arXiv:1403.6334v3 [] 12 Oct 2017

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF
FOUR QUADRICS
CRAIG HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH,
AND ALEXANDRA SECELEANU
Abstract. Motivated by Stillman’s question, we show that the projective dimension of an ideal generated by four quadric forms in a polynomial ring is at most 6; moreover, this bound is tight. We achieve this
bound, in part, by giving a characterization of the low degree generators
of ideals primary to height three primes of multiplicities one and two.

1. Introduction
Let S = k[x1 , . . . , xn ] be a polynomial ring over a field k, and let I =
(f1 , . . . , fN ) be a homogeneous ideal of S. Classical theorems of Hilbert or
Auslander and Buchsbaum give upper bounds on the projective dimension
of S/I (or, equivalently, of I) in terms of n, the number of variables of
S. Motivated by computational efficiency issues, Stillman [23] posed the
following question:
Question 1.1 (Stillman [23, Problem 3.14]). Is there a bound on pd(S/I)
depending only on d1 , . . . , dN , and N , where di = deg(fi )?
Ananyan and Hochster [1] recently proved an affirmative answer to Stillman’s Question in full generality. They prove that for any homogeneous
ideal I generated by N forms of degree at most d either the generators form
a regular sequence, implying that N is an upper bound for the projective
dimension, or else the generators of I are contained in an ideal generated
by a bounded number (in terms of N and d) of forms of strictly smaller
degree. A delicate inductive proof shows the existence of a bound, but even
in the case when I is generated by quadrics, the bounds they produce are
very large and far from optimal. Tighter bounds in more specific cases have
been given in the following situations:
(1) In a previous paper [2], Ananyan and Hochster showed that if I is
generated by N quadric forms, then pd(S/I) has an upper bound
asymptotic to 2N 2N . They also give more general result for nonhomogeneous ideals generated in degree at most 2. In [1, Theorem
4.2] they claim a new upper bound of 2N +1 (N − 2) + 4.
(2) When I is minimally generated by N quadrics and ht(I) = 2, we
previously showed [15] that pd(S/I) ≤ 2N − 2. (See Theorem 2.12.)
Moreover, for all N we construct examples where equality is achieved.
Thus the bound here is optimal.
1

CRAIG HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU
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(3) When I is minimally generated by 3 cubics, Engheta [12] showed that
pd(S/I) ≤ 36, while the largest known example satisfies pd(S/I) =
5.
(4) When k = p > 0 and S/I is an F -pure ring, De Stefani and NúñezBetancourt [7] showed that pd(S/I) ≤ µ(I), where µ(I) denotes the
minimal number of generators of I.
It remains open as to what the optimal bounds for projective dimension
are. The general optimal bound, however, must be rather large. Examples
of Beder et. al. [5] show that the projective dimension of ideals generated
by 3 degree-d forms can grow exponentially with respect to d.
Further motivating Question 1.1, Caviglia showed it was equivalent to the
following analogous question for Castelnuovo-Mumford regularity. (See [20,
Theorem 2.4].)
Question 1.2 (Stillman [23, Problem 3.15]). Is there a bound on reg(S/I)
depending only on d1 , . . . , dN , and N , where di = deg(fi )?
While (1) above shows that pd(S/I) is bounded for any ideal generated
by N quadrics, these bounds are exponential in N . It is clear already in
the case of an ideal generated by 3 quadrics that these bounds are far from
optimal (cf. [20, Proposition 24]). In this paper we give a tight upper bound
on the projective dimension of an ideal generated by 4 quadrics. Specifically,
we prove
Theorem 1.3. Let S by a polynomial ring over a field k, and let I =
(q1 , q2 , q3 , q4 ) be an ideal of S generated by 4 homogeneous polynomials of
degree 2. Then pd(S/I) ≤ 6.
Examples from [19] of ideals I generated by 4 quadrics with pd(S/I) = 6
show that the above bound is optimal. Our proof follows a technique similar
to Engheta’s in [10] and [12] in dividing into cases by height and multiplicity.
We notably rely on the tight bound for the height 2 case proved in [15] and
the height 3 multiplicity 6 case proved in [16]. The remaining height 3 cases
constitute the bulk of this paper and require a variety of approaches.
Previously the authors had posted a paper to the arXiv proving an upper
bound of 9; however, we suspected that 6 should be the tight upper bound.
We finally reduced all of the remaining cases to 6 producing this paper.
This was done at the expense of the length and complexity of the paper.
However, we do not believe these arguments can be significantly shortened
without weakening the upper bound and, along with Theorem 2.12, this
paper represents the only other nontrivial known tight bound for Stillman’s
Question. A table summarizing the many cases is given in Section 10.
The rest of the paper is organized as follows: in Section 2, we collect
many of the results that we cite and fix notation; in Section 3, we prove
some results that are used several times throughout the paper; in Sections 4,
5, we characterize certain height three primary ideals that can occur as the
unmixed part of an ideal of four quadrics or a direct link of such an ideal; in

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

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Sections 6, 7, 8, we handle the cases where our ideal has multiplicity 2, 3, 4
or 5. Section 9, handles the cases where the ideal has multiplicity 4 or 5 and
we know that any complete intersection contained in our ideal is extended
from a 6 variable polynomial ring. Finally, we collect all of the previous
results to prove our main result in Section 10. A contains a list of the many
primary ideals needed for Sections 4 and 5. It may be skipped without
affecting the logical flow of the paper, but it may also be of independent
interest.
2. Background and Notation
In this section we set the notation that will be used throughout this
manuscript and collect several results that will be used in the following
sections.
2.1. Unmixed ideals and multiplicity. We will frequently use the following well-known associativity formula (also sometimes referred to as the
linearity formula or additivity and reduction formula) to compute the multiplicity of an ideal.
Proposition 2.1 (cf. [26, Theorem 11.2.4]). If J is an ideal of S, then
X
e(S/J) =
e(S/p)λ(Sp /Jp ).
primes p⊇J
ht(p)=ht(J)

Here e(R) denotes the multiplicity of a graded ring R, and λ(M ) denotes
the length of an S-module M .
Recall that an ideal J of height h is unmixed if ht(p) = h for every
p ∈ Ass(S/J). Also, the unmixed part of J, denoted J un , is the intersection
of all the components of J of minimum height. One has J ⊆ J un , and from
the associativity formula it follows that e(S/J) = e(S/J un ). We follow [10]
in using the following notation: if J is an unmixed ideal, we say it is of type
he1 , . . . , em ; λ1 , . . . , λm i
if J has m associated prime ideals p1 , . . . , pm of equal height
P with ei =
e(S/pi ) and λi = λ(Spi /Jpi ) for all i. It follows that e(S/J) = m
i=1 ei λi .
The following simple fact also follows from the associativity formula:
Lemma 2.2 (cf. [10, Lemma 8]). Let J ⊆ S be an unmixed ideal. If I ⊆ S
is an ideal containing J such that ht(I) = ht(J) and e(S/I) = e(S/J), then
J = I.
Engheta’s finite characterization of height 2 unmixed ideals of multiplicity
2 gives a bound on projective dimension in this case.
Proposition 2.3 ([10, Proposition 11]). Let J be an unmixed ideal in a
polynomial ring S over an algebraically closed field. Suppose ht(J) = 2 and
e(S/J) = 2. Then pd(S/J) ≤ 3.

CRAIG HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU
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In particular, if J is primary to (x, y) for independent linear forms x, y,
then
(1) J = (x, y 2 ) or
(2) J = (x2 , xy, y 2 , ax + by), where ht(x, y, a, b) = 4.
In [17] it was shown that such a complete characterization is very special
and cannot be extended to higher heights or multiplicities. In fact, for
any multiplicity e ≥ 2 and any height h ≥ 2 with (e, h) 6= (2, 2) there are
infinitely many nonisomorphic primary ideals of multiplicity e and height h.
Moreover, they can be chosen to have arbitrarily large projective dimension.
However, in Sections 4 and 5 we give a characterization of ideals primary to
a height three linear prime ideal or to a height three multiplicity two prime
that is sufficient for the bounds we want.
2.2. Linkage. Two ideals J and K in a regular ring R are said to be linked,
denoted J ∼ K, if there exists a regular sequence α = α1 , . . . , αg such
that K = (α) : J and J = (α) : K. Notice that the definition forces J
and K to be unmixed, and (α) ⊆ J ∩ K. Linkage has been studied since
the nineteenth century, although its first modern treatment appeared in the
ground-breaking paper by Peskine and Szpiro [24]. We refer the interested
reader to [21] and [18] and their references.
We will need a few results regarding linkage.
Theorem 2.4. (Peskine-Szpiro, [24]) Let J be an unmixed ideal of S of
height g. Let α = α1 , . . . , αg be a regular sequence in J, and set K = (α) : J.
Then, one has
(1) J = (α) : K, that is, J ∼ K via α;
(2) S/J is Cohen-Macaulay if and only if S/K is Cohen-Macaulay;
(3) e(S/J) + e(S/K) = e(S/(α)).
It is easily checked that if α is a regular sequence of maximal length in
an ideal J, then (α) : J = (α) : J un , that is, J un is linked to (α) : J. We
will use this fact several times.
The following facts are well-known.
Lemma 2.5 (cf. [9, Lemma 2.6]). Suppose J and L are unmixed ideals both
linked to the same ideal. Then pd(S/J) = pd(S/L).
Lemma 2.6 (cf. [10, Theorem 7]). Let J be an almost complete intersection
ideal of S. If K is any ideal linked to J un , then one has
pd(S/J) ≤ pd(S/K) + 1.
2.3. Height three primes of small multiplicity. Here we recall some of
the structure theorems regarding prime ideals of small height and multiplicity. For most of this section we need the field k to be algebraically closed.
The reduction to this case occurs in the proof of Theorem 10.1. Also, recall
that a homogeneous ideal J is called degenerate if J contains at least one
linear form, otherwise J is said to be non-degenerate.

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The following result is an ideal-theoretic version of a classical result of
Samuel and Nagata.
Theorem 2.7 ( Samuel [25] , Nagata [22, Theorem 40.6]). Let J be a
homogeneous, unmixed ideal of S. If e(S/J) = 1, then J is generated by
ht(J) linear forms.
The next result we need is a classical simple lower bound for the multiplicity of non-degenerate prime ideals.
Proposition 2.8 ([14, Corollary 18.12]). Let p be a homogeneous prime
ideal of S. If p is a non-degenerate prime ideal, then e(S/p) ≥ ht(p) + 1.
The following result is a consequence of Proposition 2.8.

Corollary 2.9. Let p be a homogeneous prime ideal of S of height three.
If e(S/p) = 2, then there exist linear forms x, y and a quadric q such that
p = (x, y, q).
The next two results are classification theorems of varieties of small degree
over an algebraically closed field. We give an equation-oriented version of
these results that best fits our purpose. The first of these results follows
from a classical classification theorem for varieties of multiplicity 3. (See
[27].) By Proposition 2.8 all such primes are degenerate.
Theorem 2.10 (Anonymous [3], Swinnerton-Dyer [27, Theorem 3]). Let p
be a homogeneous prime ideal of S of height three. If e(S/p) = 3, then p is
of one of the following types:
(1) p = (x, y, c) where x and y are linear forms and c is a cubic form;
(2) p = (x) + I2 (M ), where M is a 2 × 3 matrix of linear forms and x
is a linear form that is necessarily a nonzerodivisor on I2 (M ).
The next result is a classification theorem for varieties of multiplicity
4. The non-degenerate case was done by Swinnerton-Dyer [27, Theorem
1]; the degenerate cases are derived from work of Brodmann-Schenzel [6,
Theorem 2.1]).
Theorem 2.11. Let p a homogeneous prime ideal of S of height 3. If
e(S/p) = 4, then p is of one of the following types:
(1) p = (x, y, r), where x, y are linear forms and r is a quartic;
(2) p = (x, q, q ′ ), where x is a linear form and q and q ′ are quadrics;
(3) p = I2 (M ), where M is a 2 × 4 matrix of linear forms or a 3 × 3
symmetric matrix of linear forms;
(4) either p is generated by a linear form, a quadric and three cubics, or
p is generated by a linear form and 7 cubics.
2.4. Related results. Finally we recall two results of the authors’ needed
for the main result:
Theorem 2.12 ([15, Theorem 3.5]). For any ideal I of height two generated
by n quadrics in a polynomial ring S, one has pd(S/I) ≤ 2n − 2. Moreover,
this bound is tight.

CRAIG HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU
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In particular, an ideal generated by 4 independent quadrics of height two
satisfies pd(S/I) ≤ 6.
The following is a generalization of a result of Engheta.
Theorem 2.13 ([16, Corollary 2.8], [11, Theorem 1]). Let I be an almost
complete intersection of height g > 1 generated by quadrics. Then e(S/I) ≤
2g − g + 1. Moreover, if e(S/I) = 2g − g + 1, then S/I is Cohen-Macaulay.
Engheta proved an upper bound in the case of any graded almost complete
intersection. The Cohen-Macaulayness and a more general bound are given
in [16]. In particular, this result shows that an ideal I of height 3 generated
by 4 quadrics has multiplicity at most 6, and moreover, if I has multiplicity
exactly 6, it is Cohen-Macaulay and hence pd(S/I) = 3.
3. Basic Results
In this section we collect some basic results and notation. If I is an ideal
generated by 4 quadric forms in a polynomial ring S, then ht(I) ≤ 4. If
ht(I) = 1 or 4, then pd(S/I) = 4. By Theorem 2.12, it suffices to consider
the height three case.
Assumptions 3.1. Unless otherwise specified, we use the following notation
in the remainder of the paper:
• S is a polynomial ring over an algebraically closed field k,
• I = (q1 , q2 , q3 , q4 ) is an S-ideal of height 3,
• q1 , q2 , q3 , q4 are independent, homogeneous polynomials of degree 2,
• ht(q1 , q2 , q3 ) = 3.
• L = (q1 , q2 , q3 ) : I = (q1 , q2 , q3 ) : I un .
Lemma 3.2. If I un is Cohen-Macaulay, then pd(S/I) ≤ 4.

Proof. By Proposition 2.4, (q1 , q2 , q3 ) : I = (q1 , q2 , q3 ) : I un is CohenMacaulay. By Lemma 2.6, pd(S/I) ≤ 4.

Proposition 3.3. If e(S/I) = 1, then pd(S/I) ≤ 4.

Proof. By Theorem 2.7, I un is Cohen-Macaulay, and by Lemma 3.2, pd(S/I) ≤
4.

Lemma 3.4. If I un or L = (q1 , q2 , q3 ) : I contains a linear form, then
pd(S/I) ≤ 5.

Proof. First suppose x is a linear form contained in I un . Since we have the
containments I ⊆ (I, x) ⊆ I un , it follows that ht (I, x) = 3. Hence, after
possibly relabeling the quadrics q1 , . . . , q4 , we may assume x, q1 , q2 form a
regular sequence. Let L′ := (x, q1 , q2 ) : I = (x, q1 , q2 ) : I un . We have now
reduced to the case when an ideal J (either L′ or L) directly linked to I un
contains a complete intersection of one linear form and two quadric forms.
Since J contains a complete intersection of one linear and two quadric
forms, say x, q1 , q2 , e(S/J) ≤ 4. If e(S/J) = 4, then J = (x, q1 , q2 ). In particular, J is Cohen-Macaulay and pd(S/I) ≤ 4 by Lemma 2.6. If e(S/J) = 1,

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

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then J is Cohen-Macaulay by Theorem 2.7 and again pd(S/I) ≤ 4. If
e(S/J) = 2, then we can write J = (x) + J ′ , where J ′ is unmixed, ht(J ′ ) = 2
and e(S/J ′ ) = 2 such that x is regular on J ′ . By Proposition 2.3, pd(S/J ′ ) ≤
3. This yields pd(S/J) ≤ 4 (because x is regular on S/J ′ ) which, by Jemma
2.6, gives pd(S/I) ≤ 5. Finally if e(S/J) = 3, consider K = (x, q1 , q2 ) : J.
By Theorem 2.4 we have e(S/K) = e(S/(x, q1 , q2 )) − e(S/J) = 4 − 3 = 1.
So K is Cohen-Macaulay by Theorem 2.7. By Lemma 2.5, pd(S/I un ) = 3,
and by Lemma 3.2, pd(S/I) ≤ 4.


Lemma 3.5. Suppose I ⊆ K ∩ J, where K = K ′ + (q), ht(K) = ht(J) = 3,
ht(K ′ ) = 2 and q is a quadric. Then there exists a quadric q ′ ∈ K ∩ J such
that K = K ′ + (q ′ ).

Proof. For each i = 1, . . . , 4, write qi = fi + αi q, where fi is an element of
K ′ and αi ∈ k. If αi = 0 for all i, one has I ⊆ K ′ , which implies ht K ′ ≤ 2
and gives a contradiction. Hence, one may assume αi 6= 0 for some i. Now,
take q ′ = qi .

Lemma 3.6. Suppose I1 , I2 are ideals. Then
pd(S/(I1 ∩ I2 )) ≤ max{pd(S/I1 ), pd(S/I2 ), pd(S/(I1 + I2 )) − 1}.

Proof. This follows from the long exact sequence of TorSi (−, k) applied to
the short exact sequence
0 → S/(I1 ∩ I2 ) → S/I1 ⊕ S/I2 → S/(I1 + I2 ) → 0.



Proposition 3.7. Let I be an almost complete intersection of height h. Let
F• be the minimal free resolution of the unmixed part I un of I. Let ∂i denote
the ith differential in F• . Then
∗
))}.
pd(S/I) ≤ max{h + 2, pd(Coker(∂h+1

Proof. Set I = (f1 , . . . , fh , fh+1 ), where C = (f1 , . . . , fh ) is a complete intersection. Then
pd(S/I) ≤ max{h, pd(S/(C : fh+1 )) + 1}

≤ max{h + 1, pd((C : fh+1 )/C) + 2}

∗
≤ max{h + 2, pd(Ker(∂h+1
)) + 2}

∗
))}.
= max{h + 2, pd(Coker(∂h+1

The first inequality follows from the the short exact sequence
0 → S/(C : fh+1 ) → S/C → S/I → 0.

The second inequality follows from the short exact sequence
0 → (C : fh+1 )/C → S/C → S/(C : fh+1 ) → 0.

The third inequality follows from the isomorphism (C : fh+1 )/C ∼
= ExthS (S/I un , S)
(see [17, Lemma 3.1]) and the short exact sequence
∗
) → ExthS (S/I un , S) → 0.
0 → Im(∂h∗ ) → Ker(∂h+1

CRAIG HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU
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By [17, Lemma 3.3], pd(Im(∂h∗ )) = h−1, which implies that pd(ExthS (S/I un , S)) =
∗ )). Finally, pd(Coker(∂ ∗ )) = pd(Ker(∂ ∗ ))+ 2 gives the last
pd(Ker(∂h+1
h+1
h+1
equality.

We make use of a slight generalization of an observation of AnanyanHochster:
Lemma 3.8 ([2, Lemma 3.3]). Let f1 , . . . , ft be a regular sequence of forms
in S and set A = k[f1 , . . . , ft ]. Then for any ideal I of S extended from A
one has pd(S/I) ≤ t. More generally, for any finite S-module M presented
by a matrix with entries in A, one has pd(M ) ≤ t. In particular, any ideal
I whose generators can be written in terms of at most t variables satisfies
pd(S/I) ≤ t.
In general pd(S/I) may be arbitrarily larger than pd(S/I un ). However,
for almost complete intersection ideals whose unmixed part is extended from
a smaller polynomial ring, we get the following useful corollary to Proposition 3.7.
Corollary 3.9. Let I = (g1 , . . . , gh , gh+1 ) be an almost complete intersection
of height h. Suppose I un is extended from A = K[f1 , . . . , ft ], where f1 , . . . , ft
are a regular sequence of forms. Then pd(S/I) ≤ max{h + 2, t}.

Proof. Let (Fi , ∂i ) be the minimal free resolution of I un ∩ A over A. Then
the entries of ∂h+1 lie in A. Hence
∗
pdS (S/I) ≤ max{h + 2, pdS (Coker(∂h+1
⊗ S))}
∗
))}
= max{h + 2, pdA (Coker(∂h+1

≤ max{h + 2, t}.
The first inequality follows from Proposition 3.7. The second line follows
from the faithful flatness of S over A. The last inequality follows from
Lemma 3.8.

4. (x, y, z)-primary ideals
In this section we give a characterization of the degree 2 component of an
ideal J primary to a height 3 linear prime p = (x, y, z) with e(S/J) = 2, 3
or 4. As shown in [17], there are infinitely many distinct such ideals, but
we show here that there are only a finite number of possibilities for the low
degree generators of such ideals. First we need a lemma regarding matrices
of linear forms.
For a matrix M, we set Ik (M) to be the ideal generated by the k × k
minors of M . Recall that Ik (M) is unchanged by k-linear row and column
operations. We say that M has a generalized zero if after k-linear row and
column operations M has a zero entry. A matrix which has no generalized
zeros is called 1-generic. We refer the reader to [8] for background on 1generic matricies.

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

9

Lemma 4.1. Suppose M is a 2 × 3 matrix of linear forms in S such that
ht(I2 (M)) = 1. Then, after k-linear row and column operations, M has one
of the following forms:


a 0 0
(1) M =
, where a 6= 0 and ht(e, f ) = 2;
d e f
a b 0
(2) M =
, where ae − bd 6= 0;
d e 0


a b 0
(3) M =
, where ht(a, b, d) = 3.
d 0 b
Proof. Since ht(I2 (M)) = 1, M is not 1-generic [8, Theorem 6.4]. Hence M
has a generalized zero and we may assume that it has the form


a b 0
M=
.
d e f

If ht(a, b) ≤ 1, then after a column operation, M has the form of case (1)
above. So we may assume that ht(a, b) = 2. Suppose ht(d, e, f ) = 3. Then
since I2 (M) = (af, bf, ae − bd) has height 1, we have ae − bd = xf for some
x ∈ S1 . Since d, e, f form a regular sequence, we must have a ∈ (d, f ). If
a ∈ (f ), then it follows that b ∈ (f ) as well, contradicting that ht(a, b) = 2.
Hence, after a row operation, we may now assume that ht(d, e, f ) ≤ 2. In
this case, we can do a column operation so that M now has one of the
following forms:




a b 0
a b 0
M=
or
M=
.
d 0 f
d e 0

We are done in the latter case. In the former case, note that I2 (M) =
(af, bf, bd) = (a, b)∩(b, f )∩(d, f ). Since ht(a, b) = 2, we must have ht(b, f ) ≤
1 or ht(d, f ) ≤ 1. If ht(b, f ) ≤ 1, we are in case (3). If ht(d, f ) ≤ 1, we can
perform row and column operations so that M has the form of case (1) or
case (2).
Finally note that the conditions on the entries of the matrices distinguish
the cases by the minimal number of generators of I2 (M).

Lemma 4.2. Suppose M is a 2 × 3 matrix of linear forms in S such that
ht(I2 (M)) = 2 and ht(I1 (M)) = 2. Then, after k-linear row and column
operations, M has the following form:


a b 0
M=
, where a, b are distinct linear forms.
0 a b
Proof. Since ht(I2 (M)) = 2, the entries in the first row must generate a
height 2 ideal. After column operations we may assume that the top-right
entry is 0. The bottom-right entry then cannot be 0, since ht(I2 (M)) = 2.
After further column operations we may assume M has the form


a
b
0
M=
.
αa + βb γa + δb b

CRAIG
10
HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU

After more column operations we may force β = δ = 0. After a row operation we can force α = 0. Since ht(I2 (M)) = 2, γ 6= 0. After a column
operation, M has the desired form.

Proposition 4.3. Let J be p-primary with e(S/J) = 2. Then one of the
following holds:
(1) J = (x, y, z 2 )
(2) J = (x, y 2 , yz, z 2 , ay + bz), where ht(x, y, z, a, b) = 5
(3) J = (x, y, z)2 + (ax + by + cz, dx + ey + f z), where a, b, c, d, e, f ∈ S1 ,
ht(x, y, 
z, I2 (M))≥ 5, ht(x, y, z, a, b, c, d, e, f ) ≥ 6, and M is the
a b c
matrix
.
d e f
(4) All quadrics in J can be written in terms of at most 6 variables.
In particular, pd(S/J) ≤ 4 for cases (1), (2) and (3).
Proof. Since e(S/J) = 2, we have (x, y, z)2 ⊆ J ⊆ (x, y, z). If J contains
a linear form, which we may take to be x, then J = (x) + J ′ , where J ′
is (y, z)-primary and e(S/J ′ ) = 2. By Lemma 2.3, either J = (x, y, z 2 ) or
J = (x, y 2 , yz, z 2 , ay + bz), where ht(x, y, z, a, b) = 5.
Suppose J does not contain a linear form. Since e(S/(x, y, z)2 ) = 4, J
must contain at least 2 more generators of the form ax + by + cz, dx + ey + f z
not contained in (x, y, z)2 . We may assume that both such generators are
linearly independent and degree 2 or else we are in case (4). By Lemma A.2,
the ideal (x, y, z)2 +(ax+by+cz, dx+ey+f z) is unmixed and has multiplicity
2, and hence equal to J, if 
ht(x, y, z,
I2 ) ≥ 5, where I2 denotes the 2 × 2
a b c
minors of the matrix M =
. So we may assume that ht(I2 ) =
d e f
1 modulo (x, y, z). Hence, after a linear change of variables and a linear
combination of generators, we may assume that M has one of the three
forms in Lemma 4.1 modulo (x, y, z). In case (1), ax ∈ J. Since a ∈
/ p and
since J is p-primary, x ∈ J contradicting that J does not contain a linear
form. In case (2), we see that (ae − db)y ∈ J and ae − db ∈
/ p and hence
y ∈ J, also a contradiction. In the remaining case (3), we see that
b(dy − az) = d(ax + by) − a(dx + bz) ∈ J.

Since b ∈
/ p, we have dy − az ∈ J and ht(x, y, z, a, b, d) = 6. By Lemma A.3
the ideal (x, y, z)2 + (ax + by, dx + bz, dy − az) is unmixed of multiplicity 2
and equal to J. All quadrics in this ideal can be written in terms of the 6
variables x, y, z, a, b, d leaving us in case (4). This completes the proof. 
Proposition 4.4. Let J be p-primary with e(S/J) = 3. Then one of the
following holds:
(1) J = (x, y 2 , yz, z 2 ).
(2) J = (x, y 2 , yz, z 3 , ay + z 2 ), where a ∈ S1 and ht(x, y, z, a) = 4.
(3) J = (x2 , xy, xz, y 2 , yz, ax + by + z 2 , cx + dy), where a, b, c, d ∈ S1 and
ht(x, y, z, c, d) = 5.

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

11

(4) J = (x2 , xy, xz, yz 2 , z 3 , ax + by + cz, dx + y 2 ), where a, b, c, d ∈ S1
and ht(x, y, z, b, c) = ht(x, y, z, c, d) = 5.
(5) J = (x2 , xy, xz, y 3 , z 3 , ax + by + cz, dx + yz), where a, b, c, d ∈ S1 and
ht(x, y, z, b, c) = ht(x, y, z, b, d) = ht(x, y, z, c, d) = 5.
(6) All quadrics in J can be written in terms of at most 6 variables.
(7) All quadrics in J generate an ideal of height at most 2.
In particular, pd(S/J) ≤ 4 for cases (1) - (5).
Proof. As e(S/J) = 3, e(S/(J : p)) = 1 or 2.
If e(S/(J : p)) = 1, then J : p = (x, y, z) and (x, y, z)2 ⊆ J. If J
contains two linear forms, then an ideal of the form (x, y, z 2 ) is contained
in J, contradicting that e(S/J) = 3. If J contains a linear form, then J
must be (x, y 2 , yz, z 2 ). If J does not contain any linear forms, then J must
contain at least one more generator of the form q = ax + by + cz, which we
may assume is degree 2 (or else we are in case (6)). If ht(x, y, z, a, b, c) ≥ 5,
then (x, y, z)2 +(ax+by +cz) is unmixed of multiplicity 3 by Lemma A.4 and
then equals J. This ideal fits into case (6) as well. If ht(x, y, z, a, b, c) = 4
then we can modify b, c modulo (x, y, z) to assume b and c are multiple of a.
Then q = aℓ ∈ J for some linear form ℓ and then J contains a linear form,
ℓ, a contradiction.
If e(S/(J : p)) = 2, then by the previous proposition there are 4 cases to
consider:
Case I. J : p = (x, y, z 2 )
In this case (x2 , xy, xz, y 2 , yz, z 3 ) ⊆ J ⊆ (x, y, z 2 ). If J contains 2 independent linear forms, then J = (x, y, z 3 ). If J contains exactly one linear form
up to scalar multiplication, we may assume this form to be x, thus we have
(x, y 2 , yz, z 3 ) ⊆ J. The ideal on the left has multiplicity 4, so we need at
least one additional generator of the form ay + z 2 , which we may assume is
a quadric, or else we are in case (7). By Lemma A.5, (x, y 2 , yz, z 3 , ay + z 2 )
is unmixed of multiplicity 3 and so is equal to J. If ht(x, y, z, a) = 4, then
we are in case (2); if a ∈ (x, y, z), then we are in case (1). Finally we assume that J does not contain a linear form. We may assume that J has
two additional quadric generators of the form ax + by + z 2 , cx + dy. Indeed,
if two linearly independent quadrics are of the form ax + by, cx + dy, then
(ad − bc)y ∈ J. Since J contains no linear forms, we have ad − bc ∈ p. Since
p is a linear prime, then S/p is a UFD, then either one of a, b, c, d ∈ p or
a ≡ b and c ≡ d modulo p, or else a ≡ c and b ≡ d modulo p. In the first
case, we get a linear form in J. In the latter case, after modifying a, b, c, d
modulo p we would get that ax + by can be assumed to be a scalar multiple
of cx + dy, which contradicts their linear independence. In the second case,
after modifying a, b, c, d modulo p we would obtain aℓ ∈ J and cℓ′ ∈ J for
linear forms ℓ, ℓ′ ∈ p. Since a ∈
/ p (otherwise one of the generators in not
minimal), then ℓ ∈ J, giving a contradiction.

CRAIG
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HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU

Finally, if ht(x, y, z, c, d) = 5, the ideal (x2 , xy, xz, y 2 , yz, ax+by +z 2 , cx+
dy) is equal to J by Lemma A.6; otherwise, c and d have a common factor
modulo p and so J contains a linear form - a contradiction.
Case II. J : p = (x, y 2 , yz, z 2 , ay + bz).
If x ∈ J, then (x, y 3 , y 2 z, yz 2 , z 3 , y(ay + bz), z(ay + bz)) ⊆ J and there is
at least one additional generator of the form αy 2 + βyz + γz 2 + δ(ay + bz),
which we may assume is a quadric or else we are in case (7). If δ = 0 (which
includes the case deg(a) = deg(b) ≥ 2), then one checks that (x, y, z)2p ⊆
Jp and hence (x, y 2 , yz, z 2 ) ⊆ J since J is p-primary. S these ideals have
the same multiplicity, J = (x, y 2 , yz, z 2 ). This contradicts that J : p =
(x, y 2 , yz, z 2 , ay +bz). If δ 6= 0 we may assume δ = 1; then by a linear change
of variables (a′ = a + αy + βz, b′ = b + γz), we may assume our additional
quadric has the form ay + bz. Then J = (x, ay + bz, y 3 , y 2 z, yz 2 , z 3 ) by
Lemma A.7. This ideal falls into case (7) as well.
If x ∈
/ J, then (x2 , xy, xz, y 3 , y 2 z, yz 2 , z 3 , y(ay + bz), z(ay + bz)) ⊆ J.
There are at least 2 additional quadric generators of the form cx + αy 2 +
βyz +γz 2 +δ(ay +bz), dx+ǫy 2 +ζyz +ηz 2 , where c, d ∈ S1 and the remaining
coefficients are in K, since otherwise we are in case (6). If δ = 0 for all such
quadrics, then either there are at most 3 such quadrics putting us in case
(6), or else there is a quadric of the form ex ∈ J with e ∈
/ (x, y, z). It follows
that x ∈ J, a contradiction. Hence we may assume that δ = 1 and that
deg(a) = deg(b) = 1. After a linear change of variables (a′ = a + αy + βz
and b′ = b + γz), we can assume that α = β = γ = 0. After a further linear
change of variables we can assume that the second quadric has the form
dx + y 2 or dx + yz. (If ǫ = ζ = η = 0, then x ∈ J, contradicting that J did
not contain a linear form.) Recall that we have ht(x, y, z, a, b) = 5.
Suppose the second quadric is dx + y 2 . If ht(x, y, z, b, d) = 5, then J =
2
(x , xy, xz, yz 2 , z 3 , cx + ay + bz, dx + y 2 ) by Lemma A.8 putting us in case
(4). If ht(x, y, z, b, d) = 4, then after rewriting we may assume d = 0 or
d = b. If d = 0, then one checks that J = (x, y, z)2 + (cx + ay + bz), which
is unmixed by Lemma A.4, putting us in case (6). If d = b, then
b(ax − yz) = a(bx + y 2 ) − y(cx + ay + bz) + cxy ∈ J,
and hence ax − yz ∈ J. By Lemma A.10 that J = (x2 , xy, xz, z 3 , bx +
y 2 , ax − yz, cx + ay + bz). Again this falls into case (6).
The case where the second quadric is dx+yz is similar. If ht(x, y, z, a, d) =
ht(x, y, z, b, d) = 5, then J = (x2 , xy, xz, y 3 , z 3 , cx + ay + bz, dx + yz) by
Lemma A.9. Otherwise we may assume one of the following: d = a, d = b
or d = 0. If d = a or d = b, then J has the form of Lemma A.10. If d = 0,
then x2 , y 2 ∈ J and J has the form from Lemma A.4. Both cases put us in
case (6).
Case III. J : p = (x, y, z)2 + (ax + by + cz, dx + ey + f z).
We have that (x, y, z)3 + (x, y, z)(ax + by + cz, dx + ey + f z) ⊆ J, where
ht(x, y, z, I2 (M)) = 5. The ideal on the left has multiplicity 5, so there are

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

13

at least two additional quadric generators which, after a linear change of
variables, we can take to be of the form ax + by + cz + q, dx + ey + f z + q ′ ,
where q, q ′ ∈ (x, y, z)2 . (If there are not at least 2 such quadrics, then all
quadrics in J can be written in terms of at most 6 variables.) After relabeling
a, . . . , f we can assume q, q ′ = 0 without changing ht(I2 (M)) or ht(I1 (M)).
Hence (x, y, z)3 + (ax + by + cz, dx + ey + f z) ⊆ J. If ht(x, y, z, I2 (M)) ≥ 5,
where


a b c
M=
,
d e f
then ht(x, y, z, a, b, c, d, e, f ) ≥ 5. If in addition ht(x, y, z, a, b, c, d, e, f ) ≥ 6,
then this ideal equals J by Lemma A.11. This ideal falls into case (7). If
ht(x, y, z, a, b, c, d, e, f ) = 5, then by Lemma 4.2 we may take M to be of the
form


a b 0
M=
,
0 a b
modulo (x, y, z). In other words, ax + by + q, ay + bz + q ′ ∈ J, where
q, q ′ ∈ (x, y, z)2 . In this case
a(xz − y 2 ) = z(ax + by + q) − y(ay + bz + q ′ ) − zq + yq ′ .

Since (x, y, z)3 ⊆ J, we have a(xz − y 2 ) ∈ J. Since a ∈
/ (x, y, z) and J is
(x, y, z)-primary, xz − y 2 ∈ J. Now J = (x, y, z)3 + (ax + by, ay + bz, xz + y 2 )
by Lemma A.12. This ideal falls into case (6).
Case IV. Quadrics in J : p can be written in terms of 6 variables.
Since J ⊆ J : p, all quadrics in J can be written in terms of 6 variables,
putting us in case (6).

Proposition 4.5. Let J be p-primary with e(S/J) = 4. Then at least one
of the following holds:
(1) All quadrics in J can be written in terms of at most 6 variables.
(2) All quadrics in J generate an ideal of height at most 2.
(3) J contains a linear form.
(4) pd(S/J) ≤ 5.
Proof. We can assume J does not contain a linear form, since otherwise we
are in case (3). The ideal J : p is p-primary and e(S/(J : p)) must be 1, 2
or 3.
Case I. e(S/(J : p)) = 1.
In this case J : p = (x, y, z) and hence J = (x, y, z)2 . This fall into case (1).
Case II. e(S/(J : p)) = 2.
There are 3 subcases to consider corresponding to Lemma 4.3.
Case II.A. J : p = (x, y, z 2 ).
We have (x2 , xy, xz, y 2 , yz, z 3 ) ⊆ J ⊆ J : p. There is another generator
which we may assume is a quadric of the form ax + by + αz 2 , where a, b ∈ S1
and α ∈ K. We may further assume that α = 1, otherwise all quadrics in
J are contained in (x, y), putting us in case (2). This ideal is then unmixed
of multiplicity 4 by Lemma A.13 and so equal to J. Note that all quadrics

CRAIG
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HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU

in J can be written in terms of the linear forms x, y, z, a, b.
Case II.B. J : p = (x, y 2 , yz, z 2 , ay + bz), where ht(x, y, z, a, b) = 5.
Here we have (x2 , xy, xz, y 3 , y 2 z, yz 2 , z 3 , y(ay + bz), z(ay + bz)) ⊆ J ⊆ J : p.
We may assume J contains a quadric of the form cx+αy 2 +βyz +γz 2 +(ay +
bz). After a linear change of variables we may assume (without changing
ht(x, y, z, a, b)) that the quadric has the form cx + ay + bz. By assumption
ht(x, y, z, a, b) = 5, so this ideal is unmixed and equal to J by Lemma A.14.
All quadrics in J can be written in terms of the linear forms x, y, z, a, b, c
putting us in case (1).
Case II.C. J : p = (x, y, z)2 + (ax + by + cz, dx + ey + f z), where
ht(x, y, z, I2 (M)) = 5.
Since (x, y, z)3 + (x, y, z)(ax + by + cz, dx + ey + f z) ⊆ J ⊆ J : p, it is
easy to see that either all quadrics in J can be written in terms of at most
6 variables, or else, after relabeling the coefficients (and without affecting
ht(x, y, z, I2 (M))), we may assume ax + by + cz, dx + ey + f z ⊆ J. These
two forms are linearly independent elements of pp \ p2p . Since (x, y, z)3 ⊆ J,
we have HFSp /Jp = (1, 1, 1, 0, . . .); in particular e(S/J) ≤ 3, contradicting
that e(S/J) = 4.
Case III. e(S/(J : p)) = 3.
In this case J : p is one of cases (1) - (7) from the previous proposition. We
are done if J : p is as in cases (6) or (7) or if J contains a linear form by
Lemma 3.4.
Case III.A. J : p = (x, y 2 , yz, z 2 ).
In this case (x2 , xy, xz, y 3 , y 2 z, yz 2 , z 3 ) ⊆ J ⊆ J : p and J contains additional generators of the form ax + αy 2 + βyz + γz 2 where a ∈ S1 − p. Either
all quadrics in J can be written in terms of at most 6 variables, or there
are at least 4 such generators with linearly independent x-coefficients. After
taking a linear combination of these 4 generators, we have a′ x ∈ J for some
a′ ∈
/ p and hence x ∈ J, contradicting that J does not contain a linear form.
Case III.B. J : p = (x, y 2 , yz, z 3 , ay + z 2 ), where ht(x, y, z, a) = 4.
We have (x2 , xy, xz, y 3 , y 2 z, yz 2 , ayz +z 3 ) ⊆ J ⊆ J : p. We may assume that
J has additional quadric generators of the form bi x+αi yz+βi y 2 +γi (ay+z 2 ).
We have three possibilities to consider:
(1) ht(x, y, z, a, b1 , b2 , . . .) ≤ 6, in which case all quadrics in J can be
expressed in terms of at most 6 variables.
(2) One of extra quadric generators has the form bx, with b ∈
/ (x, y, z).
Since J is (x, y, z)-primary, it follows that J contains the linear form
x, a contradiction.
(3) If neither of the above hold, we must have exactly 3 additional
quadrics. If there are fewer than 3 such quadrics, we are in case
(1). If there are more than 3 then after taking linear combinations of the generators we can produce a generator of the form
bx and are in case (2). Moreover, after taking linear combinations
of these 3 quadrics, we may assume they have the following form:

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

15

b1 x + ay + z 2 , b2 x + yz, b3 x + y 2 with ht(x, y, z, a, b1 , b2 , b3 ) = 7. In
this case, one notes that
ab3 x = a(b3 x + y 2 ) − y(b1 x + ay + z 2 ) + b1 xy + yz 2 ∈ J.
As ab3 ∈
/ (x, y, z), we again have the linear form x ∈ J.
Case III.C. J : p = (x2 , xy, xz, y 2 , yz, ax + by + z 2 , cx + dy), where
ht(x, y, z, c, d) = 5.
We have (x, y, z)(x2 , xy, xz, y 2 , yz, ax + by + z 2 , cx + dy) ⊆ J ⊆ J : p. If
it is not the case that all quadrics in J are expressible in terms of at most
6 variables, then after a relabeling of linear forms, we may assume that
ax + by + z 2 , cx + dy ∈ J. One checks then that (ad − bc)(x2 , xy, y 2 ) ⊆ J. If
ad−bc ∈ p, then ht(x, y, z, a, b, c, d) ≤ 6 putting us in case (6). If ad−bc ∈
/ p,
then since J is p-primary, we have J1 = (x2 , xy, y 2 , ax+by +z 2 , cx+dy) ⊆ J.
This ideal is unmixed of multiplicity 4 by Lemma A.15, J = J1 and we have
pd(S/J) = 4.
Case III.D. J : p = (x2 , xy, xz, yz 2 , z 3 , cx + ay + bz, dx + y 2 ), where
ht(x, y, z, b, c) = ht(x, y, z, c, d) = 5.
Once again either we are in case (6) or we may assume cx+ay +bz, dx+y 2 ⊆
J and ht(x, y, z, a, b, c, d) = 7. Since dx2 = x(dx+y 2 )−y(xy) ∈ J and d 6∈ p,
we also have x2 ∈ J. Then J1 = (x2 , xz 2 , xy 2 , y 4 , z 4 , dx + yz, cx + ay + bz) ⊆
J ⊆ J : p. One checks that J1 is unmixed of multiplicity 4 and hence equal
to J and that pd(S/J) = 4.
Case III.E. J : p = (x2 , xy, xz, y 3 , z 3 , cx + ay + bz, dx + yz),where
ht(x, y, z, b, c) = ht(x, y, z, b, d) = ht(x, y, z, c, d) = 5.
This case is nearly identical to the previous case, so we leave the proof to
the reader.
This completes the proof.

5. h2; 2i structures
The section contains our characterization of certain q-primary ideals,
where q = (x, y, q) is a height three prime generated by 2 linear forms
x, y and a quadric form q. For our purposes, we need to characterize the
possible h2; 2i structures; that is, q-primary ideals J with e(S/J) = 4.
Proposition 5.1. Let q = (x, y, q) be a height 3 multiplicity 2 prime and let
J be a q-primary ideal with e(S/J) = 4. Further assume that J contains a
height 3 ideal generated by quadrics. Then J has one of the following forms:
(1) J = (x, y 2 , q),
(2) J = (x2 , xy, y 2 , q) + L′ , where L′ is generated in degrees ≥ 3,
(3) J = (x2 , xy, y 2 , ax + by, q) + L′ , where a, b ∈ S1 , ht(x, y, a, b, q) = 5
and L′ is generated in degrees ≥ 3,
(4) J = (x2 , xy, y 2 , ax + by, cx + dy, ad − bc + ex + f y = q), where
a, b, c, d, e, f ∈ S1 , ht(x, y, a, b) = ht(x, y, c, d) = 4 and ht(x, y, ad −
bc) = 3.

CRAIG
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HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU

Proof. First, notice that after possibly modifying q modulo (x, y) we may
assume q ∈ J. Indeed, if not, every quadric in J would be contained in
(x, y), a contradiction. Notice that the Hilbert function of J locally at q
is HFSq /Jq : (1, 1, 0, . . .). In particular, q2q ⊆ Jq . Since J is q-primary, this
yields q2 ⊆ J ⊆ q. We conclude that (x2 , xy, y 2 , q) ⊆ J.
Assume J also contains a linear form. Without loss of generality we
may assume it is x. Then (x, y 2 , q) ⊆ J. However, x, y 2 , q is a complete
intersection ideal contained in the unmixed ideal J and e(S/(x, y 2 , q)) =
e(S/J) = 4; hence J = (x, y 2 , q) by Lemma 2.2, and we are in case (1).
We now assume J does not contain a linear form. Since e(S/(x2 , xy, y 2 , q)) =
6, there must be at least an additional generator for J. If all such additional generators have degree at least three, then we are in case (2)
of the statement. We may then assume J contains a quadric generator
q′ ∈
/ (x2 , xy, y 2 , q). Since q ∈ J, without loss of generality we may assume q ′
is of the form q ′ = ax + by, where a, b are linear forms. Since (x, y)2 ⊆ J we
may assume ht(x, y, a, b) ≥ 3. Also, if ht(x, y, a, b) = 3 = ht(x, y, a), then it
is easily seen that aℓ ∈ J for some ℓ ∈ (x, y), and hence ℓ ∈ J, contradicting
that J contains no linear forms. Thus we may assume ht(x, y, a, b) = 4.
If ht(x, y, a, b, q) = 5 and there are no additional quadric generators of
J, then we are in case (3). If ht(x, y, a, b, q) = 4, then we may write q =
f x + ey + da − cb for linear forms c, d, e, f . Then b(cx + dy) = −x(f x +
ey + da − cb) + f (x2 ) + e(xy) + d(ax + by) ∈ J. Since b ∈
/ q and J is qprimary, cx + dy ∈ J. If ht(x, y, c, d) = 2, then q ∈ (x, y), a contradiction. If
ht(x, y, c, d) = 3, then c, d share a common factor modulo (x, y) and again
we have a linear form in J.
Hence, we may assume there is an additional quadric generator q ′′ = cx +
dy ∈ J − (x2 , xy, y 2 , q, q ′ ), where c, d are linear forms with ht(x, y, c, d) = 4.
Note that (ad − bc)x = d(ax + by) − b(cx + dy) ∈ J. If ad − bc 6∈ q, then
x ∈ J, contradicting that J does not contain any linear forms. Hence we
can write ad − bc = ex + f y + αq for linear forms e, f and α ∈ k.
If α = 0, then ad − bc ∈ (x, y). So ad − bc = 0(modulo (x, y)). It
follows that ax + by = α(cx + dy) modulo (x, y)2 , which contradicts that
q ′′ = cx+ dy ∈
/ (x2 , xy, y 2 , q, q ′ = ax + by). Hence α 6= 0, and we may assume
that q = ad − bc + ex + f y, for linear forms e, f .
Set J ′ = (x2 , xy, y 2 , ax + by, cx + dy, ad − bc + ex + f y). By Lemma A.16,
J ′ has multiplicity 4 and is Cohen-Macaulay; in particular, J ′ is unmixed,
ht(J ′ ) = ht(J) = 3 and J ′ ⊆ J; hence J ′ = J by Lemma 2.2.

6. Multiplicity 2
In this section, we again set I = (q1 , q2 , q3 , q4 ) to be an ideal generated
by 4 quadrics with ht(I) = 3. The next proposition gives a bound on the
projective dimension when e(S/I) = 2.
Proposition 6.1. Suppose e(S/I) = 2. Then pd(S/I) ≤ 6.

Proof. There are three cases for the type of I un : h2; 1i, h1, 1; 1, 1i, and h1; 2i.

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

17

If I un is of type h2; 1i, then I un is prime and hence is Cohen-Macaulay
by Corollary 2.9. By Lemma 3.2, pd(S/I) ≤ 4.
If I un is of type h1, 1; 1, 1i, then I un = (u, v, w)∩(x, y, z), hence pd(S/I) ≤
6 by Corollary 3.9.
If I un is of type h1; 2i, then I un satisfies one of the four cases in Proposition 4.3. In the first two cases, I un contains a linear form and pd(S/I) ≤ 5
by Lemma 3.4. In the last case, we have pd(S/I) ≤ 6 by Lemma 3.8 again.
Finally, if I un = (x, y, z)2 + (ax + by + cz, dx + ey + f z), then by Lemma A.2,
pd(S/(x2 , y 2 , z 2 ) : I un ) = 4. Hence pd(S/I) ≤ 5 by Lemma 2.6.

7. Multiplicity 3 and 5
Recall that we are assuming I = (q1 , q2 , q3 , q4 ) is a height 3 ideal generated
by four quadrics and (q1 , q2 , q3 ) form a complete intersection. As the cases
e(S/I) = 3 and e(S/I) = 5 are very similar, we will handle these cases
simultaneously. When e(S/I) = 3, we show that pd(S/I) ≤ 6. When
e(S/I) = 5, we consider the ideal L = (q1 , q2 , q3 ) : I, which is directly
linked to I un and satisfies e(S/L) = 3. In this case, we show that either
pd(S/I) ≤ 6 or the quadrics in L are extended from a polynomial ring
with at most 6 variables. Thus either I un or L is of one of the following
types: h3; 1i, h1; 3i, h1, 2; 1, 1i, h1, 1; 1, 2i, h1, 1, 1; 1, 1, 1i. For several of these,
we devote a separate lemma. The final bound arguments are collected in
Proposition 7.5.
Lemma 7.1. If I un is type h1; 3i, then pd(S/I) ≤ 6. If L is of type h1; 3i,
then pd(S/I) ≤ 6 or the quadrics in L can be expressed in terms of at most
6 variables.
Proof. Note that we are done if either I un or L contains a linear form (by
Lemma 3.4), or if the quadrics in either are expressible in terms of at most
6 variables (by Lemma 3.8), or generate an ideal of height at most 2 (which
contradicts our assumptions about I). By Proposition 4.4, this leaves only
cases (3) - (5) to consider. If L is one of these ideals, then pd(S/L) ≤ 4 by
Lemmas A.6, A.8 and A.9, and so pd(S/I) ≤ 5 by Lemma 2.6. If I un is one
of these ideals, then we note that again by Lemmas A.6, A.8 and A.9, an
ideal L′ directly linked to I un satisfies pd(S/L′ ) ≤ 4. Hence pd(S/I) ≤ 5 by
Lemmas 2.5 and 2.6.

Lemma 7.2. If I un is of type h1, 2; 1, 1i, then pd(S/I) ≤ 6. If L is of type
h1, 2; 1, 1i, then pd(S/I) ≤ 6 or the quadrics in L can be expressed in terms
of at most 6 variables.
Proof. First consider the case L = (x, y, z) ∩ (u, v, q), where x, y, z, u, v ∈ S1
and q ∈ S2 . By Lemma 3.5, we may assume that q ∈ (x, y, z). Thus
(x, y, z)+(u, v, q) = (x, y, z, u, v). From Lemma 3.6 it follows that pd(S/L) ≤
4. Hence pd(S/I) ≤ 5 by Lemma 2.6.
Now if I un = (x, y, z) ∩ (u, v, q), we may again assume that q ∈ (x, y, z),
say q = ax + by + cz. If ht(x, y, z, u, v) ≤ 4, then I un contains a linear

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form and we are done. We may then assume that ht(x, y, z, u, v) = 5 and so
I un = (q, xu, xv, yu, yv, zu, zv). Let ∂i denote the ith differential map in the
minimal free resolution of I un . By Lemma A.17, we have pd(Coker(∂4 ∗)) ≤
5. Hence by Lemma 3.7, pd(S/I) ≤ 5.


Lemma 7.3. If I un is of type h1, 1; 1, 2i, then pd(S/I) ≤ 6. If L is of type
h1, 1; 1, 2i, then pd(S/I) ≤ 6 or the quadrics in L can be expressed in terms
of at most 6 variables.

Proof. Suppose that L or I un is (u, v, w) ∩ L2 , where L2 is as in Proposition 4.3. First assume L2 = (x, y, z 2 ). If L = (u, v, w) ∩ (x, y, z 2 ), then since
pd(S/(u, v, w, x, y, z 2 )) ≤ 6 then we have pd(S/L) ≤ 5 by Lemma 3.6. Thus
pd(S/I) ≤ 6 by Lemma 2.6. If I un = (u, v, w) ∩ (x, y, z 2 ), then pd(S/I) ≤ 6
by Corollary 3.9.
Assume then that L2 = (x, y 2 , yz, z 2 , ay + bz). By Lemma 3.4 we may
assume that ht(u, v, w, x) = 4 or else (u, v, w) ∩ L2 contains a linear form.
We may further assume that deg(a) = deg(b) = 1 and that ay + bz ∈
(u, v, w) ∩ L2 , or else all quadrics in (u, v, w) ∩ L2 are expressible in terms
of u, v, w, x, y, z. If ht(u, v, w, x, y, z) = 6, then (u, v, w) ∩ L2 = (ay + bz) +
(u, v, w)(x, y 2 , yz, z 2 ) all of whose quadrics are contained in the height 2
ideal (ay + bz, x), contradicting that ht(q1 , q2 , q3 ) = 3. So we must have
ht(u, v, w, x, y, z) ≤ 5 and may assume that y = u. Since au+bz ∈ (u, v, w)∩
L2 ⊆ (u, v, w), we must have b ∈ (u, v, w) or z ∈ (u, v, w). In either case
ht(u, v, w, x, y, z, a, b) ≤ 6. Hence all quadrics in (u, v, w)∩L2 are expressible
in terms of at most 6 variables.
Suppose L2 = (x, y, z)2 + (ax + by + cz, dx + ey + f z). After rewriting, we
may assume that ax + by + cz, dx + ey + f z ∈ (u, v, w), since otherwise all
quadrics in (u, v, w) ∩ L2 could be expressed in terms of at most 6 variables.
Thus (u, v, w) + L2 = (u, v, w) + (x, y, z)2 , which has projective dimension
at most 6. If L = (u, v, w) ∩ L2 , then pd(S/L) ≤ 5 by Lemma 3.6, and hence
pd(S/I) ≤ 6 by Lemma 2.6.
The last case to consider is when I un = (u, v, w)∩L2 . We may still assume
ax + by + cz, dx + ey + f z ∈ (u, v, w). If ht(u, v, w, x, y, z) = 6, it follows
that a, b, c, d, e, f ∈ (x, y, z, u, v, w). Thus all quadrics in I un are expressible
in terms of 6 variables. If ht(u, v, w, x, y, z) = 5, we may assume that u = x.
In the above situation, we can take I = (xℓ1 , xℓ2 , ax+by +cz, dx+ey +f z)
and e(S/I) = 3 for linear forms ℓ1 , ℓ2 . But clearly I + (x) = (x, by + cz, ey +
f z). Since ht(I2 ) = 2, where I2 is as in Proposition 4.3, we must have
ht(I + (x)) = 3. Therefore e(S/(I + (x))) = 4. This cannot happen since
I + (x) ⊇ I and e(S/I) = 3.
If ht(u, v, w, x, y, z) = 4, we may assume that u = x and v = y. As
above, we may assume ax + by + cz, dx + ey + f z ∈ (x, y, w), hence we
may assume c = w and f = 0, since w, x, y, z is a regular sequence (the
case where both c, f are non-zero multiples of w can be reduced to the
above, by taking a linear combination of the two generators, and the case

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19

where c = f = 0 cannot happen otherwise the height of the ideal of minors
I2 is 1 modulo (x, y, z), where I2 is as in Proposition 4.3). It follows that
I un = (x2 , xy, xz, y 2 , yz, z 2 w, ax+by +wz, dx+ey). By the height restriction
on I2 , we have ht(x, y, z, d, e) = 5. By Lemma A.18, pd(Coker(∂4 ∗)) ≤ 5,
where ∂4 is the 4th map in the free resolution of I un . By Proposition 3.7,
pd(S/I) ≤ 5.

Lemma 7.4. If I un is type h1, 1, 1; 1, 1, 1i, then pd(S/I) ≤ 6. If L is of type
h1, 1, 1; 1, 1, 1i, then pd(S/I) ≤ 6 or the quadrics in L can be expressed in
terms of at most 6 variables.
T
Proof. Write J = 3i=1 Li where Li = (xi , yi , zi ). If ht(Li + Lj ) = 6 for any
i 6= j, then all quadrics in J are expressible in terms of xi , yi , zi , xj , yj , zj .
Otherwise, ht(Li + Lj ) ≤ 5 for all pairs i 6= j. Unless there is a linear form
in J, this forces ht(L1 + L2 + L3 ) ≤ 6. Thus either I un or L contain a linear
form, yielding pd(S/I) ≤ 5 by Lemma 3.4, or all generators are expressible in
terms of at most 6 variables. This means pd(S/L) ≤ 5, L being an unmixed
ideal extended from a 6 variable polynomial ring, and hence pd(S/I) ≤ 6
by Lemma 2.6. The case for I un follows from Collolary 3.9.

Proposition 7.5. If e(S/I) = 3, then pd(S/I) ≤ 6. If e(S/I) = 5, then
pd(S/I) ≤ 6 or all quadrics in L = (q1 , q2 , q3 ) : I can be expressed in terms
of at most 6 variables.
Proof. If e(S/I) = 3, then I un or L = (q1 , q2 , q3 ) : I is of one of the five types
listed at the beginning of the section. If either I un or L is of type h3; 1i, then
it contains a linear form by Theorem 2.10. By Lemma 3.4, pd(S/I) ≤ 5.
The other four cases are covered by Lemmas 7.1, 7.2, 7.3 and 7.4.

8. Multiplicity 4
In this section we cover the case when e(S/I) = 4. Recall that I =
(q1 , q2 , q3 , q4 ), ht(I) = ht(q1 , q2 , q3 ) = 3. We again set L = (q1 , q2 , q3 ) : I and
note that L is unmixed and e(S/L) = 4. Hence our strategy for bounding
the projective dimension of I is one of the following:
(1) Show pd(S/L) ≤ 5, yielding pd(S/I) ≤ 6 by Lemma 2.6.
(2) Show that L contains a linear form and hence pd(S/I) ≤ 5 by
Lemma 3.4.
(3) Show that all quadrics in L can be written in terms of at most 6
variables.
The last case is completed in the following section where we show that if
e(S/I) = 4 or 5 and any complete intersection of quadrics in I can be written
in terms of at most 6 variables, then pd(S/I) ≤ 6.
We handle several of the cases for the type of L separately in the following
lemmas.
Lemma 8.1. If e(S/L) = 4 and L is of type h1, 3; 1, 1i, then pd(S/I) ≤ 5.

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Proof. We have L = (u, v, w) ∩ L2 , where L2 is generated by a linear form x
and the 2 × 2 minors of a 2 × 3 matrix M of linear forms. We may assume
that ht(x, u, v, w) = 4, since otherwise L contains a linear form. We may
further assume that there are two quadrics ∆1 , ∆2 ∈ I2 (M) ∩ (u, v, w) that
are linearly independent in modulo (x), since otherwise the quadrics in L
generate an ideal of height at most 2. It follows that J = (u, v, w) + L2 is
generated by 4 linear forms and a quadric form. Hence pd(S/J) ≤ 5. By
Lemma 3.6, pd(S/L) ≤ 4 and by Lemma 2.6, pd(S/I) ≤ 5.

Lemma 8.2. If e(S/L) = 4 and L is of type h1, 1; 1, 3i, then pd(S/I) ≤ 6
or the quadrics in L can be written in terms of at most 6 variables.
Proof. Here L = (u, v, w) ∩ L2 , where u, v, w ∈ S1 and L2 is as in Proposition 4.4. If L2 = (x, y 2 , yz, z 2 ), then (u, v, w) + L2 = (u, v, w, x, y 2 , yz, z 2 ).
Hence pd(S/(L2 + (u, v, w))) ≤ 6 and by Lemmas 3.6 and 2.6 we have
pd(S/I) ≤ 6.
If L2 = (x, y 2 , yz, z 3 , ay + z 2 ), then again we consider L2 + (u, v, w). If
ht(u, v, w, x, y, z, a) = 7, then the quadrics in L are contained in (x), which
is not possible. If ht(u, v, w, x, y, z, a) ≤ 6, then pd(S/((u, v, w) + L2 )) ≤ 6
and we are done similarly to the above.
If L2 = (x2 , xy, xz, y 2 , yz, ax + by + z 2 , cx + dy), then after a linear change
of coordinates we may assume that ax + by + z 2 , cx + dy ∈ (u, v, w), since
otherwise all quadrics in L are expressible in terms of at most 6 variables.
It follows that pd(S/((u, v, w) + L2 )) ≤ 6 and hence pd(S/I) ≤ 6. A similar
argument works if L2 is as in cases (4) and (5) of Proposition 4.4. In cases
(6) or (7) there is nothing to show.

Lemma 8.3. Suppose e(S/L) = 4 and L is of type h2, 2; 1, 1i, h1, 2; 2, 1i,
h1, 1; 2, 2i, h1, 1, 2; 1, 1, 1i, h1, 1, 1; 1, 1, 2i, or h1, 1, 1, 1; 1, 1, 1, 1i. Then either
pd(S/I) ≤ 6 or any three quadrics in L can be written in terms of at most
6 variables.
Proof. By the associativity formula 2.1, an unmixed height three ideal of
multiplicity 2 is either prime, primary to a linear prime, or the intersection
of two linear primes. Then, by the results of the previous section, it has one
of the following 5 forms:
• (x, y, q), where x, y ∈ S1 and q ∈ S2 (this includes the case (x, y, z) ∩
(x, y, w))
• (x, a, b) ∩ (x, c, d), where x, a, b, c, d ∈ S1 and ht(x, a, b, c, d) = 5
• (u, v, w)∩ (x, y, z), where u, v, w, x, y, z ∈ S1 and ht(u, v, w, x, y, z) =
6
• (x, y 2 , yz, z 2 , ay + bz), where x, y, z, a, b ∈ S1 , ht(x, y, z, a, b) = 5
• (x, y, z)2 +(ax+by +cz, dx+ey +f z), where x, y, z, a, b, c, d, e, f ∈ S1
and ht(x, y, z, I2 ) = 5, where I2 denotes the ideal of 2 × 2 minors as
in Proposition 4.3.
Therefore, an ideal L as in the statement can be written as L = L1 ∩ L2 ,
where L1 and L2 are ideals of one of the above 5 forms.

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21

Note that if either L1 or L2 fall into the third case, then all quadrics in L
can be expressed in terms of 6 variables. So we may ignore this case. In the
fourth case, we add the assumption that a, b ∈ S1 since if deg(a), deg(b) ≥ 2,
then q1 , q2 , q3 ∈ (x, y 2 , yz, z 2 ) and require at most 6 variables to express. We
handle each of the other 10 possible pairings separately.
Case I. L1 = (x, y, q), L2 = (w, z, q ′ ).
Here L1 + L2 is generated by at most 4 linear forms and 2 quadric forms.
Hence pd(S/(L1 + L2 )) ≤ 6. By Lemma 3.6, pd(S/L) ≤ 5. By Lemma 2.6,
pd(S/I) ≤ 6.
Case II. L1 = (x, y, q), L2 = (z, ac, ad, bc, bd).
By Lemma 3.5, we may assume that q ∈ L2 . If ht(x, y, z, a, b, c, d) = 7, then
all quadrics in L are contained in (z, q), contradicting that ht(q1 , q2 , q3 ) = 3.
If ht(x, y, z, a, b, c, d) ≤ 6, then pd(S/(L1 + L2 )) ≤ 6 and hence pd(S/I) ≤ 6
by Lemmas 3.6 and 2.6.
Case III. L1 = (x, y, q), L2 = (z, v 2 , vw, w2 , av + bw).
This is very similar to the previous case. We may again assume q ∈ L2 .
If ht(x, y, z, v, w, a, b) = 7, then all quadrics in L are contained in (z, q), a
contradiction. Otherwise ht(x, y, z, v, w, a, b) ≤ 6, in which case pd(S/(L1 +
L2 )) ≤ 6, whence pd(S/I) ≤ 6 as before.
Case IV. L1 = (u, v, q), L2 = (x, y, z)2 + (ax + by + cz, dx + ey + f z).
After a possible linear change of variables, we may assume ax + by + cz, dx +
ey + f z ∈ L1 , or else all quadrics in L can be written in terms of at most
6 variables. Then L1 + L2 = (x, y, z)2 + (u, v, q) with q ∈ (x, y, z). Write
q = a′ x + b′ y + c′ z. If ht(x, y, z, u, v, a′ , b′ , c′ ) ≤ 6, then pd(S/(L1 + L2 )) ≤
6 and we are done. If ht(x, y, z, u, v, a′ , b′ , c′ ) ≥ 7, then one checks that
pd(S/(L1 + L2 )) = 6.
Case V. L1 = (x, ac, ad, bc, bd), L2 = (y, eg, eh, f g, f h).
Here L is the intersection of four linear primes. If pairwise, the height of 6
linear generators of 2 primes is 6, then all quadrics in L could be written
in terms of those 6 linear forms only. So we may assume that every pair of
linear primes has a linear form in its intersection. If no three linear primes
have a common linear form, then we must have 4 more linear dependencies
among the above 10 linear forms. It follows that pd(S/L) ≤ 5, as L is an
unmixed, height three ideal whose generators live in a polynomial ring with
6 variables.
If three of these primes have a common linear form, we may assume
that either x = y or x = e. In the first case, L contains a linear form
and we are done by Lemma 3.4. In the second case, we still must have
ht(y, g, h, x, a, b) ≤ 5, ht(y, g, h, x, c, d) ≤ 5. This forces at least two more
linear dependencies which, after a linear change of variables, we may take
to be one of the following cases:
• a = y, c = g
• a = g, c = h
• a = g, c = y + g

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One checks that in each of these cases, if there are no remaining linear dependencies among the linear generators, then all quadrics in L are contained
in (x, y), contradicting that L contains a height 3 ideal of quadrics. Hence
there is a fourth linear dependence among the 10 variables. It follows again
that all quadrics in L can be written in terms of at most 6 variables, hence
pd(S/I) ≤ 6.
Case VI. L1 = (w, ac, ad, bc, bd), L2 = (x, y 2 , yz, z 2 , ey + f z).
If w ∈ (x, y, z), after a linear change of variables we may assume that w = x
or w = y. If w = x, then L contains a linear form. If w = y, then x ∈
(y, a, b, c, d) (otherwise all quadrics in L are contained in (x, z), a contradiction), and similarly z ∈ (y, a, b, c, d). So L1 +L2 = (y, ac, ad, bc, bd, x, z 2 , f z),
whose generators require at most 6 linear forms to express. If follows that
pd(S/(L1 + L2 )) ≤ 6, whence pd(S/I) ≤ 6.
If w ∈
/ (x, y, z) and if ht(x, y, z, w, a, b) = 6 or ht(x, y, z, w, c, d) = 6, then
all quadrics can be expressed in terms of 6 variables. Hence we may assume
that ht(x, y, z, w, a, b) ≤ 5 and ht(x, y, z, w, c, d) ≤ 5 and without loss of
generality that x = a and y = c. Then L ⊆ (w, xy, xd, by, bd) ∩ (x, y, z) ⊆
(wx, wy, wz, xy, xd, by, bd) and again all quadrics in L are expressible in
terms of at most 6 variables.
Case VII.
L1 = (s, tv, tw, uv, uw), L2 = (x, y, z)2 + (ax + by + cz, dx + ey + f z).
If ht(s, x, y, z) = 4, then all quadrics in L are in (ax + by + cz, dx + ey + f z),
a contradiction on the height of L. Hence we may assume that s = x after a linear change of variables. As usual we may assume that ax + by +
cz, dx + ey + f z ∈ L1 . Then L1 + L2 = (x, tv, tw, uv, uw, y 2 , yz, z 2 ). If
ht(t, u, v, w, x, y, z) = 7, then one checks that pd(S/(L1 + L2 )) = 6. Otherwise, one has pd(S/(L1 + L2 )) ≤ 6 and hence pd(S/I) ≤ 6 by Lemmas 3.6
and 2.6.
Case VIII. L1 = (x, y 2 , yz, z 2 , ay + bz), L2 = (u, v 2 , vw, w2 , cv + dw).
After a change of variables, we may suppose cv + dw ∈ L1 , or else any complete intersection of quadrics in L1 ∩ L2 would be expressible in terms of at
most 6 variables. Hence L1 + L2 ⊆ (x, y 2 , yz, z 2 , ay + bz, u, v 2 , vw, w2 ). By
symmetry, we can also assume ay +bz ∈ (u, v, w). If ht(a, b, y, z, u, v, w) ≥ 6,
then we have a contradiction. Therefore, ht(x, y, z, a, b, u, v, w) ≤ 6 and
pd(S/(L1 + L2 )) ≤ 6. It follows from Lemmas 3.6 and 2.6 that pd(S/I) ≤ 6.
Case IX.
L1 = (u, v 2 , vw, w2 , gv + hw), L2 = (x, y, z)2 + (ax + by + cz, dx + ey + f z).
This case is similar to the previous one. After a change of variables, we may
assume that ax + by + cz, dx + ey + f z ∈ L1 , or else all quadrics in L
are expressible in terms of at most 6 linear forms. Similarly, we may assume that gv + hw ∈ (x, y, z). This forces ht(g, h, v, w, x, y, z) ≤ 5 and so
ht(u, v, w, g, h, x, y, z) ≤ 6. Once again we have pd(S/I) ≤ 6.
Case X. L1 = (x, y, z)2 + (ax + by + cz, dx + ey + f z),
L2 = (u, v, w)2 + (a′ x + b′ y + c′ z, d′ x + e′ y + f ′ z).

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

23

As before, we may assume ht(u, v, w, x, y, z) ≤ 5 and thus that x = u. We
may also assume that ax + by + cz, dx + ey + f z ∈ L2 and symmetrically
that a′ x + b′ y + c′ z, d′ x + e′ y + f ′ z ∈ L1 , after a change of variables. Suppose
ht(u, v, w, x, y, z) = 5, then
L1 +L2 = (x2 , xy, xz, y 2 , yz, z 2 , xv, xw, v 2 , vw, w2 , ax+by +cz, dx+ey +f z).
Moreover (L1 +L2 ) : x = L1 : x+L2 : x = (x, y, z, v, w) and (L1 +L2 )+(x) =
(x, y 2 , yz, z 2 , v 2 , vw, w2 , by + cz, ey + f z). Clearly pd(S/(x, y, z, v, w)) ≤
5. As ax + by + cz, dx + ey + f z ∈ (u, v, w), it follows that b, c, e, f ∈
(u, v, w, x, y, z). Hence pd(L1 +L2 +(x)) ≤ 5. Therefore pd(S/(L1 +L2 )) ≤ 5
and pd(S/I) ≤ 5 by Lemmas 3.6 and 2.6.
If ht(u, v, w, x, y, z) = 4, we may further suppose that v = y. It follows
that we may assume c = w and f = 0 after a linear change of coordinates
and generators. Hence L1 + L2 = (x2 , xy, xz, y 2 , yz, z 2 , xw, yw, w2 , ax + by +
wz, dx + ey). If ht(x, y, z, w, d, e) ≥ 5, one checks that pd(S/(L1 + L2 )) = 5;
otherwise, L1 + L2 can be expressed in terms of at most 6 variables. In
either case, pd(S/I) ≤ 6.

Proposition 8.4. Suppose e(S/L) = 4. Then pd(S/I) ≤ 6 or any three
quadrics in L can be written in terms of at most 6 variables.

Proof. There are the following 11 possibilities for the type of L:
(1) h4; 1i
(2) h2; 2i
(3) h1; 4i
(4) h1, 3; 1, 1i
(5) h1, 1; 1, 3i
(6) h2, 2; 1, 1i
(7) h1, 2; 2, 1i
(8) h1, 1; 2, 2i
(9) h1, 1, 2; 1, 1, 1i
(10) h1, 1, 1; 1, 1, 2i
(11) h1, 1, 1, 1; 1, 1, 1, 1i
The last 6 of these cases are covered in Lemma 8.3. The cases h1, 3; 1, 1i and
h1, 1; 1, 3i are covered in Lemmas 8.1 and 8.2, respectively. If L is of type
h4; 1i, it is prime and one of the first 3 cases of Theorem 2.11. All three are
Cohen-Macaulay. Hence pd(S/I) ≤ 4 by Lemma 2.6. If L is of type h2; 2i
or h1; 4i, then by Propositions 4.5 and 5.1, either pd(S/L) ≤ 5, all quadrics
in L can be expressed in terms at most 6 variables, or L contains a linear
form.

9. The Remaining Cases of Multiplicity 4 and 5
Throughout this section I = (q1 , q2 , q3 , q4 ) is an ideal generated by 4
quadrics. By the previous sections, we have proved the main result in all
cases except when e(S/I) = 4 or 5. In those cases where we have not
already produced the desired bound on pd(S/I), we have that any complete

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intersection of quadrics in I can be written in terms of at most 6 variables.
(Different choices of the complete intersection may lead to different sets of
6 variables.) Indeed after taking a linear combinations of the generators,
we may assume that q1 , q2 , q3 is the complete intersection in question and
apply Prop. 7.5 or Prop. 8.4. While it is tempting to conclude that this
assumption should imply that all four quadrics are expressible in terms of
at most 6 linear forms, the following example shows this is not true.
Example 9.1. Consider the ideal I = (ax + y 2 , bx + yz, cx + z 2 , dx) in
S = k[a, b, c, d, x, y, z]. Note that ht(I) = 3 and I is generated by four
quadric forms. It is easy to check that any complete intersection of quadrics
q1 , q2 , q3 in I can be expressed in terms of at most 6 linear forms (x, y and z
and the three linear coefficients of x), but it requires all 7 variables to write
all four minimal generators of I.
For the remainder of this section, we assume that any complete intersection of three quadrics is extended from a polynomial ring in at most 6
variables and e(S/I) = 4 or 5.
Lemma 9.2. If I contains a product of two linear forms, then pd(S/I) ≤ 6.
Proof. Assume xy ∈ I for linear forms x, y so that I = (xy, q1 , q2 , q3 ). Then
after possibly replacing the qi with a linear combination of the qj and xy,
we may further assume ht(q1 , q2 , q3 ) = 3 and then, by the above, we may
assume q1 , q2 , q3 ∈ S ′ = k[z1 , . . . , z6 ].
Set C = (q1 , q2 , q3 )S and observe that I : x = (C : x) + (y). Then we
have the short exact sequence
(1)

0 −→ S/(I : x) −→ S/I −→ S/(I + (x)) −→ 0

If x is regular on S/C, then the first and last term of (1) are of the form
C + (ℓ) where ℓ is a linear form. Write C + (ℓ) as C1 + (ℓ), with ℓ regular on
S/C1 . Then pd(S/(C + (ℓ))) = 1 + pd(S/C1 ). The ideal C1 need not be a
complete intersection but is generated by 3 quadrics, hence by Theorem 2.12,
we have pd(S/C1 ) ≤ 4. Then the first and last term of (1) have projective
dimension at most 5, and by Depth Lemma we obtain pd(S/I) ≤ 5.
We may then assume x and y are not regular on S/C. Since C is extended from S ′ and x and y are not regular on S/C, it follows that x, y
are both contained in some associated prime of (q1 , q2 , q3 )S, all of which are
extended from S ′ by [4, Exercise 4.7]. Since x, y are linear forms we have
ht(x, y, z1 , . . . , z6 ) = 6. Then I itself is extended from S ′ = k[z1 , . . . , z6 ] and
then pd(S/I) ≤ 6.

Lemma 9.3. If I un is radical, then pd(S/I) ≤ 6.

Proof. Let q1 , q2 , q3 be a regular sequence of quadrics in I, by assumption
there is a polynomial ring S ′ = k[z1 , . . . , z6 ] such that q1 , q2 , q3 ∈ S ′ . Let
C ′ = (q1 , q2 , q3 )S ′ and C = C ′ S and observe that all the associated primes
of S/C are extended from S ′ . Since C is a complete intersection in I un it

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

25

follows that Ass(S/I un ) ⊆ Ass(S/C) yielding that all associated primes of
S/I un are extended from S ′ . Since I un is radical then I un itself is extended
from S ′ and then, by Proposition 3.9, it follows that pd(S/I) ≤ 6.

Lemma 9.4. If there exist quadrics Q1 , Q2 , Q3 and linear forms x, y such
that I ⊆ (x, Q1 , Q2 , Q3 ) or I ⊆ (x, y)2 + (Q1 , Q2 , Q3 ), then pd(S/I) ≤ 6.
Proof. In the first case, some linear combination of the 4 quadric generators
of I is xy. The statement now follows by Lemma 9.2. In the second case,
some linear combination of the 4 quadric generators of I is contained in
(x, y)2 . Since k = k, every such quadric can be written as the product of
two linear forms. We are again done by Lemma 9.2.

Lemma 9.5. Let p be a linear prime, H a p-primary ideal. If x is a linear
form in p which is not contained in H then H : x and (H + (x))un are
p-primary and e(S/H) = e(S/H : x) + e(S/(H + (x))).
Proof. Since Ass(S/(H : x)) ⊆ Ass(S/H), it is clear that H : x is p-primary;
(H+(x))un is p-primary because H is unmixed, of equal height and contained
in it. The second statement follows by additivity of e(−) applied to the
following short exact sequence where all modules have the same dimension:
0 −→ S/(H : x) −→ S/H −→ S/(H + (x)) −→ 0

together with the well-known fact that e(S/(H, x)) = e(S/(H + (x))un ).



Proposition 9.6. If e(R/I) = 4 or 5 and every complete intersection of
3 quadrics in I can be written in terms of at most 6 linear forms, then
pd(R/I) ≤ 6.

Proof. Consider the primary components of I un . The hypotheses allow us
to consider three cases: when I un is radical or has components of the form
(x, y, z 2 ), when I un has a component primary to a multiplicity 2 prime that
is not itself prime, and when I un has a component primary to a linear prime
which is neither prime nor of the form (x, y, z 2 ).
Case 1: There is a primary component J of type h2; 2i.
Then J has one of the 4 forms listed in Proposition 5.1. All four cases are
covered by Lemma 9.4.
Case 2: There is a primary component J of type h1; ei, with e ≥ 2.
First consider the case when e = 3. Then J has one of the 7 forms in
Proposition 4.4. Cases (1), (2), (4) and (5) are covered by Lemma 9.4.
Cases (6) and (7) easily follow as well. In the remaining case (3), we have
J = (x2 , xy, xz, y 2 , yz, ax+by+z 2 , cx+dy). By expressing q1 , . . . , q4 in terms
of the minimal generators of I and taking linear combinations we may ensure
that q4 ∈ (x, y, z)2 and that q1 , q2 , q3 for a complete intersection. Hence we
may assume that q1 , q2 , q3 are extended from a polynomial ring R in at
most 6 variables and since (x, y, z) minimally contains (q1 , q2 , q3 ), the prime
(x, y, z) is also extended from R. Hence q4 is extended from R and we are
done by Lemma 3.8.

CRAIG
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HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU

If e = 5, then let p = (x, y, z) be the linear prime for which J is p-primary.
Let ℓ ∈ p − J. By Lemma 9.5, J : ℓ and J + (ℓ) are p-primary and e(S/(J :
ℓ)) + e(S/(J + (ℓ))) = 5. Either one of these multiplicities is 3 and we are
done by the previous argument, or one of these multiplicities is 4 and we
reduce to the the next case.
If e = 4, we consider J : p, which contains J and is p-primary with
strictly smaller multiplicity. If J : p contains a linear form ℓ, then J : ℓ = p.
Hence e(S/(J + (ℓ))) = 3 and we are again done by a previous case. If
J : p does not contain a linear form, then either e(S/(J : p)) = 3 or all
quadrics in J : p, and hence I, can be written in terms of 6 variables,
in which case we are done again. Otherwise by Proposition 4.3, J : p =
(x, y, z)2 + (ax + by + cz, dx + ey + f z), where a, . . . , f ∈ S1 . For any ideal
I generated by 4 quadrics contained in this ideal, either the generators are
expressible in terms of 6 linear forms, or without loss of generality we can
take q1 = ax + by + cz + q1′ , q2 = dx + ey + f z + q2′ , where q1′ , q2′ , q3 , q4 ∈
(x, y, z)2 . Since (x, y, z) ∈ Ass(S/I) ⊆ Ass(S/(q1 , q2 , q3 )), and since q1 , q2 , q3
are extended from a 6 variable polynomial ring, we may take x, y, z to be 3
of these variables. It follows that all the generators of I can be written in
terms of at most 6 variables and so pd(S/I) ≤ 6.
If e = 2, then J is of one of the 4 forms in Proposition 4.3. The previous
arguments cover cases (3) and (4). In case (2), J = (x, y 2 , yz, z 2 , ay + bz)
with ht(x, y, z, a, b) = 5. Either I contains a monomial or else we can write
I = (q1 , q2 , q3 , q4 ) with q1 = c1 x + y 2 , q2 = c2 x + yz, q3 = c3 x + z 2 and
q4 = c4 x+ay+bz. If ht(c1 , c2 , c3 , c4 , x, y, z, a, b) = 5, we are done. Otherwise,
after possibly taking a linear combination of the generators we may assume
ht(c4 , x, y, z, a, b) = 6 and (q1 , q2 , q4 ) and (q1 , q3 , q4 ) form a compete intersections. It follows that all 4 quadrics are extended from k[c4 , x, y, z, a, b]
and hence pd(S/I) ≤ 6.
Case 3: Every primary component of I un is prime or is equal to (x, y, z 2 ).
Since Ass(S/I un ) ⊆ Ass(S/(q1 , q2 , q3 )), each associated prime of I un is extended from a single polynomial ring R′ in at most 6 variables by Proposition 3.9. In the case of a component of the form (x, y, z 2 ), the prime
ideal (x, y, z) is extended from R′ and hence (x, y, z 2 ) is as well. Since
all primary components of I un are extended from R′ it follows from that
pd(S/I) ≤ 6.

10. Main Result
Theorem 10.1. Let S by a polynomial ring over a field K, and let I =
(q1 , q2 , q3 , q4 ) be an ideal of S minimally generated by 4 homogeneous polynomials of degree 2. Then pd(S/I) ≤ 6. Moreover, this bound is tight.
Proof. By extension of scalars, we may assume that K is algebraically closed.
As previously noted, ht(I) ≤ 4. If ht(I) = 1 or 4, then pd(S/I) = 4. If
ht(I) = 2, then pd(S/I) ≤ 6 by Theorem 2.12. So we may assume that
ht(I) = 3.

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

e(S/I)
1
2

3

6

he; λi for I un
h1; 1i
h2; 1i
h1; 2i
h1, 1; 1, 1i
h3; 1i
h1; 3i
h1, 2; 1, 1i
h1, 1; 1, 2i
h1, 1, 1; 1, 1, 1i
any

27

Bound for pd(S/I) Justification
4
Proposition 3.3
4
Lemma 3.2
5
Proposition 6.1
5
Proposition 6.1
5
Lemma 3.4
6
Lemma 7.1
6
Lemma 7.2
6
Lemma 7.3
6
Lemma 7.4
3
Theorem 2.13

e(S/L) he; λi for L
Bound for pd(S/I)
4
h4; 1i
4
h2; 2i
5
h1; 4i
6
h1, 3; 1, 1i
5
h1, 1; 1, 3i
6
h2, 2; 1, 1i
6
h1, 2; 2, 1i
6
h1, 1; 2, 2i
6
h1, 1, 2; 1, 1, 1i
6
h1, 1, 1; 1, 1, 2i
6
h1, 1, 1, 1; 1, 1, 1, 1i 6
3
h3; 1i
5
h1; 3i
6
h1, 2; 1, 1i
5
h1, 1; 1, 2i
6
h1, 1, 1; 1, 1, 1i
6
Table 1. Bounds on pd(S/I) where I is a
generated by 4 quadrics

Justification
Lemma 2.6
Proposition 5.1*
Proposition 4.5*
Proposition 8.1
Proposition 8.2*
Proposition 8.3*
Proposition 8.3*
Proposition 8.3*
Proposition 8.3*
Proposition 8.3*
Proposition 8.3*
Lemma 3.4
Lemma 7.1*
Lemma 7.2
Lemma 7.3*
Lemma 7.4*
height 3 ideal

By Theorem 2.13, we have e(S/I) ≤ 6. Moreover, if e(S/I) = 6, then
pd(S/I) = 3. The cases where e(S/I) = 1, 2, 3, 4, 5 are proved in Propositions 3.3, 6.1, 7.5, 8.4 and 9.6.

A summary of the many cases in the proof of Theorem 10.1 is given in
Table 1. The Lemma or Proposition covering each case is referenced on the
right. Asterisks denote cases that are also covered by Proposition 9.6.
We remark that there are canonical examples of ideals I2 and I3 generated by 4 quadrics of heights 2 and 3, respectively, and with pd(S/I2 ) =
pd(S/I3 ) = 6. Namely, we can take I2 = (x2 , y 2 , ax + by, cx + dy) and
I3 = (x2 , y 2 , z 2 , ax + by + cz) in the polynomial ring S = k[a, b, c, d, x, y, z].

CRAIG
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HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU

These correspond to the examples I2,2,2 and I3,1,2 from [19]. More generally,
we previously posed the following question:
Question 10.2 ( [15, Question 6.2]). Let S be a polynomial ring and let I
be an ideal of S generated by n quadrics and having ht I = h. Is it true that
pd(S/I) ≤ h(n − h + 1)?
There are examples in [19] achieving this bound for all possible integers
h and n. This paper now answers this question affirmatively if n ≤ 4, while
our previous paper [15] gave an affirmative answer if h ≤ 2. Note that if
the question has a positive answer, it would give a bound on the projective
dimension of ideals generated by n quadrics that is quadratic in n – much
smaller than the known bounds of Ananyan-Hochster.
Appendix A. Resolutions of Primary Ideals
In this section we collect the details of the many unmixed and primary
ideals listed in the structure theorems in Sections 4 and 5. The technique is
the same for each of the following: one resolves the given ideal generically
and uses the Buchsbaum-Eisenbud exactness criteria to check the conditions that ensure the resolution is exact. If F• is the resolution of S/I
and ∂i denotes the ith differential map,
P then this amounts to checking that
ht(Irj (∂j )) ≥ j for all j. Here rj = pi=j (−1)p−i rank(Fi ) denotes the expected rank of the jth map, and Ir (∂j ) denotes the ideal of r × r minors
of the matrix associated to the jth differential. Moreover one checks that
ht(Irj (∂j )) ≥ j + 1 for j > ht(I) to ensure that the ideal is unmixed (cf. [17,
Proposition 2.4]). In the case that ht(I) = 3, this amounts to showing that
(
j
if j = 1, 2, 3
ht(Irj (∂j )) ≥
.
j + 1 if j ≥ 4
We do this explicitly for Lemmas A.1 and A.2. The reader may check
the remaining ones with the help of a computer algebra system, such as
Macaulay2 [13]. We have posted a supplementary Macaulay2 [13] file at
http://www.math.unl.edu/~aseceleanu2/research/fourQuadrics.m2
Throughout this section x, y, z represent independent linear forms so that
p = (x, y, z) is a height three linear prime ideal.
Lemma A.1. If
L = (x2 , y 2 , z 2 , xyz, Cxy + Bxz + Ayz),
where ht(x, y, z, A, B, C) ≥ 5, then L is (x, y, z)-primary, pd(S/L) = 4 and
e(S/L) = 6.
Proof. Consider the complex
So

∂1

S5 o

∂2

S9 o

∂3

S6 o

∂4

S1 o

0,

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

29


where∂1 = x2 y 2 z 2 Cxy + Bxz + Ayz xyz

−Cy − Bz
0
−y 2
0
−yz
0
−z 2
0
0

0
−Cx − Az x2
0
0
−xz
0
0
−z 2 


2

0
0
0
Bx + Ay
0
0
x
−xy y 2 
∂2 = 


x
y
0
−z
0
0
0
0
0 
−A
−B
0
C
x
y
0
z
0


−y −z
0
0
0
0
 x
0
−z
0
0
0 



C
0
0
z
0
0


 0
−x −y 0
0
0 


C
0 −y −z 0 
∂3 = 

B
−A 0
C
x
0 −z 



 0
B
0
0
y
0


 0 −A −B 0
x
y 
0
0
A
0
0
x


−z
 y 


 −x 


∂4 = 

C


−B 
A
Note that x2 ∈ I1 (∂1 ), x6 , y 6 ∈ I4 (∂2 ) and x5 , y 5 , z 5 ∈ I5 (∂3 ) and I1 (∂4 ) =
(x, y, z, A, B, C). It follows that this √
is exact and forms a resolution of L,
pd(S/L) = 4 and L is unmixed. As L = p and λ(Sp /Lp ) = 6, it follows
that L is (x, y, z)-primary and e(S/L) = 6.

Lemma A.2. If
J = (x, y, z)2 + (ax + by + cz, dx + ey + f z),



a b c
where ht (x, y, z) + I2
= 5, then J is (x, y, z)-primary, pd(S/J) =
d e f
4 and e(S/J) = 2. Moreover,
L = (x2 , y 2 , z 2 ) : J
= (x2 , y 2 , z 2 , xyz, (ae − bd)xy + (af − cd)xz + (bf − ce)yz)
and pd(S/L) = 4.
Proof. Consider the complex
So

∂1

S8 o

∂2

S 14 o

∂3

S9 o

∂4

S2 o

0,

where

∂1 = ax + by + cz dx + ey + f z x2 xy y 2 xz yz z 2

CRAIG
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HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU



−x 0 −y 0
0
0 −z 0
0
0
0
0
0
0
 0 −x 0 −y 0
0
0 −z 0
0
0
0
0
0


 a
d
0
0 −y 0
0
0 −z 0
0
0
0
0


 b

e
a
d
x
−y
0
0
0
−z
0
0
0
0

∂2 = 
 0
0
b
e
0
x
0
0
0
0
0 −z 0
0


 c
f
0
0
0
0
a
d
x
y −y 0 −z 0 


 0
0
c
f
0
0
b
e
0
0
x
y
0 −z 
0
0
0
0
0
0
c
f
0
0
0
0
x
y


y
0
z
0
0
0
0
0
0
 0
y
0
z
0
0
0
0
0 


−x 0
0
0
z
0
0
0
0 


0
0
z
0
0
0 
 0 −x 0


d
0
0
0
0
z
0
0 
 a


e
0
0
0
0
0
z
0 
 b


0 −x 0 −y 0
0
0
0 
 0
∂3 = 

0
0 −x 0 −y 0
0
0 
 0


0
a
d
0
0 −y 0
0 
 0


0
b
e
a
d
x −y 0 
 0
 c
f
b
e
0
0
0 −y z 


 0
0
0
0
b
e
0
x
0 


 0
0
c
f
0
0
0
0 −y 
0  0
0
c
f
0
0
x
 0
−z 0
 0 −z 


 y
0 


 0

y



∂4 = −x 0 

 0 −x


 a
d 


 b
e 
c
f
It is easy to check that this is a complex. We note that x2 ∈ I1 (∂1 ),
x7 , y 7 ∈ I7 (∂2 ), x7 , y 7 , z 7 ∈ I7 (∂3 ) and x2 , y 2 , z 2 , ae − bd, af − cd, bf − ce ∈
I2 (∂4 ). It follows that the complex is exact and resolves
J. Moreover, we
√
have that pd(S/J) = 4 and J is unmixed. As J = (x, y, z) = p and
λ(Sp /Jp ) = 2, it follows that J is (x, y, z)-primary and e(S/J) = 2. The
inclusion L ⊆ (x2 , y 2 , z 2 ) : I is clear. Moreover L is (x, y, z)-primary by
Lemma A.1. Since both ideals are height 3 unmixed of multiplicity 6, they
must be equal by Lemma 2.2.

Lemma A.3. If
J = (x, y, z)2 + (bx − ay, cx − az, cy − bz),
where ht(x, y, z, a, b, c) = 6, then J is (x, y, z)-primary, pd(S/J) = 5 and
e(S/J) = 2.

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

31

Lemma A.4. If
J = (x, y, z)2 + (ax + by + cz),
where ht(x, y, z, a, b, c) ≥ 5, then J is (x, y, z)-primary, pd(S/J) = 4 and
e(S/J) = 3.
Lemma A.5. If
J = (x, y 2 , yz, z 3 , ay + z 2 ),
then J is (x, y, z)-primary, pd(S/J) = 3 and e(S/J) = 3.
Lemma A.6. If
J = (x2 , xy, xz, y 2 , yz, ax + by + z 2 , cx + dy),
where ht(x, y, z, c, d) = 5, then J is (x, y, z)-primary, pd(S/J) = 4 and
e(S/J) = 3. Moreover,
L = (x2 , y 2 , ax + by + z 2 ) : J = (x2 , xy, y 2 , ax + by + z 2 , cxz − dyz)
and pd(S/L) = 4.
Lemma A.7.
J = (x, y 3 , y 2 z, yz 2 , z 3 , ay + bz),
where ht(x, y, z, a, b) = 5, then J is (x, y, z)-primary, pd(S/J) = 4 and
e(S/J) = 4.
Lemma A.8. If
J = (x2 , xy, xz, yz 2 , z 3 , ax + by + cz, dx + y 2 ),
where ht(x, y, z, b, c) = ht(x, y, z, c, d) = 5, then J is (x, y, z)-primary, pd(S/J) =
4 and e(S/J) = 3.
Lemma A.9. If
J = (x2 , xy, xz, y 3 , z 3 , ax + by + cz, dx + yz),
where ht(x, y, z, b, c) = ht(x, y, z, b, d) = ht(x, y, z, c, d) = 5, then J is (x, y, z)primary, pd(S/J) = 4 and e(S/J) = 3.
Lemma A.10. If
J = (x2 , xy, xz, z 3 , cx + y 2 , bx − yz, ax + by + cz),
where ht(x, y, z, b, c) = 5, then J is (x, y, z)-primary, pd(S/J) = 4 and
e(S/J) = 3.
Lemma A.11. If
J = (x, y, z)3 + (ax + by + cz, dx + ey + f z),


a b c
where ht (x, y, z) + I2
= 5 and ht(x, y, z, a, b, c, d, e, f ) ≥ 6,
d e f
then J is (x, y, z)-primary, pd(S/J) = 5 and e(S/J) = 3.


CRAIG
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HUNEKE, PAOLO MANTERO, JASON MCCULLOUGH, AND ALEXANDRA SECELEANU

Lemma A.12. If
J = (x, y, z)3 + (ax + by + q, ay + bz + q ′ , xz − y 2 ),
where ht(x, y, z, a, b) = 5 and q, q ′ ∈ (x, y, z)2 , then J is (x, y, z)-primary,
pd(S/J) = 4 and e(S/J) = 3.
Lemma A.13. If
J = (x2 , xy, xz, y 2 , yz, z 3 , ax + by + z 2 ),
where ht(x, y, z) = 3, then J is (x, y, z)-primary, pd(S/J) = 3 and e(S/J) =
4.
Lemma A.14. If
J = (x2 , xy, xz, y 3 , y 2 z, yz 2 , z 3 , ax + by + cz),
where ht(x, y, z, b, c) = 5, then J is (x, y, z)-primary, pd(S/J) = 4 and
e(S/J) = 4.
Lemma A.15. If
J = (x2 , xy, y 2 , ax + by + z 2 , cx + dy),
where ht(x, y, z, c, d) = 5, then J is (x, y, z)-primary, pd(S/J) = 4 and
e(S/J) = 4.
This lemma gives the only nontrivial h2; 2i structure we need to consider.
Lemma A.16. If
J = (x2 , xy, y 2 , ax + by, cx + dy, ad − bc + ex + f y),
where (x, y, ad−bc) is a height 3 prime ideal, then J is (x, y, ad−bc)-primary,
pd(S/J) = 3 and e(S/J) = 4.
The last two lemmas are referenced in the body of the paper.
Lemma A.17. If
J = (ax + by + cz) + (x, y, z)(u, v),
where ht(u, v, x, y, z) = 5 and ht(u, v, a, b, c) ≥ 3, then J is unmixed and has
resolution


So

∂1

S7 o

∂2

S 11 o

∂3

0



 −z 




 y 


∂4 =

−x


 −v 

S6 o

u

S1 o

0,

A TIGHT BOUND ON THE PROJECTIVE DIMENSION OF FOUR QUADRICS

33

Lemma A.18. If
J = (x2 , xy, xz, y 2 , yz, ax + by + wz, cx + dy),
where ht(x, y, w) = 3 and ht(c, d, x, y, z) = 5, then J is unmixed and has
resolution
0
 −z 




 y 

∂4 =
−x




 c 
d
S6 o
S1 o


So

∂1

S7 o

∂2

S 11 o

∂3



0,

Acknowledgements
This work was started while all the authors were members of the MSRI
program on Commutative Algebra. We thank MSRI for its support and
great mathematical environment. We thank the referee for several suggestions that improved this paper. We also thank Chuck Weibel for handling
our paper through the later editorial stages.
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University of Virginia, Department of Mathematics, 141 Cabell Drive, Kerchof Hall, P.O. Box 400137, Charlottesville, VA 22904-4137
E-mail address: huneke@virginia.edu
Department of Mathematical Sciences, 309 SCEN - 1, University of Arkansas
- Fayetteville, AR 72701
E-mail address: pmantero@uark.edu
Department of Mathematics, Iowa State University, Ames, IA 50011
E-mail address: jmccullo@iastate.edu
University of Nebraska, Department of Mathematics, 203 Avery Hall, Lincoln, NE 68588
E-mail address: aseceleanu@unl.edu

