SURJECTIVE AND SPLITTING CAPACITIES

arXiv:1710.06010v2 [] 8 Jan 2018

ROBIN BAIDYA
Abstract. Let R be a commutative ring, S a module-finite R-algebra, M a right Smodule, and N a finitely generated right S-module such that Max(R) ∩ SuppR (N ) is finitedimensional and Noetherian. Working under various combinations of additional hypotheses
on R, M , and N , we give lower bounds on the global surjective capacity of M with respect
to N over S, that is, the supremum of the nonnegative integers t such that N ⊕t is a direct
summand of a quotient S-module of M . We express our lower bounds in terms of local analogues of global surjective capacity and topological properties of Spec(R). Assuming that N
is finitely presented over S, we also give lower bounds on the global splitting capacity of M
with respect to N over S, that is, the supremum of the nonnegative integers t such that N ⊕t
is a direct summand of M . In the process, we generalize Serre’s Splitting Theorem from
algebraic K-theory and a theorem from stable algebra due to Bass. We also generalize a
theorem on Noetherian modules by De Stefani, Polstra, and Yao that serves as an analogue
of an older result by Stafford. To close, we consider the case of finitely generated modules
over a Dedekind domain; in this case, we show that we can provide conditions equivalent to
having a given global surjective or splitting capacity.

0. Introduction
In this paper, every ring has a 1, and every module is unital.
Let R be a commutative ring, and let M and N be R-modules. It is well known that an
R-linear map f : M → N is surjective if and only if fp is surjective for every p ∈ Spec(R).
Suppose now that, for every p ∈ Spec(R), there is a surjective Rp -linear map from Mp to
Np . Can we conclude that there is a surjective R-linear map from M to N?
In general, the answer is no. For example, suppose that R is a Dedekind domain with a
nonprincipal ideal I. Then Rp ∼
= Ip for every p ∈ Spec(R), but there is neither a surjective
R-linear map from R to I nor a surjective R-linear map from I to R.
How might we strengthen our hypothesis on R, M, and N, then, to be able to conclude
that there is a surjective R-linear map from M to N or, more generally, from M to N ⊕t
for a given positive integer t? Before considering this question further, we introduce some
definitions and notation:
Date: January 9, 2018.
2010 Mathematics Subject Classification. Primary 13E05; Secondary 13C05, 13D15, 13E15, 13F05.
Key words and phrases. basic set, splitting capacity, surjective capacity.
This paper constitutes a portion of a dissertation [3] that will be submitted to the Department of Mathematics and Statistics at Georgia State University in partial fulfillment of the requirements for the degree of
Doctor of Philosophy. The author would like to thank his advisor Yongwei Yao for suggesting the problems
considered in this paper, for proposing several definitions that ultimately led to solutions of these problems,
and for reviewing many early drafts of this work. The author would also like to thank Florian Enescu for
his helpful comments during a presentation of some of these results at the algebra seminar at Georgia State
University.
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Definition 0.1. Let R be a commutative ring, S an R-algebra, and M and N right Smodules. We let surS (M, N) denote the supremum of the nonnegative integers t such that
there exists a surjective S-linear map from M to N ⊕t , and we refer to surS (M, N) as the
global surjective capacity of M with respect to N over S.
Let p ∈ Spec(R). We refer to surSp (Mp , Np ) as the local surjective capacity of M with
respect to N over S at p.
Under the hypotheses of the previous definition, we observe that we can always get an
upper bound on a global surjective capacity in terms of local surjective capacities:
surS (M, N) 6 inf{surSm (Mm , Nm ) : m ∈ Max(R) ∩ SuppR (N)}.

Can we also get a lower bound on a global surjective capacity using local surjective capacities?
Bass gives an answer to this question in his work on stable algebra [4, Theorem 8.2]. In the
sequel, we endow every subset Y of Spec(R) with the subspace topology induced by the
Zariski topology on Spec(R), and we define dim(Y ) to be the Krull dimension of Y as a
topological space. We define dim(∅) = −∞.

Theorem 0.2 (Bass [4, Theorem 8.2]). Let R be a commutative ring, S a module-finite Ralgebra, and M a direct summand of a direct sum of finitely presented right S-modules. Suppose that Y := Max(R) is Noetherian with dim(Y ) < ∞. Suppose also that surSm (Mm , Sm ) >
1 + dim(Y ) for every m ∈ Y . Then surS (M, S) > 1.

We generalize this theorem in two ways: We replace the module S with an arbitrary
finitely generated right S-module N, and we change the number 1 to an arbitrary positive
integer t. With the help of these modifications, we get two new conclusions.

Theorem 0.3. Let R be a commutative ring, S a module-finite R-algebra, M a direct summand of a direct sum of finitely presented right S-modules, and N a finitely generated right
S-module. Suppose that Y := Max(R) ∩ SuppR (N) is Noetherian with dim(Y ) < ∞. Then
the following statements hold:
(1) Let t be a positive integer, and suppose that surSm (Mm , Nm ) > t + dim(Y ) for every
m ∈ Y . Then surS (M, N) > t.
(2) surS (M, N) = ∞ if and only if surSm (Mm , Nm ) = ∞ for every m ∈ Y .
(3) Suppose that surSn (Mn , Nn ) < ∞ for some n ∈ Y . Then
surS (M, N) > min{surSm (Mm , Nm ) : m ∈ Y } − dim(Y ).

Notice that Theorem 0.2 is concerned with the existence of a surjective S-linear map from
M to S. If such a map exists, then it must split. This observation motivates the following
definition:
Definition 0.4. Let R be a commutative ring, S an R-algebra, and M and N right Smodules. We let splS (M, N) denote the supremum of the nonnegative integers t such that
there exists a split surjective S-linear map from M to N ⊕t , and we refer to splS (M, N) as
the global splitting capacity of M with respect to N over S.
Let p ∈ Spec(R). We refer to splSp (Mp , Np ) as the local splitting capacity of M with respect
to N over S at p.
As with surjective capacities, we can always get an upper bound on a global splitting
capacity in terms of local splitting capacities:
splS (M, N) 6 inf{splSm (Mm , Nm ) : m ∈ Max(R) ∩ SuppR (N)}.

SURJECTIVE AND SPLITTING CAPACITIES

3

Next, we notice that, in Theorem 0.2, we can replace every instance of the symbol sur
with the symbol spl to get a lower bound on a global splitting capacity using local splitting
capacities. Our next theorem ensures that we can make the same modifications to Theorem 0.3 if we assume that N is finitely presented over S. For emphasis, we mention that N
need not be flat over S for the following theorem to hold.
Theorem 0.5. Let R be a commutative ring, S a module-finite R-algebra, M a direct summand of a direct sum of finitely presented right S-modules, and N a finitely presented right
S-module. Suppose that Y := Max(R) ∩ SuppR (N) is Noetherian with dim(Y ) < ∞. Then
the following statements hold:
(1) Let t be a positive integer, and suppose that splSm (Mm , Nm ) > t + dim(Y ) for every
m ∈ Y . Then splS (M, N) > t.
(2) splS (M, N) = ∞ if and only if splSm (Mm , Nm ) = ∞ for every m ∈ Y .
(3) Suppose that splSn (Mn , Nn ) < ∞ for some n ∈ Y . Then
splS (M, N) > min{splSm (Mm , Nm ) : m ∈ Y } − dim(Y ).
Now suppose that R is a commutative Noetherian ring and that M and N are finitely
generated R-modules. In this more restricted setting, we can improve our lower bounds on
global surjective and splitting capacities. Our starting point for this work is Serre’s Splitting
Theorem from algebraic K-theory [14, Théorème 1 ]. Below, dimX (p) refers to the Krull
dimension of Var(p) ∩ X, where p ∈ X ⊆ Spec(R) and Var(p) := {q ∈ Spec(R) : p ⊆ q}.
The symbol j-Spec(R) refers to the set of all p ∈ Spec(R) such that p can be expressed as
an intersection of maximal ideals of R. The set j-Spec(R) was introduced by Swan in [16]
and was used by Eisenbud and Evans in [9] and by De Stefani, Polstra, and Yao in [6].
In the corollary following Proposition 1 in [16], Swan notes that j-Spec(R) is Noetherian
if and only if Max(R) is Noetherian. In the same corollary, Swan indicates that dim(jSpec(R)) = dim(Max(R)).
Theorem 0.6 (Serre’s Splitting Theorem [14, Théorème 1 ]). Let R be a commutative Noetherian ring, and let P be a finitely generated projective R-module. Let X := j- Spec(R),
and suppose that dim(X) < ∞. Then
surR (P, R) > inf{surRp (Pp , Rp ) − dimX (p) : p ∈ X}.
The following result by De Stefani, Polstra, and Yao generalizes Serre’s Splitting Theorem
by replacing P with an arbitrary finitely generated R-module M:
Theorem 0.7 (De Stefani, Polstra, and Yao [6, Theorem 3.12]). Let R be a commutative
Noetherian ring, and let M be a finitely generated R-module. Let X := j- Spec(R), and
suppose that dim(X) < ∞. Then
surR (M, R) > inf{surRp (Mp , Rp ) − dimX (p) : p ∈ X}.
Theorem 0.7 can be compared with another extension of Serre’s Splitting Theorem due to
Stafford [15, Theorem 5.7].
We generalize Theorem 0.7 in the following manner: We let S be a module-finite R-algebra;
we allow M to be a right S-module; and we replace the module R with an arbitrary finitely
generated S-module N:

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Theorem 0.8. Let R be a commutative Noetherian ring, S a module-finite R-algebra, and
M and N finitely generated right S-modules. Let X := j- Spec(R) ∩ SuppR (N), and suppose
that dim(X) < ∞. Then
surS (M, N) > inf{surSp (Mp , Np ) − dimX (p) : p ∈ X}.

As with Theorem 0.2, we can replace every instance of the symbol sur with the symbol
spl in Theorems 0.6 and 0.7. Our next theorem indicates that we can revise Theorem 0.8 in
the same way:
Theorem 0.9. Let R be a commutative Noetherian ring, S a module-finite R-algebra, and
M and N finitely generated right S-modules. Let X := j- Spec(R) ∩ SuppR (N), and suppose
that dim(X) < ∞. Then
splS (M, N) > inf{splSp (Mp , Np ) − dimX (p) : p ∈ X}.

We now describe the structure of our paper. In Section 1, we prove Theorem 0.8, modulo
a lemma (Lemma 1.12). We call this lemma the Surjective Lemma, and we prove it in several
stages: Over the course of Sections 2 and 3, we first reduce the proof of the Surjective Lemma
to a verification that a certain condition holds on a finite set Λ of prime ideals of a given
commutative ring R. Along the way, we prove Theorem 0.3, modulo the Surjective Lemma.
In Section 4, we study the maximal ideals of R in Λ and continue working toward a proof of
the Surjective Lemma. In Section 5, we address the remaining members of Λ and complete
a proof of the Surjective Lemma. In Section 6, we show that Theorems 0.5 and 0.9 can be
proved by making a few key modifications to the proofs of Theorems 0.3 and 0.8. Finally,
in Section 7, we consider the case of finitely generated modules over a Dedekind domain; in
this case, we show that we can give conditions equivalent to having a given global surjective
or splitting capacity.
In Sections 1–5, we abide by the following conventions: We let R be a commutative ring;
we let S be a module-finite R-algebra; we let M denote a right S-module; and we let N
denote a finitely generated right S-module. We view every left and right S-module as a
standard R-module in the natural way.
1. A Proof of Theorem 0.8, modulo the Surjective Lemma
In this section, we prove Theorem 0.8, modulo the Surjective Lemma (Lemma 1.12). The
statement of the Surjective Lemma requires the following definition, which establishes analogues of local and global surjective capacities for arbitrary R-submodules of HomS (M, N):
Definition 1.1. Let F be an R-submodule of HomS (M, N), and let p ∈ Spec(R). We let
∂(F ) denote the supremum of the nonnegative integers t such that there exists a surjective
f ∈ F ⊕t ⊆ HomS (M, N ⊕t ). We let ∂p (F ) denote the supremum of the nonnegative integers
t such that there exists f ∈ F ⊕t with the property that fp is surjective.
Let n be a positive integer. We will think of a member of HomS (M, N ⊕n ) as a column
 
f1
 ...  ,
fn
where f1 , . . . , fn ∈ HomS (M, N). When we refer to such an n-tuple without using a display,
we write (f1 , . . . , fn )⊤ to denote the transpose of a row of functions.

SURJECTIVE AND SPLITTING CAPACITIES

5

Remark 1.2. Let F be a finitely generated R-submodule of HomS (M, N). We observe that
∂(F ) = ∞ if and only if N = 0: Certainly, if N = 0, then ∂(F ) = ∞. Suppose that ∂(F ) =
∞, and let n := 1 + µR (F ). Then there exist f1 , . . . , fn ∈ F such that F = Rf1 + · · · + Rfn .
Let f := (f1 , . . . , fn )⊤ . Since ∂(F ) = ∞, there exists an (n + 1) × n matrix B with entries
in R such that Bf is surjective. Hence B represents a surjective R-linear map from N ⊕n to
N ⊕(n+1) . Now, for every m ∈ Max(R), we see that B ⊗ 1R/m is surjective, which implies that
Nm /mm Nm ∼
= N/mN = 0 since N is finitely generated over R. Hence, for every m ∈ Max(R),
Nakayama’s Lemma tells us that Nm = 0 since Nm is finitely generated over Rm . Thus N = 0.
Let p ∈ Spec(R). Then, as a result of the discussion above, ∂p (F ) = ∞ if and only if
p 6∈ SuppR (N).

Using the symbol ∂, we now describe a condition weaker than surjectivity that we can
impose on a member of HomS (M, N ⊕n ).

Definition 1.3. Let n, t be positive integers with n > t; let p ∈ X ⊆ Spec(R); and let
f := (f1 , . . . , fn )⊤ ∈ HomS (M, N ⊕n ). We say that f is (t, X, p)-surjective if ∂p (Rf1 + · · · +
Rfn ) > min{n, t + dimX (p)}.
Let Y ⊆ X. We say that f is (t, X, Y )-surjective if f is (t, X, q)-surjective for every q ∈ Y .
When t and X are understood, we use the less cumbersome terms p-surjective and Y surjective in place of (t, X, p)-surjective and (t, X, Y )-surjective, respectively.
Remark 1.4. Maintaining the hypotheses in the previous definition, we see that f is (n, X, p)surjective if and only if fp is surjective: Suppose that f is (n, X, p)-surjective. Then there
exists an n × n matrix B with entries in R such that Bp fp = (Bf )p is surjective. Now Bp
represents a surjective Sp -linear map from Np⊕n to itself. Since N is finitely generated over
R, we see that Np is finitely generated over Rp . Hence Bp is bijective, and so fp is surjective.
Conversely, if fp is surjective, then ∂p (Rf1 + · · · + Rfn ) = n, and so f is (n, X, p)-surjective.
We need one more definition to state the Surjective Lemma, and this concerns the notion
of a basic set for R. De Stefani, Polstra, and Yao use basic sets to prove [6, Theorems 3.9
and 4.5]. These theorems have conclusions that are weaker than those of [6, Theorems 3.12
and 4.8], respectively, but the hypotheses of [6, Theorems 3.9 and 4.5] are more general.
Later in this section, we state an analogue of [6, Theorems 3.9 and 4.5] and a generalization
of [6, Theorems 3.12 and 4.8] in Theorem 1.13. We generalize [6, Theorems 3.9 and 4.5] in
a separate paper [2] where our goal is to extend Bass’s Cancellation Theorem.
Here is the definition of a basic set for R, along with a few examples:
Definition 1.5. Let X ⊆ Spec(R). We say that X is a basic set for R if X is Noetherian
and if, for every p ∈ Spec(R) that can be written as an intersection of members of X, it is
the case that p ∈ X.

Example 1.6. Every finite subset of Spec(R) is a basic set for R.

Example 1.7. If R is Noetherian, then Spec(R) is a basic set for R. If R is a Jacobson ring,
then j-Spec(R) = Spec(R). Hence, if R is a Noetherian Jacobson ring, then j-Spec(R) is a
basic set for R. In fact, a more general statement holds, as the next example shows.
Example 1.8. Let X := j-Spec(R) ∩ SuppR (N), and suppose that X is Noetherian. Then
X is a basic set for R: Let p ∈ Spec(R) such that p is an intersection of members of X.
Since every member of j-Spec(R) is an intersection of maximal ideals of R, so is p. Hence
p ∈ j-Spec(R). Since N is finitely generated over R, we see that SuppR (N) = Var(AnnR (N)).

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Hence p is an intersection of prime ideals that contain AnnR (N), and so p must itself contain
AnnR (N). Thus p ∈ Var(AnnR (N)) = SuppR (N). We have proved then that p ∈ X. Thus
X is a basic set for R.
Example 1.9. By Example 1.8, if j-Spec(R) is Noetherian, then j-Spec(R) is a basic set
for R. If R is Artinian, then Max(R) = j-Spec(R) = Spec(R). If R is semilocal (Noetherian
with only finitely many maximal ideals) but not Artinian, then Max(R) = j-Spec(R) (
Spec(R), and dim(j-Spec(R)) = 0 < dim(Spec(R)).
Example 1.10. Suppose that R is a one-dimensional Noetherian domain with infinitely
many maximal ideals. Then Max(R) is not a basic set for R: If Max(R) is a basic set for R,
then Jac(R) 6= 0, and so Min(Jac(R)) = Max(R) is a finite set, a contradiction.
Notice that, in this example, Max(R) ( j-Spec(R) = Spec(R).
Example 1.11. Let d ∈ Z with d > 2. Let T be the domain
K[x1 , . . . , xd , y1 , y2, y3 , . . .],

where K is a field and x1 , . . . , xd , y1 , y2 , y3 , . . . are indeterminates. Invert every element of T
outside of the set
(T x1 + · · · + T xd ) ∪ T y1 ∪ T y2 ∪ T y3 ∪ · · ·
to form a new domain, and suppose that R is this new domain. Example 3.3 in [13] then
indicates that R is a d-dimensional Noetherian ring such that
Max(R) = {Rx1 + · · · + Rxd , Ry1 , Ry2, Ry3 , . . .}

and such that every nonzero prime ideal of R is contained a unique maximal ideal. We will
prove that Max(R) ( j-Spec(R) ( Spec(R).
To prove that the first inclusion is strict, we will show that 0 ∈ j-Spec(R) − Max(R).
Of course, 0 6∈ Max(R). Suppose that 0 6∈ j-Spec(R). Then Jac(R) 6= 0. Since R is
Noetherian, Min(Jac(R)) is a finite set. On the other hand, Min(Jac(R)) contains the
infinite set {Ry1 , Ry2 , Ry3, . . .}, a contradiction. Hence 0 ∈ j-Spec(R).
To prove that j-Spec(R) 6= Spec(R), we will first show that j-Spec(R) = {0} ∪ Max(R).
We have already shown that 0 ∈ j-Spec(R), and Max(R) ⊆ j-Spec(R) by definition, so it
remains to show that every nonzero member p of j-Spec(R) is in Max(R). As we mentioned
above, p is contained in a unique maximal ideal of R. Since p is an intersection of maximal
ideals of R, it must then be the case that p ∈ Max(R). Hence j-Spec(R) = {0} ∪ Max(R),
and so Rx1 ∈ Spec(R) − j-Spec(R). Thus j-Spec(R) 6= Spec(R).
Notice that, in this example, dim(j-Spec(R)) = 1 < d = dim(Spec(R)). In particular, if
X := Spec(R) and Y := j-Spec(R), then dimY (0) = 1 < d = dimX (0).
We are now prepared to state the Surjective Lemma. This lemma is similar in spirit to [4,
Section 8, Lemma II]; [5, Theorem 2.4]; [6, Lemmas 3.8 and 4.4]; [9, Lemma 3, Lemma 5,
and Theorem B]; [11, Theorem 2.1]; [14, Théorème 2 ]; and [15, Lemma 5.4], although several
techniques that we use to prove the Surjective Lemma, as we will see, are decidedly distinct
from previous work.
Lemma 1.12 (Surjective Lemma). Let n, t be positive integers with n > 1 + t, and let X
be a subset of SuppR (N) that is a basic set for R. Let f := (f1 , . . . , fn )⊤ ∈ HomS (M, N ⊕n ),
′
and suppose that f is (t, X, X)-surjective. Then there exist f1′ , . . . , fn−1
∈ Rf1 + · · · + Rfn
′
′
′
⊤
such that f := (f1 , . . . , fn−1) is (t, X, X)-surjective.

SURJECTIVE AND SPLITTING CAPACITIES

7

We prove this lemma in Section 5. Given this lemma, we can prove Theorem 1.13 below.
Part (1) of Theorem 1.13 provides an analogue of [6, Theorems 3.9 and 4.5]; and Part (2)
generalizes [6, Theorems 3.12 and 4.8] and [14, Théorème 1 ]. Theorems that extend [14,
Théorème 1 ] in other directions include [5, Corollary 3.2]; [9, Theorem A]; [11, Theorem 2.5];
and [15, Theorem 5.7].
Theorem 1.13. Let L be an S-submodule of M; let F be a finitely generated R-submodule
of HomS (L, N); and let G be an R-submodule of HomS (M, N). Suppose that every member
of F can be extended to a member of G. Let X be a subset of SuppR (N) that is a basic set
for R, and suppose that dim(X) < ∞. Then the following statements hold:
(1) Let t be a positive integer, and suppose that ∂p (F ) > t + dimX (p) for every p ∈ X.
Then there exists g ∈ G⊕t such that gp is surjective for every p ∈ X.
(2) Suppose that Max(R) ∩ SuppR (N) ⊆ X. Then
∂(G) > inf{∂p (F ) − dimX (p) : p ∈ X}.
Proof of Theorem 1.13, modulo the Surjective Lemma. (1) We may assume that X 6= ∅. Let
n := µR (F ). Since t is a positive integer, n is a positive integer. Let f1 , . . . , fn ∈ F such
that F = Rf1 + · · · + Rfn ; let f := (f1 , . . . , fn )⊤ ; and let q ∈ X. Then there exists a t × n
matrix B with entries in R such that (Bf )q = Bq fq is surjective. Hence B ⊗1κ(q) represents a
surjective κ(q)-linear map from N ⊕n ⊗κ(q) to N ⊕t ⊗κ(q). Since N ⊗κ(q) is finitely generated
over κ(q), we see that n > t. Thus f is (t, X, X)-surjective. Now, after n − t applications
of the Surjective Lemma, we obtain f1′ , . . . , ft′ ∈ F such that f ′ := (f1′ , . . . , ft′ )⊤ is (t, X, X)surjective. By Remark 1.4, we see that fp′ is surjective for every p ∈ X. By assumption,
the members f1′ , . . . , ft′ of F can be extended to members g1 , . . . , gt of G, respectively. Let
g := (g1 , . . . , gt )⊤ ∈ G⊕t . Since fp′ is surjective for every p ∈ X, we see that gp is surjective
for every p ∈ X.
(2) Let
t := inf{∂p (F ) − dimX (p) : p ∈ X}.
If t 6 0, then there is nothing to prove. If t = ∞, then N = 0 by Remark 1.2, and so
∂(G) = t. Suppose then that t is a positive integer. By Part (1) of this theorem, there exists
g ∈ G⊕t such that gp is surjective for every p ∈ X. Since Max(R) ∩ SuppR (N) ⊆ X, we see
that g is surjective. Hence ∂(G) > t.

We now show that Theorem 0.8 is an easy corollary of Theorem 1.13.
Proof of Theorem 0.8, modulo the Surjective Lemma. Let L = M, and let F = G denote
HomS (M, N). By our hypotheses, F is finitely generated over R. By Example 1.8, we see
that X is a subset of SuppR (N) that is a basic set for R. Also, Max(R) ∩ SuppR (N) ⊆ X.
Hence, by Part (2) of Theorem 1.13, we get
surS (M, N) = ∂(G)
> inf{∂p (F ) − dimX (p) : p ∈ X}
= inf{surSp (Mp , Np ) − dimX (p) : p ∈ X}. 
With a little more work, we can also prove Theorem 0.3, modulo the Surjective Lemma.
We accomplish this goal in the next section.

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ROBIN BAIDYA

2. A Proof of Theorem 0.3, modulo the Surjective Lemma
Under the hypotheses of Theorem 0.8, we see that HomS (M, N) is finitely generated over
R, and we see that [HomS (M, N)]p and HomSp (Mp , Np ) are isomorphic as Rp -modules for
every p ∈ Spec(R). Hence, a finite number of global S-linear maps govern all local surjective
capacities.
This property does not follow from the more general hypotheses of Theorem 0.3. As a
result, we are not immediately in a position to apply the Surjective Lemma (Lemma 1.12)
when attempting to prove Theorem 0.3.
In this section, we show that we can circumvent the issue by obtaining a finitely generated
R-submodule F of HomS (M, N) such that ∂m (F ) is sufficiently large for every m ∈ Max(R)∩
SuppR (N). We can then prove Theorem 0.3, modulo the Surjective Lemma.
With this goal in mind, we first show that, for every R-submodule F of HomS (M, N),
the function on Spec(R) taking p to ∂p (F ) is lower semicontinuous. This lemma can be
compared to [4, Lemma 7.2]; [6, Lemmas 3.5 and 4.1]; [11, Theorem 1.5]; [14, Lemme 3 ];
and [16, Proposition 3].
Lemma 2.1. Let F be an R-submodule of HomS (M, N), and let t be a nonnegative integer.
Then the set {p ∈ Spec(R) : ∂p (F ) > t} is open, and so the set {p ∈ Spec(R) : ∂p (F ) 6 t} is
closed. Hence, for every subspace X of Spec(R), the set Yt := {p ∈ X : ∂p (F ) 6 t} is closed
in X.
Proof. Let p ∈ Spec(R) such that ∂p (F ) > t; let f ∈ F ⊕(t+1) such that fp is surjective; and
let C := coker f . Since Cp = 0 and since C is finitely generated over R, there is an element
s ∈ R − p such that sC = 0. Let U := {q ∈ Spec(R) : s 6∈ q}. Then Cq = 0 for every q ∈ U.
Hence U is an open neighborhood of p such that ∂q (F ) > t for every q ∈ U. Thus the set
{p ∈ Spec(R) : ∂p (F ) > t} is open. This proves the first claim of the lemma. The last two
claims of the lemma follow from the first claim.

Using the previous lemma, we can show that, under the hypotheses of Part (1) of Theorem 0.3, there exists a finitely generated R-submodule F of HomS (M, N) such that ∂m (F ) >
t + dim(Y ) for every m ∈ Y . This is, in fact, a consequence of a more general statement:

Lemma 2.2. Suppose that M is a direct summand of a direct sum of finitely presented right
S-modules. Let X be a Noetherian subspace of SuppR (N), and suppose that dim(X) < ∞.
Let t be a positive integer, and suppose that surSp (Mp , Np ) > t + dim(X) for every p ∈ X.
Then there exists a finitely generated R-submodule F of HomS (M, N) such that ∂p (F ) >
t + dim(X) for every p ∈ X.

Proof. Let u := (t − 1) + dim(X), and let G denote the collection of all finitely generated
R-submodules of HomS (M, N). For every G ∈ G , let Y (G) := {p ∈ X : ∂p (G) 6 u},
and let Y := {Y (G) : G ∈ G }. We aim to prove that ∅ ∈ Y . Suppose not. Then X
is nonempty. By Lemma 2.1, we see that Y (G) is closed in X for every G ∈ G . Since
X is Noetherian and nonempty, there is G′ ∈ G such that Y (G′ ) is a minimal member
of Y . By assumption, Y (G′ ) 6= ∅, so let q ∈ Y (G′ ). By our hypothesis on M, there
exist h1 , . . . , hu+1 ∈ HomS (M, N) such that h := (h1 , . . . , hu+1 )⊤ has the property that hq is
surjective. Let H := G′ + Rh1 + · · · + Rhu+1 . Then H ∈ G , and ∂q (H) > u + 1 = t + dim(X).
Thus q ∈ Y (G′ ) −Y (H), and so Y (H) ( Y (G′ ), contradicting the minimality of Y (G′ ) in Y .
Thus ∅ ∈ Y , and so there exists a finitely generated R-submodule F of HomS (M, N) such
that Y (F ) = ∅. Consequently, ∂p (F ) > u + 1 = t + dim(X) for every p ∈ X.


SURJECTIVE AND SPLITTING CAPACITIES

9

We now prove Theorem 0.3, modulo the Surjective Lemma.
Proof of Theorem 0.3, modulo the Surjective Lemma. Let X := j-Spec(R) ∩ SuppR (N). By
the corollary following Theorem 1 in [16], we see that dim(X) = dim(Y ) < ∞ and that X
is Noetherian since Y is Noetherian. By Example 1.8, we see that X is a basic set for R.
(1) Since surSm (Mm , Nm ) > t + dim(Y ) for every m ∈ Y , we see that surSp (Mp , Np ) > t +
dim(X) for every p ∈ X. By Lemma 2.2, there exists a finitely generated R-submodule F of
HomS (M, N) such that ∂p (F ) > t + dim(X) > t + dimX (p) for every p ∈ X. Now, let L = M
and G = HomS (M, N). Then Part (2) of Theorem 1.13 tells us that surS (M, N) = ∂(G) > t.
(2) Certainly, if surS (M, N) = ∞, then surSm (Mm , Nm ) = ∞ for every m ∈ Y . Suppose
then that surSm (Mm , Nm ) = ∞ for every m ∈ Y . Then, surSm (Mm , Nm ) > t+dim(Y ) for every
m ∈ Y and for every positive integer t. Hence, by Part (1) of this theorem, surS (M, N) > t
for every positive integer t. Thus surS (M, N) = ∞.
(3) Let
t := min{surSm (Mm , Nm ) : m ∈ Y } − dim(Y ).
Then surSm (Mm , Nm ) > t + dim(Y ) for every m ∈ Y . If t 6 0, then certainly surS (M, N) > t.
Suppose then that t is a positive integer. Then, by Part (1) of this theorem, surS (M, N) > t.

We can also prove the following variations of Lemma 2.2 and Theorems 1.13 and 0.3.
These variations are noteworthy in the sense that they do not require M to be a direct
summand of a direct sum of finitely presented right S-modules. We omit the proofs since
they are similar to the proofs of Lemma 2.2 and Theorems 1.13 and 0.3.
Lemma 2.3. Let F be an R-submodule of HomS (M, N). Let X be a Noetherian subspace
of SuppR (N), and suppose that dim(X) < ∞. Let t be a positive integer, and suppose that
∂p (F ) > t + dim(X) for every p ∈ X. Then there exists a finitely generated R-submodule F ′
of F such that ∂p (F ′ ) > t + dim(X) for every p ∈ X.
Theorem 2.4. Let L be an S-submodule of M; let F be an R-submodule of HomS (L, N);
and let G be an R-submodule of HomS (M, N). Suppose that every member of F can be
extended to a member of G. Then the following statements hold:
(1) Let X be a subset of SuppR (N) that is a basic set for R with dim(X) < ∞. Let t be
a positive integer, and suppose that ∂p (F ) > t + dim(X) for every p ∈ X. Then there
exists g ∈ G⊕t such that gp is surjective for every p ∈ X.
(2) Suppose that Y := Max(R) ∩ SuppR (N) is Noetherian with dim(Y ) < ∞. Then the
following statements hold:
(a) Let t be a positive integer, and suppose that ∂m (F ) > t+dim(Y ) for every m ∈ Y .
Then ∂(G) > t.
(b) If ∂m (F ) = ∞ for every m ∈ Y , then ∂(G) = ∞. Hence ∂(F ) = ∞ if and only
if ∂p (F ) = ∞ for every m ∈ Y .
(c) Suppose that ∂n (F ) < ∞ for some n ∈ Y . Then
∂(G) > min{∂m (F ) : m ∈ Y } − dim(Y ).
In the next section, we continue working toward a proof of the Surjective Lemma.

10

ROBIN BAIDYA

3. The Set Λ
Assuming the hypotheses of the Surjective Lemma (Lemma 1.12), we show in this section
that there is a finite subset Λ of X with a crucial property: If there exists an invertible
n × n matrix A with entries in R such that the first n − 1 components of Af := (f1′ , . . . , fn′ )⊤
′
form a map f ′ := (f1′ , . . . , fn−1
)⊤ that is (t, X, Λ)-surjective, then f ′ is (t, X, X)-surjective.
In other words, we show that proving the Surjective Lemma, which may involve infinitely
many prime ideals of R, reduces to proving a statement about Λ, a finite set of prime ideals.
We begin by covering two properties of basic sets for R with Proposition 3.1. Property (1),
which holds not only for every basic set but also for every Noetherian topological space, is
proved, for example, in [10, Proposition 1.5]. Property (2) is proved for the special case of
X := j-Spec(R) in [16, Proposition 2], although the more general result here can be proved
by similar means.
Proposition 3.1. Let X be a basic set for R. Then X has the following properties:
(1) Every closed subset of X is a union of finitely many closed irreducible subsets of X.
(2) Every closed irreducible subset of X has a unique generic point.
In the next lemma, we define the set Λ mentioned earlier in this section. We omit the
proof of this lemma since it is similar to the proofs of [6, Lemmas 3.6 and 4.2].
Lemma 3.2. Let F be a finitely generated R-submodule of HomS (M, N), and let X be a
subset of SuppR (N) that is a basic set for R. Then there exists a finite subset Λ of X
such that, for every p ∈ X − Λ, there exists q ∈ Λ with the properties that q ( p and
∂q (F ) = ∂p (F ).
Assume the hypotheses of Part (2) of Theorem 1.13, and define Λ as in Lemma 3.2 with
respect to F and X. In Part (2) of Theorem 1.13, we give a lower bound on ∂(G) whose
expression involves all of the members of X, where X could be an infinite set. In Part (1)
of the following corollary, we express this same lower bound using only the members of the
finite set Λ. As a result, we can improve our expression of the lower bound on surS (M, N)
in Theorem 0.8 as well, and this is the content of Part (2) of the following corollary. Of
course, since Theorems 1.13 and 0.8 rely on the Surjective Lemma, we still need to assume
the truth of the Surjective Lemma for the following corollary.
Corollary 3.3. We make the following improvements to Theorems 1.13 and 0.8:
(1) Assume the hypotheses of Part (2) of Theorem 1.13, and let Λ be defined as in
Lemma 3.2 with respect to F and X. Then
∂(G) > inf{∂p (F ) − dimX (p) : p ∈ Λ}.

(2) Assume the hypotheses of Theorem 0.8, and let Λ be defined as in Lemma 3.2 with
respect to F := HomS (M, N) and X. Then
surS (M, N) > inf{surSp (Mp , Np ) − dimX (p) : p ∈ Λ}.
Proof of Corollary 3.3, modulo the Surjective Lemma. (1) Let
and let

t := inf{∂p (F ) − dimX (p) : p ∈ X},
u := inf{∂p (F ) − dimX (p) : p ∈ Λ}.

SURJECTIVE AND SPLITTING CAPACITIES

11

We will show that t = u. Certainly t 6 u, so it remains to show that t > u. If t = ∞,
then certainly t > u. Suppose then that t is an integer so that X 6= ∅. Let p0 ∈ X such
that t = ∂p0 (F ) − dimX (p0 ). Suppose, by way of contradiction, that p0 6∈ Λ. By Lemma 3.2,
there exists q0 ∈ Λ such that q0 ( p0 and ∂q0 (F ) = ∂p0 (F ). Hence
t = ∂p0 (F ) − dimX (p0 ) > ∂q0 (F ) − dimX (q0 ) > t,
a contradiction. Thus p0 ∈ Λ, and so t > u. Thus t = u. Now, by Part (2) of Theorem 1.13,
we see that ∂(G) > t = u.
(2) Let L = M, and let G = F . Since F is finitely generated over R, Part (1) of this
corollary tells us that
surS (M, N) = ∂(G)
> inf{∂p (F ) − dimX (p) : p ∈ Λ}
= inf{surSp (Mp , Np ) − dimX (p) : p ∈ Λ}. 
We now return to the task of reducing the proof of the Surjective Lemma to the study of
a finite subset Λ of X. The following definition will be useful in subsequent work:
Definition 3.4. Let n be a positive integer. The symbol GL(n, R) refers to the group of
all invertible n × n matrices with entries in R. This group is called the general linear group
of degree n over R.
Remark 3.5. With respect to the previous definition, an n × n matrix with entries in R is in
GL(n, R) if and only if its determinant is a unit of R.
In the following definition, we establish a large amount of notation that we will use in the
remainder of this section and in the next three sections:
Definition 3.6. Let n ∈ Z with n > 2; let f := (f1 , . . . , fn )⊤ ∈ HomS (M, N ⊕n ); and let
p ∈ SuppR (N). Fix the following notation relative to n, f , and p: Let F := Rf1 + · · · + Rfn ;
let − denote the functor
− ⊗R κ(p); and, for every matrix Ξ := (ξi,j ) with entries in R or

Rp , let Ξ := ξi,j . If a matrix A ∈ GL(n, R) is given, then let f ∗ := Af := (f1′ , . . . , fn′ )⊤ ;
′
′
let f ′ := (f1′ , . . . , fn−1
)⊤ ; and let F ′ := Rf1′ + · · · + Rfn−1
.
Upon proving the following lemma, we will be prepared to achieve the goal of this section.
Lemma 3.7. Let n ∈ Z with n > 2; let f := (f1 , . . . , fn )⊤ ∈ HomS (M, N ⊕n ); let A ∈
GL(n, R); and let p ∈ SuppR (N). Then, relative to Definition 3.6, we have ∂p (F ′) >
∂p (F ) − 1.
Proof. If ∂p (F ) 6 1, then ∂p (F ′ ) > 0 > ∂p (F ) − 1, and so we are done. Suppose then, for
the rest of the proof, that ∂p (F ) > 2.
Let d := ∂p (F ), and let B be a d ×n matrix with entries in R such that (Bf )p is surjective.
Let C ∈ GL(d, Rp ) such that CBA−1 can be represented as a matrix (bi,j ) with entries in
R, where b1,n , . . . , bd−1,n ∈ p. Hence


b1,1 · · · b1,n−1
0
 ..
.. 
..
..

.
. 
.
CBA−1 =  .
.
bd−1,1 · · · bd−1,n−1 0 
bd,1 · · · bd,n−1 bd,n

12

ROBIN BAIDYA

Let

b1,1 · · · b1,n−1
..  .
..
B ′ :=  ...
.
.
bd−1,1 · · · bd−1,n−1


Then B ′ f ′ ∈ (F ′ )⊕(d−1) , and B ′ f ′ is surjective. Nakayama’s Lemma then tells us that (B ′ f ′ )p
is surjective. Hence ∂p (F ′ ) > d − 1 = ∂p (F ) − 1.

Lemma 3.8. Assume the hypotheses of the Surjective Lemma, and define Λ as in Lemma 3.2
with respect to F := Rf1 +· · ·+Rfn and X. Let A ∈ GL(n, R), and suppose that, with respect
to Definition 3.6, we have that f ′ is (t, X, Λ)-surjective. Then f ′ is (t, X, X)-surjective.
Proof. Since t and X are understood, we can use the terms p-surjective and Y -surjective for
any p ∈ X and for any Y ⊆ X without the risk of confusion. Hence, we aim to prove that
f ′ is X-surjective.
Let p ∈ X − Λ. By Lemma 3.7, we have ∂p (F ′ ) > ∂p (F ) − 1. By Lemma 3.2, there exists
q ∈ Λ such that q ( p and ∂q (F ) = ∂p (F ). Since f is q-surjective by assumption, we now
have
∂p (F ′ ) > ∂p (F ) − 1
= ∂q (F ) − 1
> min{n, t + dimX (q)} − 1
> min{n − 1, t + dimX (p)},

and so f ′ is p-surjective. Thus f ′ is (X −Λ)-surjective. Since f ′ is Λ-surjective by assumption,
we conclude that f ′ is X-surjective.

Assume the hypotheses of the Surjective Lemma. By Lemma 3.8, proving the Surjective
Lemma now reduces to finding a matrix V ∈ GL(n, R) such that the first n − 1 components
of V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1 )⊤ that is (t, X, Λ)-surjective, where Λ is
defined as in Lemma 3.2 with respect to F := Rf1 + · · · + Rfn and X. We compute such
a matrix V over the course of the next two sections, completing a proof of the Surjective
Lemma in Section 5.
4. The Maximal Ideals of R in Λ
Assume the hypotheses of the Surjective Lemma, and define Λ as in Lemma 3.2 with
respect to F := Rf1 + · · · + Rfn and X. In this section, we find a matrix V ∈ GL(n, R)
such that the first n − 1 components of V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1 )⊤ that
is (t, X, m)-surjective for every m ∈ Λ ∩ Max(R). We accomplish this goal over the course of
three lemmas. The following definition will be useful in the work to come:
Definition 4.1. Let n ∈ Z with n > 2. For every i ∈ {1, . . . , n − 1}, let Pi be the n × n
permutation matrix obtained by switching the ith row and the nth row of the n × n identity
matrix, and let Pn denote the n × n identity matrix itself.
We begin with the following lemma:
Lemma 4.2. Assume the hypotheses of the Surjective Lemma; define Λ as in Lemma 3.2
with respect to F := Rf1 + · · · + Rfn and X; and let m ∈ Λ ∩ Max(R). Then there exist

SURJECTIVE AND SPLITTING CAPACITIES

13

elements rm,1 , . . . , rm,n−1 ∈ R and a number Lm ∈ {1, . . . , n} with the following property:
For all s1 , . . . , sn ∈ R − m and for every n × n matrix V with entries in R such that

1

0

V ≡  ...

0
0



s1
0 · · · 0 rm,1
.. 
. . . . ..

.
. .
. 
0
.
.. 
.. ..
P
L  .
.
. 0
. 
 m.
0
· · · 0 1 rm,n−1 
0
··· ··· 0
1


0 ···
0
0
.. 
..
.. ..
.
.
.
.
.. 
.. ..
(mod m),
.
.
.
0

· · · 0 sn−1 0 
··· ···
0
sn

the first n−1 components of V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1)⊤ that is (t, X, m)surjective.
Proof. For all s1 , . . . , sn ∈ R − m, if G := Rs1 f1 + · · · + Rsn fn , then ∂m (G) = ∂m (F ). Hence,
by Nakayama’s Lemma, it suffices to prove the case in which s1 = · · · = sn = 1.
Define every object in Definition 3.6 with respect to our current hypotheses, with m
taking the place of p and with A defined as the n × n identity matrix. Then we have
f ′ := (f1 , . . . , fn−1 )⊤ . Let d := ∂m (F ), and let B be a d × n matrix with entries in R such
that (Bf )m is surjective. Let C ∈ GL(d, Rm ) such that CB can be represented by a matrix
(bi,j ) with entries in R and such that CB is in the following reduced row echelon form, where
the nonzero entries are clustered toward the top right corner of the matrix:


.
.
.
0 · · · 0 1 .. 0 .. 0 ..


0 · · · 0 0 ... 1 ... 0 ...
.

CB = 

. . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
0 · · · 0 0 .. 0 .. 1 ..
Here, the vertical and horizontal ellipses denote possible omissions of entries, and the zero
columns on the left may not be present. For every i ∈ {1, . . . , d}, let ji be the smallest
number in the set {1, . . . , n} such that bi,ji 6= 0. We assume that, for every i ∈ {1, . . . , d},
the entry bi,ji is the only nonzero entry in the (ji )th column of CB.
Below, we consider several cases and prove the lemma in each case. Since t and X are
understood, we may use the term m-surjective for the rest of the proof without the risk of
ambiguity.
First suppose that f ′ is m-surjective. Then, by Nakayama’s Lemma, we may let Lm = n
and rm,1 = · · · = rm,n−1 = 0.
Suppose next that jd 6 n − 1. Then, by Nakayama’s Lemma, we may take Lm = n, and
we may define rm,j for every j ∈ {1, . . . , n − 1} as follows: If j = ji for some i ∈ {1, . . . , d},
then let rm,j = bi,n ; otherwise, let rm,j = 0.
Now suppose, for the rest of the proof, that f ′ is not m-surjective and that jd = n. If
∂m (F ) = n, then ∂m (F ′ ) > ∂m (F ) − 1 = n − 1 by Lemma 3.7, and so f ′ is m-surjective, a
contradiction. Hence ∂m (F ) 6 n − 1.
Since jd = n and since d = ∂m (F ) 6 n − 1, there exists k ∈ {1, . . . , n − 1} − {j1 , . . . , jd−1 }.
Accordingly, by Nakayama’s Lemma, we may let Lm be any such k, and we may define
rm,j for every j ∈ {1, . . . , n − 1} as follows: If j = ji for some i ∈ {1, . . . , d − 1}, then let
rm,j = bi,k ; otherwise, let rm,j = 0.


14

ROBIN BAIDYA

We use the next lemma to find a matrix Q ∈ GL(n, R) and elements
[
s1 , . . . , sn ∈ R −
m
m∈Λ∩Max(R)

such that


s1 0 · · ·
0
0
..
.. 
..
..

.
.
.
.
0
. .

.
.
Q ≡ PLm  ..
..
..
..  (mod m)
0


 0 · · · 0 sn−1 0 
0 ··· ···
0
sn

for every m ∈ Λ ∩ Max(R). We phrase the next lemma using language more general than
this because we will use the lemma again under different circumstances in Sections 5 and 6
of this paper and in a separate paper on cancellation properties of modules [2].
Lemma 4.3. Let R be a commutative ring, and let n ∈ Z with n > 2. Let Λ1 , . . . , Λn be
finite, pairwise disjoint subsets of Max(R). (Here, we allow some, or even all, of these sets
to be empty.) Then there exist a matrix Q ∈ GL(n, R) and elements
[
m
s1 , . . . , sn ∈ R −
m∈Λ1 ∪···∪Λn

such that, for every i ∈ {1, . . . , n} and for every m ∈ Λi , the matrix Q satisfies the congruence


s1 0 · · ·
0
0
..
.. 
.. ..

.
.
.
.
0

.. 
Q ≡ Pi  ... . . . . . .
(mod m).
0
.


 0 · · · 0 sn−1 0 
0 ··· ···
0
sn
Moreover, for any

a∈R−

[

m,

m∈Λ1

we can arrange for the first row of Q to be of the form

1 − ab 0 · · · 0 ab

for some b ∈ R.

Remark 4.4. We do not use the last claim of Lemma 4.3 in this paper, but we use it in a
paper on the cancellation of modules [2].
Proof of Lemma 4.3. Let
a∈R−

[

m.

m∈Λ1

For every i ∈ {1, . . . , n}, let
Ii :=

\

m∈Λi

m

and

Ji :=

\

j∈{1,...,n}−{i}




\

m∈Λj



m ,

SURJECTIVE AND SPLITTING CAPACITIES

15

and let
Ui :=

[

m

and

m∈Λi

Vi :=

[

j∈{1,...,n}−{i}

We would like to prove that there exist




[

m∈Λj



m .

a1 ∈ I1 − V1 , a2 ∈ I2 − V2 , . . . , an−1 ∈ In−1 − Vn−1 ,
b1 ∈ aJ1 − U1 , b2 ∈ J2 − U2 , . . . , bn−1 ∈ Jn−1 − Un−1 ,
c1 ∈ J1 − U1 , c2 ∈ J2 − U2 , . . . , cn−1 ∈ Jn−1 − Un−1 , cn ∈ Jn − Un
such that a1 = 1 − b1 and such that the n × n matrix


a1 0 · · ·
0
b1
.. 
..
..
..

.
.
. 
.
0

.. 
Q :=  ... . . . . . .
0
. 


 0 · · · 0 an−1 bn−1 
c1 · · · · · · cn−1 cn
has determinant 1 and is thus invertible. After we have accomplished this goal, we can
appeal to the Chinese Remainder Theorem to produce
[
m
s1 , . . . , sn ∈ R −
m∈Λ1 ∪···∪Λn

such that the following conditions hold:
(1) For every i ∈ {1, . . . , n − 1} and for every m ∈ (Λ1 ∪ · · · ∪ Λn ) − Λi , the element si
satisfies the congruence si ≡ ai (mod m).
(2) For every i ∈ {1, . . . , n − 1} and for every m ∈ Λi , the element sn satisfies the
congruence sn ≡ bi (mod m).
(3) For every i ∈ {1, . . . , n} and for every m ∈ Λi , the element si satisfies the congruence
si ≡ ci (mod m).
Finally, since b1 ∈ aJ1 , we can choose b ∈ J1 such that b1 = ab. Hence a1 = 1 − b1 = 1 − ab.
The matrix Q and the elements s1 , . . . , sn of R will then jointly satisfy all of the conditions
described in the lemma.
We note that, for any comaximal ideals
K and β ∈ L such
p for any α ∈ √
√ K, L of R and
2
2
that α + β = 1, it is the case that R = αR + βR = α R + β R ⊆ αK + βL, and so
αK + βL = R. We will use this observation shortly.
Next, we prove that aJ1 + J2 + · · · + Jn = R. Suppose not. Then there exists n ∈ Max(R)
such that aJ1 + J2 + · · · + Jn ⊆ n, and so aJ1 , J2 , . . . , Jn ⊆ n. Since aJ1 ⊆ n, we see that
a ∈ n or J1 ⊆ n. Either way, n 6∈ Λ1 . On the other hand, J2 , . . . , Jn ⊆ n, and so n ∈ Λ1 , a
contradiction. Hence aJ1 + J2 + · · · + Jn = R, and so J1 + J2 + · · · + Jn = R as well.
Let i ∈ {1, . . . , n − 1}, and suppose that we have defined
a1 ∈ I1 − V1 , a2 ∈ I2 − V2 , . . . , ai−1 ∈ Ii−1 − Vi−1 ,
b1 ∈ aJ1 − U1 , b2 ∈ J2 − U2 , . . . , bi−1 ∈ Ji−1 − Ui−1
and the ideals
Ki−1 := (b1 a2 · · · ai−2 J1 ) + · · · + (a1 · · · ai−3 bi−2 Ji−2 )
+(a1 · · · ai−2 Ji ) + · · · + (a1 · · · ai−2 Jn )

16

ROBIN BAIDYA

and
of R so that ai−1 Ki−1 + bi−1 Li−1

Li−1 := a1 · · · ai−2 Ji−1
= R. Let

Ki := (b1 a2 · · · ai−1 J1 ) + · · · + (a1 · · · ai−2 bi−1 Ji−1 )
+(a1 · · · ai−1 Ji+1 ) + · · · + (a1 · · · ai−1 Jn )

and

Li := a1 · · · ai−1 Ji
so that Ki + Li = ai−1 Ki−1 + bi−1 Li−1 = R. If i = 1, then let a1 ∈ K1 and b1 ∈ aJ1 ⊆ L1 with
a1 + b1 = 1 so that R = a1 K1 + b1 aJ1 ⊆ a1 K1 + b1 L1 and, hence, so that a1 K1 + b1 L1 = R.
If i > 2, then simply let ai ∈ Ki and bi ∈ Li with ai + bi = 1 so that ai Ki + bi Li = R.
We will prove that ai ∈ Ii − Vi and that bi ∈ Ji − Ui . Certainly ai ∈ Ki ⊆ Ii , and
bi ∈ Li ⊆ Ji . Suppose that ai ∈ Vi . Then there is n ∈ (Λ1 ∪ · · · ∪ Λn ) − Λi such that ai ∈ n.
Since bi ∈ Ji ⊆ n, we have 1 = ai + bi ∈ n, a contradiction. Hence ai ∈ Ii − Vi . Similarly,
bi ∈ Ji − Ui .
By induction on i, we can thus define
a1 ∈ I1 − V1 , a2 ∈ I2 − V2 , . . . , an−1 ∈ In−1 − Vn−1 ,
b1 ∈ aJ1 − U1 , b2 ∈ J2 − U2 , . . . , bn−1 ∈ Jn−1 − Un−1

and ideals
and

Kn−1 := (b1 a2 · · · an−2 J1 ) + · · · + (a1 · · · an−3 bn−2 Jn−2 ) + (a1 · · · an−2 Jn )

Ln−1 := a1 · · · an−2 Jn−1
of R so that an−1 Kn−1 + bn−1 Ln−1 = R. Hence
(b1 a2 · · · an−1 J1 ) + · · · + (a1 · · · an−2 bn−1 Jn−1 ) + (a1 · · · an−1 Jn ) = R.

Accordingly, we can choose
such that the determinant
of the matrix

c1 ∈ J1 , . . . , cn−1 ∈ Jn−1 , cn ∈ Jn

−(b1 a2 · · · an−1 c1 ) − · · · − (a1 · · · an−2 bn−1 cn−1 ) + (a1 · · · an−1 cn )


a1 0 · · ·
0
b1
.. 
..
..
..

.
.
. 
.
0

.. 
Q :=  ... . . . . . .
. 
0


 0 · · · 0 an−1 bn−1 
c1 · · · · · · cn−1 cn

is equal to 1.
It remains to show that ci 6∈ Ui for every i ∈ {1, . . . , n}. Suppose that c1 ∈ U1 . Then
there is n ∈ Λ1 such that c1 ∈ n. Since J2 , . . . , Jn ⊆ n, we have c2 , . . . , cn ∈ n, and so
1 = det(Q) ∈ Rc1 + · · · + Rcn ⊆ n, a contradiction. Hence c1 6∈ U1 . Similarly, ci 6∈ Ui for
every i ∈ {2, . . . , n}.

We combine the results of the last two lemmas to achieve the goal of this section:

SURJECTIVE AND SPLITTING CAPACITIES

17

Lemma 4.5. Assume the hypotheses of the Surjective Lemma, and define Λ as in Lemma 3.2
with respect to F := Rf1 + · · · + Rfn and X. Then there exists a matrix V ∈ GL(n, R) such
that the first n − 1 components of V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1 )⊤ that is
(t, X, m)-surjective for every m ∈ Λ ∩ Max(R).
Proof. For every m ∈ Λ ∩ Max(R), choose rm,1 , . . . , rm,n−1 ∈ R and Lm ∈ {1, . . . , n} so that
they jointly satisfy the conclusion of Lemma 4.2. For every i ∈ {1, . . . , n}, let
Λi = {m ∈ Λ ∩ Max(R) : Lm = i}.
Then Λ1 , . . . , Λn are finite, pairwise disjoint subsets of Max(R). Hence, by Lemma 4.3, there
exist a matrix Q ∈ GL(n, R) and elements
[
m
s1 , . . . , sn ∈ R −
m∈Λ1 ∪···∪Λn

such that, for every i ∈ {1, . . . , n} and for every m ∈ Λi , we have


s1 0 · · ·
0
0
..
.. 
..
..

.
.
.
.
0
. .
.. 
.
Q ≡ Pi  ..
(mod m).
..
..
0
.


 0 · · · 0 sn−1 0 
0 ··· ···
0
sn

Next, we use the Chinese Remainder Theorem to find r1 , . . . , rn−1 ∈ R such that ri ≡ rm,i
(mod m) for every i ∈ {1, . . . , n − 1} and for every m ∈ Λ ∩ Max(R), and we define


1 0 · · · 0 r1
.. 
 . . . . . . ..
.
. 
0
. .
.. 
.
U :=  .. . .
∈ GL(n, R).
.. 0
. 


0 · · · 0 1 rn−1 
0 ··· ··· 0 1
Let V := UQ. Then




1 0 · · · 0 rm,1
s1 0 · · ·
0
0
.. 
..
.. 
..
..
 . . . . . . ..

.
.
.
. 
.
.
0
0

. .

.. 
.
.
V ≡  ... . . . . . . 0
P
..
..
..  (mod m)
L  .
. 
0

 m.

0 · · · 0 1 rm,n−1 
 0 · · · 0 sn−1 0 
0 ··· ··· 0
1
0 ··· ···
0
sn

for every m ∈ Λ ∩ Max(R). Now, Lemma 4.2 tells us that the first n − 1 components of
V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1 )⊤ that is (t, X, m)-surjective for every m ∈
Λ ∩ Max(R).

In the next section, we complete our proof of the Surjective Lemma.
5. A Proof of the Surjective Lemma
Throughout this section, we assume the hypotheses of the Surjective Lemma (Lemma 1.12),
and we let Λ be defined as in Lemma 3.2 with respect to F := Rf1 + · · · + Rfn and X. Since

18

ROBIN BAIDYA

t and X are understood, we may use the terms p-surjective and Y -surjective for any p ∈ X
and for any Y ⊆ X without the risk of confusion.
In this section, we find a matrix V ∈ GL(n, R) such that the first n − 1 components of
V f := (g1 , . . . , gn )⊤ form a map g := (g1 , . . . , gn−1 )⊤ that is Λ-surjective. Lemma 3.8 will
then tell us that g is X-surjective and, hence, that we have proved the Surjective Lemma.
Proof of the Surjective Lemma. Let q1 , . . . , qm be the distinct members of Λ − Max(R), and
arrange q1 , . . . , qm so that, for every ℓ ∈ {1, . . . , m}, the prime qℓ is a minimal member of
the set {q1 , . . . , qℓ }. We prove, by induction on ℓ > 0, that there exists V ∈ GL(n, R) such
that the first n − 1 components of V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1)⊤ that is
p-surjective for every p ∈ Λ − {qℓ+1 , . . . , qm }.
Lemma 4.5 proves the case in which ℓ = 0. Suppose then that 1 6 ℓ 6 m and that there
exists A ∈ GL(n, R) such that the first n − 1 components of f ∗ := Af := (f1′ , . . . , fn′ )⊤ form
′
a map f ′ := (f1′ , . . . , fn−1
)⊤ that is p-surjective for every p ∈ Λ − {qℓ , . . . , qm }. If f ′ happens
to be qℓ -surjective as well, then we may set V = A to finish the inductive step. Suppose
then that f ′ is not qℓ -surjective. Define every object in Definition 3.6 with respect to our
current hypotheses, with qℓ taking the place of p.
Let
\
J :=
p.
p∈Λ−{qℓ ,...,qm }

It suffices to find r1 , . . . , rn−1 ∈ J such that, if

1

0

U :=  ...

0
0


0 · · · 0 r1
.. 
..
. . ..
.
. .
. 
.. 
..
..
.
. 0
. 

· · · 0 1 rn−1 
··· ··· 0 1

and if Uf ∗ := (g1 , . . . , gn )⊤ , then G := Rg1 +· · ·+Rgn−1 satisfies ∂qℓ (G) = ∂qℓ (F ): Given such
r1 , . . . , rn−1 ∈ J, we will see that the first n − 1 components of Uf ∗ = UAf := (g1 , . . . , gn )⊤
form a map (g1 , . . . , gn−1 )⊤ that is not only qℓ -surjective but, by Nakayama’s Lemma, also
p-surjective for every p ∈ Λ − {qℓ , . . . , qm }. Thus we will be able to take V := UA to finish
the inductive step and, thus, the proof overall. Before we find such r1 , . . . , rn−1 ∈ J, though,
we must complete some more preparatory work.
To simplify notation, let q := qℓ from now on. First we show that ∂q (F ′ ) = ∂q (F ) − 1 and
that ∂q (F ) 6 n − 1. By Lemma 3.7 and by our assumption that f ′ is not q-surjective, we
have
∂q (F ) − 1 6 ∂q (F ′ ) < min{n − 1, t + dimX (q)} 6 ∂q (F ),
and so ∂q (F ′) = ∂q (F ) − 1. Now, if ∂q (F ) = n, then ∂q (F ′) = ∂q (F ) − 1 = n − 1, and so f ′
is q-surjective, a contradiction. Hence ∂q (F ) 6 n − 1.
Let d := ∂p (F ), and let B be a d×n matrix with entries in R such that (Bf ∗ )q is surjective.
Let C ∈ GL(d, Rq ) such that CB can be represented by a matrix (bi,j ) with entries from R
and such that CB is in the following row echelon form with the nonzero entries clustered

SURJECTIVE AND SPLITTING CAPACITIES

19

toward the top right corner of the matrix and with s ∈ R − q:


.
.
.
0 · · · 0 s .. 0 .. 0 ..


0 · · · 0 0 ... s ... 0 ...
.

CB = 

. . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
0 · · · 0 0 .. 0 .. s ..

Here, the vertical and horizontal ellipses denote possible omissions of entries, and the zero
columns on the left may not be present. Now, for every i ∈ {1, . . . , d}, let ji be the smallest
number in the set {1, . . . , n} such that bi,ji 6= 0. We assume that, for every i ∈ {1, . . . , d},
the entry bi,ji is the only nonzero entry in the (ji )th column of CB. Let


..
..
..
0 ··· 0 s . 0 . 0 .

..
..
..

∗
∗
0
·
·
·
0
0
.
s
.
0
.


B := (bi,j ) := 

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.


..
..
..
0 ··· 0 0 . 0 . s .
be a d × n matrix with entries in R that satisfies the following conditions:
(1) B ∗ = CB.
(2) For every i ∈ {1, . . . , d} and for every j ∈ {1, . . . , n}, if bi,j = 0, then b∗i,j = 0.
(3) For every i ∈ {1, . . . , d}, we have b∗i,ji = s.

Hence B ∗ f ∗ ∈ F ⊕d , and B ∗ f ∗ is surjective. Nakayama’s Lemma then tells us that (B ∗ f ∗ )q
is surjective. Thus, we assume, without loss of generality, that B = CB = (bi,j ) and that B
already has the desirable form of B ∗ .
Since (Bf ∗ )q is surjective, there exists a finitely generated R-submodule L of M such that
the restriction of (Bf ∗ )q to Lq is surjective. We may assume, then, without loss of generality,
that Mq is a finitely generated Rq -module.
Let µ := µRq (Mq ), and let ν := µRq (Nq ). Since q ∈ SuppR (N), we see that ν > 1. Since
f is q-surjective, d > t > 1, and so Nakayama’s Lemma tells us that µ > dν > 1. In fact,
without loss of generality, we may assume that µ = dν.
Let E := (ε1 , . . . , εdν )⊤ be an ordered dν-tuple of elements of Mq such that {ε1 , . . . , εdν }
is a minimal generating set for Mq over Rq , and let Z := (ζ1 , . . . , ζν )⊤ be an ordered ν-tuple
of elements of Nq such that {ζ1 , . . . , ζν } is a minimal generating set for Nq over Rq . For
every i ∈ {1, . . . , n}, let ϕ′i := (fi′ )q , and let Φi′ be a ν × dν matrix with entries in Rq that
represents ϕ′i with respect to E and Z in the following sense: For every j ∈ {1, . . . , dν}, if
θ1,j , . . . , θν,j ∈ Rq such that ϕ′i (εj ) = θ1,j ζ1 + · · · + θν,j ζν , then we may define the jth column
of Φi′ to be
 
θ1,j
 ...  .
θν,j
Now let Φ ∗ be the nν × dν matrix whose ith ν × dν block is Φi′ . Hence
 ′
Φ1
∗
Φ =  ...  .
Φn′

20

ROBIN BAIDYA

Finally, we let rank(Ξ ) denote the rank of a matrix Ξ with entries in κ(q).
We now return to the task of finding r1 , . . . , rn−1 ∈ J that satisfy the criteria described
earlier. We consider two cases.
Case 1: jd 6 n − 1. In this case, B has the following form:


..
.
.
.
. s .. 0 .. 0 .. b1,n
.

.. 0 ... s ... 0 ... b 
2,n 
B=
. . . . . . . . . . . . . . . . . . . . . . .


..
..
..
..
. 0 . 0 . s . bd,n
Let rj = 0 ∈ J for every j ∈ {1, . . . , n − 1} − {j1 , . . . , jd }.
Let i ∈ {1, . . . , d}, and suppose that we have defined rj1 , . . . , rj(i−1) ∈ J. Let


..
..
..
..
..
. s . 0 . 0 . 0 . srj1 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.
..
..
..
..

.
. 0 . s . 0 . 0 . srj(i−1) 
Bi := .

.. 0 ... 0 ... s ... 0 ...
0 



..
.
.
.
.
.
.
.
.
. 0 . 0 . 0 . s . bi+1,n 
...............................

be the d×n matrix obtained from B by replacing b1,n , . . . , bi−1,n , bi,n with srj1 , . . . , srj(i−1) ,
0, respectively. Let Ωi := (Bi ⊗ Iν )Φ ∗ , and let
 
0
 .. 
 . 
 
0
 
′
Ωi := Φn′ 
 
0
 . 
 .. 
0
be the dν × dν matrix obtained by replacing the
 ith ν × dν block of the zero dν × dν
matrix with Φn′ . Suppose that rank Ωi + bi,n Ωi′ = dν. We will prove that there exists

rji ∈ J such that rank Ωi + srji Ωi′ = dν.
Let I denote the ideal (sJ + q)/q of R/q. Since q is a nonmaximal prime ideal of R,
we see that R/q is an infinite domain. Since s ∈ R − q and since J 6⊆ q, the ideal I is
nonzero, hence infinite.
Let


Si := σ ∈ κ(q) : rank Ωi + σΩi′ 6 dν − 1 .

We will show that I contains an element ρi that avoids Si . Let Di(x) denote the
determinant of Ωi + xΩi′ , where x is a variable. Since rank Ωi + bi,n Ωi′ = dν, we see

that Di bi,n 6= 0. Hence Di (x) is a nonzero polynomial. Since the degree of Di (x) is at
most ν, we see that |Si | 6 ν. Since I is infinite, I must then contain an element ρi
that avoids Si .

SURJECTIVE AND SPLITTING CAPACITIES

21


Now let rji ∈ J such that srji = ρi . Then rank Ωi + srji Ωi′ = dν, as promised.
By induction, then, we can define
matrices B1 , Ω1 , Ω1′ , . . . , Bd , Ωd, Ωd′ and rj1 , . . . , rjd ∈

′
J such that rank Ωd + srjd Ωd = dν. Now, let B ′ be the d × (n − 1) matrix obtained by
deleting the nth column of B; let


1 0 · · · 0 r1
.. 
 . . . . . . ..
.
. 
0

.. 
;
U :=  ... . . . . . . 0
. 


0 · · · 0 1 rn−1 
0 ··· ··· 0 1
and let Γ denote the (n − 1)ν × dν matrix obtained by deleting
h the nth νi × dν block
∗
′
′
of (U ⊗ Iν )Φ . Then (B ⊗ Iν )Γ = Ωd + srjd Ωd , and so rank (B ′ ⊗ Iν )Γ = dν. Let

Uf ∗ := (g1 , . . . , gn )⊤ , and let G := Rg1 + · · · + Rgn−1 . Then, by Nakayama’s Lemma,
⊕d
(B ′ ⊗ Iν )Γ represents a surjection in G⊕d
q from Mq to Nq , and so ∂q (G) = d = ∂q (F ),
as desired.
Case 2: jd = n. In this case, B has the following form:


..
.
.
. s .. 0 .. 0
.

.. 0 ... s ... 0

B=
. . . . . . . . . . . . . . . .


..
..
..
. 0 . 0 . s

Since d 6 n − 1 and since jd = n, there is k ∈ {1, . . . , n − 1} − {j1 , . . . , jd−1 }. Let
Ω := (B ⊗ Iν )Φ ∗ , and let
 
0
.. 

. 
Ω ′ := 
0
Φk′

be the dν × dν matrix obtained by replacing the dth ν × dν block of the zero dν × dν
matrix with Φk′ .
Let J denote the ideal (J + q)/q of R/q. Since I ⊆ J , we see that J is infinite.
Let


S := σ ∈ κ(q) : rank Ω + σΩ ′ 6 dν − 1 .

We will show that J contains a nonzero element ρ such that ρ−1 avoids
be
 S . Let D(x)

′
′
the determinant of
 Ω +xΩ , where x is a variable. Since rank Ω + 0Ω = rank Ω = dν,
we see that D 0 6= 0. Hence D(x) is a nonzero polynomial. Since the degree of D(x)
is at most ν, we see that |S | 6 ν. Since J is infinite, J must then contain a nonzero
element ρ such that ρ−1 avoids S .
Now let r ∈ J − q such that r = ρ, and let 1/r denote the multiplicative inverse of the
element r/1 of Rq so that (1/r) = ρ−1 .

22

ROBIN BAIDYA

Let


..
..
. b1,k . 0
. 
..
. b2,k .. 0


B1 = . . . . . . . . . . . . . 
.

.. bd−1,k ... 0


..
..
. 1/r . s

be the d × n matrix obtained from B by replacing bd,kh = 0 with 1/r.
i
∗
′
∗
Note that (B1 ⊗ Iν )Φ = Ω + (1/r)Ω so that rank (B1 ⊗ Iν )Φ = dν.
Next, let


1 0 · · · 0 −rb1,k
 . . . . . . ..

. −rb2,k 
0


..
B2 :=  ... . . . . . . 0
 ∈ GL(d, Rq ).
.


0 · · · 0 1 −rbd−1,k 
0 ··· ··· 0
rs
i
h
Then rank (B2 B1 ⊗ Iν )Φ ∗ = dν. Also,


..
..
. 0 . s(−rb1,k ) 
.

..
. 0 .. s(−rb2,k ) 


B2 B1 = . . . . . . . . . . . . . . . . . . . ,

.

.. 0 ... s(−rb
d−1,k )

..
.
. s ..
s(rs)
where the column (0, 0, . . . , 0, s)⊤ displayed above is the kth column of B2 B1 . Now permutehthe rows of B
i 2 B1 to yield a matrix B3 such that B3 is in row echelon form. Then
∗
rank (B3 ⊗ Iν )Φ = dν, and so we have reduced to Case 1.
This completes the inductive step of our proof.



Now that we have worked through our proof of the Surjective Lemma, we can reveal why
we address the members of Λ ∩ Max(R) separately in Section 4. Let q be defined as in the
main inductive step of this section. Since q is a nonmaximal prime ideal of R, we see that
R/q is an infinite domain and, hence, that the nonzero ideals I and J of R/q are also
infinite. In Case 1, we find that, for every i ∈ {1, . . . , d}, there is an element of I that avoids
Si since |Si | 6 µRq (Nq ) < ∞. In Case 2, we find that J must have a nonzero element
whose multiplicative inverse avoids S since |{0} ∪ S | 6 1 + µRq (Nq ) < ∞. These lines of
reasoning, mutatis mutandis, are not necessarily available for a given m ∈ Λ ∩ Max(R). In
particular, it is not always the case that R/m is infinite or even that |R/m| > 2 + µRm (Nm ).
Hence, in general, we cannot mimic the method that we use on the members of Λ − Max(R)
to treat the members of Λ ∩ Max(R). The possibility that |R/m| 6 1 + µRm (Nm ) for some
m ∈ Λ ∩ Max(R) is what compelled us to find a special method for dealing with the members
of Λ ∩ Max(R), and it is this method that we present in Section 4.

SURJECTIVE AND SPLITTING CAPACITIES

23

This is not the only method that works. In fact, there is an alternative to the method of
Sections 4 and 5 that automatically simplifies our proof of the Surjective Lemma in a special
case: With respect to Lemma 4.2, we can define
Λi := {m ∈ Λ ∩ Max(R) : Lm = i and |R/m| 6 1 + µRm (Nm )}

for every i ∈ {1, . . . , n}, use Lemmas 4.2 and 4.3 to account for the members of Λ1 ∪ · · · ∪ Λn
only, and then proceed by induction on the remaining members of Λ as in Section 5. The
benefit of this approach is that, if |R/m| > 2 + µRm (Nm ) for every m ∈ Λ ∩ Max(R), then
Λ1 , . . . , Λn are all empty, and so it suffices to use the method of Section 5 for the entirety
of Λ.
In another special case, the proof of the Surjective Lemma does not go through as quickly,
but the result of the inductive step is still comparable to that of Section 5: If µRm (Nm ) = 1
for every m ∈ Λ ∩ Max(R), then we can mimic the method used in the inductive step of
the proof of [6, Lemma 3.8] for the members of Λ ∩ Max(R), and we can apply the method
of Section 5 of this paper for the remaining members of Λ. This case is noteworthy in the
following sense: Let q1 , . . . , qm be the distinct members of Λ, and arrange q1 , . . . , qm so that,
for every ℓ ∈ {1, . . . , m}, the prime qℓ is a minimal member of the set {q1 , . . . , qℓ }. Also, list
the members of Λ ∩ Max(R) first. Let ℓ ∈ {1, . . . , m}, and let
J :=

ℓ−1
\

qi .

i=1

Suppose that there exists A ∈ GL(n, R) such that the first n − 1 components of f ∗ := Af :=
′
(f1′ , . . . , fn′ )⊤ form a map (f1′ , . . . , fn−1
)⊤ that is qi -surjective for every i ∈ {1, . . . , ℓ − 1}.
Then there exist r1 , . . . , rn−1 ∈ J such that, if


1 0 · · · 0 r1
.. 
 . . . . . . ..
.
. 
0
. .
.. 
.
,
U :=  .. . .
.. 0
. 


0 · · · 0 1 rn−1 
0 ··· ··· 0 1
then the first n − 1 components of Uf ∗ := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1 )⊤ that is
qi -surjective for every i ∈ {1, . . . , ℓ}.
We can combine the two special cases that we have mentioned to yield the following
corollary of the Surjective Lemma:

Corollary 5.1. Assume the hypotheses of the Surjective Lemma. Define Λ as in Lemma 3.2
with respect to F := Rf1 + · · · + Rfn and X. Suppose that, for every m ∈ Λ ∩ Max(R), one
of the following conditions holds:
(1) |R/m| > 2 + µRm (Nm ).
(2) µRm (Nm ) = 1.
(For example, we may suppose that every residue field of R is infinite or that N is a locally
cyclic R-module.) Then there exist r1 , . . . , rn−1 ∈ R such that (f1 + r1 fn , . . . , fn−1 + rn−1 fn )⊤
is (t, X, X)-surjective.
The conclusion of this corollary can be compared to that of [9, Theorem B]. We leave it to
the reader to spell out the ramifications of this corollary for Theorems 0.3, 0.8, 1.13, and 2.4.

24

ROBIN BAIDYA

Despite their benefits, the alternative approaches to the Surjective Lemma have an obvious
drawback: Conditions on the sizes of residue fields and minimal generating sets do not receive
proper context until the middle of Section 5. For this reason, we decided to present a method
that avoids specific reference to the sizes of residue fields and minimal generating sets when
handling the members of Λ∩Max(R). We remain faithful to this decision in our presentation
of the analogous results of the next section.
6. Proofs of Theorems 0.5 and 0.9
Throughout this section, let R denote a commutative ring; let S denote a module-finite
R-algebra; let M denote a right S-module; and let N denote a finitely presented right Smodule. As before, we view every left and right S-module as a standard R-module in the
natural way.
In this section, we prove Theorems 0.5 and 0.9. Since many of the techniques here are
similar to those that we use in Sections 1–5, we do not provide as much detail here as
before. Still, we state all of the necessary definitions and lemmas, and we indicate the major
differences between the proofs here and their earlier analogues.
Of note is the fact that there are only four results in this section that do not require N to
be finitely presented over S: For Remark 6.2, Remark 6.4, Lemma 6.13, and Lemma 6.15,
it suffices for N to be finitely generated over S. Every other result in this section ultimately
relies on Lemma 6.7, and Lemma 6.7 relies on the finite presentation of N over S.
We begin with the following definitions and remarks:
Definition 6.1. Let F be an R-submodule of HomS (M, N), and let p ∈ Spec(R). We let
δ(F ) denote the supremum of the nonnegative integers t such that there exists f ∈ F ⊕t ⊆
HomS (M, N ⊕t ) that is split surjective over S. We let δp (F ) denote the supremum of the
nonnegative integers t such that there exists f ∈ F ⊕t with the property that fp is split
surjective over Sp .
Remark 6.2. Let F be a finitely generated R-submodule of HomS (M, N). We observe that
δ(F ) = ∞ if and only if N = 0: Certainly, if N = 0, then δ(F ) = ∞. On the other hand, if
δ(F ) = ∞, then ∂(F ) = ∞, and so N = 0 by Remark 1.2.
Let p ∈ Spec(R). Then, by the preceding discussion, δp (F ) = ∞ if and only if p 6∈
SuppR (N).
Definition 6.3. Let n, t be positive integers with n > t; let p ∈ X ⊆ Spec(R); and let
f := (f1 , . . . , fn )⊤ ∈ HomS (M, N ⊕n ). We say that f is (t, X, p)-split if δp (Rf1 + · · · + Rfn ) >
min{n, t + dimX (p)}.
Let Y ⊆ X. We say that f is (t, X, Y )-split if f is (t, X, q)-split for every q ∈ Y .
When t and X are understood, we use the terms p-split and Y -split in place of (t, X, p)-split
and (t, X, Y )-split, respectively.
Remark 6.4. Maintaining the hypotheses in the previous definition, we see that f is (n, X, p)split if and only if fp is split surjective over Sp . The reasoning is basically the same as in
Remark 1.4.
We now state an analogue of the Surjective Lemma (Lemma 1.12).
Lemma 6.5 (Splitting Lemma). Let n, t be positive integers with n > 1 + t, and let X be a
subset of SuppR (N) that is a basic set for R. Let f := (f1 , . . . , fn )⊤ ∈ HomS (M, N ⊕n ), and

SURJECTIVE AND SPLITTING CAPACITIES

25

′
suppose that f is (t, X, X)-split. Then there exist f1′ , . . . , fn−1
∈ Rf1 + · · · + Rfn such that
′
′
′
⊤
f := (f1 , . . . , fn−1 ) is (t, X, X)-split.

We defer the proof of the Splitting Lemma to the end of this section. Assuming the truth
of the Splitting Lemma, we could prove the following theorem at this point, but we omit
the proof since it is basically the same as the proof of Theorem 1.13. Still, we would like to
make one note about the proof. We use the finite presentation of N over S more than just
through the use of Lemma 6.7: When applying Part (1) of Theorem 6.6 to prove Part (2),
we use the fact that a map g ∈ HomS (M, N ⊕t ) is split surjective over S if and only if gm is
split surjective over Sm for every m ∈ Max(R) ∩ SuppR (N).
Theorem 6.6. Let L be an S-submodule of M; let F be a finitely generated R-submodule of
HomS (L, N); and let G be an R-submodule of HomS (M, N). Suppose that every member of
F can be extended to a member of G. Let X be a subset of SuppR (N) that is a basic set for
R, and suppose that dim(X) < ∞. Then the following statements hold:
(1) Let t be a positive integer, and suppose that δp (F ) > t + dimX (p) for every p ∈ X.
Then there exists g ∈ G⊕t such that gp is split surjective over Sp for every p ∈ X.
(2) Suppose that Max(R) ∩ SuppR (N) ⊆ X. Then
δ(G) > inf{δp (F ) − dimX (p) : p ∈ X}.
We could prove Theorem 0.9 at this point, but we omit the proof since it is similar to the
proof of Theorem 0.8.
As with the Surjective Lemma, we can reduce the proof of the Splitting Lemma to a consideration of a finite subset Λ of X. The following lemma, which is analogous to Lemma 2.1,
helps us reach this goal. As we mention above, this lemma marks the main point in this
section that relies on the finite presentation of N over S.
Lemma 6.7. Let F be an R-submodule of HomS (M, N), and let t be a nonnegative integer.
Then the set {p ∈ Spec(R) : δp (F ) > t} is open, and so the set {p ∈ Spec(R) : δp (F ) 6 t} is
closed. Hence, for every subspace X of Spec(R), the set Yt := {p ∈ X : δp (F ) 6 t} is closed
in X.
Proof. Let p ∈ Spec(R) such that δp (F ) > t; let f ∈ F ⊕(t+1) such that fp is split surjective over Sp ; and let L = N ⊕(t+1) . Since N is finitely presented over S, there exist
g ∈ HomS (L, M) and s ∈ R − p such that f ◦ g = s · 1L . Let U := {q ∈ Spec(R) : s 6∈ q}.
Then fq is split surjective over Sq for every q ∈ U. Hence U is an open neighborhood of p
such that δq (F ) > t for every q ∈ U. Thus the set {p ∈ Spec(R) : δp (F ) > t} is open. This
proves the first claim of the lemma. The last two claims of the lemma follow from the first
claim.

We state the following lemma with an eye toward Theorem 0.5. We omit the proof of this
lemma since it is similar to the proof of Lemma 2.2.
Lemma 6.8. Suppose that M is a direct summand of a direct sum of finitely presented right
S-modules. Let X be a Noetherian subspace of SuppR (N), and suppose that dim(X) < ∞.
Let t be a positive integer, and suppose that splSp (Mp , Np ) > t + dim(X) for every p ∈ X.
Then there exists a finitely generated R-submodule F of HomS (M, N) such that δp (F ) >
t + dim(X) for every p ∈ X.

26

ROBIN BAIDYA

We would now be in a position to prove Theorem 0.5, modulo the Splitting Lemma. We
omit the proof since it is similar to the proof of Theorem 0.3.
We would now also be able to prove the following variations of Lemma 6.8 and Theorems 6.6 and 0.5, but we omit the proofs. As with Lemma 2.3 and Theorem 2.4, the
following variations are noteworthy in the sense that they do not require M to be a direct
summand of a direct sum of finitely presented right S-modules.
Lemma 6.9. Let F be an R-submodule of HomS (M, N). Let X be a Noetherian subspace
of SuppR (N), and suppose that dim(X) < ∞. Let t be a positive integer, and suppose that
δp (F ) > t + dim(X) for every p ∈ X. Then there exists a finitely generated R-submodule F ′
of F such that δp (F ′ ) > t + dim(X) for every p ∈ X.
Theorem 6.10. Let L be an S-submodule of M; let F be an R-submodule of HomS (L, N);
and let G be an R-submodule of HomS (M, N). Suppose that every member of F can be
extended to a member of G. Then the following statements hold:
(1) Let X be a subset of SuppR (N) that is a basic set for R with dim(X) < ∞. Let t be
a positive integer, and suppose that δp (F ) > t + dim(X) for every p ∈ X. Then there
exists g ∈ G⊕t such that gp is split surjective over Sp for every p ∈ X.
(2) Suppose that Y := Max(R) ∩ SuppR (N) is Noetherian with dim(Y ) < ∞. Then the
following statements hold:
(a) Let t be a positive integer, and suppose that δm (F ) > t+dim(Y ) for every m ∈ Y .
Then δ(G) > t.
(b) If δm (F ) = ∞ for every m ∈ Y , then δ(G) = ∞. Hence δ(F ) = ∞ if and only if
δp (F ) = ∞ for every m ∈ Y .
(c) Suppose that δn (F ) < ∞ for some n ∈ Y . Then
δ(G) > min{δm (F ) : m ∈ Y } − dim(Y ).

Let F be a finitely generated R-submodule of HomS (M, N), and let X be a subset of
SuppR (N) that is a basic set for R. The next lemma, which can be compared to Lemma 3.2,
shows that there is a finite subset Λ of X that completely determines the function on X
taking p to δp (F ). We omit the proof since it is basically the same as the ones for [6,
Lemmas 3.6 and 4.2].
Lemma 6.11. Let F be a finitely generated R-submodule of HomS (M, N), and let X be a
subset of SuppR (N) that is a basic set for R. Then there exists a finite subset Λ of X such
that, for every p ∈ X −Λ, there exists q ∈ Λ with the properties that q ( p and δq (F ) = δp (F ).
Using this lemma, we can prove the following analogue of Corollary 3.3. We omit the
proof on account of its similarity with the proof of Corollary 3.3.
Corollary 6.12. We make the following improvements to Theorems 6.6 and 0.9:
(1) Assume the hypotheses of Part (2) of Theorem 6.6, and let Λ be defined as in
Lemma 6.11 with respect to F and X. Then
δ(G) > inf{δp (F ) − dimX (p) : p ∈ Λ}.

(2) Assume the hypotheses of Theorem 0.9, and let Λ be defined as in Lemma 6.11 with
respect to F := HomS (M, N) and X. Then
splS (M, N) > inf{splSp (Mp , Np ) − dimX (p) : p ∈ Λ}.

SURJECTIVE AND SPLITTING CAPACITIES

27

We now return to the task of reducing the proof of the Splitting Lemma to the study of a
finite subset Λ of X. To this end, we present the following analogues of Lemmas 3.7 and 3.8.
We omit the proofs.
Lemma 6.13. Let n ∈ Z with n > 2; let p ∈ SuppR (N); let f := (f1 , . . . , fn )⊤ ∈
HomS (M, N ⊕n ); and let A ∈ GL(n, R). Then, with respect to Definition 3.6, we have
δp (F ′ ) > δp (F ) − 1.
Lemma 6.14. Assume the hypotheses of the Splitting Lemma, and define Λ as in Lemma 6.11
with respect to F := Rf1 + · · · + Rfn and X. Let A ∈ GL(n, R), and suppose that, with
respect to Definition 3.6, we have that f ′ is (t, X, Λ)-split. Then f ′ is (t, X, X)-split.
For the rest of this section, we assume the hypotheses of the Splitting Lemma, and we let
Λ be defined as in Lemma 6.11 with respect to F := Rf1 + · · · + Rfn and X. Since t and X
are understood, we can use the terms p-split and Y -split for any p ∈ X and for any Y ⊆ X
without the risk of confusion.
Given the next two lemmas, we can find a matrix V ∈ GL(n, R) such that the first n − 1
components of V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1 )⊤ that is m-split for every
m ∈ Λ ∩ Max(R). The proofs of the following two lemmas are basically the same as the
proofs of Lemmas 4.2 and 4.5, and so we omit them. We make one note, however: In the
proof of Lemma 6.16 below, we use Lemma 4.3 by defining
Λi := {m ∈ Λ ∩ Max(R) : Lm = i}
for every i ∈ {1, . . . , n}, where we can find an appropriate choice of Lm for every m ∈
Λ ∩ Max(R) by using Lemma 6.15.
Lemma 6.15. Let m ∈ Λ ∩ Max(R). Then there exist elements rm,1 , . . . , rm,n−1 ∈ R and
a number Lm ∈ {1, . . . , n} with the following property: For all s1 , . . . , sn ∈ R − m and for
every n × n matrix V with entries in R such that




s1 0 · · ·
0
0
1 0 · · · 0 rm,1
..
.. 
.. 
.. ..

 . . . . . . ..
.
.
.
.
.
. 
0
0


. .
. .
.
.
.
.
V ≡  .. . . . . 0
.. ..
..  (mod m),
..  PLm  ..
0




 0 · · · 0 sn−1 0 
0 · · · 0 1 rm,n−1 
0 ··· ···
0
sn
0 ··· ··· 0
1
the first n − 1 components of V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1 )⊤ that is m-split.
Lemma 6.16. There exists a matrix V ∈ GL(n, R) such that the first n − 1 components of
V f := (g1 , . . . , gn )⊤ form a map (g1 , . . . , gn−1)⊤ that is m-split for every m ∈ Λ ∩ Max(R).
Lastly, we must find a matrix V ∈ GL(n, R) such that the first n − 1 components of
V f := (g1 , . . . , gn )⊤ form a map g := (g1 , . . . , gn−1 )⊤ that is Λ-split. Lemma 6.14 will then
tell us that g is X-split and, hence, that we have proved the Splitting Lemma.
Proof of the Splitting Lemma. The beginning of the proof is basically the same as the proof
of the Surjective Lemma up to, and including, the point where we reduce to the case in
which a matrix B has a certain desirable form with (Bf ∗ )q surjective. Of course, here, we
need (Bf ∗ )q to be not only surjective but also split surjective over Sq .

28

ROBIN BAIDYA

Let d := δq (F ). Since (Bf ∗ )q is split surjective over Sq , there exists an S-submodule L
of M such that the restriction of (Bf ∗ )q to Lq is an isomorphism. We may assume, then,
without loss of generality, that Mq = Nq⊕d .
It remains to find r1 , . . . , rn−1 ∈ J such that, if


1 0 · · · 0 r1
.. 
 . . . . . . ..
.
. 
0
. .
.. 
.
U :=  .. . .
.. 0
. 


0 · · · 0 1 rn−1 
0 ··· ··· 0 1
and if Uf ∗ := (g1 , . . . , gn )⊤ , then G := Rg1 + · · · + Rgn−1 satisfies δq (G) = d. On the
other hand, since we have reduced to the case in which Mq = Nq⊕d , it suffices to verify that
∂q (G) = d, for this will imply that δq (G) = d. Thus, from here, we may proceed once more
as in the proof of the Surjective Lemma.

As with the Surjective Lemma, there is a special case that admits a stronger version
of the Splitting Lemma. The reasoning is virtually identical to the discussion preceding
Corollary 5.1, so we simply state the result here.
Corollary 6.17. Assume the hypotheses of the Splitting Lemma. Define Λ as in Lemma 6.11
with respect to F := Rf1 + · · · + Rfn and X. Suppose that, for every m ∈ Λ ∩ Max(R), one
of the following conditions holds:
(1) |R/m| > 2 + µRm (Nm ).
(2) µRm (Nm ) = 1.
(For example, we may suppose that every residue field of R is infinite or that N is a locally
cyclic R-module.) Then there exist r1 , . . . , rn−1 ∈ R such that (f1 + r1 fn , . . . , fn−1 + rn−1 fn )⊤
is (t, X, X)-split.
It would now be possible to state improved versions of Theorems 0.5, 0.9, 6.6, and 6.10 in
the special case acknowledged by the previous corollary, but we omit the details.
In the next section, we consider another special case in which we can improve upon our
previous results.
7. Finitely Generated Modules over Dedekind Domains
In this section, we characterize global surjective and splitting capacities of finitely generated modules over Dedekind domains. We define a Dedekind domain to be an integral
domain in which every ideal is projective. Hence we consider a field to be a Dedekind domain. Since every ideal of a Dedekind domain is projective, every ideal is finitely generated [7,
pages 760–762], and so a Dedekind domain is always Noetherian.
A fractional ideal of a Dedekind domain R is an R-submodule I of the fraction field of
R such that there exists a nonzero a ∈ R with aI an ideal of R. Hence every fractional
ideal of R is isomorphic to an ideal of R. We define an equivalence relation ∼ on the set
F of all nonzero fractional ideals of R by letting I ∼ J if and only if there exist nonzero
a, b ∈ R such that aI = bJ. The set of all equivalence classes of F with respect to ∼ forms
an abelian group under multiplication with [I][J] := [IJ] for all nonzero ideals I and J of
R. This group, denoted Pic(R), is called the Picard group of R or the class group of R, and

SURJECTIVE AND SPLITTING CAPACITIES

29

its identity is [R], the class of all principal fractional ideals of R. See [1, pages 457–460]; [7,
pages 760–762]; or [8, pages 253–258] for more details.
For every module M over a Dedekind domain R, we define TorR (M) to be the torsion
R-submodule of M. We omit the subscript R if the underlying ring is understood. Indeed,
in this section, we will often omit subscripts referring to rings in notation such as AssR (N),
SuppR (N), surR (M, N), and splRp (Mp , Np ) since, in many cases, the underlying ring will be
understood. In particular, we do not consider arbitrary module-finite algebras over Dedekind
domains here.
We now recall the structure theorem for finitely generated modules over Dedekind domains.
The various parts of this theorem can be found in [7, pages 771 and 774]. Alternatively,
Parts (1), (3), and (4) can be found in [1, page 463] or [8, pages 484–485], and Part (2) can
be deduced by applying the Chinese Remainder Theorem to results from [1, page 458] or [8,
page 258].
Theorem 7.1. Let M be a finitely generated module over a Dedekind domain R. Then the
following statements hold:
(1) M ∼
= Tor(M) ⊕ M/ Tor(M).
(2) Tor(M) is a direct sum of R-modules, each of which has the form R/mi for some
m ∈ Spec(R) − {0} and some positive integer i. This decomposition is unique up to
a permutation of factors.
(3) There is an alternative decomposition of Tor(M) as (R/I1 ) ⊕ · · · ⊕ (R/Iu ) for some
nonzero proper ideals I1 , . . . , Iu of R such that I1 ⊆ · · · ⊆ Iu . This decomposition is
unique.
(4) Suppose that M 6= Tor(M). Then there is a nonzero ideal I of R such that M/ Tor(M)
∼
= R⊕(r−1) ⊕ I, where r := rank(M). The ideal I is unique up to isomorphism.
Part (2) of the preceding theorem gives the elementary divisor decomposition of Tor(M),
and Part (3) gives the invariant factor decomposition of Tor(M). If M 6= Tor(M), then the
class [I] of the ideal I from Part (4) is called the Steinitz class of M and is denoted by [M].
See [7, page 773].
We collect a few more properties of Dedekind domains in the following lemma.
Lemma 7.2. Let R be a Dedekind domain. Then R satisfies the following properties:
(1) Let I and J be nonzero ideals of R. Then I ⊕ J ∼
= R ⊕ IJ.
(2) Let I and J be nonzero ideals of R. Then there is a nonzero ideal K of R such that
I∼
= KJ.
(3) Let I be a nonzero ideal of R, and let M be a cyclic torsion R-module. Then there
is a surjective R-linear map from I to M.
(4) Let M be a finitely generated torsion R-module, and let u be a positive integer. Suppose that µRm (Mm ) 6 u for every m ∈ Ass(M). Then µR (M) 6 u.
Proof. (1) See [1, pages 461–462]; [7, page 769]; or [8, page 484].
(2) Since Pic(R) is a group, there is an ideal K of R such that [I] = [K][J] = [KJ]. Hence
I∼
= KJ.
(3) The result is obvious if M = 0, so suppose that M 6= 0. Then there is a nonzero proper
ideal J of R such that M ∼
= R/J. From [7, page 765], we learn that I and J can be written as
v1
vu
wu
1
I = m1 · · · mu and J = mw
1 · · · mu , where m1 , . . . , mu are distinct nonzero prime ideals of R
and where v1 , . . . , vu , w1 , . . . , wu are nonnegative integers. Now let K := mv11 +w1 · · · mvuu +wu .

30

ROBIN BAIDYA

i
By results from [7, page 768], we see that mvi i /mvi i +wi ∼
= R/mw
i for every i ∈ {1, . . . , u} and,
hence, that I/K ∼
= R/J.
(4) This follows from the invariant factor decomposition of Tor(M) in Theorem 7.1. 

Here is our first main result on Dedekind domains:
Proposition 7.3. Let M and N be finitely generated modules over a Dedekind domain R;
let X := Ass(N) − {0}; and let t be a positive integer. Then sur(M, N) > t if and only if
sur(Mm , Nm ) > t for every m ∈ X and one of the following conditions holds:
(1) rank(N) = 0.
(2) rank(M) > 1 + t · rank(N).
(3) rank(M) = t · rank(N) > t, and [M] = [N]t .
Moreover, if sur(M, N) > t and (3) holds, then we have that sur(Tor(M), Tor(N)) > t and
sur(M, N) = t.
Proof. Let r := rank(M), and let s := rank(N).
Suppose first that sur(M, N) > t. Certainly sur(Mm , Nm ) > t for every m ∈ X. Suppose
that neither (1) nor (2) holds. Then r = st > t, and so sur(M, N) = t. Let I, J be
ideals of R that represent [M], [N], respectively. Then N ⊕t / Tor(N ⊕t ) ∼
= R⊕(st−1) ⊕ J t by
Part (1) of Lemma 7.2. Let e be a surjective R-linear map from M to N ⊕t . Note that
HomR (Tor(M), R⊕(st−1) ⊕ J t ) = 0. Hence there exist R-linear maps
f : Tor(M)
→
Tor(N)⊕t ,
g : R⊕(st−1) ⊕ I →
Tor(N)⊕t ,
⊕(st−1)
⊕(st−1)
h : R
⊕I → R
⊕ Jt

such that the following matrix represents e:

Tor(M) R⊕(st−1) ⊕ I 

Tor(N)⊕t
f
g
.
⊕(st−1)
t
R
⊕J
0
h

Clearly, h is surjective. Since R⊕(st−1) ⊕ I is torsion-free, ker h is torsion-free. Since
rank(ker h) = rank(M) − rank(N ⊕t ) = 0, we see that ker h = 0. Hence h is an isomorphism, and so Theorem 7.1 tells us that I ∼
= J t . Thus [M] = [N]t , proving (3). As a result,
the Snake Lemma tells us that f is surjective. Hence sur(Tor(M), Tor(N)) > t.
Next suppose that sur(Mm , Nm ) > t for every m ∈ X. If (1) holds, then clearly it is the
case that sur(M, N) > t.
Suppose then that (1) does not hold but that (2) does hold. Let I, J be ideals of R that
represent [M], [N], respectively. By Part (2) of Lemma 7.2, there exists a nonzero ideal K
of R such that I ∼
= KJ t . By Part (1) of Lemma 7.2, we may then write M/ Tor(M) as
R⊕(r−1) ⊕ I ∼
= R⊕(r−st−1) ⊕ K ⊕ R(st−1) ⊕ J t .
If X = ∅, then immediately we see that sur(M, N) > t since N ⊕t / Tor(N ⊕t ) ∼
= R⊕(st−1) ⊕ J t
by Part (1) of Lemma 7.2. Suppose then that X 6= ∅. Let m ∈ X, and let e(m) be a
⊕st
surjective R-linear map from Mm to Nm⊕t . Note that HomR (Tor(Mm ), Rm
) = 0. Hence
there exist R-linear maps
f (m) : Tor(Mm ) → Tor(Nm )⊕t ,
⊕r
g(m) : Rm
→ Tor(Nm )⊕t ,
⊕r
⊕st
h(m) : Rm
→
Rm

SURJECTIVE AND SPLITTING CAPACITIES

31

such that the following matrix represents e(m):
Tor(Nm )
⊕st
Rm


⊕t

⊕r
Tor(Mm ) Rm

f (m)
g(m)
.
0
h(m)

⊕(r−st)
⊕st
. As
Clearly, h(m) is surjective. Since Rm
is free over Rm , we see that ker h(m) ∼
= Rm
a result, the Snake Lemma tells us that µRm (coker f (m)) 6 r − st. Lift a generating set
of coker f (m) to a subset C (m) of Tor(Nm )⊕t ⊆ Tor(N)⊕t . Let m1 , . . . , mu be the distinct
members of X, and let C be the R-module generated by C (m1 ) ∪ · · · ∪ C (mu ). Then, by
Part (4) of Lemma 7.2, we see that µR (C) 6 r − st. Hence, by Part (3) of Lemma 7.2, there
is a surjective R-linear map from R⊕(r−st−1) ⊕ K to C. As a result, there is a surjective
R-linear map from Tor(M) ⊕ R⊕(r−st−1) ⊕ K to im f (m1 ) + · · · + im f (mu ) + C = Tor(N)⊕t .
Altogether, then, we find that sur(M, N) > t.
Finally, suppose that (3) holds. If X = ∅, then immediately we see that sur(M, N) = t.
Suppose then that X 6= ∅. Let m ∈ X, and let e(m) be a surjective R-linear map from Mm
⊕st
to Nm⊕t . Note that HomR (Tor(Mm ), Rm
) = 0. Hence there exist R-linear maps

f (m) : Tor(Mm ) → Tor(Nm )⊕t ,
⊕st
g(m) : Rm
→ Tor(Nm )⊕t ,
⊕st
⊕st
h(m) : Rm
→
Rm

such that the following matrix represents e(m):
Tor(Nm )
⊕st
Rm


⊕t

⊕st
Tor(Mm ) Rm

f (m)
g(m)
.
0
h(m)

⊕st
Clearly, h(m) is surjective. Since Rm
is finitely generated over Rm , we see that h(m) is
an isomorphism. As a result, the Snake Lemma tells us that f (m) is surjective. Hence
sur(Tor(Mm ), Tor(Nm )) > t, and so sur(Tor(M), Tor(N)) > t. Moreover, since rank(M) =
rank(N ⊕t ) and since [M] = [N]t , Theorem 7.1 and Part (1) of Lemma 7.2 tell us that

M/ Tor(M) ∼
= N ⊕t / Tor(N ⊕t ). Altogether then, sur(M, N) = t.

Using the previous proposition, we can characterize the global surjective capacity of a
finitely generated module M with a respect to a finitely generated module N over a Dedekind
domain R: To see this, let u be a positive integer. Then sur(M, N) = u if and only if
sur(M, N) > u and sur(M, N) < u + 1. We can apply the previous proposition with t = u to
characterize the statement that sur(M, N) > u, and we can apply the previous proposition
with t = u + 1 to characterize the statement that sur(M, N) < u + 1. We also have the
following corollary:
Corollary 7.4. Let M and N be finitely generated modules over a Dedekind domain R, and
let X := Ass(N) − {0}. Then sur(M, N) = 0 if and only if one of the following conditions
holds:
(1) sur(Mm , Nm ) = 0 for some m ∈ X.
(2) rank(N) > 1 + rank(M).
(3) rank(M) = rank(N) > 1, and [M] 6= [N].
Now recall the example given in the introduction to this paper: If R is a Dedekind domain
with a nonprincipal ideal I, then sur(Rp , Ip ) = sur(Ip , Rp ) = 1 for every p ∈ Spec(R), and yet

32

ROBIN BAIDYA

sur(R, I) = sur(I, R) = 0. Corollary 7.4 extends this example to a complete characterization
of the condition that sur(M, N) = 0 when M and N are finitely generated modules over a
Dedekind domain R.
We can give a result analogous to Proposition 7.3 for splitting capacities:
Proposition 7.5. Let M and N be finitely generated modules over a Dedekind domain R;
let X := Ass(N) − {0}; and let t be a positive integer. Then spl(M, N) > t if and only if
spl(Mm , Nm ) > t for every m ∈ X and one of the following conditions holds:
(1) rank(N) = 0.
(2) rank(M) > 1 + t · rank(N).
(3) rank(M) = t · rank(N) > t, and [M] = [N]t .
Moreover, if spl(M, N) > t and (3) holds, then spl(Tor(M), Tor(N)) > t, and spl(M, N) = t.
Proof. Let r := rank(M), and let s := rank(N).
Suppose first that spl(M, N) > t. Certainly spl(Mm , Nm ) > t for every m ∈ X. Suppose
that neither (1) nor (2) holds. Then r = st > t, and so spl(M, N) = t. Let I, J be
ideals of R that represent [M], [N], respectively. Then, by Theorem 7.1 and Part (1) of
Lemma 7.2, we see that I ∼
= J t , and so [M] = [N]t , proving (3). Theorem 7.1 also implies
that spl(Tor(M), Tor(N)) > t.
Next suppose that spl(Mm , Nm ) > t for every m ∈ X. If (1) holds, then it must be the
case that spl(M, N) > t.
Suppose then that (1) does not hold but that (2) does hold. Let I, J be ideals of R that
represent [M], [N], respectively. By Part (2) of Lemma 7.2, there exists a nonzero ideal K
of R such that I ∼
= KJ t . By Part (1) of Lemma 7.2, we may write M/ Tor(M) as
R⊕(r−1) ⊕ I ∼
= R⊕(r−st−1) ⊕ K ⊕ R(st−1) ⊕ J t .
Now Theorem 7.1 implies that spl(M, N) > t since N ⊕t / Tor(N ⊕t ) ∼
= R⊕(st−1) ⊕J t by Part (1)
of Lemma 7.2.
Finally, suppose that (3) holds. Then Theorem 7.1 and Part (1) of Lemma 7.2 immediately
imply that spl(Tor(M), Tor(N)) > t and that spl(M, N) = t.

In light of the previous proposition, we could now characterize global splitting capacities
of finitely generated modules over a Dedekind domain in a manner similar to the case for
global surjective capacities. We omit the details.
Given the wealth of results that we have obtained on surjective and splitting capacities,
we propose the following injective analogue of these concepts:
Definition 7.6. Let R be a commutative ring, S an R-algebra, and M and N right Smodules. We let injS (M, N) denote the supremum of the nonnegative integers t such that
there exists an injective S-linear map from N ⊕t to M (the order of the modules here is
correct), and we refer to injS (M, N) as the global injective capacity of M with respect to N
over S.
Let p ∈ Spec(R). We refer to injSp (Mp , Np ) as the local injective capacity of M with respect
to N over S at p.
As with surjective and splitting capacities, we can always get an upper bound on a global
injective capacity in terms of local injective capacities:
injS (M, N) 6 inf{injSm (Mm , Nm ) : m ∈ Max(R) ∩ SuppR (N)}.

SURJECTIVE AND SPLITTING CAPACITIES

33

Can we also get a lower bound in terms of local injective capacities? Can we characterize a
given global injective capacity using local injective capacities? We have not been able to find
any answers to these questions in the existing literature, and we cannot provide results that
are as far-reaching as our results regarding surjective and splitting capacities. However, in
the case of finitely generated modules over a Dedekind domain, we can provide an answer.
As with surjective and splitting capacities, we can write inj(M, N) and inj(Mp , Np ) for every
p ∈ Spec(R) without causing confusion concerning the underlying ring.
Proposition 7.7. Let M and N be finitely generated modules over a Dedekind domain R,
and let X := Ass(N). Then
inj(M, N) = inf{inj(Mp , Np ) : p ∈ X}.
Proof. Let r := rank(M); let s := rank(N); and let
t := inf{inj(Mp , Np ) : p ∈ X}.

It is clear that inj(M, N) 6 t, and so it remains to prove that inj(M, N) > t. If t = 0, then
there is nothing to prove. If t = ∞, then X = ∅, and so inj(M, N) = t. Suppose then that
t is a positive integer so that X 6= ∅. If s = 0, then Theorem 7.1 immediately tells us that
inj(M, N) > t. Suppose then that s > 1 so that r > st > 1. Let I, J be ideals of R that
represent [M], [N], respectively. Let a ∈ I − {0}, and let f : J t → I be the R-linear map
defined by letting f (x) = ax for every x ∈ J t . Then f is injective, and so inj(I, J t ) = 1.
Thus Theorem 7.1 and Part (1) of Lemma 7.2 tell us that inj(M, N) > t.

To close, we mention a few more cases in which we can characterize global surjective,
splitting, and injective capacities. We have already covered some of these cases. For example, a characterization of an infinite global surjective capacity can be found in Part (2)
of Theorem 0.3, and Part (2) of Theorem 0.5 offers a characterization of an infinite global
splitting capacity. Under the hypotheses of Part (3) of Theorem 0.3, if dim(Y ) = 0, then
since we always have

surS (M, N) = min{surSm (Mm , Nm ) : m ∈ Y }
surS (M, N) 6 inf{surSm (Mm , Nm ) : m ∈ Y }.

We can draw an analogous conclusion from Part (3) of Theorem 0.5 when dim(Y ) = 0: In
this case,
splS (M, N) = min{splSm (Mm , Nm ) : m ∈ Y }.
In particular, if our underlying commutative ring is quasisemilocal (the ring has only finitely
many maximal ideals), then global surjective capacities are completely determined by local
surjective capacities, and we can say the same for splitting capacities.
We now observe that, whenever we have a result concerning surjective, splitting, or injective capacities over finitely many commutative rings R1 , . . . , Ru , we have an analogous
result for the direct product R1 × · · · × Ru of these rings. For example, since a commutative
Noetherian hereditary ring is a direct product of finitely many Dedekind domains, we can
characterize global surjective, splitting, and injective capacities of finitely generated modules
over any commutative Noetherian hereditary ring, given Propositions 7.3, 7.5, and 7.7. In
light of our additional results on global surjective and splitting capacities over commutative
quasisemilocal rings, we can characterize global surjective and splitting capacities over any
direct product of finitely many commutative quasisemilocal rings and Dedekind domains.

34

ROBIN BAIDYA

One historically significant example of such a direct product is given by Hungerford’s Theorem [12, Theorem 1]: A commutative principal ideal ring (that is, a commutative ring in
which every ideal is principal) is a direct product of finitely many quotients of principal ideal
domains. Direct products thus provide a way to extend some of the results of this paper to
larger classes of rings.
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E-mail address: rbaidya1@gsu.edu
Department of Mathematics and Statistics, Georgia State University, Atlanta, Georgia
30303

