤ i, j ≤ m}.Moreover,

• eEuler(T/RLP(I)) = 2∏mi=1

(deg(fi) + 1

),

• pdT (T/RLP(I)) = pd(S/I) +m− 1,• ht(RLP(I)) = m,

and in particular, T/RLP(I) is Cohen-Macaulay if and only if m = 1.

In the previous theorem, eEuler(M) denotes the Euler multiplicity of the posi-tively graded T -moduleM defined as follows. LetEM (u) =

∑i

∑j(−1)iβTi,j(M)uj ∈

Z[u] denote the Euler polynomial of M . After factoring out a maximal possiblepower of (1 − u) we write EM = (1 − u)chM (u). Finally we define the Eulermultiplicity of M to be eEuler(M) = hM (1). When T is a standard graded poly-nomial ring, this is the usual degree or multiplicity of M . See [1, Theorem 2.5] forfurther details.

3. THE CANONICAL MODULE

We start this section with a brief summary of the proof of the main theoremconcerning the structure of the canonical module. Recall, the Rees-like algebraS[It, t2] is a quotient of a polynomial ring T . Set Q := RLP(I). As T/Q is notCohen-Macaulay if ht(I) ≥ 2, we take as our definition of the canonical moduleωT/Q := ExtcT (T/Q, T ), where c = codimQ. To calculate the canonical module,


our approach is based on linkage. Two ideals I and J in S are said to be linked pro-vided there is a complete intersection C ⊂ I ∩ J so that J = C : I and I = C : J .Many nice properties of ideals persist on linkage, namely if I and J are linked,then I defines a Cohen-Macaulay quotient if and only if J does. The applicationfor us is to compute the canonical module via the following well-known result; fora proof we refer the reader to [14, Thm. 6.25].

Theorem 3.1. For a polynomial ring T , a prime idealQ of heightm, and aC ⊂ Qa complete intersection of height m,

ωT/Q ∼= (C : Q)/Q.

The key observation is that among the generators of Q we find a natural com-plete intersection C to work with. We determine the primary decomposition ofC in an explicit manner and provide a Rees-like algebra interpretation for it, seeLemma 3.2(4). We then compute the minimal generators for C : Q, which alsoform a Gröbner basis. These generators allow one to relate this calculation ofthe canonical module to an interesting chain complex, obtained by combining twoKoszul complexes, which then serves as the claimed explicit minimal free resolu-tion.

We assume k is a field with char(k) 6= 2. Let S = k[x1, . . . , xn] and f1, . . . , fmminimal generators of a homogeneous ideal I . We also assume that ht(I) ≥ 2.Denote by RLP(f1, . . . , fm) the Rees-like prime defined in Section 2. There is adistinguished complete intersection inRLP(f1, . . . , fm), namely,

C =(y21 − zf21 , y22 − zf22 , . . . , y2m − zf2m

).

Note that a different choice of minimal generating set g1, . . . , gm of I givesa different but isomorphic Rees–like prime in the same polynomial ring T =S[y1, . . . , ym, z]. For instance, RLP(f1,−f2, f3, . . . , fm) 6= RLP(f1, . . . , fm),whileRLP(f1,−f2, f3, . . . , fm) ∼= RLP(f1, . . . , fm).

Lemma 3.2. With the the notation above, we have the following:(1) For any choice of +− signs, C ⊂ RLP(+−f1,+−f2, . . . ,+−fm).(2) RLP(f1, f2, . . . , fm) = RLP(−f1,−f2, . . . ,−fm).(3) If m ≥ 2, then for any choice of +− sign as indicated

RLP(f1, f2, . . . , fm) 6= RLP(f1,−f2,+−f3,+−f4, . . . ,+−fm).(4) The complete intersection ideal C defined above is radical and has the

following primary decomposition

C =⋂RLP(f1,+−f2,+−f3, . . . ,+−fm),

where the intersection is taken over all possible choices of +− sign.

Proof. (1) One simply observes that when we replace yi by +−fit and z by t2, we

see that y2i − zf2i becomes (+−fit)2 − t2f2i = 0.

(2) Let φ : T → S[t] be the map sending yi 7→ fit and z 7→ t2. Then clearly


RLP(f1, f2, . . . , fm) = Ker(φ) = Ker(−φ) = RLP(−f1,−f2, . . . ,−fm).(3) The element y1y2 − zf1f2 is in the left-hand ideal but not the right-hand one.(4) By (3), there are 2m−1 distinct primes in the intersection above, let us writethem Q1, . . . , Q2m−1 . By (1), C is a subset of the ideal H =

⋂2m−1j=1 Qj .

Both C and Q are unmixed homogeneous ideals with the grading deg(xj) = 1,deg(yi) = di + 1 and deg(z) = 2. Since y2i − zf2i is homogeneous of degree2(deg(fi)+1), we have eEuler(T/C) = 2mD, whereD =

∏mi=1(di+1). By The-

orem 2.1, eEuler(T/Qi) = 2D for every i = 1, . . . , 2m−1. Then eEuler(T/C) =eEuler(T/H) = 2

mD. Since C ⊆ H are unmixed ideals of the same Euler multi-plicity and height, we have C = H . �

Next, we want to obtain an explicit description of the link L = C : RLP(I),where RLP(I) = RLP(f1, f2, . . . , fm). To do this, we identify interest-ing candidate generators which posses remarkable symmetries. For any subsetA ⊆ [m] := {1, 2, . . . ,m} we define the elements gevenA and goddA as follows. Fora subset S ⊆ A, let yS denote

∏i∈S yi and set S = A \S. We define two elements

of T ,

gevenA :=

b#A/2c∑i=0

∑S⊆A#S=2i

ySfSzi,

goddA :=

b(#A−1)/2c∑i=0

∑S⊆A

#S=2i+1

ySfSzi,

where #A denotes the cardinality of A. For example, when m = 4 we get

geven[4] = y1y2y3y4 + y1y2f3f4z + y1f2y3f4z + · · ·+ f1f2y3y4z + f1f2f3f4z2,

godd[4] = y1y2y3f4 + y1y2f3y4 + · · ·+ f1y2y3y4 + y1f2f3f4z + · · ·+ f1f2f3y4z.

The elements geven[j] and godd[j] are invariant under an Sj-action which permutes

the variables yi, and they satisfy the following useful identities.

Lemma 3.3. For 1 ≤ j ≤ m and 1 ≤ h ≤ j, we have

godd[j] = yhgodd[j]r{h} + fhg

even[j]r{h},

geven[j] = yhgeven[j]r{h} + zfhg

odd[j]r{h},

yhgeven[j] = zfhg

odd[j] +

(y2h − zf2h

)geven[j]r{h},

fhgeven[j] = yhg

odd[j] −

(y2h − zf2h

)godd[j]r{h}.


Proof. The proof of the first two identities are similar to each other as are the proofsof the last two. We provide the reasoning for the first and third identities and leavethe other two for the interested reader.

To prove the first identity, we fix h and isolate the terms involving yh to obtain

godd[j] =

b(j−1)/2c∑i=0

∑S⊆{1,...,j}#S=2i+1

ySfSzi.

= yh

b(j−1)/2c∑i=0

∑S⊆{1,...,ĥ,...,j}

#S=2i+1

ySfSzi + fh

b(j−1)/2c∑i=0

∑S⊆{1,...,ĥ,...,j}

#S=2i

ySfSzi

= yh

b(j−1)/2c∑i=0

∑S⊆{1,...,ĥ,...,j}

#S=2i+1

ySfSzi + fhgeven[j]r{h}

= yhgodd[j]r{h} + fhg

even[j]r{h}.

To see the last equality holds note the following observations.• If j is even, then b(j − 1)/2c = b(j − 2)/2c, so∑b(j−1)/2c

i=0

∑S⊆{1,...,ĥ,...,j}

#S=2i+1

ySyhfSzi = godd[j]r{h}.

• If j is odd, for i = b(j − 1)/2c there is only one subset S ⊆ [j] with#S = 2i + 1, namely S = [j]. For this value of i and the only possibleassociated S, the variable yh does not divide ySfSzi = f1f2 · · · fjzi. Thus

b(j−1)/2c∑i=0

∑S⊆{1,...,ĥ,...,j}

#S=2i+1

ySfSzi =

b(j−2)/2c∑i=0

∑S⊆{1,...,ĥ,...,j}

#S=2i+1

ySfSzi = godd[j]r{h}.

Thus the first identity holds.

As for the third identity, we have

yhgeven[j] = y

2hg

even[j]r{h} + yhzfhg

odd[j]r{h}

= y2hgeven[j]r{h} + zfh

(godd[j] − fhg

even[j]r{h}

)=(y2h − zf2h

)geven[j]r{h} + zfhg

odd[j] ,

where the first equality follows from the second identity, and the middle equalityfrom the first identity and the last simply rearranges the terms.

�


To simplify the notation, in what follows we simply write goddj for godd[j] and

gevenj for geven[j] .

Lemma 3.4. If Q = RLP(f1,−f2,+−f3,+−f4, . . . ,+−fm), then gevenm , g

oddm ∈ Q.

Proof. We show that gevenj , goddj ∈ Q by induction on 2 ≤ j ≤ m. First note

that geven2 = y1y2 + zf1f2 = y1y2 − zf1(−f2) ∈ Q and, similarly, godd2 =y1f2 + y2f1 = y2f1 − y1(−f2) ∈ Q,

Now let j > 2 and suppose gevenj−1 , goddj−1 ∈ Q. Then, by Lemma 3.3, gevenj =

yjgevenj−1 + zfjg

oddj−1 ∈ Q and, similarly, goddj = yjgoddj−1 + fjgevenj−1 ∈ Q. �

Corollary 3.5. If Q = RLP(+−f1,+−f2, . . . ,+−fm), then gevenm , g

oddm ∈ Q for any

choice of +− signs except for Q = RLP(f1, . . . , fm) = RLP(−f1, . . . ,−fm).

Proof. By the symmetry of gevenm , goddm , we can assume that the signs on f1 and f2

are different. Then the statement follows from Lemma 3.4 and Lemma 3.2(2). �

Our next goal is to prove thatC : RLP(f1, . . . , fm) = C+(gevenm , g

oddm

). From

now on we adopt the following notation

Notation 3.6. Let I = (f1, . . . , fm) ⊆ S, and let Q = RLP(f1, . . . , fm) ⊆ T beits Rees-like prime. We set L := C : Q ⊆ T , and J := C +

(gevenm , g

oddm

)⊆ T .

Proving L = J will require a sequence of lemmas. First we construct two usefulshort exact sequences.

Lemma 3.7. With Notation 3.6, we have short exact sequences

0→ T/Q ·goddm−−−→ T/C → T/(C + (goddm ))→ 0,

and

0→ T/(IT + (y1, . . . , ym))·gevenm−−−→ T/(C + (goddm ))→ T/J → 0.

In particular, Q = C : (goddm ) and IT + (y1, . . . , ym) = (C + (goddm )) : (g

evenm ).

Proof. The first short exact sequence is explained by the equality C : (goddm ) = Q,which follows by Lemma 3.2(4) and Corollary 3.5.Analogously, for the second sequence we need to show (C + (goddm )) : (g

evenm ) =

IT + (y1, . . . , ym). First note that by the third and fourth equalities in Lemma 3.3we see that fh and yh lie in (C + (goddm )) : (g

evenm ) for every 1 ≤ h ≤ m, so

IT + (y1, . . . , ym) ⊆ (C + (goddm )) : (gevenm ).Since IT ⊂ (C + (goddm )) : (gevenm ), it suffices to consider the reverse inclusion

modulo IT . Let a ∈ T be such that a · gevenm ∈ (C + (goddm )) modulo IT . Sincegevenm ≡ y1y2 · · · ym modulo IT and (C + (goddm )) ≡ (y21, . . . , y2m) modulo IT ,we have ay1 · · · ym ∈ (y21, . . . , y2m) in T/IT . Because y1, . . . , ym is a regularsequence on T/IT , we get a ∈ (y1, . . . , ym)+IT . Therefore IT+(y1, . . . , ym) =(C + (goddm )) : (g

evenm ).

�

Next, we compute the initial ideal of J .


Lemma 3.8. Fix y1 > y2 > · · · > ym > z > x1 > · · · > xn and let < be the lexorder < on T . Then y21 − zf21 , . . . , y2m − zf2m, gevenm , goddm form a Gröbner basis ofJ with respect to


As a step toward proving J is unmixed, we next show that (y1, y2, . . . , ym, z) isnot an associated prime of T/J .

Lemma 3.10. Let p = (y1, y2, . . . , ym, z). Then p /∈ Ass(T/J).

Proof. First we show that Qp is a complete intersection. Recall that we have adecomposition Q = RLP(I)syz + RLP(I)gen as in Theorem 2.1. The idealRLP(I)syz is generated by elements of the form

∑i siyi such that

∑i sifi = 0 in

S. In particular, the following elements corresponding to Koszul syzygies of I arein (RLP(I)syz)p: y1 − f1fm ym, y2 −

f2fmym, . . . , ym−1 − fm−1fm ym. For brevity, set

y′i = yi −fifmym. Since y2m − zf2m ∈ RLP(I)gen, it follows that Qp is generated

by the regular sequence y′1, y′2, . . . , y

′m−1, y

2m − zf2m. (These elements, along with

ym, form a regular system of parameters of the regular local ring Sp.)Now we compute the link Lp = Cp : Qp. Set yi = yi + fifm ym, so that

y2i − zf2i = yiy′i +f2if2m

(y2m − zf2m).

Therefore[y21 − zf21 , . . . , y2m − zf2m

]= D

[y′1, . . . , y

′m−1, y

2m − zf2m

]T, where

D =

y1 0 · · · 0 f21 /f2m0 y2 · · · 0 f22 /f2m0 0

. . . 0...

0 0 0 ym−1 f2m−1/f

2m

0 0 0 0 1

.By [13, Theorem A.140], Lp = (C + (detD))p. Note that

det(D) =m−1∏i=1

yi

=m−1∏i=1

(yi +fifm

ym)

=∑

S⊆{1,...,m−1}

ySfS

f|S|m

y|S|m

=

b(m−1)/2c∑i=0

∑S⊆{1,...,m−1}

#S=2i

ySfS

f2imy2im +

∑S⊆{1,...,m−1}

#S=2i+1

ySfS

f2i+1my2i+1m

≡b(m−1)/2c∑

i=0

∑S⊆{1,...,m−1}

#S=2i

ySfSzi +∑

S⊆{1,...,m−1}#S=2i+1

ySfSziymfm

(mod Cp)= gevenm−1 +

ymfm

goddm−1,


where the third line follows from expanding the product, the fourth line separatesthe even and odd terms, and the fifth line follows since z − y

2mf2m∈ Cp. Finally note

thatfm det(D) ≡ fmgevenm−1 + ymgoddm−1 ≡ goddm (mod Cp).

It follows that Lp = (C + goddm )p. Since

fmgevenm = ymg

oddm − (y2m − zf2m)goddm−1 ∈ C + (goddm ),

we haveLp = (C + (g

oddm ))p = Jp.

Since Qp is a complete intersection, in particular Tp/Qp is Cohen-Macaulay.Since Jp = Cp : Qp is a link of Qp, then by linkage, e.g., [11, Prop. 2.6]. AlsoTp/Jp is Cohen-Macaulay. In particular, Jp is unmixed of height m; thereforepTp /∈ Ass(Tp/Jp) and so p /∈ Ass(T/J). �

We can now prove the following:

Proposition 3.11. In Notation 3.6, one has L = J , i.e. C : Q = C+(goddm , gevenm ).

Proof. The containment L ⊇ J follows from Lemma 3.2 and Corollary 3.5. Nextwe show Jun = L. Since C = Q ∩ L ⊆ J ⊆ L, since all these ideals have heightm, and since Q,L are unmixed, we have C ⊆ Jun ⊆ L. Since C ⊆ Jun areunmixed of the same height, Ass(T/Jun) ⊆ Ass(T/C), so, by Lemma 3.2(4), allassociated primes of T/Jun have the formRLP(f1,±f2, . . . ,±fm). By Theorem2.1 (or the proof of Lemma 3.10) they are all contained in p = (y1, . . . , ym, z).Since Jp = Lp, by Lemma 3.10, then JQi = LQi for eachQi ∈ Ass(T/Jun). Thisproves Jun = L.

It then suffices to prove that J is unmixed. We observe that for any associatedprime q of T/J we have ht(q) ≤ m+ 1, because

ht(q) ≤ pd(T/J) ≤ pd(T/in 1, the only

possibility is that z ∈ p, and therefore p = (y1, . . . , ym, z). But this possibility isruled out by Lemma 3.10. �

Claim 2. We may assume ym is regular on T/J .


Proof of Claim 2. By Claim 1 there is a linear form 0 6= ` ∈ k[y1, . . . , ym] thatis regular on T/J . By possibly multiplying by a unit and permuting the variables,we may assume that ` = ym +

∑m−1i=1 αiyi, where αi ∈ k. We consider the

automorphism ψ of T that fixes all variables except it sends ym 7→ `. It is easycheck that ψ−1(J) has the same generators as J except that every instance of fmis replaced by fm +

∑m−1i=1 αifi. This then corresponds to choosing a different

minimal set of generators of I before constructing the Rees-like prime. Since `is not in any associated prime of J , ym is not in any associated prime of ψ−1(J). �

We now conclude the proof of Proposition 3.11. Since ym is regular onT/J and y2m − zf2m ∈ J , then also fm is regular on T/J . To prove J is unmixedit then suffices to show Jfm is unmixed in the localization Tfm . Since fm is a unitin Tfm and fmg

evenm = ymg

oddm − (y2m − zf2m)goddm−1 ∈ (C + (goddm ))fm , the ideal

Jfm = (C + (goddm ))fm is an almost complete intersection of height m.

Now, in the ring Tfm we have(2)

J + (ym) = (y21 − zf21 , . . . , y2m−1 − zf2m−1, y2m − zf2m, ym, goddm )

= (y21 − zf21 , . . . , y2m−1 − zf2m−1, zf2m, ym, goddm )(because fm is a unit) = (y21 − zf21 , . . . , y2m−1 − zf2m−1, z, ym, goddm )

= (y21, y22, . . . , y

2m−1, ym, z, g

oddm )

(by definition of goddm ) = (y21, y

22, . . . , y

2m−1, ym, z, y1 · · · ym−1).

Since M = (y21, . . . , y2m−1, y1y2 · · · ym−1) is (y1, . . . , ym−1)-primary and

extended from k[y1, . . . , ym−1], then M is Cohen-Macaulay of heightm − 1. Since ym, z is a regular sequence on (T/M)fm , the ideal(y21, y

22, . . . , y

2m−1, y1 · · · ym−1, ym, z)fm = (J+(ym))fm is Cohen-Macaulay too.

Since ym is regular on T/J and fm is regular on T/J , ym is also regular on(T/J)fm , and thus (T/J)fm is Cohen-Macaulay. In particular, Jfm is unmixedand then so is J . �

We are now able to construct a finite T -free resolution of the canonical moduleof any Rees-like algebraRL(I) = S[It, t2] = T/RLP(I), assuming char(k) 6= 2and I has height at least 2. It is built from an amalgamation of the Koszul com-plexes on the generators f1, . . . , fm of I and the variables y1, . . . , ym.

Theorem 3.12. Suppose k is a field with char(k) 6= 2. Let S = k[x1, . . . , xn]and let I = (f1, . . . , fm) be an ideal of S with ht(I) ≥ 2. Then the canonicalmodule ωRL(I) of the Rees-like algebra RL(I) is a maximal Cohen-MacaulayRL(I)-module. In particular, if M is the matrix

M =

[y1 y2 · · · ym f1 f2 · · · fmzf1 zf2 · · · zfm y1 y2 · · · ym

],

then the canonical module of the Rees-like algebraRL(I) isωRL(I) ∼= coker(M),

as T -modules, and thus type(RL(I)) = 2.


Proof. As usual let T = S[y1, . . . , ym, z]. Let K•(y) denote the Koszul complexon y1, . . . , ym over T with differential maps d

y

i : Ki(y) → Ki−1(y), and letK•(f) denote the Koszul complex on f1, . . . , fm over T with differential maps

df

i : Ki(f) → Ki−1(f). Define a new complex of free T -modules D• withDi = T

2(mi ) for 0 ≤ i ≤ m with differential given as a matrix by

dDi =

dyi dfiz ·dfi d

y

i

.It is easy to check that dDi−1 ◦ dDi = 0 and thus D• is a complex. We also have thefollowing short exact sequences of complexes

0→ D•z−→ D• → D•/zD• → 0.

and0→ K•(y)⊗T T/zT → D•/zD• → K•(y)⊗T T/zT → 0.

Because K•(y)⊗T T/zT is acyclic, it follows from the long exact sequence of ho-mology associated to the second short exact sequence that D•/zD• is also acyclic.Now from the long exact sequence associated to the first short exact sequence wesee that multiplication by z induces an isomorphism on Hi(D•) for i > 0; then byNakayama’s Lemma we get Hi(D•) = 0 for i > 0. Note that dD1 = M.

Now define dD0 : D0 →C:QC as follows. By Proposition 3.11,

C:QC is minimally

generated by gevenm and −goddm . Since D0 = T 2, we map the first basis element togevenm and the second basis element to −goddm . By Lemma 3.3, we have

ymgevenm + zfm(−goddm ) = gevenm−1

(y2m − zf2m

)∈ C

andfmg

evenm + ym(−goddm ) = −goddm−1

(y2m − zf2m

)∈ C.

Therefore Im(dD1 ) = Im(M) ⊆ Ker(dD0 ). To show the reverse inclusion, supposethat a, b ∈ T such that dD0 [a, b]T = 0 ∈

C:QC ; that is,

a · gevenm + b(−goddm ) ∈ C.

Then by Lemma 3.7, a ∈ (C + (goddm )) : (gevenm ) = IT + (y1, . . . , ym). Since theentries in the first row of M generate IT + (y1, . . . , ym), we can use the columnsof M to rewrite a and b and we may assume that a = 0. But then b ∈ C : (goddm ) =Q. By Theorem 2.1, every element of Q is a linear combination of the elementsyiyj − zfifj , where 1 ≤ i ≤ j ≤ m and

∑j cjyj , where

∑j cjfj = 0. Note that[

0yiyj − zfifj

]= yj

[fiyi

]− fi

[yjzfj

]∈ Im(dD1 ),

and [0∑j cjyj

]=∑j

cj

[fjyj

]∈ Im(dD1 ),


where∑

j cjfj = 0. Therefore [0, b]T ∈ Im(dD1 ), for any b ∈ Q. It follows that

Im(dD1 ) = Ker(dD0 ), and that D• is a minimal T -free resolution of

C:QC . Finally,

we have ωRL(I) ∼= C:QC , e.g. by [6, Lemma 3.1].�

In retrospect, the fact that the canonical module is Cohen-Macaulay should notbe surprising since the integral closure of S[It, t2] is a polynomial ring, and thusa finite Cohen-Macaulay module over the non-Cohen-Macaulay Rees-like algebraRL(I). Yet, we find the self-dual nature of the T -free resolution of the canon-ical module in the previous theorem interesting, especially given that one of theconstituent Koszul complexes, K•(f) need not be exact.

As a corollary, we get the following surprising self-duality statement:

Corollary 3.13. Using the notation above,

ωRL(I) ∼= ExtmT (T/Q, T ) ∼= ExtmT(ωRL(I), T ).

Proof. Because K•(y) and K•(f) are self-dual, it follows from the definition thatD• is self-dual as well, i.e. D• ∼= HomT (D•, T ).

�

Example 3.14. Let S = k[x1, x2] and set I = (x1, x2)2. We construct the res-olution of the canonical module of the Rees-like algebra RL(I). As such, setT = S[y1, y2, y3, z] and let Q = RLP(x21, x1x2, x22). By the previous theorem,ωRL(I) ∼= C:QC , where C = (y

21 − zx41, y22 − zx21x22, y23 − zx42) and

C : Q = C + (godd3 , geven3 ),

where

geven3 = y1y2y3 + x1x32y1z + x

21x

22y2z + x

31x2y3z,

godd3 = x22y1y2 + x1x2y1y3 + x

21y2y3 + x

31x

32z.

Moreover, as a T -module, ωRL(I) has T -free resolution:

T 2 T 6d1oo T 6

d2oo T 2d3oo 0,oo


where

d1 =

[y1 y2 y3 x

21 x1x2 x

22

zx21 zx1x2 zx22 y1 y2 y3

]

d2 =

−y2 −y3 0 −x1x2 −x22 0y1 0 −y3 x21 0 −x220 y1 y2 0 x

21 x1x2

−zx1x2 −zx22 0 −y2 −y3 0zx21 0 −zx22 y1 0 −y30 zx21 zx1x2 0 y1 y2

d3 =

−y3 −x22y2 x1x2−y1 −x21−zx22 −y3zx1x2 y2−zx21 −y1

.

4. CLASS GROUPS

We now turn our attention to the investigation of class groups of Rees-like alge-bras. A main complication comes from the fact that Rees-like algebras are nevernormal. On the other hand, the integral closure of S[It, t2] is the UFD S[t], and,when the height of I is at least 2, S[It, t2] satisfies Serre’s (R1) condition by [7,Thm. 5.1]. We leverage these two facts in our computations.

4.1. Divisor class group. We first review class groups in the generality we con-sider; for details we refer the reader to [3, Section 11.5]. Denote for a ringR the setof height 1 primes by Spec1R. Let R be a Noetherian domain. A Weil divisor isa formal finite Z-linear combination

∑p∈Spec1(R) np[p] of height 1 primes. These

naturally form an abelian group Div(R).If R is normal, then Rq would be a DVR for all height 1 primes q, leading

to the usual notion of linear equivalence. Note however that for any (possiblynon-normal) domain R, the ring Rq is still a one dimensional domain. Thusfor any nonzero x ∈ R, the Rq-module Rq/xRq has finite length which we de-note ordq(x) := λ (Rq/xRq). When Rq is a DVR, ordq(x) agrees with theq-adic valuation of x and so this recovers the more familiar definition of classgroup. This extends in the natural fashion to Frac(R) and yields a well-definedmap divR : Frac(R) → Div(R) sending x/y ∈ Frac(R) with x, y ∈ R to∑

q∈Spec1R (ordq(x)− ordq(y)) [q]. Elements in the image Prin(R) of this mapare called principal divisors and the divisor class group or codimension-1 Chowgroup is the quotient

Cl(R) := Div(R)/Prin(R).

There are few computations in the literature of class groups of non-normal do-mains.

To compute the class group of a Rees-like algebra we prove a much more generaltheorem providing sufficient conditions under which the class group of an algebra


is isomorphic to the one of its integral closure (Theorem 4.1). Since the integralclosure of a Rees-like algebra is a polynomial ring, it follows that the class groupof a Rees-like algebra is trivial, under mild hypotheses.

Theorem 4.1 is likely unsurprising for experts, but we could not locate its state-ment or proof in the literature, so we provide a proof along with examples to il-lustrate the necessity of the hypothesis. The proof of [16, Chapter V, Section 5,Remark, p. 269] makes essentially similar claims and one could deduce a quickerargument accepting those, but we opted to provide a more detailed argument.

We work first in the following general setup. Let A be a Noetherian integraldomain, and let A denote its integral closure.

Theorem 4.1. Let A be a Noetherian, universally catenary, integral domain satis-fying Serre’s condition (R1). Let A denote the integral closure of A. Then

Cl(A) ∼= Cl(A).

Proof. The proof follows by showing that contraction of primes along the inclu-sion A → A induces a bijection between the sets of height one primes Spec1(A)and Spec1(A). Let ϕ : Div(A) → Div(A) be the function obtained by lin-early extending ϕ(P ) := P ∩ A. This map is clearly a group homomorphism.In the following, we will demonstrate an equality of rings AP = Aϕ(P ). AsFrac(A) = Frac(A), any principal divisor divA(f) =

∑ai[Pi] in Div(A) has

image∑ai[Pi ∩ A] = divA(f), which then will guarantee that Cl(A) ∼= Cl(A).

We establish these in the following claims.

Claim 1. If P ∈ Spec1(A), then p := P ∩A ∈ Spec1(A).

Since dim(A) = dim(A), then trdegA(A) = 0. Also, this forcestrdegκ(p)(κ(P )) = 0. Finally, the dimension equality [8, Theorem 15.6] holds,so one has

ht(P ) = ht(p) + trdegA(A)− trdegκ(p)(κ(P )).It follows that ht(P ) = ht(p) = 1.

Claim 2. For every p ∈ Spec1(A), there is P ∈ Spec1(A) with p = P ∩A.

The existence of a prime P ∈ Spec(A) contracting to p is guaranteed by thelying-over property of integral extensions. By the dimension formula

ht(P ) ≤ ht(P ) + trdegκ(p)(κ(P )) = ht(p) + trdegA(A) = ht(p) = 1.

As P is a nonzero ideal of the domain A, ht(P ) = 1.

Claim 3. We have an equality of rings Ap = AP inside their common fractionfield.

First, observe that Ap is a DVR, so we can write pAp = fAp for some f ∈ Ap,and every element a ∈ Ap has the form a = wf t for some unitw ∈ Ap and t ∈ N0.If the equality does not hold, then there exists x ∈ AP with x /∈ Ap. Since A ⊆ A


is birational, then AP ⊆ Frac(AP ) = Frac(Ap), so we can write x = a1/a2 witha1, a2 ∈ Ap. By the above, there exist r1, r2 ∈ N0 and units u1, u2 ∈ Ap suchthat ai = uf ri for i = 1, 2, so x = uf r for some unit u = u1/u2 ∈ Ap andr = r1 − r2 ∈ Z. Since x /∈ Ap, then r < 0, and since u−1 and f lie in Ap ⊆ AP ,then also f−1 ∈ AP . We use it to prove that AP is a field: any non-zero elementy ∈ AP can be written, as above, in the form y = vfs, where v is a unit in Ap ands ∈ Z. By the above, both v and f s are units in AP , thus y is a unit, and thereforeAP is a field. This is a contradiction, so Ap = AP as claimed.

By the above, for every prime p ∈ Spec1(A), there is a prime P ∈ Spec1(A)lying over p. We now show that P is unique. Let P ′ ∈ Spec(A) be another heightone prime with P ′ ∩ A = p, then by the arguments above AP ′ = Ap = AP . Lety ∈ P ′, and write y = ab , with a, b ∈ AP and b /∈ P . Since y ∈ P

′AP ′ = PAP ,we have by = a ∈ P and thus y ∈ P . It follows that P ′ ⊆ P and then, bysymmetry, P ′ = P , which concludes the proof. �

In particular, we obtain that any birational, integral extension of a k-algebra Asatisfying (R1) has the same class group as A.

Corollary 4.2. Let k be a field and letA ⊆ B be a birational, integral extension offinitely generated k-algebra domains such that A satisfies Serre’s condition (R1).Then

Cl(A) ∼= Cl(B).

Proof. Being k-algebras, both A and B are universally catenary and, by assump-tion, they have the same integral closureA. We then prove thatB satisfies the (R1)condition. Since the proof of Claim 2 in the previous theorem does not requireA tobe (R1), one has the every prime ideal p′ ofB is contracted from a height one primeideal P ofA. One then has natural inclusionsAP∩A ⊆ Bp ⊆ AP . SinceA is (R1),the proof of Claim 3 of the previous theorem implies that AP∩A = Bp = AP , soB is (R1).

The conclusion now follows from the previous theorem, because Cl(A) andCl(B) are both isomorphic to Cl(A). �

The next examples show the necessity of each assumption in the previous result.

Example 4.3 (Necessity of birationality). Let A = k[x3, x2y, xy2, y3] be the thirdVeronese of B = k[x, y]. The ring B is regular, A is (R1), and A → B is anintegral but not birational extension. One has Cl(B) = 0, but Cl(A) ∼= Z/3Z isnon-zero.

Example 4.4 (Necessity of integrality). LetA = k[x, y, xt, yt] be the Rees algebraof (x, y) in k[x, y] and B = k[x, y, t]. The ring B is regular and hence a UFD,A satisfies Serre’s (R1) property, and A → B is a birational extension but not anintegral extension. Thus all assumptions of Corollary 4.2 apply except integrality.

Clearly Cl(B) = 0 but Cl(A) 6= 0 as A is an integrally-closed non-UFD. Infact, Cl(A) ∼= Z. Thus we cannot remove the integral extension hypothesis inCorollary 4.2.


Example 4.5 (Necessity of (R1)). Let A = k[x, xt, t2] be the Rees-like algebra of(x) in K[x], and let B = k[x, t] be its integral closure. Then the ring B is regular,A→ B is an integral, birational extension, but A is not (R1). A is the coordinatering of a Whitney Umbrella variety which is a semi-normal hypersurface that is notnormal and thus not (R1). Here we verify that Cl(A) 6= 0.

Since A is not (R1), some care must be taken with computing the class group.Consider the height one prime ideal P = (x, xy) = xB ∩ A of A. If Cl(A) = 0,then ordP (f) = 1 for some f ∈ Frac(A) = Frac(B). Writing f = aa′ fora, a′ ∈ A, we have 1 = ordP (f) = λ(AP /aAP )− λ(AP /a′AP ).

We now find a contradiction by proving that λ(AP /cAP ) is an even integer forall c ∈ A.

Observe that A ∼= k[u, v, w]/(v2 − u2w), where we identify x ↔ u, xy ↔ v,and y2 ↔ w. It is easy to see that the multiplicity of the one-dimensional ringAP is 2 and AP has Hilbert-Samuel function λ(AP /P i+1AP ) = 1 + 2i. Let0 6= c ∈ AP , and write c = αud + βud−1v + higher order terms in u and vfor some integer d ≥ 0 and some α, β units in AP not both 0. Thenλ(AP /cAP ) = e(AP /cAP ) = e(grPAP (AP /cAP )). Since grPAP (AP /cAP )

∼=K(W )[U, V ]/(V 2−U2W,αUd+ βUd−1V ) is defined by a complete intersectionof degrees 2 and d, it follows that λ(AP /cAP ) = 2d. This gives a contradiction,so Cl(A) 6= 0.

We now prove that Rees-like algebras have trivial class groups.

Corollary 4.6. Let S = k[x1, . . . , xn]. If I ⊆ S is an ideal of height at least two,then Cl(S[It, t2]) = 0.

Proof. By [7, Theorem 6], S[It, t2] satisfies Serre’s condition (R1). It is easy tocheck that the hypothesis char(k) 6= 2 stated in [7] is not necessary. Its integralclosure S[t] is a UFD and so has Cl(A) = 0. Thus Cl(S[It, t2]) = 0 by Theo-rem 4.1. �

4.2. Picard group. Finally we consider the Picard group, i.e., the group Pic(R)of invertible fractional ideals modulo principal fractional ideals ofR. In the normalcase, the Picard group is a subgroup of the divisor class group, and so if we werein a normal setting it would be reasonable to expect the Picard group of a Rees-likealgebra S[It, t2] to also be trivial. However, in our setting the situation is moreinteresting, as we show that Pic(S[It, t2]) = 0 if and only if I is radical.

Our approach uses Milnor squares; see e.g. [15, Ex. 2.6]. We use the followingsetup. Suppose A → B is an inclusion of rings and let c := AnnA(B/A) be theconductor ideal. The ideal c is the largest ideal of A that is also an ideal of B. Inthis situation, A is the pullback of the diagram

A �

//

��

B

��

A/c �

// B/c


Spec(S[It, t2]) An+1

V (I)× A12 : 1 map

V (I)× A1

FIGURE 1. Pushout diagram for Spec(S[It, t2]) - Pictured withI = (x) ⊂ k[x]

It is easy to verify that the conductor ideal for the Rees-like algebra S[It, t2] isAnnS[It,t2](S[t]/S[It, t

2]) = I + It and the corresponding Milnor square is:

S[It, t2] �

//

��

S[t]

��

(S/I)[t2] �

// (S/I)[t].

Dual to this diagram is a pushout of affine varieties, which provides some per-spective on the geometry of Rees-like algebras. Fix an ideal I in S = k[x1, . . . , xn]and consider the cylinder V (I) × A1k inside A

n+1k . Identifying (a, b) ∼ (a,−b) ∈

V (I)× A1K creates a 2 : 1 gluing which, when extended to all of An+1k , yields the

affine variety Spec(S[It, t2]). The pinch point, or Whitney umbrella, is the varietyassociated to the Rees-like algebra of the ideal (x) ⊆ k[x]. Thus one can view va-rieties defined by Rees-like algebras as higher dimensional analogues of the pinchpoint surface. See Figure 1.

To compute the Picard group of a Rees-like algebra, we apply the Units-Picexact sequence associated to the Milnor square for Rees-like algebra.

Theorem 4.7. Let I ⊆ S = k[x1, . . . , xn] be a ideal. The Picard groupPic(S[It, t2]) = 0 if and only if I is radical.


Proof. Given the Milnor square above, the Units-Pic exact sequence [15, Thm.3.10] is as follows

1 // S[It, t2]× // S[t]× × (S/I)[t2]× // (S/I)[t]×∂

// Pic(S[It, t2]) // Pic(S[t])× Pic((S/I)[t2]) // Pic((S/I)[t]).

As S[It, t2] and S[t] are standard graded domains, both have units groups iso-morphic to k×. If I is not radical, then S/I has a nonzero nilpotent element, sayη ∈ S/I . Since 1 + ηt ∈ (S/I)[t]× r (S/I)[t2]×, it follows that coker(∂) 6= 0,whence Pic(S[It, t2]) 6= 0.

If on the other hand I is radical, then (S/I)[t]× = (S/I)[t2]× = k× and so∂ = 0. As S is regular, Pic(S[t]) = 0. The inclusion (S/I)[t2] → (S/I)[t]is a free extension and hence the natural map Pic((S/I)[t2]) → Pic((S/I)[t]) isinjective. It follows from the above sequence that Pic(S[It, t2]) = 0. �

Remark 4.8. The Rees-like algebra S[It, t2] is seminormal if and only if I is radi-cal, by [7, Corollary 4]. The proof cited states that I is homogeneous, however thiscondition is not needed. In general, it is not true that the Picard group of everyseminormal ring is trivial. Clearly, any number ring with class number greaterthan 1 is a counterexample. Specifically, the Dedekind domain R = Z[

√−5] R is

of course normal, whence seminormal, but Pic(R) = Cl(R) ∼= Z/2Z 6= 0.

Remark 4.9. When I is not radical, Pic(S[It, t2]) is not just nonzero but infinite.Here we consider the case I = (x2) ⊂ k[x] again. By the proof of Theorem 4.7,Pic(S[It, t2]) ∼= Im(∂) = coker((S/I)[t2]× → (S/I)[t]×). The units group of(S/I)[t2] decomposes as k× ⊕

⊕i≥1 k with (α0, α1, α2, . . .) ∈ k× ⊕

⊕i≥1 k

corresponding to α0(1 + α1xt+ α2xt2 + · · · ) ∈ (S/I)[t]×. A similar calculationworks for (S/I)[t2]×, with the copies of k appearing in even degrees only. Itfollows that Pic(S[It, t2]) ∼=

⊕i∈N k.

The same computation works for I = (x2, y) ⊂ k[x, y], when ht(I) = 2 andCl(S[It, t2]) = 0.

ACKNOWLEDGEMENTS

The second author was supported by a grant from the Simons Foundation(576107, JGM) and NSF grant DMS-1900792.

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UNIVERSITY OF ARKANSAS, DEPARTMENT OF MATHEMATICAL SCIENCES, FAYETTEVILLE,AR 72701

E-mail address: [email protected]

IOWA STATE UNIVERSITY, DEPARTMENT OF MATHEMATICS, AMES, IA 50011E-mail address: [email protected]

UNIVERSITY OF ARKANSAS, DEPARTMENT OF MATHEMATICAL SCIENCES, FAYETTEVILLE,AR 72701

E-mail address: [email protected]

1. Introduction2. Preliminaries3. The canonical module 4. Class groups4.1. Divisor class group4.2. Picard group

AcknowledgementsReferences

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