NEW CLASSES OF EXAMPLES SATISFYING THE THREE MATRIX
ANALOG OF GERSTENHABER’S THEOREM

arXiv:1711.10109v1 [] 28 Nov 2017

JENNA RAJCHGOT AND MATTHEW SATRIANO
Abstract. In 1961, Gerstenhaber proved the following theorem: if k is a field and X and Y are
commuting d × d matrices with entries in k, then the unital k-algebra generated by these matrices
has dimension at most d. The analog of this statement for four or more commuting matrices is
false. The three matrix version remains open. We use commutative-algebraic techniques to prove
that the three matrix analog of Gerstenhaber’s theorem is true for some new classes of examples.
In particular, we translate this three commuting matrix statement into an equivalent statement
about certain maps between modules, and prove that this commutative-algebraic reformulation is
true in special cases. We end with ideas for an inductive approach intended to handle the three
matrix analog of Gerstenhaber’s theorem more generally.

Contents
1. Introduction and statement of results
2. Reformulating Statement 1.2 in terms of module morphisms
3. Reducing Theorem 1.5 to a special case
4. Completing the proof of Theorem 1.5
5. Proof of Theorem 1.6
6. Some other instances where Statement 1.2 holds
7. An inductive approach to Statement 1.2
References

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1. Introduction and statement of results
1.1. An overview of Gerstenhaber’s theorem and its three matrix analog. Let k be a
field and let Md (k) denote the space of d × d matrices with entries in k.
Question 1.1. Let X1 , . . . , Xn ∈ Md (k) be pairwise commuting matrices. Must the unital k-algebra
that they generate be a finite-dimensional vector space of dimension at most d?

When n = 1, the answer to Question 1.1 is “yes” by the Cayley-Hamilton theorem: X ∈ Md (k)
satisfies its characteristic polynomial, thus I, X, X 2 , . . . , X d−1 is a vector space spanning set for the
algebra generated by X.
When n ≥ 4, the answer is “no” in general. The standard n = 4 counter-example is given by the
matrices E13 , E14 , E23 , E24 ∈ M4 (k), where Eij denotes the matrix with a 1 in position (i, j) and
0s elsewhere. These 4 matrices generate a 5-dimensional algebra.
The first interesting case is n = 2. Here the answer to Question 1.1 is “yes.” This result is
often called Gerstenhaber’s theorem and was proved in [Ger61]. Gerstenhaber’s proof was
algebro-geometric, and relied on the irreducibility of the commuting scheme C(2, d) of pairs of
d × d commuting matrices (a fact also proved in the earlier paper [MT55]). Some years later,
J.R. is partially supported by NSERC grant RGPIN-2017-05732.
M.S. is partially supported by NSERC grant RGPIN-2015-05631.
1

linear algebraic proofs (see [BH90, LL91]) and commutative algebraic proofs (see [Wad90, Ber13])
of Gerstenhaber’s theorem were found. More detailed summaries on the history and approaches to
Gerstenhaber’s theorem can be found in [Set11, HO15].
The case n = 3 is still open, and is the subject of this paper. We refer to the following statement
as the three matrix analog of Gerstenhaber’s theorem.
Statement 1.2. If X, Y, Z ∈ Md (k) are matrices which pairwise commute, then the unital k-algebra
generated by X, Y , and Z is a finite-dimensional vector space of dimension at most d.
To prove Statement 1.2, one might try to mimic the algebro-geometric proof of Gerstenhaber’s
theorem. This approach succeeds whenever the affine scheme of triples of commuting d×d matrices,
denoted C(3, d), is irreducible. Consequently, Statement 1.2 is true when d ≤ 10 and k is of
characteristic 0 (see [Š12] and references therein). However, since C(3, d) has multiple irreducible
components when d ≥ 30 [Gur92, HO01], a different approach is necessary to handle the general
case.
1.2. Summary of the main results. In this paper, we use commutative-algebraic methods to
study the three matrix analog of Gerstenhaber’s theorem. We do so by reformulating Statement
1.2 in terms of morphisms of modules, our key technical tools being Propositions 1.8 and 1.10.
Although this is nothing more than a simple reformulation, an approach along these lines appears
to be new.
We work over an arbitrary field k and let S = k[x1 , . . . , xn ] denote a polynomial ring in n
variables. One can easily rephrase Question 1.1 on n commuting matrices in terms of S-modules.
We provide a proof of this in Section 2 (see also [Ber13, §5]).
Proposition 1.3. Question 1.1 has answer “yes” if and only if for all S-modules N which are
finite-dimensional over k, we have
(1.4)

dim S/ Ann(N ) ≤ dim N.

Thus, Statement 1.2 is true if and only if inequality (1.4) holds for all finite-dimensional k[x, y, z]modules N .
From the perspective of Proposition 1.3, one can approach Statement 1.2 by successively considering modules of increasing complexity. The simplest modules to consider are cyclic ones: when
N = S/I it is obvious that equation (1.4) holds. The next simplest case to consider is extensions
0 → S/I → N → S/m → 0,
where m ⊆ S is a maximal ideal. This case is much less obvious, and is the central focus of this
paper. Our main result is that Statement 1.2 holds for such modules:
Theorem 1.5. Let k be an infinite field, S = k[x, y, z], I ⊆ S an ideal of finite colength, and
m ⊆ S a maximal ideal. If N is an extension of S-modules
0 → S/I → N → S/m → 0,
then the inequality (1.4) holds for N .
Using Theorem 1.5, we also obtain the following more general result.
Theorem 1.6. Let k be an infinite field, S = k[x, y, z], and N an S-module which is finite dimension over k. Suppose N = N1 ⊕ · · · ⊕ Nr where for each i, one of the following holds:
(1) Ni is a cyclic module, or
(2) Ni is local Gorenstein, or
2

(3) there is an extension
0 → Mi → Ni → S/mi → 0
L
where mi ⊆ S is a maximal ideal and Mi =
ℓ Mi,ℓ with each Mi,ℓ a cyclic or local
Gorenstein module.
Then the inequality (1.4) holds for N .
Before discussing the technique of proof, it is worth remarking why Theorems 1.5 and 1.6 are
more difficult than the cyclic case N = S/I. One reason is that the cyclic case holds over polynomial
rings S = k[x1 , . . . , xn ] in arbitrarily many variables, whereas Theorem 1.5 is specific to 3 variables.
Indeed, the technique of proof must be specific to 3 variables since there are counter-examples of
this form in 4 variables:
Example 1.7 (Theorems 1.5 and 1.6 are false for 4 variables). Let S = k[x, y, z, w] and m = (x, y, z, w).
Recall the standard 4 variable counter-example is given by the matrices E13 , E14 , E23 , E24 ∈ M4 (k).
Via the proof of Proposition 1.3, this corresponds to the S-module N given as follows: we have an
extension
π
0 → S/((x, y) + m2 ) → N → S/m → 0
so that N is generated by the elements of M := S/((x, y) + m2 ) and an additional element f ∈ N
such that π(f ) = 1 ∈ S/m. The S-module structure is given by
zf = wf = 0,

One checks Ann(N ) = m2 so that

xf = z ∈ M,

yf = w ∈ M.

dim S/ Ann(N ) = 5 > 4 = dim N
which violates the inequality of Proposition 1.3.

⋄

The proof of Theorem 1.5 consists of two steps. The first (see Theorem 3.2) is showing that if a
counter-example of this form exists, then we can reduce to the case where I = (F12 , F13 , F23 , g), the
Fij are the maximal minors of a specific 2 × 3 matrix, g is a non-zero divisor, and N satisfies certain
additional properties. The second step (see §4) consists of showing that no such counter-example
exists.
The case of Gorenstein modules, and extensions of S/m by Gorenstein modules, are handled in
Section 5. This, together with some preliminary results in Section 2, quickly reduces Theorem 1.6
to the case of Theorem 1.5.
As mentioned above, the following is a key tool we use in the proofs of Theorems 1.5 and 1.6.
Proposition 1.8. Let S = k[x, y, z] be a polynomial ring and m √
= (x, y, z). Then Statement 1.2
is true if and only if for all finite-dimensional S-modules M with Ann M = m, and all S-module
maps β : m → M , we have

(1.9)

dim S/ Ann(M ) + dim β(Ann M ) ≤ dim M + 1.

From the perspective of Proposition 1.8, Statement 1.2 is an assertion about bounding the
dimension of the image of Ann(M ) under a module map. Along these lines, we obtain an inductive
approach to Statement 1.2:
Proposition 1.10. Let S = k[x, y, z] and m = (x, y, z). Then Statement
1.2 is equivalent to the
√
following assertion: for all finite-dimensional S-modules M with Ann M = m, all submodules
M ′ ⊆ M with dim(M/M ′ ) = 1, all ideals J ⊆ m, and all S-module maps β : J → M ′ , if
then

dim J/(J ∩ Ann M ′ ) + dim β(J ∩ Ann M ′ ) ≤ dim M ′
dim J/(J ∩ Ann M ) + dim β(J ∩ Ann M ) ≤ dim M.
3

We prove Proposition 1.8 in Section 2, and Proposition 1.10 in Section 7.
Notation. All rings in this paper are commutative with unit. We will let Supp(M ) and Soc(M )
denote the support, respectively the socle, of a module M . Unless otherwise specified, dim(M ) and
the word dimension will refer to the dimension of a module M as a vector space over a given base
field.
Acknowledgments. It is a pleasure to thank Jason Bell, Mel Hochster, and Steven Sam for
many enlightening conversations. We thank the Casa Mathemática Oaxaca for the wonderful
accommodations where part of this work took place. We are especially grateful to Matt Kennedy
for suggesting this problem. We performed many computations in Macaulay2 [GS], which inspired
several ideas.
2. Reformulating Statement 1.2 in terms of module morphisms
Throughout this section, we fix a field k, let S = k[x1 , . . . , xn ] and m = (x1 , . . . , xn ). Although
our primary focus is the three matrix analogue of Gerstenhaber’s theorem, our proofs work for
arbitrary fields and arbitrarily many matrices. For convenience, consider the following:
Statement 2.1. The algebra generated by commuting matrices X1 , X2 , . . . , Xn ∈ Md (k) has dimension at most d.
For n > 3 this statement is false, n < 3 it is true, and the case when n = 3 is Statement 1.2.
Our goal in this section is to prove the following version of Proposition 1.8 which is valid for n
commuting matrices.
Proposition
2.2. Statement 2.1 is true if and only if for all finite-dimensional S-modules M with
√
Ann M = m, and S-module maps β : m → M , we have

(2.3)

dim S/ Ann(M ) + dim β(Ann M ) ≤ dim M + 1.

We begin with a different module-theoretic reformulation of Statement 2.1. This appeared as
Proposition 1.3 in the introduction.
Proposition 2.4. Statement 2.1 holds if and only if for all S-modules N which are finite-dimensional
over k, we have
dim S/ Ann(N ) ≤ dim N.
Proof. Given commuting matrices X1 , . . . , Xn ∈ Md (k), we obtain an S-module structure on k⊕d
where we define multiplication by xi to be the action of Xi . Conversely, given any S-module N
of dimension d over k, after fixing a basis we have N ≃ k⊕d as k-vector spaces; multiplication
by x1 , . . . , xn on N can then be viewed as matrices X1 , . . . , Xn ∈ Md (k) which commute since
xi xj = xj xi .
Let A denote the unital k-algebra generated by our commuting matrices. We then have a surjection π : S → A given by π(xi ) = Xi , and so A ≃ S/ ker π as S-modules. Under the correspondence
in the previous paragraph, we have ker π = Ann(N ) and so A = S/ Ann(N ). Therefore, the
inequality dim A ≤ d is equivalent to the inequality dim S/ Ann(N ) ≤ dim N .

Example 2.5. Following the above proof, we describe some triples of commuting matrices obtained
from our main theorem, Theorem 1.5. Let S = k[x, y, z] and m = (x, y, z). Let I ⊆ S be an ideal
of finite colength. Each module N that fits into a short exact sequence
i

π

0 → S/I −
→N −
→ S/m → 0

has an ordered basis of the form i(m1 ), . . . , i(md−1 ), f where m1 , . . . , md−1 is a basis of S/I and
π(f ) = 1 in S/m. Let nj = i(mj ). Let X, Y, Z ∈ Md (k) be determined by multiplication of x, y, z
4

on N and the given basis. To describe the first d − 1 columns of X, observe that each xnj is a
linear combination of n1 , . . . , nd−1 , and this linear combination is determined by multiplication by
x in S/I. Thus, if S/I with basis m1 , . . . , md−1 is given, the first d − 1 columns of X are fixed
and do not depend on the choice of extension of S/I by S/m. On the other hand, the last column
of X may be chosen almost arbitrarily. Indeed, xf is a linear combination of n1 , . . . , nd−1 (since
π(xf ) = 0 in S/m), but there are no restrictions on which linear combinations may occur: given
any linear combination of n1 , . . . , nd−1 , one can construct an extension N such that xf is the given
linear combination. The matrices Y and Z are described analogously. In other words, the matrices
X, Y , and Z take the form






a1
b.1
c.1
..
.
.
′
′
′
. 
. 



. 
X= X
Y = Y
Z= Z
bd−1  ,
cd−1  ,
ad−1  ,
0···0
0
0···0
0
0···0
0

where X ′ , Y ′ , Z ′ ∈ Md−1 (k) are determined by multiplication, in S/I, by x, y, and z respectively,
and the entries aj , bj , cj can be chosen arbitrarily by choosing an appropriate extension of S/I by
S/m.
⋄
The next observation allows us to handle the case of direct sums of modules and in particular,
reduce to the local case.
Lemma 2.6. Let N = N1 ⊕ · · · ⊕ Nr where each Ni is an S-module which is finite-dimensional
over k. If dim S/ Ann(Ni ) ≤ dim Ni for all i, then dim S/ Ann(N ) ≤ dim N .
T
Proof. We have Ann(N ) = i Ann(Ni ), so the diagonal map
M
S/ Ann(N ) →
S/ Ann(Ni )
i

is injective. As a result,
dim S/ Ann(N ) ≤

X
i

dim S/ Ann(Ni ) ≤

X

dim Ni = dim N.



i

Corollary 2.7. If N is an S-module that is finite-dimensional over k, and dim S/ Ann(N ′ ) ≤
dim N ′ for all localizations N ′ of N , then dim S/ Ann(N ) ≤ dim N . In particular, Statement 2.1 is
true if and only if dim S/ Ann(N ) ≤ dim N for all S-modules N which are finite-dimensional over
k and such that Supp N is a point.
Proof. Since N is finite-dimensional as a k-vector space, its support consists of finitely many points,
and N can be written as a direct sum N1 ⊕ · · · ⊕ Nr where Ni are the localizations of N . By
Lemma 2.6, the desired inequality for N is implied by that for each of the Ni . In particular,
dim S/ Ann(N ) ≤ dim N for all N if and only if it holds for those N which are supported at a
point. The corollary then follows from Proposition 2.4.

Remark 2.8. Corollary 2.7 is equivalent to the following statement about commuting matrices:
“Statement 2.1 holds for all choices of pairwise commuting matrices X1 , X2 , . . . , Xn ∈ Md (k) if and
only if it holds for all choices of n pairwise commuting nilpotent matrices.” To see this, we follow
the proof of Proposition 2.4, using the same notation: if the matrices Xi are nilpotent then one
easily checks that S/ Ann N is a local ring with maximal ideal m. Conversely, if S/ Ann N is local
with maximal ideal m then Ann N ⊇ mc for some large enough c. Identifying the variable xi ∈ S
with the matrix Xi as in the proof of Proposition 2.4, and recalling that S/ Ann N ∼
= A, we see
c
that each Xi satisfies Xi = 0, hence is nilpotent.
We next prove Proposition 2.2, and hence Proposition 1.8.
5

Proof of Proposition 2.2. By Corollary 2.7, we need only show that the inequality in the statement
of the theorem is equivalent to dim S/ Ann(N ) ≤ dim N for all S-modules N which are finitedimensional over k and such that Supp N is a point. Let N be such a module.
By translation, we
√
can assume without loss of generality that its support is the origin, i.e. Ann N = m. Then the
Jordan-Hölder filtration yields a short exact sequence
0 → M → N → S/m → 0

and this corresponds to a class α ∈ Ext1 (S/m, M ).
From the short exact sequence
we have a long exact sequence

0 → m → S → S/m → 0

0 → Hom(S/m, M ) → M → Hom(m, M ) → Ext1 (S/m, M ) → 0.

So, Ext1 (S/m, M ) ≃ Hom(m, M )/M . Thus, our class α lifts to an S-module map β : m → M . It is
easy to check that Ann(N ) = Ann(M ) ∩ ker β, so
dim S/ Ann(N ) = dim S/ Ann(M ) + dim Ann(M )/(Ann(M ) ∩ ker β)
= dim S/ Ann(M ) + dim β(Ann M ).

Since dim N = dim M + 1, the inequality
dim S/ Ann(M ) + dim β(Ann M ) ≤ dim M + 1

is equivalent to dim S/ Ann(N ) ≤ dim N .



Remark 2.9. The proof of Proposition 2.2 shows that if
0 → M → N → S/m → 0

is the extension corresponding to the map β : m → M , then the inequality (2.3) holds if and only
if the inequality (1.4) holds for N . Note that N only depends on the class [β] ∈ Ext1 (S/m, M ).
Consequently, whether or not the inequality (2.3) holds depends only on the class [β] rather than
the map β itself.
In light of Proposition 2.2, we make the following definition:
Definition 2.10. Let M be an S-module which is finite-dimensional as a k-vector space with
√
Ann M = m. If β : m → M is an S-module map, we say that (β, M ) is a counter-example if
dim S/ Ann(M ) + dim β(Ann M ) > dim M + 1.
Corollary 2.11. If I ⊆ S is an ideal with S/I finite-dimensional over k, then (β, S/I) is a counterexample if and only if dim β(I) ≥ 2.

Proof. Let M = S/I and notice that Ann(S/I) = I. By definition, (β, S/I) is a counter-example if
and only if
dim S/I + dim β(I) = dim S/ Ann(M ) + dim β(Ann M ) > dim M + 1 = dim S/I + 1.
In other words, it is a counter-example if and only if dim β(I) > 1.



We end with the following result which reduces our search for counter-examples (β, M ) to the
case where M is indecomposable. Note the distinction from Lemma 2.6: the lemma concerns
the inequality dim S/ Ann(M ) ≤ dim M whereas the proposition below concerns the inequality
f) ≤ dim M
f, where M
f is the extension of S/m by M defined by β.
dim S/ Ann(M
Proposition
√ 2.12. Let β : m → M be an S-module map with M finite-dimensional over k and
satisfying Ann M = m. Suppose M = N1 ⊕ N2 and let πj : M → Nj be the two projections. If
neither (π1 β, N1 ) nor (π2 β, N2 ) is a counter-example, then (β, M ) is also not a counter-example.
6

f be the extension defined by β, and let M
fj be the extension defined by πj β. We must
Proof. Let M
f
f
show dim S/ Ann(M ) ≤ dim M . We know that
\
\
\
f) = ker β ∩ Ann(M ) =
fj ),
Ann(M
ker(πj β) ∩
Ann(Nj ) =
Ann(M
j

so we have
f) =
dim S/ Ann(M

X
j

j

j

fj ) − dim S/(Ann(M
f1 ) + Ann(M
f2 )).
dim S/ Ann(M

fj ) ≤ dim M
fj . Also notice that
Since (πj β, Nj ) is not a counter-example, we have dim S/ Ann(M
√
f1 ) + Ann(M
f2 ) ⊆ m. Hence, dim S/(Ann(M
f1 ) + Ann(M
f2 )) ≥ 1. Since
Ann M = m implies Ann(M
X
X
f=1+
fj ,
dim M
dim Nj = −1 +
dim M
j

j

f) ≤ dim M
f.
we have our desired inequality dim S/ Ann(M



3. Reducing Theorem 1.5 to a special case

Throughout this section, we fix an infinite field k, let S = k[x1 , x2 , x3 ] and m = (x1 , x2 , x3 ). The
goal of the next two sections is to prove Theorem 1.5. By Lemma 2.6, we reduce immediately to
the case where Supp(S/I) = m, so we must prove:
Theorem 3.1. If M is cyclic and Supp M = m, then (β, M ) is not a counter-example in the sense
of Definition 2.10.
The focus of this section is to prove the following theorem, which reduces Theorem 3.1 to a
special case:
√
Theorem 3.2. Suppose (β, M ) is a counter-example over k with M cyclic and Ann M = m.
Then there exist f1 , f2 , f3 ∈ m and g, h ∈ Ann(M ) with the following properties:
(1) letting Fij = xi fj − xj fi and J = (F12 , F13 , F23 , g), we have h ∈
/ J, the module S/J is
finite-dimensional over k, and m ∈ Supp(S/J),
(2) letting β ′ : m → S/J be the S-module map defined by β ′ (xi ) = fi , the elements β ′ (g) and
β ′ (h) are linearly independent in the localization of S/(J + hhi) at m, and
(3) dim Soc(Sm /Jm ) = 2.

This theorem is proved over the course of §§3.1–3.2. We begin with
√ some preliminary results.
Since M is cyclic, it is of the form S/I where I is an ideal of S with I = m. By Corollary 2.11,
we know dim β(I) ≥ 2. Furthermore, we can make a minimality assumption: we may assume that
S/I is the cyclic module of smallest dimension for which there exists a counter-example (β, S/I),
i.e. for all cyclic modules S/K with dim S/K < dim S/I and all pairs (γ, S/K), we can assume
dim γ(K) ≤ 1.
Let fi ∈ S such that β(xi ) = fi mod I. Letting Fij = xi fj − xj fi , we see Fij ∈ I. So, we can
write
I = (F12 , F13 , F23 , g1 , . . . , gs )
for some polynomials g1 , . . . , gs ∈ m.
Lemma 3.3. β(I) ⊆ Soc(S/I) is the vector space spanned by the β(gi ). Moreover, there exist i 6= j
such that β(gi ) and β(gj ) are linearly independent.
7

Proof. Notice that if p ∈ I, then xi β(p) = pβ(xi ) = 0 in S/I. This shows that β(I) ⊆ Soc(S/I).
Next observe that
β(Fij ) = β(xi )fj − β(xj )fi = fi fj − fj fi = 0.
As a result, β(I) is the ideal generated by the β(gj ) and since the β(gj ) are contained in Soc(S/I),
this ideal is nothing more than the vector space they span. Since dim β(I) ≥ 2, there must be i 6= j
such that β(gi ) and β(gj ) are linearly independent.

Proposition 3.4. For each 1 ≤ i ≤ 3, we have fi ∈ m.

√
Proof. Without loss of generality, assume that f1 ∈
/ m. Since I = m, we see S/I is an Artin local
ring, and so f1 is a unit in S/I. Since x2 f1 − x1 f2 = F12 is 0 in S/I, we see x2 = x1 f2 f1−1 ∈ (x1 ).
Similarly for x3 .
Thus, the maximal ideal m/I of S/I is principally generated by x1 . Any Artin local ring with
principal maximal ideal has the property that all ideals are of the form (m/I)n , hence principal.
Since Soc(S/I) is an ideal, it is principal and so 1-dimensional. This contradicts Lemma 3.3 which
shows that dim Soc(S/I) ≥ 2.

3.1. Showing the existence of a non-zero divisor. In this subsection, we show the existence
of g and h in the statement of Theorem 3.2. Let us give an outline of how we proceed. We begin by
showing that S/(F12 , F13 , F23 ) is Cohen-Macaulay of Krull dimension 1. Now there are of course
many choices of g such that S/(F12 , F13 , F23 , g) has Krull dimension 0 (i.e. finite dimensional over
k), however to prove Theorem 3.2, we need to guarantee that dim Soc(Sm /Jm ) = 2 and that β ′ (g)
and β ′ (h) are linearly independent. This is accomplished by choosing g to be a suitable non-zero
divisor in S/(F12 , F13 , F23 ).
Proposition 3.5. Every minimal prime of S over (F12 , F13 , F23 ) has height 2.
Proof. For convenience, let L = (F12 , F13 , F23 ). As F12 , F13 , F23 are the minors of the 2 × 3 matrix


x1 x2 x3
,
f1 f2 f3
the height of each minimal prime over L is at most 2. So assume that there is a minimal prime p
over L of height ≤ 1.
If p has height 0, then p = {0} and so L = {0}. Then Fij = 0, so xi fj = xj fi for each i, j.
From this it follows that there exists q ∈ S such that fi = xi q for all i. Thus, given any h ∈ m, we
have β(h) = hq. In particular, for h ∈ I we see β(h) = hq ∈ I, so dim β(I) = 0, which contradicts
dim β(I) ≥ 2.
If p has height 1, then there exists p ∈ S irreducible
with p = (p). By Bertini’s theorem, we
P
know that for a generic linear combination y = i λi xi , the ideal (p, y) is prime. Choose yP
so that
(p, y) is prime and so that (the open conditions) λ3 6= 0 and y ∈
/ (p) are satisfied. Let fy = i λi fi ,
and let F1y = x1 fy − yf1 , F2y = x2 fy − yf2 . Observe that L = (F12 , F1y , F2y ).
Next, since F1y , F2y ∈ L ⊆ (p) ( (p, y), we have x1 fy , x2 fy ∈ (p, y). Recalling that (p, y) is prime,
we see fy ∈ (p, y), or both x1 , x2 ∈ (p, y). In the latter case, we have that (x1 , x2 , x3 ) = (x1 , x2 , y) ⊆
(p, y), which is impossible as the vanishing locus V (p, y) ⊆ A3 is irreducible of dimension at least
1. We must therefore have fy ∈ (p, y).
Since fy ∈ (p, y), we have fy = qy + ry p for some q, ry ∈ S. Using that p divides F1y =
x1 fy − yf1 = (x1 q − f1 )y + x1 ry p, we see that p divides (x1 q − f1 )y. Since (p) is prime and y ∈
/ p,
we have x1 q − f1 ∈ (p) and so f1 = qx1 + r1 p for some r1 . Similarly, f2 = qx2 + r2 p for some
r2 . As a result, β = β ′ + β ′′ where β ′ (h) = qh for all h ∈ m, and β ′′ (xi ) = pri . By Remark 2.9,
whether or not (β, S/I) is a counter-example depends only on the value of [β] ∈ Ext1 (S/m, S/I)
and since [β ′ ] = 0, we can assume β = β ′′ . As a result, we can assume the image of β factors
through (p) ( S/I. Since (p) is generated by a single element, we have (p) ≃ S/J where J =
8

Ann(p). Since dim S/J < dim S/I, by our minimality assumption at the start of §3, we know that
β : m → (p) = S/J is not a counter-example, and so dim β(J) ≤ 1. But, I ⊆ J because I kills p.
So, dim β(I) ≤ dim β(J) ≤ 1.

Corollary 3.6. S/(F12 , F13 , F23 ) is Cohen-Macaulay of Krull dimension 1.
Proof. By Proposition 3.5, the variety cut out by (F12 , F13 , F23 ) has codimension 2 in A3 , so
S/(F12 , F13 , F23 ) has Krull dimension 1. Since F12 , F13 , F23 are the 2 × 2 minors of a 2 × 3 matrix,
we may apply [Eis95, Theorem 18.18] (originally proven in [HE71]) to see that S/(F12 , F13 , F23 ) is
Cohen-Macaulay.

The following proposition establishes the existence of our desired g, h ∈ m.
Proposition 3.7. There exist g, h ∈ I such that g is a non-zero divisor in S/(F12 , F13 , F23 ) and
β(g) and β(h) are linearly independent. Furthermore, we necessarily have h ∈
/ (F12 , F13 , F23 , g).
Proof. Recall our notation I = (F12 , F13 , F23 , g1 , . . . , gs ). Our first goal is to show that I contains
a non-zero divisor in S/(F12 , F13 , F23 ).
We know from Corollary 3.6 that S/(F12 , F13 , F23 ) is Cohen-Macaulay, so its associated primes
are its minimal primes. Consequently, the set of zero divisors of S/(F12 , F13 , F23 ) is the union of
minimal primes over (F12 , F13 , F23 ). If I is contained in this union of minimal primes then, by
the prime avoidance lemma, I is contained in one of these minimal primes. But this is impossible as S/(F12 , F13 , F23 ) is Cohen-Macaulay of Krull dimension 1 by Corollary 3.6, and S/I has
Krull dimension 0 by assumption. Thus, I is not contained in the union of minimal primes over
(F12 , F13 , F23 ) and so I contains a non-zero divisor of S/(F12 , F13 , F23 ).
Next, by Lemma 3.3, there exist i 6= j such that β(gi ) and β(gj ) are linearly independent in
S/I. Let q ∈ I be a non-zero divisor of S/(F12 , F13 , F23 ). If c, d ∈ k r {0} with c 6= d, and if
q + cgi and q + dgi are in the same minimal prime over (F12 , F13 , F23 ), then q and gi are also in that
minimal prime, a contradiction to q being a non-zero divisor. Thus, since there are only finitely
many minimal primes, we see that for all but finitely many c ∈ k, the polynomial q + cgi is a
non-zero divisor.
Since q ∈ I, we know from Lemma 3.3 that β(q) ∈ Soc(S/I). Then β(gi ) and β(q) are elements
of the vector space Soc(S/I) which has dimension at least 2 and β(gi ) 6= 0, so for infinitely many
c ∈ k, we see β(q + cgi ) = β(q) + cβ(gi ) 6= 0. Combining this with our conclusion from the previous
paragraph that q + cgi is a non-zero divisor for all but finitely many c ∈ k, we see we can find
a non-zero divisor g := q + cgi ∈ I such that β(g) 6= 0. Now choose h ∈ m to be any linear
combination of gi and gj such that β(h) is not a scalar multiple of β(g); this is possible by Lemma
3.3 as β(gi ) and β(gj ) span a 2-dimensional subspace of Soc(S/I).
Lastly, we show h ∈
/ J := (F12 , F13 , F23 , g). From Lemma 3.3, we know 0 6= β(g) ∈ Soc(S/I) and
we see that β(Fij ) = 0. Thus, β(J) is 1-dimensional, generated by β(g). Since β(g) and β(h) are
linearly independent, it follows that h ∈
/ J.

Remark 3.8 (Hypothesis that k is infinite). Proposition 3.7 is the only step in the proof of Theorem
3.1 that assumes that k is infinite.
With the choice of g and h from Proposition 3.7, we have:
Corollary 3.9. Conditions (1) and (2) of Theorem 3.2 hold.
Proof. From Proposition 3.7, g is a non-zero divisor of S/(F12 , F13 , F23 ), which is Cohen-Macaulay of
Krull dimension 1 by Corollary 3.6. Thus, S/J = S/(F12 , F13
√, F23 , g) has Krull dimension 0, so it is
finite-dimensional over k. Since S/J surjects onto S/I and I = m, we know that m ∈ Supp(S/J),
which proves condition (1).
9

Next since h ∈ I, we have
√ surjections S/J → S/(J + hhi) → S/I. After localizing at m, these
remain surjections. Since I = m, we know Sm /Im = S/I and so
β′

m −→ S/J → Sm /Jm → Sm /(J + hhi)m

is a lift of β, meaning that after post-composing the above map by Sm /(J + hhi)m → Sm /Im = S/I,
we obtain β. Since β(g) and β(h) are linearly independent in S/I, it must also be the case that
β ′ (g) and β ′ (h) are linearly independent in Sm /(J + hhi)m , proving condition (2).

3.2. Computing the socle. In this subsection we compute the socle of Sm /Jm , thereby showing
condition (3) of Theorem 3.2 and finishing the proof.
Proposition 3.10. dim Soc(Sm /Jm ) = 2.
Proof. From Corollary 3.6 we know that S/(F12 , F13 , F23 ) is Cohen-Macaulay of dimension 1. Thus,
the Eagon-Northcott complex yields a minimal free resolution [Eis05, Theorem A2.60]:


x3 f 3

B
A
→ S → S/hF12 , F13 , F23 i → 0, A = x2 f2  , B = F12 −F13 F23
→ S3 −
0 → S2 −
x1 f 1

is an exact sequence.
Since g ∈ m is a nonzerodivisior in S/(F12 , F13 , F23 ), we can use the above resolution to obtain
the minimal free resolution of S/J:
A′



g
0

A′ = 
x3
x2
x1
Localizing at m, we

C′

B′

0 → S 2 −→ S 5 −→ S 4 −→ S → S/J → 0,




0
x
f
−g
0
0
3
3
g
 x2 f 2 0


−g
0 
′

 , C ′ = F12 −F13 F23 g
f3 
 , B =  x1 f 1 0
0
−g 
f2 
0 0 F12 −F13 F23
f1
obtain the minimal free resolution
2
5
4
0 → Sm
→ Sm
→ Sm
→ Sm → (S/J)m → 0.

Consequently, dim Soc(Sm /Jm ) = 2, as the left-most term in the above resolution is of rank 2.



4. Completing the proof of Theorem 1.5
Having now proved Theorem 3.2, we have reduced Theorem 3.1 to showing the following. Let
f1 , f2 , f3 , g ∈ m, Fij = xi fj − xj fi , and J = (F12 , F13 , F23 , g) such that S/J is finite-dimensional
with dim Soc(Sm /Jm ) = 2. Let β : m → S/J be defined by β(xi ) = fi . Then it is impossible to find
h ∈ m r J such that β(g) and β(h) are linearly independent in Sm /(Jm + hhi).
We begin with two well-known lemmas.
Lemma 4.1. If R is an Artinian Gorenstein local ring, and I1 and I2 are ideals of R, then
Ann(I1 ) + Ann(I2 ) = Ann(I1 ∩ I2 ).
Proof. In any commutative ring, we have the equality Ann(K1 + K2 ) = Ann(K1 ) ∩ Ann(K2 ) for
all ideals K1 and K2 . For Artinian Gorenstein local rings, we have Ann(Ann(K)) = K for all
ideals K, see e.g. [BH98, Exercise 3.2.15]. So letting Kj = Ann(Ij ), we see in our case that
Ann(Ann(I1 ) + Ann(I2 )) = I1 ∩ I2 . Taking annihilators of both sides then proves the result.

Lemma 4.2. Let R be any Artinian local ring and 0 6= r ∈ R. If s1 , . . . , sm form a basis for
Soc(R), then there is a linear dependence relation among the si in R/r.
10

Proof. Since every non-zero ideal of R intersects
the socle non-trivially, (r) ∩ Soc(R) contains a
P
non-trivial element s. We can write
s
=
a
s
with
(a1 , . . . , am ) ∈ km r 0. Then in R/r we have
i i
P
the linear dependence relation
ai si = 0.


Proposition 4.3. Let K ⊆ S = k[x1 , x2 , x3 ] be an ideal with S/K an Artinian Gorenstein local
ring, and γ : m → S/K an S-module map. If γ(q) is divisible by q for all q ∈ m, then there exists
r ∈ S/K such that for all q ∈ m, we have γ(q) = qr.
Proof. For every r ∈ S/K, let δr : m → S/K be the map δr (q) = qr. To prove the result, it suffices
to replace γ by γ − δr for any r. Let γ(xi ) = xi pi . Replacing γ by γ − δp3 , we can assume p3 = 0,
i.e. γ(x3 ) = 0. Then x1 x3 p1 = x3 γ(x1 ) = x1 γ(x3 ) = 0. In other words,
p1 ∈ Ann(x1 x3 ).
Similarly,
p2 ∈ Ann(x2 x3 ) and p1 − p2 ∈ Ann(x1 x2 ).

Our first goal is to show that p2 ∈ Ann(x2 ) + Ann(x3 ). By Lemma 4.1, we have Ann(x2 ) +
Ann(x3 ) = Ann((x2 ) ∩ (x3 )). So, let q ∈ (x2 ) ∩ (x3 ) and we must show p2 q = 0. Since q is
divisible by both x2 and x3 , we can write q = x3 q ′ and q = x2 q ′′ . We can further write q ′′ =
a(x2 ) + x3 b(x2 , x3 ) + x1 c(x1 , x2 , x3 ) where a ∈ k[x2 ], b ∈ k[x2 , x3 ], and c ∈ k[x1 , x2 , x3 ]. Then using
p2 ∈ Ann(x2 x3 ) and p1 − p2 ∈ Ann(x1 x2 ), we have
p2 q = p2 x2 a(x2 ) + p2 x1 x2 c(x1 , x2 , x3 ) = p2 x2 a(x2 ) + p1 x1 x2 c(x1 , x2 , x3 )

= γ(x2 a(x2 ) + x1 x2 c(x1 , x2 , x3 ) + x3 d)
for any choice of d. Choosing d = x2 b(x2 , x3 ) − q ′ , we see

s := x2 a(x2 ) + x1 x2 c(x1 , x2 , x3 ) + x3 d = q − x3 q ′ = 0.

But, by assumption p2 q = γ(s) is divisible by s = 0, and hence p2 q = 0. We have therefore shown
p2 ∈ Ann(x2 ) + Ann(x3 ).
Thus, we can write p2 = (p2 − r)+ r with r ∈ Ann(x3 ) and p2 − r ∈ Ann(x2 ). Then (γ − δr )(x2 ) =
x2 (p2 − r) = 0 and (γ − δr )(x3 ) = −x3 r = 0, and so we can assume
p2 = p3 = 0.
Then
p1 ∈ Ann(x1 x2 ) ∩ Ann(x1 x3 ).

To finish the proof we need only show that p1 ∈ Ann(x1 ) + Ann(x2 , x3 ). Indeed, upon doing so,
we can write p1 = (p1 − r) + r with r ∈ Ann(x2 , x3 ) and p1 − r ∈ Ann(x1 ). Then (γ − δr )(x1 ) =
x1 (p1 − r) = 0, (γ − δr )(x2 ) = −x2 r = 0, and (γ − δr )(x3 ) = −x3 r = 0. In other words, we will
have found r ∈ S such that γ − δr = 0, i.e. γ(q) = qr for all q ∈ m.
To show that p1 ∈ Ann(x1 ) + Ann(x2 , x3 ), and thereby finish the proof, we again note by Lemma
4.1 that Ann(x1 ) + Ann(x2 , x3 ) = Ann((x1 ) ∩ (x2 , x3 )). We let q ∈ (x1 ) ∩ (x2 , x3 ) and must show
that p1 q = 0. We can then write q = x1 (a(x1 ) + x2 b(x1 , x2 ) + x3 c(x1 , x2 , x3 )) and q = x2 q ′ + x3 q ′′ .
Then using that p1 ∈ Ann(x1 x2 ) ∩ Ann(x1 x3 ), we have
p1 q = p1 x1 a(x1 ) = γ(x1 a(x1 ) + x2 d + x3 e)
for any choice of d and e. As before, choosing d = x1 b(x1 , x2 ) − q ′ and e = x1 c(x1 , x2 , x3 ) − q ′′
yields s := x1 a(x1 ) + x2 d + x3 e = 0, and since p1 q = γ(s) is divisible by s = 0, we have p1 q = 0. 
Finally, we turn to the proof of the main theorem.
11

Proof of Theorem 3.1. Since S/J is finite-dimensional over k, we know that Sm /Jm = S/K for some
ideal K. Recall that dim Soc(S/K) = 2 and our goal is to show that for all h ∈ m r J there is a
linear dependence relation between β(g) and β(h) in Sm /(Jm + hhi) = S/(K + hhi). In particular,
we may assume β(g) 6= 0 in S/(K + hhi).
To begin, we show β(h) ∈
/ Soc(S/K). If β(h) were in the socle, then since dim Soc(S/K) = 2
and β(g) ∈ Soc(S/K), either we have our desired linear dependence relation between β(g) and β(h)
in S/K (and hence in S/(K + hhi)), or β(g) and β(h) form a basis for Soc(S/K). In the latter
case, Lemma 4.2 shows there is a linear dependence relation between β(g) and β(h) in S/(K + hhi).
So, we have shown our claim that β(h) ∈
/ Soc(S/K). Further note that h ∈ Soc(S/K) implies
β(h) ∈ Soc(S/K), since xi β(h) = hβ(xi ) ∈ hm = 0. So, we conclude
h, β(h) ∈
/ Soc(S/K).

Next, notice that β(h) does not divide β(g) in S/K. Indeed, suppose to the contrary that
β(g) = qβ(h) with q ∈ S. If q ∈ k, then we have a linear dependence relation in S/K and hence
in S/(K + hhi). If q ∈ m, then β(g) = qβ(h) = hβ(q) ∈ hhi so β(g) = 0 in S/(K + hhi), which is
again a linear dependence relation. This shows our claim that
β(g) ∈
/ Sβ(h).

Now, Sβ(h) is an ideal of S/K, so it intersects Soc(S/K) non-trivially. Since dim Soc(S/K) = 2,
we know Sβ(h) ∩ Soc(S/K) has dimension 1 or 2, but β(g) ∈
/ Sβ(h) ∩ Soc(S/K), and so Sβ(h) ∩
Soc(S/K) is 1-dimensional. Let q0 ∈ S such that q0 β(h) is a basis vector for Sβ(h) ∩ Soc(S/K).
Then
β(g) and q0 β(h) form a basis for Soc(S/K).
Since h ∈
/ Soc(S/K), we can induct on the smallest ℓ for which mℓ h ∈ Soc(S/K). That is, we
can assume the result for qh for all q ∈ m, i.e. we can assume that β(g) and β(qh) are linearly
dependent in S/(K + hqhi) for all q ∈ m. So for all q ∈ m, there exists p ∈ S/K and a, b ∈ k such
that (a, b) 6= (0, 0) and
aβ(g) + bβ(qh) = pqh.
Note that β(qh) = hβ(q) ∈ hhi, so the above equality shows aβ(g) ∈ hhi. This yields our desired
linear dependence relation among β(g) and β(h) in S/(K + hhi) unless a = 0, in which case after
rescaling we can assume b = 1. We can therefore assume that
Next, let

β(qh) ∈ Sqh ∀q ∈ m.

γ : m → hhi ⊆ S/K, γ(q) = β(qh).
Since hhi ∩ Soc(S/K) is non-trivial, it has dimension 1 or 2. If β(g) is in this intersection, then
we have our desired linear dependence relation among β(g) and β(h) in S/(K + hhi), so we can
assume this is not the case. Thus, hhi ∩ Soc(S/K) does not contain β(g), so it is 1-dimensional, and
hence hhi is Gorenstein. Notice that hhi ≃ S/ AnnS/K (h) and via this identification, the condition
β(qh) ∈ hqhi is equivalent to the condition that q divides γ(q). Applying Proposition 4.3, there is
r ∈ S/ AnnS/K (h) such that γ(q) = qr. Translating this back into a statement about hhi via our
identification with S/ Ann(h), this says
∃ r ∈ S such that β(qh) = qhr ∀q ∈ m.

As a result, β(h) − hr ∈ Ann(m) = Soc(S/K), and so there exist a, b ∈ k such that
β(h) − hr = aβ(g) + bq0 β(h).

If q0 ∈ m, then we see β(h) − aβ(g) = h(bβ(q0 ) + r) = 0 in S/(K + hhi) and gives our linear
dependence relation. So, q0 ∈
/ m, i.e. q0 is a unit. By construction β(g) and q0 β(h) form a basis
12

for Soc(S/K), and q0 is a unit, so β(g) and β(h) form a basis for Soc(S/K). Then by Lemma 4.2,
they have a linear dependence relation in S/(K + hhi).

5. Proof of Theorem 1.6
We begin with the following result which holds for arbitrarily many variables:
Proposition 5.1. Let S = k[x1 , . . . , xn ] and let m = (x1 , . . . , xn ). Suppose that M is a finitedimensional S-module with Supp M = m and dim Soc(M ) = 1. Then the following hold:
(1) dim S/ Ann M ≤ dim M
(2) for every short exact sequence
0 → M → N → S/m → 0,

we have dim S/ Ann N ≤ dim N .

Proof. We first prove (1) by induction on the dimension of M . If dim M = 1 then M ≃ S/m and
the result holds trivially. For higher dimensional M , recall that M has a composition series
0 ⊆ M1 ⊆ · · · ⊆ Mr−1 ⊆ M

where Mi /Mi−1 ≃ S/m. Since the socle of Mr−1 is contained in the socle of M , we also have that
dim Soc(Mr−1 ) = 1, and we may apply induction to see that dim S/ Ann Mr−1 ≤ dim Mr−1 . Then
we have a short exact sequence
0 → Mr−1 → Mr → S/m → 0

which corresponds to a map β : m → Mr−1 . By Remark 2.9, we need only show
dim S/ Ann Mr−1 + dim β(Ann Mr−1 ) ≤ dim Mr−1 + 1.

By induction, we know dim S/ Ann Mr−1 ≤ dim Mr−1 . Furthermore, β(Ann Mr−1 ) ⊆ Soc(Mr−1 ),
so has dimension at most 1. This proves the desired inequality.
The proof of (2) is entirely analogous. We know that the short exact sequence defining N
corresponds to a map β : m → M . By (1), we know dim S/ Ann(M ) ≤ dim M . Since β(Ann M ) ⊆
Soc(M ), we have dim β(Ann M ) ≤ 1. Combining these two statements, inequality (2.3) holds for
β, and so dim S/ Ann N ≤ dim N by Remark 2.9.

We now turn to the proof of the second main result of this paper.
Proof of Theorem 1.6. By Lemma 2.6, we reduce immediately to the case where: (i) N is either
cyclic, or (ii) N is local Gorenstein, or (iii) there is an extension
(5.2)

0 → M → N → S/m → 0
L
where m ⊆ S is a maximal ideal and M = ℓ Mℓ with each Mℓ a cyclic or local Gorenstein module.
For case (i), inequality (1.4) holds trivially. Case (ii) is handled by Proposition 5.1 (1).
It remains to handle case (iii). Further decomposing if necessary, we can assume each
L Mℓ is
local. Next, letting L be the set of ℓ for which Supp(Mℓ ) = m, we can write N = N ′ ⊕ ℓ∈L
/ Mℓ
where we have a short exact sequence
M
0→
Mℓ → N ′ → S/m → 0.
ℓ∈L

Since Mℓ is cyclic or local Gorenstein for every ℓ ∈
/ L, another application of Lemma 2.6 combined
with cases (i) and (ii) above allows us to assume N = N ′ , i.e. we can assume that Supp(Mℓ ) = m
for all ℓ. Proposition 2.12 then reduces
√ us to the case where there is only one ℓ; that is, we need
only consider extensions (5.2) where Ann M = m and M itself is cyclic or Gorenstein. If M is
Gorenstein, then inequality (1.4) holds by Proposition 5.1 (2). If M is cyclic, then the inequality
holds by Theorem 1.5.

13

6. Some other instances where Statement 1.2 holds
In this short section we record a couple of additional situations where a finite-dimensional module
M over S = k[x1 , . . . , xn ] satisfies dim S/ Ann M ≤ dim M .
Proposition 6.1. If β : m → M is surjective, then the pair (β, M ) is not a counterexample.
Moreover, if N is the extension of M defined by β then
dim S/ Ann(N ) = dim N.
Proof. Since Ann(N ) = Ann(M ) ∩ ker(β), we see
dim S/ Ann(N ) = dim S/m + dim m/ ker(β) + dim ker(β)/(Ann(N ) ∩ ker(β))
= 1 + dim M + dim ker(β)/(Ann(N ) ∩ ker(β)).

Since dim N = 1 + dim M , we must show ker(β) ⊆ Ann(M ). Given m ∈ M and f ∈ ker(β),
we know β is surjective, so m = β(g) for some g ∈ m. Then f m = f β(g) = gβ(f ) = 0, and so
f ∈ Ann(M ).

Proposition 6.2. If there exists m ∈ N such that Ann(m) = Ann(N ), then
dim S/ Ann N ≤ dim N.
Proof. In this case, dim S/ Ann(N ) = dim S/ Ann(m) and since S/ Ann(m) is isomorphic to the
cyclic submodule Sm ⊆ N , we necessarily have dim S/ Ann(m) = dim Sm ≤ dim N .

We end this section with an example where Theorem 1.5 applies but Proposition 6.2 does not.
Example 6.3. Let I = (x, y 2 , z) ⊆ k[x, y, z]. Let N = (xy, z)/(yz, x2 , z 2 , xy 2 − xz) and observe that
N fits into a short exact sequence
0 → S/I → N → S/m → 0
where the injective map sends 1 to xy. We know by Theorem 1.5 that inequality (1.4) holds
for N . Proposition 6.2, however does not apply here: every element of N can be represented as
m = az + bxy + cxz, for some a, b, c ∈ k, and one checks that there is no choice of a, b, c ∈ k such
that Ann(m) agrees with Ann N = (z, y 2 , xy, x2 ). Indeed, if a 6= 0, then y − (b/a)x ∈ Ann(m) and
if a = 0, then x ∈ Ann(m).
⋄
7. An inductive approach to Statement 1.2
Our goal is to prove Proposition 1.10 stated in the introduction. We do so after a preliminary
lemma.
Lemma 7.1. Let S = k[x1 , . . . , xn ] and m = (x1 , . . . , xn ). Then Statement
2.1 is true if and only
√
if for all ideals J ⊆ m, all finite-dimensional S-modules M with Ann M = m, and all S-module
morphisms β : J → M , we have
(7.2)

dim J/(J ∩ Ann M ) + dim β(J ∩ Ann M ) ≤ dim M.

Proof. Notice that if J = m, then J ∩ Ann M = Ann M , and so the inequalities (2.3) and (7.2) are
equivalent. So by Proposition 2.2, the inequality (7.2) implies Statement 2.1.
Thus, it remains to show that Statement 2.1 implies inequality (7.2). To see this, fix J ⊆ m and
an S-module map β : J → M . As in the proof of Proposition 2.2, the map β defines an extension
0 → M → N → S/J → 0
14

with Ann(N ) = Ann(M ) ∩ ker β. By Proposition 2.4, we know Statement 2.1 implies inequality
(1.4) for N . Now note that
dim S/ Ann(N ) = dim S/(J ∩ Ann M ) + dim(J ∩ Ann M )/(ker β ∩ Ann M )
= dim S/(J ∩ Ann M ) + dim β(J ∩ Ann M ).

Since dim N = dim M + dim S/J, we obtain inequality (7.2) by subtracting dim S/J from both
sides of the inequality (1.4).

Remark 7.3. The proof of Lemma 7.1 shows that if
0 → M → N → S/J → 0
is the extension corresponding to the map β : J → M , then the inequality (7.2) holds if and only if
the inequality (1.4) holds for N .
Proof of Proposition 1.10. Let S = k[x, y, z] be a polynomial ring and m = (x, y, z). We know
by Lemma 7.1 that Statement 1.2 is true if and only if inequality (7.2) holds for all J, M , and
maps β : J → M . So, if Statement 1.2 is true, then both of the inequalities in the statement of
Proposition 1.10 are true, hence the first inequality implies the second.
We now show that if the implication of inequalities in the statement of Proposition 1.10 holds,
then Statement 1.2 is true. By virtue of Lemma 7.1, we need only show that inequality (7.2) holds
for all J, M , and β : J → M . We prove this latter statement by induction on dim M , the base case
being trivial. So, we need only handle the induction step. For this, we can choose a submodule
M ′ ⊆ M such that dim(M/M ′ ) = 1. Then M/M ′ ≃ S/m and we let π : M → S/m be the quotient
map.
Suppose first that πβ : J → S/m is surjective. Then letting I = ker(πβ), we have a map of short
exact sequences
0

// I

// J
α



0

// M ′

// S/m
β



// M

// 0

≃
π


// S/m

// 0

The maps α and β define extensions
0 → M ′ → N ′ → S/I → 0 and

0 → M → N → S/J → 0,

respectively, and one checks that N ′ ≃ N . Since dim M ′ < dim M , we can assume by induction
that
dim I/(I ∩ Ann M ′ ) + dim α(I ∩ Ann M ′ ) ≤ dim M ′ .
By Remark 7.3, this is equivalent to the inequality dim S/ Ann(N ′ ) ≤ dim N ′ . Since N ≃ N ′ we
obtain the inequality dim S/ Ann(N ) ≤ dim N , and applying Remark 7.3 again, we have
dim J/(J ∩ Ann M ) + dim β(J ∩ Ann M ) ≤ dim M.
It remains to handle the case when πβ : J → S/m is not surjective. In this case πβ = 0 and so β
factors through M ′ . We are thus in the situation β : J → M ′ ⊆ M . By induction, we can assume
dim J/(J ∩ Ann M ′ ) + dim β(J ∩ Ann M ′ ) ≤ dim M ′

and we want to show
dim J/(J ∩ Ann M ) + dim β(J ∩ Ann M ) ≤ dim M.
This is precisely the implication of inequalities in the hypothesis of Proposition 1.10.
15



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