# Tensor-Hom formalism for modules
over nonunital rings

Leonid Positselski  Institute of Mathematics, Czech Academy of Sciences
Žitná 25, 115 67 Prague 1
Czech Republic<EMAIL_ADDRESS>
###### Abstract.

We say that a ring $R$ is t-unital if the natural map
$R\otimes_{R}R\longrightarrow R$ is an isomorphism, and a left $R$-module $P$
is c-unital if the natural map $P\longrightarrow\operatorname{Hom}_{R}(R,P)$
is an isomorphism. For a t-unital ring $R$, the category of t-unital left
$R$-modules is a unital left module category over an associative, unital
monoidal category of t-unital $R$-$R$-bimodules, while the category of
c-unital left $R$-modules is opposite to a unital right module category over
the same monoidal category. Any left s-unital ring $R$, as defined by Tominaga
in 1975, is t-unital; and a left $R$-module is s-unital if and only if it is
t-unital. For any (nonunital) ring $R$, the full subcategory of s-unital
$R$-modules is a hereditary torsion class in the category of nonunital
$R$-modules; and for rings $R$ arising from small preadditive categories, the
full subcategory of c-unital $R$-modules is closed under kernels and
cokernels. However, over a t-unital ring $R$, the full subcategory of t-unital
modules need not be closed under kernels, and the full subcategory of c-unital
modules need not be closed under cokernels in the category of nonunital
modules. Nevertheless, the categories of t-unital and c-unital left
$R$-modules are abelian and naturally equivalent to each other; they are also
equivalent to the quotient category of the abelian category of nonunital
modules by the Serre subcategory of modules with zero action of $R$. This is a
particular case of the result from a 1996 manuscript of Quillen. We also
discuss related flatness, projectivity, and injectivity properties; and study
the behavior of t-unitality and c-unitality under the restriction of scalars
for a homomorphism of nonunital rings. Our motivation comes from the theory of
semialgebras over coalgebras over fields.

###### Contents

  1. 1 t-Unital Rings and Modules
  2. 2 s-Unital Rings and Modules
  3. 3 c-Unital Modules
  4. 4 Counterexamples
  5. 5 The Main Abelian Category Equivalence
  6. 6 Rings Arising from Small Preadditive Categories
  7. 7 More on t-Unital and c-Unital Modules
  8. 8 t-Flat, c-Projective, and t-Injective Modules
  9. 9 s-Unital and t-Unital Homomorphisms of Rings

## Introduction

Let $A$ be an associative, unital ring. Then the category of unital
$A$-$A$-bimodules $A{\operatorname{\mathsf{--Bimod--}}}A$ is an associative,
unital monoidal category with respect to the operation $\otimes_{A}$ of tensor
product over $A$. The diagonal $A$-$A$-bimodule $A$ is the unit object of this
monoidal category. Furthermore, the category of unital left $A$-modules
$A{\operatorname{\mathsf{--Mod}}}$ is a unital left module category over the
monoidal category $A{\operatorname{\mathsf{--Bimod--}}}A$, while the category
of unital right $A$-modules ${\operatorname{\mathsf{Mod--}}}A$ is a unital
right module category over $A{\operatorname{\mathsf{--Bimod--}}}A$. The same
tensor product operation $\otimes_{A}$ provides these module category
structures. Finally, the tensor product of right and left $A$-modules defines
a pairing between the right module category ${\operatorname{\mathsf{Mod--}}}A$
and the left module category $A{\operatorname{\mathsf{--Mod}}}$ over the
monoidal category $A{\operatorname{\mathsf{--Bimod--}}}A$, taking values in
the category of abelian groups $\mathsf{Ab}$. This structure is what we mean
by the _tensor formalism for $A$-modules_. We refer to the introduction to the
book [17] for a much more general and abstract discussion of this formalism.

Explicitly, the tensor formalism for $A$-modules is expressed in three natural
isomorphisms of tensor products. For any right $A$-module $N$, any
$A$-$A$-bimodule $K$, and any left $A$-module $M$, one has a natural
isomorphism of abelian groups

(1) $(N\otimes_{A}K)\otimes_{A}M\simeq N\otimes_{A}(K\otimes_{A}M).$

For any left $A$-module $M$, the natural left $A$-module morphism

(2) $a\otimes m\longmapsto am\colon A\otimes_{A}M\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip 1.5muM$

is an isomorphism. For any right $A$-module $N$, the natural right $A$-module
morphism

(3) $n\otimes a\longmapsto na\colon N\otimes_{A}A\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip 1.5muN$

is an isomorphism.

Now let $R$ be a possibly nonunital, associative ring. Then the category of
nonunital $R$-$R$-bimodules $R{\operatorname{\mathsf{--{}^{n}Bimod^{n}--}}}R$
is still an associative, unital monoidal category with respect to the
operation $\otimes_{R}$ of tensor product over $R$. However, the
$R$-$R$-bimodule $R$ is _not_ the unit object of this monoidal category.

To see what is going on, it suffices to consider the ring
$\widetilde{R}=\mathbb{Z}\oplus R$ obtained by adjoining formally to $R$ a
(new) unit together with its integer multiples. So the elements of
$\widetilde{R}$ are the formal expressions $n+r$, where $n\in\mathbb{Z}$ and
$r\in R$; and the multiplication in $\widetilde{R}$ is given by the obvious
rule making $\mathbb{Z}$ a subring in $\widetilde{R}$ and $R$ a two-sided
ideal in $\widetilde{R}$. Then the category of nonunital $R$-$R$-bimodules is
naturally equivalent to the category of unital
$\widetilde{R}$-$\widetilde{R}$-bimodules, and similarly for the categories of
left and right modules:

${\operatorname{\mathsf{Mod^{n}--}}}R\simeq{\operatorname{\mathsf{Mod--}}}\widetilde{R},\quad
R{\operatorname{\mathsf{--{}^{n}Bimod^{n}--}}}R\simeq\widetilde{R}{\operatorname{\mathsf{--Bimod--}}}\widetilde{R},\quad
R{\operatorname{\mathsf{--{}^{n}Mod}}}\simeq\widetilde{R}{\operatorname{\mathsf{--Mod}}}.$

The tensor product operation $\otimes_{R}$ over the nonunital ring $R$ agrees
with the tensor product $\otimes_{\widetilde{R}}$ over the unital ring
$\widetilde{R}$ under these category equivalences. The bottom line is that the
$R$-$R$-bimodule $\widetilde{R}$ (rather than the $R$-$R$-bimodule $R$) is the
unit object of the monoidal category of nonunital bimodules
$R{\operatorname{\mathsf{--{}^{n}Bimod^{n}--}}}R$.

In this approach, the theory of nonunital rings and modules becomes a
particular case of the theory of unital ones. It is well-known, however, that
there is a different point of view, making the theory of modules over
nonunital rings quite substantial.

What one wants to do is to impose some kind of weak unitality condition on the
ring $R$, and particularly on the $R$-modules. So, we are not primarily
interested in arbitrary nonunital $R$-modules, but rather in $R$-modules that
are “weakly unital” in one sense or another. In this paper, we develop this
approach aiming to obtain an _associative, unital monoidal category of weakly
unital $R$-$R$-bimodules_ in which _the $R$-$R$-bimodule $R$ would be the unit
object_.

Alongside with the tensor formalism for modules, the _Hom formalism_ plays an
important role. Returning to our associative, unital ring $A$, let us consider
the opposite category $A{\operatorname{\mathsf{--Mod}}}^{\mathsf{op}}$ to the
category of left $A$-modules, and endow it with the right action of the
monoidal category $A{\operatorname{\mathsf{--Bimod--}}}A$ defined by the rule

(4) $P^{\mathsf{op}}*K=\operatorname{Hom}_{A}(K,P)^{\mathsf{op}}$

for all $P\in A{\operatorname{\mathsf{--Mod}}}$ and $K\in
A{\operatorname{\mathsf{--Bimod--}}}A$. This rule makes
$A{\operatorname{\mathsf{--Mod}}}^{\mathsf{op}}$ an associative, unital right
module category over $A{\operatorname{\mathsf{--Bimod--}}}A$. Furthermore, the
functor $(\operatorname{Hom}_{A})^{\mathsf{op}}\colon
A{\operatorname{\mathsf{--Mod}}}\times
A{\operatorname{\mathsf{--Mod}}}^{\mathsf{op}}\longrightarrow\mathsf{Ab}^{\mathsf{op}}$
defines a pairing between the right module category
$A{\operatorname{\mathsf{--Mod}}}^{\mathsf{op}}$ and the left module category
$A{\operatorname{\mathsf{--Mod}}}$ over
$A{\operatorname{\mathsf{--Bimod--}}}A$, taking values in the opposite
category $\mathsf{Ab}^{\mathsf{op}}$ to the category of abelian groups. This
is what we call the _Hom formalism for $A$-modules_. Once again, we refer to
the introduction to [17] for an abstract discussion.

Explicitly, the Hom formalism for $A$-modules is expressed in two natural
isomorphisms of Hom modules/groups. For any left $A$-module $M$, any
$A$-$A$-bimodule $K$, and any left $A$-module $P$, one has a natural
isomorphism of abelian groups

(5)
$\operatorname{Hom}_{A}(K\otimes_{A}M,\>P)\simeq\operatorname{Hom}_{A}(M,\operatorname{Hom}_{A}(K,P)),$

which can be written as
$P^{\mathsf{op}}*(K\otimes_{A}M)\simeq(P^{\mathsf{op}}*K)*M$ in the
$*$-notation of (4). For any left $A$-module $P$, the natural left $A$-module
morphism

(6) $p\longmapsto(a\mapsto ap)\colon P\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip
1.5mu\operatorname{Hom}_{A}(A,P)$

is an isomorphism.

In addition to the tensor formalism, one would like also to extend the Hom
formalism to suitably defined “weakly unital” modules $P$ over some nonunital
rings. Once again, both the tensor associativity isomorphism (1) and the Hom
associativity isomorphism (5) hold without any unitality assumptions on the
rings and modules. It is the unitality isomorphisms (2–3) and (6) in the
tensor-Hom formalism that depend on the unitality assumptions on the ring and
modules and require some nontrivial generalizations in the contexts of “weakly
unital” modules.

We refer to the very nice recent survey paper [16] for a general discussion of
various classes of (what we call) “weakly unital” rings and modules and
relations between them, containing many examples and counterexamples. It is
from the survey [16] that we learned about Tominaga’s 1975 definition of _s-
unital_ rings and modules [24, 25, 26] playing an important role in this
paper.

Before we finish this introduction, let us briefly explain our motivation. Let
$k$ be a field, $R$ be an associative $k$-algebra, and $K\subset R$ be a
subalgebra. Given a coassociative coalgebra $\mathcal{C}$ over $k$ which is
“dual to the algebra $K$” in some weak sense, under certain assumptions, the
construction of [17, Chapter 10] (for the base ring $A=k$) produces a
(semiassociative, semiunital) _semialgebra_
$\boldsymbol{\mathcal{S}}=\mathcal{C}\otimes_{K}R$ over the coalgebra
$\mathcal{C}$. The algebras $K$ and $R$ were, of course, assumed to be unital
in [17]. In a forthcoming paper [19], we intend to generalize the construction
of [17, Chapter 10] to certain nonunital (but “weakly unital”) algebras $K$
and $R$. In this generalization, the coalgebra $\mathcal{C}$ is still presumed
to be counital. The point is that the semialgebra $\boldsymbol{\mathcal{S}}$
produced as the output of the generalized construction remains semiunital. The
present paper is intended to serve as a background reference source for [19].

By far the most relevant reference in the context of the present paper is an
unpublished 1996 manuscript of Quillen [22] (see also [21]). The author
learned about its existence on an advanced stage of the present project; as a
consequence, this paper contains our own proofs of many results from [22]
(with references given for comparison wherever relevant). What we call _t-
unital_ modules are called “firm modules” in [22], and what we call _c-unital_
modules are called “closed modules” in [22]. Another important and relevant
reference is the _almost ring theory_ [7].

There is a difference in motivation between [22] and the present paper. It
appears from [22, §1] that Quillen’s aim was to construct an abelian module
category over a nonunital ring $R$ such that, when $R$ is a unital ring, the
category of unital $R$-modules is produced as the output. This problem is
solved in [22] with the most well-behaved solution obtained in the great
generality of _idempotent_ rings $R$, i. e., rings $R$ such that $R^{2}=R$. On
the other hand, the almost ring theory seems to be primarily interested in
commutative rings. Our task is to obtain a well-behaved associative,
noncommutative tensor-Hom formalism with the $R$-$R$-bimodule $R$ appearing as
the unit object of the monoidal category of “weakly unital” bimodules; and we
solve this problem under more restrictive assumptions of, at least, a _t-
unital_ ring $R$.

### Acknowledgement

I am grateful to Jan Št’ovíček and Michal Hrbek for helpful discussions. The
author is supported by the GAČR project 23-05148S and the Czech Academy of
Sciences (RVO 67985840).

##  1\. t-Unital Rings and Modules

All _rings_ and _modules_ in this paper are associative, but nonunital by
default. The unitalization of a ring $R$ is denoted by
$\widetilde{R}=\mathbb{Z}\oplus R$. So $R$ is a two-sided ideal in the unital
ring $\widetilde{R}$ with the unit element $1+0\in\mathbb{Z}\oplus R$.

The categories of nonunital left and right $R$-modules are denoted by
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$ and
${\operatorname{\mathsf{Mod^{n}--}}}R$. Given two rings $R^{\prime}$ and
$R^{\prime\prime}$, the notation
$R^{\prime}{\operatorname{\mathsf{--{}^{n}Bimod^{n}--}}}R^{\prime\prime}$
stands for the category of nonunital
$R^{\prime}$-$R^{\prime\prime}$-bimodules. The undecorated notation
$A{\operatorname{\mathsf{--Mod}}}$, ${\operatorname{\mathsf{Mod--}}}A$, and
$A^{\prime}{\operatorname{\mathsf{--Bimod--}}}A^{\prime\prime}$ refers to the
categories of unital (bi)modules and presumes the rings $A$, $A^{\prime}$,
$A^{\prime\prime}$ to be unital. So there are natural equivalences (in fact,
isomorphisms) of abelian categories

(7)
${\operatorname{\mathsf{Mod^{n}--}}}R\simeq{\operatorname{\mathsf{Mod--}}}\widetilde{R},\quad
R^{\prime}{\operatorname{\mathsf{--{}^{n}Bimod^{n}--}}}R^{\prime\prime}\simeq\widetilde{R}^{\prime}{\operatorname{\mathsf{--Bimod--}}}\widetilde{R}^{\prime\prime},\quad
R{\operatorname{\mathsf{--{}^{n}Mod}}}\simeq\widetilde{R}{\operatorname{\mathsf{--Mod}}}.$

Given a nonunital right $R$-module $N$ and a nonunital left $R$-module $M$,
the tensor product $N\otimes_{R}M$ is defined in the same way as for unital
modules. Then there is a natural (identity) isomorphism of abelian groups

(8) $N\otimes_{R}M\simeq N\otimes_{\widetilde{R}}M.$

For any right $R^{\prime}$-module $N$, any
$R^{\prime}$-$R^{\prime\prime}$-bimodule $B$, and any left
$R^{\prime\prime}$-module $M$, there is a natural isomorphism of abelian
groups

(9) $(N\otimes_{R^{\prime}}B)\otimes_{R^{\prime\prime}}M\simeq
N\otimes_{R^{\prime}}(B\otimes_{R^{\prime\prime}}M).$

The associativity isomorphism (9) can be obtained from the similar isomorphism
for unital rings $\widetilde{R}^{\prime}$ and $\widetilde{R}^{\prime\prime}$
and unital (bi)modules over them using the category equivalences (7) and the
isomorphism (8).

For any left $R$-module $M$, there is a natural left $R$-module morphism

(10) $r\otimes m\longmapsto rm\colon R\otimes_{R}M\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip 1.5muM.$

Similarly, for any right $R$-module $N$, there is a natural right $R$-module
morphism

(11) $n\otimes r\longmapsto nr\colon N\otimes_{R}R\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip 1.5muN.$

For the $R$-modules $M=R=N$, the maps (10) and (11) agree with each other, and
provide an $R$-$R$-bimodule morphism

(12) $r^{\prime}\otimes r^{\prime\prime}\longmapsto
r^{\prime}r^{\prime\prime}\colon R\otimes_{R}R\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip 1.5muR.$

######  Definition 1.1.

We will say that a left $R$-module $M$ is _t-unital_ if the natural map
$R\otimes_{R}M\longrightarrow M$ (10) is an isomorphism. Similarly, a right
$R$-module $N$ is said to be _t-unital_ if the natural map
$N\otimes_{R}R\longrightarrow N$ (11) is an isomorphism. An
$R^{\prime}$-$R^{\prime\prime}$-bimodule is said to be _t-unital_ if it is
t-unital both as a left $R^{\prime}$-module and as a right
$R^{\prime\prime}$-module. (What we call “t-unital modules” are called “firm
modules” in [22, Definition 2.3].)

It is clear from the preceding discussion of the map (12) that the left
$R$-module $R$ is t-unital if and only if the right $R$-module $R$ is t-unital
(and if and only if the $R$-$R$-bimodule $R$ is t-unital). Any one of these
equivalent conditions holds if and only if the natural map
$R\otimes_{R}R\longrightarrow R$ (12) is an isomorphism. If this is the case,
we will say that _the ring $R$ is t-unital_.

Clearly, the map $R\otimes_{R}R\longrightarrow R$ is surjective if and only if
the ring $R$ is idempotent, i. e., $R^{2}=R$. In Example 4.3 below we
construct a counterexample of an idempotent, but not t-unital ring $R$, i. e.,
a ring $R$ for which the map $R\otimes_{R}R\longrightarrow R$ is surjective,
but not injective.

######  Lemma 1.2.

(a) Let $B$ be an $R^{\prime}$-$R^{\prime\prime}$-bimodule and $M$ be an left
$R^{\prime\prime}$-module. Assume that $B$ is t-unital as a left
$R^{\prime}$-module. Then the left $R^{\prime}$-module
$B\otimes_{R^{\prime\prime}}M$ is also t-unital.

(b) Let $N$ be a right $R^{\prime}$-module and $B$ be an
$R^{\prime}$-$R^{\prime\prime}$-bimodule. Assume that $B$ is t-unital as a
right $R^{\prime\prime}$-module. Then the right $R^{\prime\prime}$-module
$N\otimes_{R^{\prime}}B$ is also t-unital.

###### Proof.

Follows from the associativity isomorphism (9) and commutativity of the
suitable triangular diagram formed by the associativity isomorphism together
with the unitality comparison maps (10) or (11). ∎

The following corollary is a particular case of [22, Proposition 4.4].

######  Corollary 1.3.

Let $R$ be a t-unital ring. Then

(a) for any left $R$-module $M$, the left $R$-module $R\otimes_{R}M$ is
t-unital;

(b) for any right $R$-module $N$, the right $R$-module $N\otimes_{R}R$ is
t-unital.

###### Proof.

Take $B=R=R^{\prime}=R^{\prime\prime}$ in Lemma 1.2. ∎

Given three rings $R$, $R^{\prime}$, $R^{\prime\prime}$, we will denote by
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$,
${\operatorname{\mathsf{Mod^{t}--}}}R$, and
$R^{\prime}{\operatorname{\mathsf{--{}^{t}Bimod^{t}--}}}R^{\prime\prime}$ the
full subcategories of t-unital left $R$-modules, t-unital right $R$-modules,
and t-unital $R^{\prime}$-$R^{\prime\prime}$-bimodules in the abelian
categories $R{\operatorname{\mathsf{--{}^{n}Mod}}}$,
${\operatorname{\mathsf{Mod^{n}--}}}R$, and
$R^{\prime}{\operatorname{\mathsf{--{}^{n}Bimod^{n}--}}}R^{\prime\prime}$,
respectively.

######  Corollary 1.4.

Let $R$ be a t-unital ring. Then the additive category of t-unital
$R$-$R$-bimodules $R{\operatorname{\mathsf{--{}^{t}Bimod^{t}--}}}R$ is an
associative and unital monoidal category, with the unit object $R\in
R{\operatorname{\mathsf{--{}^{t}Bimod^{t}--}}}R$, with respect to the tensor
product operation $\otimes_{R}$. The additive category of t-unital left
$R$-modules $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is an associative and
unital left module category over
$R{\operatorname{\mathsf{--{}^{t}Bimod^{t}--}}}R$, and the additive category
of t-unital right $R$-modules ${\operatorname{\mathsf{Mod^{t}--}}}R$ is an
associative and unital right module category over
$R{\operatorname{\mathsf{--{}^{t}Bimod^{t}--}}}R$, with respect to the tensor
product operation $\otimes_{R}$.

###### Proof.

Follows from the associativity isomorphism (9) and Lemma 1.2 (for
$R^{\prime}=R=R^{\prime\prime}$). ∎

######  Lemma 1.5.

Let $R$ be an arbitrary (nonunital) ring. Then the full subcategory of
t-unital left $R$-modules $R{\operatorname{\mathsf{--{}^{t}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is closed under extensions and all
colimits (including cokernels, direct sums, and direct limits) in the abelian
category of nonunital modules $R{\operatorname{\mathsf{--{}^{n}Mod}}}$.

###### Proof.

This is our version of [22, Proposition 4.2]. Notice that it is _not_ claimed
in the lemma that $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is closed under
quotient modules in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$, but only under
cokernels. The closedness under arbitrary colimits follows immediately from
the fact that the tensor product functor $R\otimes_{R}{-}$ preserves colimits
(which can be proved in the way similar to the unital case, or deduced from
the unital case using the isomorphism (8)).

To prove the closedness under extensions, consider a short exact sequence
$0\longrightarrow K\longrightarrow L\longrightarrow M\longrightarrow 0$ in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$ with $K$, $M\in
R{\operatorname{\mathsf{--{}^{t}Mod}}}$. We have a commutative diagram

$\textstyle{R\otimes_{R}K\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R\otimes_{R}L\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R\otimes_{R}M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

where the sequence in the upper row is right exact and the sequence in the
lower row is short exact. The rightmost and leftmost vertical maps are
isomorphisms by assumption. Now the composition $R\otimes_{R}K\longrightarrow
K\longrightarrow L$ is injective, since the map $K\longrightarrow L$ is
injective. It follows that the map $R\otimes_{R}K\longrightarrow
R\otimes_{R}L$ is injective as well. So the upper row is, in fact, a short
exact sequence, and the middle vertical map $R\otimes_{R}L\longrightarrow L$
is an isomorphism by the 5-lemma. ∎

One shortcoming of the theory developed in this section is that the full
subcategory $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ _need not_ be closed
under kernels in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$, as explained in the
following remark and example. A way to resolve this problem will be suggested
in the next Section 2, where we will consider the more restrictive _s-unital_
setting.

######  Remark 1.6.

The construction of the full subcategory
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ can be viewed as a particular case of
the construction of the _$\operatorname{Tor}_{0,1}$ -perpendicular
subcategory_ in the spirit of [9, Proposition 1.1] or [18, Theorem 1.2(b)].

Let us consider the ring $R$ as an ideal in the unital ring
$\widetilde{R}=\mathbb{Z}\oplus R$, as above. Then the quotient ring
$\mathbb{Z}=\widetilde{R}/R$ becomes a unital left and right
$\widetilde{R}$-module, or equivalently, a nonunital left and right $R$-module
with the zero action of $R$. Given a nonunital left $R$-module $M$, the
natural map
$R\otimes_{R}M=R\otimes_{\widetilde{R}}M\longrightarrow\widetilde{R}\otimes_{\widetilde{R}}M=M$
is an isomorphism _if and only if_ one has
$\widetilde{R}/R\otimes_{\widetilde{R}}M=0=\operatorname{Tor}_{1}^{\widetilde{R}}(\widetilde{R}/R,M)$.
(This observation is a particular case of the discussion in [22, Definition
2.3].)

Now the mentioned results from [9, 18] tell that the full subcategory
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is closed under kernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$ (and, consequently, abelian) whenever
the flat dimension of the unital right $\widetilde{R}$-module
$\widetilde{R}/R$ does not exceed $1$, i. e., in other words, the unital
$\widetilde{R}$-module $R$ is flat. This can be also easily seen directly from
the definition: if the functor $R\otimes_{R}{-}\,\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is exact (preserves kernels), then the
full subcategory $R{\operatorname{\mathsf{--{}^{t}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is closed under kernels. The next
example shows that this does _not_ hold in general.

######  Example 1.7.

Let $\widetilde{R}=\mathbb{Z}[x,y]$ be the commutative ring of polynomials in
two variables $x$, $y$ with integer coefficients. Let $R\subset\widetilde{R}$
be the ideal $R=(x,y)\subset\widetilde{R}$ spanned by the elements $x$,
$y\in~{}\widetilde{R}$. Then it is clear that the ring $\widetilde{R}$ is the
unitalization of the ring $R$, just as the notation suggests. The flat
dimension of the unital $\widetilde{R}$-module $\mathbb{Z}=\widetilde{R}/R$ is
equal to $2$, and the flat dimension of the unital $\widetilde{R}$-module $R$
is equal to $1$. We claim that the full subcategory of t-unital $R$-modules
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is _not_ closed under kernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}=\widetilde{R}{\operatorname{\mathsf{--Mod}}}$.

Indeed, consider the generalized Prüfer module
$P_{x}=\mathbb{Z}[x,x^{-1}]/\mathbb{Z}[x]$ over the ring $\mathbb{Z}[x]$, and
the similar module $P_{y}=\mathbb{Z}[y,y^{-1}]/\mathbb{Z}[y]$ over the ring
$\mathbb{Z}[y]$. Furthermore, consider the $\mathbb{Z}[x,y]$-module
$P_{x}\otimes_{\mathbb{Z}}P_{y}$. Then one can easily see that
$\operatorname{Tor}^{\mathbb{Z}[x]}_{0}(\mathbb{Z},P_{x})=0$ and
$\operatorname{Tor}^{\mathbb{Z}[x]}_{1}(\mathbb{Z},P_{x})=\mathbb{Z}$ (where
$x$ acts by zero in the $\mathbb{Z}[x]$-module $\mathbb{Z}$). Consequently, by
the Künneth formula (see Lemma 4.4 below),
$\operatorname{Tor}_{n}^{\widetilde{R}}(\mathbb{Z},\>P_{x}\otimes_{\mathbb{Z}}P_{y})=0$
for $n=0$, $1$, and $\mathbb{Z}$ for $n=2$. Following Remark 1.6, we have
$P_{x}\otimes_{\mathbb{Z}}P_{y}\in R{\operatorname{\mathsf{--{}^{t}Mod}}}$.

On the other hand, denote by $Q_{y}$ the $\mathbb{Z}[y]$-module $P_{y}$ viewed
as a $\mathbb{Z}[x,y]$-module with the given (Prüfer) action of $y$ and the
zero action of $x$. Then
$\operatorname{Tor}_{1}^{\widetilde{R}}(\mathbb{Z},Q_{y})=\operatorname{Tor}_{1}^{\mathbb{Z}[y]}(\mathbb{Z},P_{y})=\mathbb{Z}$.
So $Q_{y}\notin R{\operatorname{\mathsf{--{}^{t}Mod}}}$. Nevertheless, $Q_{y}$
is the kernel of the (surjective) $R$-module (or $\widetilde{R}$-module) map
$x\colon P_{x}\otimes_{\mathbb{Z}}P_{y}\longrightarrow
P_{x}\otimes_{\mathbb{Z}}P_{y}$.

Notice, however, that the ring $R$ in this example is _not_ t-unital. In fact,
$R$ is not even an idempotent ideal in the unital ring $\widetilde{R}$, i. e.,
$R^{2}\neq R$. An example of t-unital ring $T$ for which the full subcategory
$T{\operatorname{\mathsf{--{}^{t}Mod}}}$ is not closed under kernels in
$T{\operatorname{\mathsf{--{}^{n}Mod}}}$ will be given below in Example 4.7.

For any t-unital ring $R$, Theorem 5.8(a) below tells that the category
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is abelian. On the other hand, again
over a t-unital ring $R$, according to Corollary 7.4, the full subcategory of
t-unital modules $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is closed under
kernels in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$ _if and only if_ $R$ is a
flat unital right $\widetilde{R}$-module.

######  Remark 1.8.

Speaking of unital modules over a unital ring $A$ (such as $A=\widetilde{R}$),
we often emphasize that we work in the category of unital $A$-modules (e. g.,
that particular modules are flat or projective _as unital modules_), but it
actually does not matter that much. Nonunital modules over any ring $R$ are
the same things as unital modules over the unitalization $\widetilde{R}$ of
the ring $R$, and the unitalization $\widetilde{A}=\mathbb{Z}\oplus A$ of a
_unital_ ring $A$ is naturally isomorphic to the Cartesian product of the
rings $\mathbb{Z}$ and $A$, that is $\widetilde{A}\simeq\mathbb{Z}\times A$.
The ring isomorphism $\mathbb{Z}\oplus A\longrightarrow\mathbb{Z}\times A$
takes a formal sum $n+a$ with $n\in\mathbb{Z}$ and $a\in A$ to the pair
$(n,\>ne+a)$, where $e\in A$ is the unit element. Accordingly, the abelian
category of nonunital $A$-modules is just the Cartesian product of the abelian
categories of abelian groups and unital $A$-modules,

$A{\operatorname{\mathsf{--{}^{n}Mod}}}\simeq\widetilde{A}{\operatorname{\mathsf{--Mod}}}\simeq(\mathbb{Z}\times
A){\operatorname{\mathsf{--Mod}}}\simeq\mathbb{Z}{\operatorname{\mathsf{--Mod}}}\times
A{\operatorname{\mathsf{--Mod}}}.$

So there is not much of a difference whether to consider unital modules over
unital rings as objects of the category of unital or nonunital modules over
such ring. (Cf. the discussion in [22, §1], which emphasizes a different
point.)

In particular, all the groups/modules $\operatorname{Tor}$ and
$\operatorname{Ext}$ in this paper are taken over unital rings and in the
categories of unital modules.

##  2\. s-Unital Rings and Modules

The following definition is due to Tominaga [24, 25, 26]. The exposition in
the survey paper [16] is very much recommended.

######  Definition 2.1.

Let $R$ be a (nonunital) ring. A left $R$-module $M$ is said to be _s-unital_
if for every element $m\in M$ there exists an element $e\in R$ such that
$em=m$ in $M$. (Notice that _no_ assumption of idempotency of $e$ is made
here!) Similarly, a right $R$-module $N$ is said to be _s-unital_ if for every
element $n\in N$ there exists an element $e\in R$ such that $ne=n$ in $N$.
Given two rings $R^{\prime}$ and $R^{\prime\prime}$, an
$R^{\prime}$-$R^{\prime\prime}$-bimodule is said to be _s-unital_ if it is
s-unital both as a left $R^{\prime}$-module and as a right
$R^{\prime\prime}$-module.

Given three rings $R$, $R^{\prime}$, $R^{\prime\prime}$, we will denote by
$R{\operatorname{\mathsf{--{}^{s}Mod}}}$,
${\operatorname{\mathsf{Mod^{s}--}}}R$, and
$R^{\prime}{\operatorname{\mathsf{--{}^{s}Bimod^{s}--}}}R^{\prime\prime}$ the
full subcategories of s-unital left $R$-modules, s-unital right $R$-modules,
and s-unital $R^{\prime}$-$R^{\prime\prime}$-bimodules in the abelian
categories $R{\operatorname{\mathsf{--{}^{n}Mod}}}$,
${\operatorname{\mathsf{Mod^{n}--}}}R$, and
$R^{\prime}{\operatorname{\mathsf{--{}^{n}Bimod^{n}--}}}R^{\prime\prime}$,
respectively.

######  Proposition 2.2.

Let $R$ be an arbitrary (nonunital) ring. Then the full subcategory of
s-unital $R$-modules $R{\operatorname{\mathsf{--{}^{s}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is closed under submodules, quotients,
extensions, and all colimits (including direct sums and direct limits) in the
abelian category of nonunital modules
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. In other words,
$R{\operatorname{\mathsf{--{}^{s}Mod}}}$ is a _hereditary torsion class_ in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$ in the sense of [5], [23, Sections
VI.2–3].

###### Proof.

Closedness under submodules, quotients, and direct limits (indexed by directed
posets) is obvious immediately from the definition. All direct sums are direct
limits of finite direct sums, and finite direct sums are finitely iterated
extensions. So it suffices to prove the closedness under extensions.

Let $0\longrightarrow K\longrightarrow L\longrightarrow M\longrightarrow 0$ be
a short exact sequence of nonunital left $R$-modules. Assume that the
$R$-modules $K$ and $M$ are s-unital. Given an element $l\in L$, we need to
find an element $g\in R$ such that $gl=l$ in $L$. Since the $R$-module $M=L/K$
is s-unital, there exists an element $e\in R$ such that $e(l+K)=l+K$ in $M$.
So we have $el-l\in K$. Since the $R$-module $K$ is s-unital, there exists an
element $f\in R$ such that $f(el-l)=el-l$ in $K$. Now we have $(e+f-fe)l=l$ in
$L$, and it remains to put $g=e+f-ef\in R$. ∎

The following result goes back to Tominaga [25]. It can be also found in [16,
Proposition 2.8 of the published version or Proposition 8 of the arXiv
version].

######  Corollary 2.3 (Tominaga [25, Theorem 1]).

Let $R$ be a ring and $M$ be an s-unital left $R$-module. Then, for any finite
collection of elements $m_{1}$, …, $m_{n}\in M$, there exists an element $e\in
R$ such that $em_{i}=m_{i}$ for all $1\leq i\leq n$.

###### Proof.

By Proposition 2.2, the $R$-module $M^{n}=\bigoplus_{i=1}^{n}M$ is s-unital.
By the definition, this means existence of the desired element $e\in R$ such
that $em=m$ for the element $m=(m_{1},\dotsc,m_{n})\in M^{n}$. ∎

We will not use the following lemma in this paper, but we include it for the
sake of completeness of the exposition.

######  Lemma 2.4.

Let $R$ be a ring and $B$ be an s-unital $R$-$R$-bimodule. Then, for any
finite collection of elements $b_{1}$, …, $b_{n}\in B$, there exists an
element $e\in R$ such that $eb_{i}=b_{i}=b_{i}e$ for all $1\leq i\leq n$.

###### Proof.

This is [16, Proposition 2.10 of the published version or Proposition 10 of
the arXiv version]. ∎

######  Definition 2.5 (Tominaga [24, 25]).

A ring $R$ is said to be _left s-unital_ if $R$ is an s-unital left
$R$-module. Similarly, $R$ is said to be _right s-unital_ if it is s-unital as
a right $R$-module. A ring $R$ is _s-unital_ if it is both left and right
s-unital.

The following theorem is due to Tominaga [25, Remark (2) in Section 1 on p.
121–122].

######  Theorem 2.6 ([25]).

(a) Let $R$ be a left s-unital ring. Then, for any s-unital left $R$-module
$M$, the natural map $R\otimes_{R}M\longrightarrow M$ (10) is an isomorphism.
In other words, any s-unital left $R$-module is t-unital.

(b) Let $R$ be a right s-unital ring. Then, for any s-unital right $R$-module
$N$, the natural map $N\otimes_{R}R\longrightarrow N$ (11) is an isomorphism.
In other words, any s-unital right $R$-module is t-unital.

###### Proof.

Let us prove part (a). For this purpose, we will construct an inverse map

$\phi_{M}\colon M\mskip 1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip
1.5muR\otimes_{R}M$

to the multiplication map (10). Given an element $m\in M$, pick an arbitrary
element $e\in R$ such that $em=m$ in $M$, and put $\phi(m)=e\otimes m\in
R\otimes_{R}M$.

The key step is to check that the map $\phi$ is well-defined. The assumption
of left s-unitality of the ring $R$ is used here.

Let $f\in R$ be another element such that $fm=m$ in $M$. Since the left
$R$-module $R$ is s-unital, Corollary 2.3 (for $n=2$) tells that there exists
an element $g\in R$ such that $ge=e$ and $gf=f$ in $R$. Now we have

$e\otimes m=ge\otimes m=g\otimes em=g\otimes m=g\otimes fm=gf\otimes
m=f\otimes m$

in $R\otimes_{R}M$. Here the first and sixth equations hold by the choice of
the element $g$, the second and fifth equations are provided by the definition
of the tensor product, and the third and fourth equations hold by the
conditions imposed on the elements $e$ and $f$. Thus the map $\phi$ is well-
defined.

To check that the map $\phi$ is additive (i. e., an abelian group
homomorphism), we apply Corollary 2.3 (for $n=2$) to the s-unital $R$-module
$M$. Given two elements $m^{\prime}$ and $m^{\prime\prime}\in M$, the
corollary tells that there exists an element $e\in R$ such that
$em^{\prime}=m^{\prime}$ and $em^{\prime\prime}=m^{\prime\prime}$ in $M$.
Hence $e(m^{\prime}+m^{\prime\prime})=m^{\prime}+m^{\prime\prime}$. Now it is
clear that
$\phi(m^{\prime}+m^{\prime\prime})=e\otimes(m^{\prime}+m^{\prime\prime})=e\otimes
m^{\prime}+e\otimes m^{\prime\prime}=\phi(m^{\prime})+\phi(m^{\prime\prime})$.

One can also check directly that $\phi$ is a left $R$-module map (though we
will not need this). Given any elements $r\in R$ and $m\in M$, choose an
element $f\in R$ such that $fr=r$ in $R$. Then $frm=rm$ in $M$. Also, choose
an element $e\in R$ such that $em=m$ in $M$. Now we have $\phi(rm)=f\otimes
rm=fr\otimes m=r\otimes m=r\otimes em=re\otimes m=r\phi(m)$ in
$R\otimes_{R}M$.

What we need to check, though, is that both the compositions $M\longrightarrow
R\otimes_{R}M\longrightarrow M$ and $R\otimes_{R}M\longrightarrow
M\longrightarrow R\otimes_{R}M$ are identity maps. In the case of the former
composition, this is immediate from the constructions. Concerning the latter
composition, we know that both the maps being composed are abelian group
homomorphisms; so it suffices to compute the action of the composition in
question on decomposable tensors in $R\otimes_{R}M$.

Once again, given any elements $r\in R$ and $m\in M$, let us choose an element
$f\in R$ such that $fr=r$ in $R$. Then $frm=m$ in $M$. Finally, we have
$r\otimes m\longmapsto rm\longmapsto f\otimes rm=fr\otimes m=r\otimes m$ in
$R\otimes_{R}M$. ∎

######  Corollary 2.7.

Any left s-unital ring is t-unital. Similarly, any right s-unital ring is
t-unital.

###### Proof.

Follows from Theorem 2.6 and the definition of a t-unital ring. ∎

In Corollary 7.5 below, we will show that, for any left s-unital ring $R$, the
unital right $\widetilde{R}$-module $R$ is flat.

The next lemma is a version of Lemma 1.2 for the s-unitality property.

######  Lemma 2.8.

Let $R^{\prime}$ and $R^{\prime\prime}$ be arbitrary (nonunital) rings.

(a) Let $B$ be an $R^{\prime}$-$R^{\prime\prime}$-bimodule and $M$ be an left
$R^{\prime\prime}$-module. Assume that $B$ is s-unital as a left
$R^{\prime}$-module. Then the left $R^{\prime}$-module
$B\otimes_{R^{\prime\prime}}M$ is also s-unital.

(b) Let $N$ be a right $R^{\prime}$-module and $B$ be an
$R^{\prime}$-$R^{\prime\prime}$-bimodule. Assume that $B$ is s-unital as a
right $R^{\prime\prime}$-module. Then the right $R^{\prime\prime}$-module
$N\otimes_{R^{\prime}}B$ is also s-unital.

###### Proof.

Let us explain part (a). The point is that the left $R^{\prime}$-module
$B\otimes_{R^{\prime\prime}}M$ is a quotient module of an infinite direct sum
of copies of the left $R^{\prime}$-module $B$. Therefore, it remains to refer
to Proposition 2.2. ∎

######  Corollary 2.9.

(a) Over a left s-unital ring, a left module is t-unital if and only if it is
s-unital.

(b) Over a right s-unital ring, a right module is t-unital if and only if it
is s-unital.

###### Proof.

Let us prove part (a). Let $R$ be a left s-unital ring. If a left $R$-module
$M$ is s-unital, then $M$ is t-unital by Theorem 2.6(a). Conversely, for any
left $R$-module $M$, the left $R$-module $R\otimes_{R}M$ is s-unital by Lemma
2.8(a). Hence, if $M$ is t-unital, then $M$ is s-unital. ∎

######  Corollary 2.10.

Let $R$ be an s-unital ring. Then the abelian category of s-unital
$R$-$R$-bimodules $R{\operatorname{\mathsf{--{}^{s}Bimod^{s}--}}}R$ is an
associative and unital monoidal category, with the unit object $R\in
R{\operatorname{\mathsf{--{}^{s}Bimod^{s}--}}}R$, with respect to the tensor
product operation $\otimes_{R}$. The abelian category of s-unital left
$R$-modules $R{\operatorname{\mathsf{--{}^{s}Mod}}}$ is an associative and
unital left module category over
$R{\operatorname{\mathsf{--{}^{s}Bimod^{s}--}}}R$, and the abelian category of
s-unital right $R$-modules ${\operatorname{\mathsf{Mod^{s}--}}}R$ is an
associative and unital right module category over
$R{\operatorname{\mathsf{--{}^{s}Bimod^{s}--}}}R$, with respect to the tensor
product operation $\otimes_{R}$.

###### Proof.

The three categories are abelian by Proposition 2.2. The assertions of the
corollary follow from Corollaries 1.4, 2.7, and 2.9. ∎

######  Example 2.11.

Here is an example of a finitely generated free abelian group $R$ with a
structure of associative ring which is t-unital but _neither_ left _nor_ right
s-unital.

(0) We start from the observation that any homomorphic image of a left (right)
s-unital ring is left (right) s-unital. This follows immediately from the
definitions.

(1) Another simple observation is that the direct product of any two t-unital
rings is t-unital. Let $R$ and $S$ be t-unital rings, and let $T=R\times S$ be
their direct product (with componentwise addition and multiplication). Let us
show that the ring $T$ is t-unital. It is clear that
$R\otimes_{T}R=R\otimes_{R}R=R$ and $S\otimes_{T}S=S\otimes_{S}S=S$. It
remains to check that $R\otimes_{T}S=0$ and $S\otimes_{T}R=0$. Here we notice
that any t-unital ring is idempotent, i. e., $R^{2}=R$. Now we have
$r^{\prime}r^{\prime\prime}\otimes_{T}s=r^{\prime}\otimes_{T}r^{\prime\prime}s=0$
for any $r^{\prime}$, $r^{\prime\prime}\in R$ and $s\in S$, and similarly
$s\otimes_{T}r^{\prime}r^{\prime\prime}=0$.

(2) The following example of a left unital, but not right s-unital associative
ring $D$ can be found in [12, Solution to Exercise 1.10(b)] and [16, Example
2.6 of the published version or Example 6 of the arXiv version]. Consider the
matrix ring $D=\left(\begin{smallmatrix}\mathbb{Z}&\mathbb{Z}\\\
0&0\end{smallmatrix}\right)$. Explicitly, $D\simeq\mathbb{Z}\oplus\mathbb{Z}$
is the free abelian group with two generators endowed with the multiplication
$(a,b)(c,d)=(ac,ad)$ for all $a$, $b$, $c$, $d\in\mathbb{Z}$. Then the ring
$D$ has a left unit $(1,0)$, but is not right s-unital, as $(0,1)D=0$ in $D$.

(3) The ring $D$ is left s-unital, so it is t-unital by Corollary 2.7.
Similarly, the opposite ring
$D^{\mathrm{op}}=\left(\begin{smallmatrix}0&\mathbb{Z}\\\
0&\mathbb{Z}\end{smallmatrix}\right)$ is right s-unital (hence t-unital), but
not left s-unital. Now the ring $R=D\times D^{\mathrm{op}}$ is t-unital by
(1). But both the rings $D$ and $D^{\mathrm{op}}$ are homomorphic images of
$R$, so $R$ is neither left nor right s-unital by (0).

An example of _commutative_ associative ring $R$ which is t-unital but not
s-unital will be given below in Example 4.1.

##  3\. c-Unital Modules

Let $R$, $R^{\prime}$, $R^{\prime\prime}$ be (nonunital) rings. Given two
(nonunital) left $R$-modules $L$ and $M$, we denote by
$\operatorname{Hom}_{R}(L,M)$ the abelian group of morphisms $L\longrightarrow
M$ in the category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}\simeq\widetilde{R}{\operatorname{\mathsf{--Mod}}}$.
So there is a natural (identity) isomorphism of abelian groups

(13)
$\operatorname{Hom}_{R}(L,M)\simeq\operatorname{Hom}_{\widetilde{R}}(L,M).$

For any left $R^{\prime\prime}$-module $M$, any
$R^{\prime}$-$R^{\prime\prime}$-bimodule $B$, and any left $R^{\prime}$-module
$P$, there is a natural isomorphism of abelian groups

(14)
$\operatorname{Hom}_{R^{\prime}}(B\otimes_{R^{\prime\prime}}M,\>P)\simeq\operatorname{Hom}_{R^{\prime\prime}}(M,\operatorname{Hom}_{R^{\prime}}(B,P)).$

The adjunction/associativity isomorphism (14) can be obtained from the similar
isomorphism for unital rings $\widetilde{R}^{\prime}$ and
$\widetilde{R}^{\prime\prime}$ and unital (bi)modules over them using the
category equivalences (7) and the isomorphisms (8) and (13).

For any left $R$-module $P$, there is a natural left $R$-module morphism

(15) $p\longmapsto(r\mapsto rp):P\longrightarrow\operatorname{Hom}_{R}(R,P).$

######  Definition 3.1.

We will say that a left $R$-module $P$ is _c-unital_ if the natural map
$P\longrightarrow\operatorname{Hom}_{R}(R,P)$ (15) is an isomorphism. (The
letter “c” here means “contra” or “contramodule”. What we call “c-unital
modules” are called “closed modules” in [22, Definition 5.2].)

######  Lemma 3.2.

Let $B$ be an $R^{\prime}$-$R^{\prime\prime}$-bimodule and $P$ be a left
$R^{\prime}$-module. Assume that $B$ is t-unital as a right
$R^{\prime\prime}$-module. Then the left $R^{\prime\prime}$-module
$\operatorname{Hom}_{R^{\prime}}(B,P)$ is c-unital.

###### Proof.

This is the Hom version of Lemma 1.2. The assertion follows from the
associativity isomorphism (14) and commutativity of the suitable triangular
diagram formed by the associativity isomorphism together with (the maps
induced by) the unitality comparison maps (11) and (15). ∎

######  Corollary 3.3.

Let $R$ be a t-unital ring. Then, for any left $R$-module $P$, the left
$R$-module $\operatorname{Hom}_{R}(R,P)$ is c-unital.

###### Proof.

This is the Hom version of Corollary 1.3. To prove the assertion, take
$B=R=R^{\prime}=R^{\prime\prime}$ in Lemma 3.2. ∎

Given a ring $R$, we will denote by $R{\operatorname{\mathsf{--{}^{c}Mod}}}$
the full subcategory of c-unital $R$-modules in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$.

######  Corollary 3.4.

Let $R$ be a t-unital ring. Then the opposite category
$R{\operatorname{\mathsf{--{}^{c}Mod}}}^{\mathsf{op}}$ to the additive
category of c-unital left $R$-modules $R{\operatorname{\mathsf{--{}^{c}Mod}}}$
is an associative and unital right module category over the monoidal category
$R{\operatorname{\mathsf{--{}^{t}Bimod^{t}--}}}R$, with respect to the Hom
operation

$P^{\mathsf{op}}*B=\operatorname{Hom}_{R}(B,P)^{\mathsf{op}}\quad\text{for all
$P\in R{\operatorname{\mathsf{--{}^{c}Mod}}}$ and $B\in
R{\operatorname{\mathsf{--{}^{t}Bimod^{t}--}}}R$}.$

###### Proof.

Follows from the associativity isomorphism (14) and Lemma 3.2. ∎

######  Lemma 3.5.

Let $R$ be an arbitrary (nonunital) ring. Then the full subcategory of
c-unital left $R$-modules $R{\operatorname{\mathsf{--{}^{c}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is closed under extensions and all
limits (including kernels, direct products, and inverse limits) in the abelian
category of nonunital modules $R{\operatorname{\mathsf{--{}^{n}Mod}}}$.

###### Proof.

This is our version of [22, Proposition 5.5]. Notice that it is _not_ claimed
in the lemma that $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is closed under
submodules in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$, but only under
kernels. The proof is dual-analogous to that of Lemma 1.5. The closedness
under limits follows from the fact that the Hom functor
$\operatorname{Hom}_{R}(R,{-})\simeq\operatorname{Hom}_{\widetilde{R}}(R,{-})$
preserves limits.

To prove the closedness under extensions, consider a short exact sequence
$0\longrightarrow K\longrightarrow L\longrightarrow M\longrightarrow 0$ in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$ with $K$, $M\in
R{\operatorname{\mathsf{--{}^{c}Mod}}}$. The assertion follows from
commutativity of the diagram

---
$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{L\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Hom}_{R}(R,K)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Hom}_{R}(R,L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Hom}_{R}(R,M)}$

with a short exact sequence in the upper row and a left exact sequence in the
lower row. ∎

######  Remark 3.6.

Similarly to Remark 1.6, the construction of the full subcategory
$R{\operatorname{\mathsf{--{}^{c}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ can be viewed as a particular case of
the construction of the _$\operatorname{Ext}^{0,1}$ -perpendicular
subcategory_ from [9, Proposition 1.1] or [18, Theorem 1.2(a)]. Given a
nonunital left $R$-module $P$, the natural map
$P=\operatorname{Hom}_{\widetilde{R}}(\widetilde{R},P)\longrightarrow\operatorname{Hom}_{\widetilde{R}}(R,P)=\operatorname{Hom}_{R}(R,P)$
is an isomorphism _if and only if_ one has
$\operatorname{Hom}_{\widetilde{R}}(\widetilde{R}/R,P)\allowbreak=0=\operatorname{Ext}^{1}_{\widetilde{R}}(\widetilde{R}/R,P)$.
(This observation is a particular case of the discussion in [22, Definition
5.2].) Therefore, Lemma 3.5 becomes a particular case of [9, first paragraph
of Proposition 1.1].

Furthermore, the full subcategory $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is
closed under cokernels in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$ (and,
consequently, abelian) whenever the projective dimension of the unital left
$\widetilde{R}$-module $\widetilde{R}/R$ does not exceed $1$, i. e., in other
words, the unital $\widetilde{R}$-module $R$ is projective. This can be also
seen directly from the definition: if the functor
$\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is exact (preserves cokernels), then
the full subcategory $R{\operatorname{\mathsf{--{}^{c}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is closed under cokernels. The
following example shows that this does _not_ hold in general.

######  Example 3.7.

Continuing the discussion from Example 1.7, let $R$ be the ideal
$R=(x,y)\subset\widetilde{R}=\mathbb{Z}[x,y]$. Then the projective dimension
of the unital $\widetilde{R}$-module $\mathbb{Z}=\widetilde{R}/R$ is equal to
$2$, and the projective dimension of the unital $\widetilde{R}$-module $R$ is
equal to $1$. We claim that the full subcategory of c-unital $R$-modules
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is _not_ closed under cokernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}=\widetilde{R}{\operatorname{\mathsf{--Mod}}}$,
_not_ even under the cokernels of monomorphisms.

Indeed, consider the free $\widetilde{R}$-module with one generator
$F=\mathbb{Z}[x,y]$ and its quotient $\widetilde{R}$-module
$G=F/xF=\mathbb{Z}[y]$. Then one can easily see that $F\in
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ but $G\notin
R{\operatorname{\mathsf{--{}^{c}Mod}}}$. Indeed, the $R$-module map
$r\longmapsto\frac{r}{y}\bmod\frac{x}{y}\colon R\longrightarrow G$ does not
arise from any element of $G$.

The ring $R$ is this example is _not_ t-unital; in fact, $R^{2}\neq R$.
Examples of t-unital rings $T$ for which the full subcategory
$T{\operatorname{\mathsf{--{}^{c}Mod}}}$ is not closed under cokernels in
$T{\operatorname{\mathsf{--{}^{n}Mod}}}$ will be given below in Examples 4.2
and 4.8. It would be interesting to find a counterexample of an s-unital ring
$S$ for which the full subcategory $S{\operatorname{\mathsf{--{}^{c}Mod}}}$ is
not closed under cokernels in $S{\operatorname{\mathsf{--{}^{n}Mod}}}$.

For any t-unital ring $R$, Theorem 5.8(b) below tells that the category
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is abelian. On the other hand, again
over a t-unital ring $R$, according to Corollary 7.6, the full subcategory of
c-unital modules $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is closed under
cokernels in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$ _if and only if_ $R$ is
a projective unital left $\widetilde{R}$-module.

In the more restrictive setting of _rings with enough idempotents_ considered
in Section 6 below, we will show that both the full subcategories
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ and
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ are closed under both the kernels and
cokernels in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$.

##  4\. Counterexamples

In this section we provide some more advanced versions of counterexamples from
Examples 1.7, 2.11, and 3.7. All the examples in this section are commutative,
associative rings. We start with an example of commutative t-unital but not
s-unital ring, continue with an example of idempotent but not t-unital ring,
and finally come to an example of t-unital ring for which the class of
t-unital modules is not closed under kernels and the class of c-unital modules
is not closed under cokernels. We also provide an example of a t-unital ring
for which $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is closed under kernels,
but $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is not closed under cokernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$.

######  Example 4.1.

Here is an example of a _commutative_ associative ring $R$ which is t-unital
but _not_ s-unital (see Example 2.11 for an easier noncommutative example).
Let $\widetilde{R}=\mathbb{Z}[z^{q}\mid q\in\mathbb{Q}_{\geq 0}]$ be the free
abelian group spanned by the formal expressions $z^{q}$, where $z$ is a
variable and $q$ ranges over the nonnegative rational numbers. The
multiplication given by the obvious rule
$z^{q^{\prime}}z^{q^{\prime\prime}}=z^{q^{\prime}+q^{\prime\prime}}$ makes
$\widetilde{R}$ an associative, commutative, unital ring. Let
$R\subset\widetilde{R}$ be the subgroup spanned by $z^{q}$ with $q>0$. Then
$R$ is an ideal in $\widetilde{R}$, and $\widetilde{R}=\mathbb{Z}\oplus R$ is
the unitalization of the nonunital ring $R$. We claim that $R$ is a t-unital,
but not s-unital ring.

Indeed, for any nonzero element $r\in R$ one has $r\notin Rr$, so $R$ is not
s-unital. In other words, the ring $R$ is not s-unital, since $R\neq 0$ and
the ring $\widetilde{R}$ has no zero-divisors. On the other hand, the
multiplication map $R\otimes_{R}R\longrightarrow R$ is surjective, since one
has $z^{q/2}\otimes z^{q/2}\longmapsto z^{q}$ for every positive rational
number $q$. To show that the multiplication map is also injective, consider
any quadruple of positive rational numbers $a$, $b$, $c$, $d$ such that
$a+b=c+d$ in $\mathbb{Q}$. We have to show that $z^{a}\otimes
z^{b}=z^{c}\otimes z^{d}$ in $R\otimes_{R}R$. Indeed, if $a=c$ and $b=d$, then
there is nothing to prove. Otherwise, assume without loss of generality that
$a>c$ and $b<d$. Then we have $z^{a}\otimes z^{b}=z^{c}z^{a-c}\otimes
z^{b}=z^{c}\otimes z^{a-c}z^{b}=z^{c}\otimes z^{d}$ in $R\otimes_{R}R$, as
$a-c+b=d$ in $\mathbb{Q}$.

######  Example 4.2.

Let $R=\mathbb{Z}[z^{q}\mid q\in\mathbb{Q}_{>0}]$ be the ring from Example
4.1. Notice that $R=\bigcup_{n\geq 1}z^{1/n}\widetilde{R}$ is a flat unital
$\widetilde{R}$-module. According to Remark 1.6, it follows that the full
subcategory of t-unital $R$-modules $R{\operatorname{\mathsf{--{}^{t}Mod}}}$
is closed under kernels in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$. So
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is an abelian category with an exact
fully faithful inclusion functor
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ (even though the ring $R$ is not
s-unital).

On the other hand, the unital $\widetilde{R}$-module $R$ is _not_ projective.
Indeed, the telescope construction provides a short exact sequence of unital
$\widetilde{R}$-modules $0\longrightarrow\bigoplus_{n\geq
1}z^{1/n}\widetilde{R}\longrightarrow\bigoplus_{n\geq
1}z^{1/n}\widetilde{R}\longrightarrow R\longrightarrow 0$. If $R$ were a
projective unital $\widetilde{R}$-module, it would follow that this short
exact seqeunce splits; so there is a section $s\colon
R\longrightarrow\bigoplus_{n\geq 1}z^{1/n}\widetilde{R}$. The image of the
element $z^{1}\in R$ would have to belong to a finite subsum
$\bigoplus_{i=1}^{m}z^{1/n_{i}}\widetilde{R}$ of the direct sum in the right-
hand side. Since $z^{1/n}\widetilde{R}$ are torsion-free
$\widetilde{R}$-modules, it would follow the image of the whole map $s$ is
contained in the same finite subsum
$\bigoplus_{i=1}^{m}z^{1/n_{i}}\widetilde{R}\subset\bigoplus_{n\geq
1}z^{1/n}\widetilde{R}$; a contradiction.

According to Corollary 7.6 below, we can conclude that the full subcategory
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is _not_ closed under cokernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. (Still, the category
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is abelian by Theorem 5.8(b).) A more
explicit example of such behavior, with a discussion not relying on references
to later material, will be given in Example 4.8.

######  Example 4.3.

Here is an example of an associative, commutative ring $S$ which is
idempotent, but not t-unital; in other words, the natural map
$S\otimes_{S}S\longrightarrow S$ is surjective, but not injective. Let
$R\subset\widetilde{R}=\mathbb{Z}[z^{q}\mid q\in\mathbb{Q}_{\geq 0}]$ be the
rings from Example 4.1. Denote by $J\subset R$ the additive subgroup spanned
by $z^{q}$ with $q\geq 1$. Then $J$ is an ideal in $\widetilde{R}$. Put
$S=R/J$ and $\widetilde{S}=\widetilde{R}/J$. Clearly,
$\widetilde{S}=\mathbb{Z}\oplus S$ is the unitalization of the nonunital ring
$S$. Denote by $\bar{z}^{q}=z^{q}+J\in\widetilde{S}$ the images of the basis
vectors $z^{q}\in\widetilde{R}$ under the surjective ring homomorphism
$\widetilde{R}\longrightarrow\widetilde{S}$.

We have $S^{2}=S$, since $S$ is a homomorphic image of $R$ and $R^{2}=R$. Let
us show that the map $S\otimes_{S}S\longrightarrow S$ is _not_ injective. For
any positive rational numbers $a$ and $b$ such that $a+b=1$ we have
$\bar{z}^{a}\otimes\bar{z}^{b}=\bar{z}^{1/2}\otimes\bar{z}^{1/2}$ in
$S\otimes_{S}S$. The image of this element under the multiplication map
$S\otimes_{S}S\longrightarrow S$ is $\bar{z}^{1}=0$ in $S$. Now we claim that
the element $\bar{z}^{1/2}\otimes\bar{z}^{1/2}$ is nonzero in $S\otimes_{S}S$.

One can argue as follows. Endow the rings $\widetilde{R}$ and $\widetilde{S}$
with a grading indexed by the abelian group of rational numbers, given by the
rule $\deg z^{q}=q=\deg\bar{z}^{q}$. Then there is the induced
$\mathbb{Q}$-valued grading on the tensor products $R\otimes_{R}R$ and
$S\otimes_{S}S$. The nonunital rings $R$ and $S$ are concentrated in positive
degrees, and the surjective ring homomorphism $R\longrightarrow S$ is an
isomorphism in the degrees $<\nobreak 1$. It follows easily that the induced
surjective map $R\otimes_{R}R\longrightarrow S\otimes_{S}S$ is an isomorphism
in the degrees $\leq\nobreak 1$. So $z^{1/2}\otimes z^{1/2}\neq 0$ in
$R\otimes_{R}R$ implies $\bar{z}^{1/2}\otimes\bar{z}^{1/2}\neq 0$ in
$S\otimes_{S}S$.

The following lemma is standard.

######  Lemma 4.4 (Künneth formula).

Let $A$ and $B$ be associative, unital rings. Let $K$ be a right $A$-module,
$N$ be a right $B$-module, $L$ be a left $A$-module, and $M$ be a left
$B$-module. Assume that all the abelian groups $A$, $B$, $K$, $N$, $L$, $M$
have no torsion; and assume further that all the abelian groups
$\operatorname{Tor}_{i}^{A}(K,L)$ and $\operatorname{Tor}_{j}^{B}(N,M)$, $i$,
$j\geq 0$, have no torsion, either. Then there is a natural isomorphism of
abelian groups

$\operatorname{Tor}_{n}^{A\otimes_{\mathbb{Z}}B}(K\otimes_{\mathbb{Z}}N,\>L\otimes_{\mathbb{Z}}M)\simeq\bigoplus\nolimits_{i+j=n}\operatorname{Tor}_{i}^{A}(K,L)\otimes_{\mathbb{Z}}\operatorname{Tor}_{j}^{B}(N,M)$

for every $n\geq 0$.

###### Proof.

Pick a flat resolution $F_{\text{\smaller\smaller$\scriptstyle\bullet$}}$ for
the left $A$-module $L$ and a flat resolution
$G_{\text{\smaller\smaller$\scriptstyle\bullet$}}$ for the left $B$-module
$M$. Then
$F_{\text{\smaller\smaller$\scriptstyle\bullet$}}\otimes_{\mathbb{Z}}G_{\text{\smaller\smaller$\scriptstyle\bullet$}}$
is a flat resolution for the left $(A\otimes_{\mathbb{Z}}B)$-module
$L\otimes_{\mathbb{Z}}M$, and one can compute
$(K\otimes_{\mathbb{Z}}N)\otimes_{A\otimes_{\mathbb{Z}}B}(F_{\text{\smaller\smaller$\scriptstyle\bullet$}}\otimes_{\mathbb{Z}}G_{\text{\smaller\smaller$\scriptstyle\bullet$}})=(K\otimes_{A}F_{\text{\smaller\smaller$\scriptstyle\bullet$}})\otimes_{\mathbb{Z}}(N\otimes_{B}G_{\text{\smaller\smaller$\scriptstyle\bullet$}})$.
Finally, for any (bounded above) complexes of torsion-free abelian groups
$C_{\text{\smaller\smaller$\scriptstyle\bullet$}}$ and
$D_{\text{\smaller\smaller$\scriptstyle\bullet$}}$ with torsion-free homology
groups, one has
$H_{n}(C_{\text{\smaller\smaller$\scriptstyle\bullet$}}\otimes_{\mathbb{Z}}D_{\text{\smaller\smaller$\scriptstyle\bullet$}})=\bigoplus_{i+j=n}H_{i}(C_{\text{\smaller\smaller$\scriptstyle\bullet$}})\otimes_{\mathbb{Z}}H_{j}(D_{\text{\smaller\smaller$\scriptstyle\bullet$}})$
for all $n\in\mathbb{Z}$. ∎

The following remark implies that t-unital rings $R$ are defined by _a low-
degree part of_ the set of conditions expressed by saying that the unital ring
homomorphism $\widetilde{R}\longrightarrow\widetilde{R}/R=\mathbb{Z}$ is a
_homological ring epimorphism_ [9, Section 4]. We are grateful to Jan
Št’ovíček for suggesting this interpretation.

######  Remark 4.5.

Let $R$ be a (nonunital) associative ring. As a particular case of Remark 1.6,
we know that the ring $R$ is t-unital if and only if
$\widetilde{R}/R\otimes_{\widetilde{R}}R=0=\operatorname{Tor}_{1}^{\widetilde{R}}(\widetilde{R}/R,R)$.
Notice that the map
$\widetilde{R}/R\otimes_{\widetilde{R}}R\longrightarrow{\widetilde{R}/R}\otimes_{\widetilde{R}}\widetilde{R}=\widetilde{R}/R=\mathbb{Z}$
induced by the inclusion $R\longrightarrow\widetilde{R}$ vanishes.
Consequently,
$\widetilde{R}/R\otimes_{\widetilde{R}}R\simeq\operatorname{Tor}_{1}^{\widetilde{R}}(\widetilde{R}/R,\widetilde{R}/R)$.
Also we have
$\operatorname{Tor}_{1}^{\widetilde{R}}(\widetilde{R}/R,R)\simeq\operatorname{Tor}_{2}^{\widetilde{R}}(\widetilde{R}/R,\widetilde{R}/R)$.
Thus a ring $R$ is t-unital if and only if
$\operatorname{Tor}_{1}^{\widetilde{R}}(\widetilde{R}/R,\widetilde{R}/R)=0=\operatorname{Tor}_{2}^{\widetilde{R}}(\widetilde{R}/R,\widetilde{R}/R)$,
or in other notation, if and only if
$\operatorname{Tor}_{1}^{\widetilde{R}}(\mathbb{Z},\mathbb{Z})=0=\operatorname{Tor}_{2}^{\widetilde{R}}(\mathbb{Z},\mathbb{Z})$.

######  Example 4.6.

Let $\widetilde{R}=\mathbb{Z}[x^{p}\mid p\in\mathbb{Q}_{\geq 0}]$ and
$\widetilde{S}=\mathbb{Z}[y^{q}\mid q\in\mathbb{Q}_{\geq 0}]$ be two copies of
the unital ring from Example 4.1. Consider the ring
$\widetilde{T}=\widetilde{R}\otimes_{\mathbb{Z}}\widetilde{S}=\mathbb{Z}[x^{p}y^{q}\mid
p,q\in\mathbb{Q}_{\geq 0}]$. So $\widetilde{T}$ is an associative,
commutative, unital ring. Let $T\subset\widetilde{T}$ be the subgroup spanned
by $x^{p}y^{q}$ with $(p,q)\neq(0,0)$. Then $T$ is an ideal in
$\widetilde{T}$, and $\widetilde{T}=\mathbb{Z}\oplus T$ is the unitalization
of the nonunital ring $T$. We claim that $T$ is a t-unital ring.

Indeed, following Remark 4.5, it suffices to check that
$\operatorname{Tor}_{1}^{\widetilde{T}}(\mathbb{Z},\mathbb{Z})=0=\operatorname{Tor}_{2}^{\widetilde{T}}(\mathbb{Z},\mathbb{Z})$.
One easily computes that
$\operatorname{Tor}^{\widetilde{R}}_{i}(\mathbb{Z},\mathbb{Z})=\operatorname{Tor}^{\widetilde{R}}_{i-1}(\mathbb{Z},R)=0$
for all $i\geq 1$ (in fact, $R=\bigcup_{n\geq 1}x^{1/n}\widetilde{R}$ is a
flat $\widetilde{R}$-module and $R^{2}=R$). Similarly,
$\operatorname{Tor}^{\widetilde{S}}_{j}(\mathbb{Z},\mathbb{Z})=0$ for all
$j\geq 1$. On the other hand,
$\operatorname{Tor}_{0}^{\widetilde{R}}(\mathbb{Z},\mathbb{Z})=\mathbb{Z}=\operatorname{Tor}_{0}^{\widetilde{S}}(\mathbb{Z},\mathbb{Z})$.
By the Künneth formula (Lemma 4.4), it follows that
$\operatorname{Tor}^{\widetilde{T}}_{n}(\mathbb{Z},\mathbb{Z})=0$ for all
$n\geq 1$.

######  Example 4.7.

This is an improved version of Example 1.7, providing a commutative t-unital
ring $T$ for which the full subcategory of t-unital $T$-modules
$T{\operatorname{\mathsf{--{}^{t}Mod}}}$ is _not_ closed under kernels (_not_
even under the kernels of epimorphisms) in the abelian category of nonunital
$T$-modules $T{\operatorname{\mathsf{--{}^{n}Mod}}}$. We keep the notation of
Example 4.6 for the rings $R$, $S$, and $T$.

Consider the localizations $\widetilde{R}[x^{-1}]=\mathbb{Z}[x^{p}\mid
p\in\mathbb{Q}]$ and $\widetilde{S}[y^{-1}]=\mathbb{Z}[y^{q}\mid
q\in\mathbb{Q}]$ of the rings $R$ and $S$. Let
$\widetilde{P}_{x}=\widetilde{R}[x^{-1}]/\widetilde{R}$ and
$\widetilde{P}_{y}=\widetilde{S}[y^{-1}]/\widetilde{S}$ be the related
generalized Prüfer modules. Then
$\operatorname{Tor}_{0}^{\widetilde{R}}(\mathbb{Z},\widetilde{P}_{x})=0$ and
$\operatorname{Tor}_{1}^{\widetilde{R}}(\mathbb{Z},\widetilde{P}_{x})=\mathbb{Z}$.
By the Künneth formula, it follows that
$\operatorname{Tor}_{0}^{\widetilde{T}}(\mathbb{Z},\>\widetilde{P}_{x}\otimes_{\mathbb{Z}}\widetilde{P}_{y})=0=\operatorname{Tor}_{1}^{\widetilde{T}}(\mathbb{Z},\>\widetilde{P}_{x}\otimes_{\mathbb{Z}}\widetilde{P}_{y})$.
Thus $\widetilde{P}_{x}\otimes_{\mathbb{Z}}\widetilde{P}_{y}$ is a t-unital
$T$-module (by Remark 1.6).

On the other hand, denote by
$\widetilde{Q}_{y}=\mathbb{Z}\otimes_{\mathbb{Z}}\widetilde{P}_{y}$ the
$\widetilde{S}$-module $\widetilde{P}_{y}$ viewed as a $\widetilde{T}$-module
with the given (Prüfer) action of the elements $y^{q}$ and the zero action of
the elements $x^{p}y^{q}\in T$ for $p>0$. Then, again by the Künneth formula,
$\operatorname{Tor}_{1}^{\widetilde{T}}(\mathbb{Z},\widetilde{Q}_{y})=\operatorname{Tor}_{1}^{\widetilde{S}}(\mathbb{Z},\widetilde{P}_{y})=\mathbb{Z}$.
So Remark 1.6 tells that $\widetilde{Q}_{y}\notin
T{\operatorname{\mathsf{--{}^{t}Mod}}}$.

Consider another version of generalized Prüfer module over $\widetilde{R}$:
namely, put $P_{x}=\widetilde{R}[x^{-1}]/R$. Then $P_{x}=\varinjlim_{n\geq
1}\widetilde{R}[x^{-1}]/x^{1/n}\widetilde{R}\simeq\varinjlim_{n\geq
1}\widetilde{P}_{x}$ is a direct limit of copies of the $\widetilde{R}$-module
$\widetilde{P}_{x}$. Furthermore, we have a short exact sequence of
$\widetilde{R}$-modules $0\longrightarrow\mathbb{Z}\longrightarrow
P_{x}\longrightarrow\widetilde{P}_{x}\longrightarrow 0$. The
$\widetilde{T}$-module $P_{x}\otimes_{\mathbb{Z}}\widetilde{P}_{y}$ is a
direct limit of copies of the $\widetilde{T}$-module
$\widetilde{P}_{x}\otimes_{\mathbb{Z}}\widetilde{P}_{y}$; so Lemma 1.5 tells
that $P_{x}\otimes_{\mathbb{Z}}\widetilde{P}_{y}$ is a t-unital T-module.
Finally, in the short exact sequence of $T$-modules
$0\longrightarrow\widetilde{Q}_{y}\longrightarrow
P_{x}\otimes_{\mathbb{Z}}\widetilde{P}_{y}\longrightarrow\widetilde{P}_{x}\otimes_{\mathbb{Z}}\widetilde{P}_{y}\longrightarrow
0$ the middle term and the cokernel are t-unital, but the kernel is not.

######  Example 4.8.

This is an improved version of Example 3.7, providing a commutative t-unital
ring $T$ for which the full subcategory of c-unital $T$-modules
$T{\operatorname{\mathsf{--{}^{c}Mod}}}$ is _not_ closed under cokernels
(_not_ even under the cokernels of monomorphisms) in the abelian category of
nonunital $T$-modules $T{\operatorname{\mathsf{--{}^{n}Mod}}}$. We still keep
the notation of Example 4.6 for the rings $R$, $S$, and $T$.

Consider the flat $\widetilde{T}$-module
$\widetilde{F}=\widetilde{R}\otimes_{\mathbb{Z}}S\subset T$ and its flat
$\widetilde{T}$-submodule $F=R\otimes_{\mathbb{Z}}S\subset\widetilde{F}$. So
the free abelian group $\widetilde{F}$ is spanned by the elements $x^{p}y^{q}$
with $p\geq 0$, $q>0$, while the free abelian group $F$ is spanned by the
elements $x^{p}y^{q}$ with $p>0$, $q>0$. Then both the $T$-modules $F$ and
$\widetilde{F}$ are c-unital. In other words, this means that any
$\widetilde{T}$-module morphism $T\longrightarrow F$ can be uniquely extended
to a $\widetilde{T}$-module morphism $\widetilde{T}\longrightarrow F$, and
similarly for morphisms into $\widetilde{F}$.

Indeed, let $h\colon T\longrightarrow F$ be a $T$-module morphism. Consider
the element $h(xy)\in F$. Then we have $h(xy)=xh(y)=yh(x)$ in $F$, and so the
element $h(xy)$ is divisible _both_ by $x$ and by $y$ in $F$. It follows
easily from the construction of $F$ that the element $h(xy)$ must be divisible
by $xy$ in $F$, i. e., there exists an element $f\in F$ such that $h(xy)=xyf$.
Hence the morphism $h\colon T\longrightarrow F$ comes from the element $f\in
F$ via the natural map $F\longrightarrow\operatorname{Hom}_{T}(T,F)$. The
argument for $\widetilde{F}$ is exactly the same. Arguing in this way, one can
show that any flat unital $\widetilde{T}$-module is a c-unital $T$-module.

On the other hand, notice that the $S$-module $S$ is _not_ c-unital, as the
identity morphism ${\mathrm{id}}_{S}\colon S\longrightarrow S$ does not belong
to the image of the natural map $S\longrightarrow\operatorname{Hom}_{S}(S,S)$.
In other words, the morphism ${\mathrm{id}}_{S}$ cannot be extended to an
$S$-module morphism $\widetilde{S}\longrightarrow S$. Essentially for this
reason, the quotient $T$-module $\widetilde{F}/F=S$ (with the zero action of
the elements $x^{p}y^{q}\in T$ for $p>0$) is also _not_ c-unital. Indeed, the
surjective morphism $T\longrightarrow
T/(R\otimes_{\mathbb{Z}}\widetilde{S})=S$ does not belong to the image of the
natural map $S\longrightarrow\operatorname{Hom}_{T}(T,S)$. So, in the short
exact sequence of $T$-modules $0\longrightarrow
F\longrightarrow\widetilde{F}\longrightarrow S\longrightarrow 0$ the middle
term and the kernel are c-unital, but the cokernel is not c-unital.

Notice that the ring $T$ in this example is t-unital, but _not_ s-unital. It
would be interesting to find an example of s-unital ring for which the full
subcategory of c-unital modules is not closed under cokernels in the ambient
abelian category of nonunital modules.

##  5\. The Main Abelian Category Equivalence

Let $R$ be a (nonunital) ring. We will say that a left $R$-module $M$ is a
_null-module_ if the action of $R$ in $M$ is zero, i. e., $rm=0$ in $M$ for
all $r\in R$ and $m\in M$. Right null-modules are defined similarly. We will
denote the full subcategories of null-modules by
$R{\operatorname{\mathsf{--{}^{0}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ and
${\operatorname{\mathsf{Mod^{0}--}}}R\subset{\operatorname{\mathsf{Mod^{n}--}}}R$.
(Cf. [22, Definitions 2.1 and 7.1], [7, Section 2.1.3].)

Notice that the full subcategory of null-modules
$R{\operatorname{\mathsf{--{}^{0}Mod}}}$ is closed under submodules,
quotients, infinite direct sums, and infinite products (hence under all limits
and colimits) in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$. The following easy
lemma says more.

######  Lemma 5.1.

Let $R$ be an idempotent ring, i. e., $R^{2}=R$. Then the full subcategory
$R{\operatorname{\mathsf{--{}^{0}Mod}}}$ is closed under extensions in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$.

###### Proof.

Let $0\longrightarrow K\longrightarrow L\longrightarrow M\longrightarrow 0$ be
a short exact sequence in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$ with $K$,
$M\in R{\operatorname{\mathsf{--{}^{0}Mod}}}$. Given an element $l\in L$, it
suffices to show that $r^{\prime}r^{\prime\prime}l=0$ in $L$ for all
$r^{\prime}$, $r^{\prime\prime}\in R$. Now we have $r^{\prime\prime}(l+K)=0$
in $M$, since $M$ is a null-module. Hence $r^{\prime\prime}l\in K\subset L$.
Finally $r^{\prime}(r^{\prime\prime}l)=0$ in $K$, since $K$ is a null-module.
∎

Thus, over an idempotent ring $R$, the full subcategory of null-modules is a
thick subcategory, or in a different terminology, a _Serre subcategory_ in the
abelian category $R{\operatorname{\mathsf{--{}^{n}Mod}}}$ [10, Section 1.11],
[8, Section III.1]. Therefore, the abelian Serre quotient category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
is defined. Ignoring set-theoretical issues, one can say that the quotient
category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
can be obtained from $R{\operatorname{\mathsf{--{}^{n}Mod}}}$ by inverting all
the morphisms whose kernel and cokernel belong to
$R{\operatorname{\mathsf{--{}^{0}Mod}}}$. (See [10, 8] for a more concrete
construction.)

The aim of this section is to explain that, for a t-unital ring $R$, both the
categories of t-unital and c-unital left $R$-modules are actually abelian, and
naturally equivalent to each other and to the Serre quotient category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$,

$R{\operatorname{\mathsf{--{}^{t}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{c}Mod}}}.$

A more general version of this result is due to Quillen [22, Theorems 4.5 and
5.6].

The following lemma can be compared to [1, Proposition 3.1(b)] (cf. the
paragraph preceding Remark 4.5 above). What we call t-unital modules are
called “comodules” in the context of [1].

######  Lemma 5.2.

Let $R$ be a t-unital ring. Then the functor $R\otimes_{R}{-}\,\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is right adjoint to the fully faithful
inclusion functor $R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$. For any left $R$-module $M$, the
natural map $R\otimes_{R}M\longrightarrow M$ is the adjunction counit.

###### Proof.

For any left $R$-module $M$, the left $R$-module $R\otimes_{R}M$ is t-unital
by Corollary 1.3(a). Now let $L$ be a t-unital left $R$-module. We need to
construct a natural isomorphism of abelian groups
$\operatorname{Hom}_{R}(L,M)\simeq\operatorname{Hom}_{R}(L,\>R\otimes_{R}M)$.
For this purpose, to any $R$-module morphism $f\colon L\longrightarrow M$ we
assign the morphism $R\otimes_{R}f\colon L\simeq R\otimes_{R}L\longrightarrow
R\otimes_{R}M$. To any $R$-module morphism $g\colon L\longrightarrow
R\otimes_{R}M$ we assign the composition
$L\overset{g}{\longrightarrow}R\otimes_{R}M\longrightarrow M$.

Let us check that these two maps between the two groups of morphisms are
inverse to each other. Starting with a morphism $f\colon L\longrightarrow M$,
it is clear that the composition $L\simeq R\otimes_{R}L\longrightarrow
R\otimes_{R}M\longrightarrow M$ is equal to $f$. Starting with a morphism
$g\colon L\longrightarrow R\otimes_{R}M$, producing the related morphism
$f\colon L\longrightarrow M$, and coming back to a new morphism
$g^{\prime}\colon L\longrightarrow R\otimes_{R}M$ via the constructions above,
in order to check that $g^{\prime}=g$ one needs to observe that the two
multiplication maps $R\otimes_{R}R\otimes_{R}M\rightrightarrows R\otimes_{R}M$
are equal to each other. The latter property holds by the definition of the
tensor product of $R$-modules $R\otimes_{R}M$. ∎

The next lemma is our version of [1, Proposition 3.2(b)]. What we call
c-unital modules are called “contramodules” in the context of [1].

######  Lemma 5.3.

Let $R$ be a t-unital ring. Then the functor
$\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is left adjoint to the fully faithful
inclusion functor $R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$. For any left $R$-module $P$, the
natural map $P\longrightarrow\operatorname{Hom}_{R}(R,P)$ is the adjunction
unit.

###### Proof.

For any left $R$-module $P$, the left $R$-module $\operatorname{Hom}_{R}(R,P)$
is t-unital by Corollary 3.3. Now let $Q$ be a c-unital left $R$-module. We
need to construct a natural isomorphism of abelian groups
$\operatorname{Hom}_{R}(P,Q)\simeq\operatorname{Hom}_{R}(\operatorname{Hom}_{R}(R,P),Q)$.
For this purpose, to any $R$-module morphism $f\colon P\longrightarrow Q$ we
assign the morphism
$\operatorname{Hom}_{R}(R,f)\colon\operatorname{Hom}_{R}(R,P)\longrightarrow\operatorname{Hom}_{R}(R,Q)\simeq
Q$. To any $R$-module morphism
$g\colon\operatorname{Hom}_{R}(R,P)\longrightarrow Q$ we assign the
composition $P\longrightarrow\operatorname{Hom}_{R}(R,P)\longrightarrow Q$.

Let us check that these two maps between the two groups of morphisms are
inverse to each other. Starting with a morphism $f\colon P\longrightarrow Q$,
it is clear that the composition
$P\longrightarrow\operatorname{Hom}_{R}(R,P)\longrightarrow\operatorname{Hom}_{R}(R,Q)\simeq
Q$ is equal to $f$. Starting with a morphism
$g\colon\operatorname{Hom}_{R}(R,P)\longrightarrow Q$, producing the related
morphism $f\colon P\longrightarrow Q$, and coming back to a new morphism
$g^{\prime}\colon\operatorname{Hom}_{R}(R,P)\longrightarrow Q$ via the
constructions above, in order to check that $g^{\prime}=g$ one needs to
observe that the two maps
$\operatorname{Hom}_{R}(R,P)\rightrightarrows\operatorname{Hom}_{R}(R\otimes_{R}R,\>P)\simeq\operatorname{Hom}_{R}(R,\operatorname{Hom}_{R}(R,P))$
induced by the maps $R\otimes_{R}R\longrightarrow R$ and
$P\longrightarrow\operatorname{Hom}_{R}(R,P)$ are equal to each other. The
latter property holds by the definition of the group of $R$-module morphisms
$\operatorname{Hom}_{R}(R,P)$. ∎

The following lemma is due to Quillen [22, discussions of formulas (2.2) and
(5.1)].

######  Lemma 5.4 ([22]).

Let $R$ be an arbitrary (nonunital) ring. Then

(a) for any left $R$-module $M$, the kernel and cokernel of the natural
$R$-module morphism $R\otimes_{R}M\longrightarrow M$ are null-modules;

(b) for any left $R$-module $P$, the kernel and cokernel of the natural
$R$-module morphism $P\longrightarrow\operatorname{Hom}_{R}(R,P)$ are null-
modules.

###### Proof.

Part (a): the cokernel of the map $R\otimes_{R}M\longrightarrow M$ is
obviously the maximal null quotient module of $M$. Concerning the kernel, let
$\sum_{i=1}^{n}r_{i}\otimes m_{i}\in R\otimes_{R}M$ be an element annihilated
by the map $R\otimes_{R}M\longrightarrow M$. Then, for any $r\in R$, one has
$r\sum_{i=1}^{n}r_{i}\otimes m_{i}=\sum_{i=1}^{n}rr_{i}\otimes
m_{i}=\sum_{i=1}^{n}r\otimes r_{i}m_{i}=r\otimes\sum_{i=1}^{n}r_{i}m_{i}=0$ in
$R\otimes_{R}M$. Part (b): the kernel of the map
$P\longrightarrow\operatorname{Hom}_{R}(R,P)$ is obviously the maximal null
submodule of $M$. Concerning the cokernel, let
$f\in\operatorname{Hom}_{R}(R,P)$ be a morphism and $r\in R$ be an element.
Then the $R$-module morphism $rf\colon R\longrightarrow P$, given by the rule
$(rf)(r^{\prime})=f(r^{\prime}r)=r^{\prime}f(r)$ for all $r^{\prime}\in R$,
comes from the element $f(r)\in P$ via the natural map
$P\longrightarrow\operatorname{Hom}_{R}(R,P)$. So $R$ acts by zero on the
coset of the element $f$ in the cokernel of the map
$P\longrightarrow\operatorname{Hom}_{R}(R,P)$. ∎

In the proof of part (b) of the next lemma, the following well-known fact is
used. Let $A$ be a unital ring, $E$ be a right $A$-module, and $M$ be a left
$A$-module. Then, for every $n\geq 0$, there is a natural isomorphism of
abelian groups

$\operatorname{Ext}^{n}_{A}(M,\operatorname{Hom}_{\mathbb{Z}}(E,\mathbb{Q}/\mathbb{Z}))\simeq\operatorname{Hom}_{\mathbb{Z}}(\operatorname{Tor}_{n}^{A}(E,M),\mathbb{Q}/\mathbb{Z}).$

######  Lemma 5.5.

Let $R$ be a t-unital ring, $\widetilde{R}=\mathbb{Z}\oplus R$ be its
unitalization, and $N$ be a left null-module over $R$. Then

(a)
$R\otimes_{R}N=R\otimes_{\widetilde{R}}N=0=\operatorname{Tor}^{\widetilde{R}}_{1}(R,N)$;

(b)
$\operatorname{Hom}_{R}(R,N)=\operatorname{Hom}_{\widetilde{R}}(R,N)=0=\operatorname{Ext}_{\widetilde{R}}^{1}(R,N)$.

###### Proof.

Part (a): one has $R\otimes_{R}N=0$ for any null-module $N$ over an idempotent
ring $R$, as $r^{\prime}r^{\prime\prime}\otimes n=r^{\prime}\otimes
r^{\prime\prime}n=0$ in $R\otimes_{R}N$ for any $r^{\prime}$,
$r^{\prime\prime}\in R$ and $n\in N$. To prove the $\operatorname{Tor}_{1}$
vanishing, notice that the category of (left) null-modules over $R$ is
isomorphic to the category of abelian groups: any abelian group can be
uniquely endowed with a null-module structure. Consider a short exact sequence
of abelian groups $0\longrightarrow G\longrightarrow F\longrightarrow
N\longrightarrow 0$ with a free abelian group $F$, and view it as a short
exact sequence of left null-modules over $R$. We have already seen that
$R\otimes_{\widetilde{R}}G=0$. Following Remark 1.6, we have
$\operatorname{Tor}^{\widetilde{R}}_{1}(R,\mathbb{Z})=\operatorname{Tor}^{\widetilde{R}}_{1}(R,\widetilde{R}/R)=0$,
since $R$ is a t-unital right $R$-module. Hence
$\operatorname{Tor}^{\widetilde{R}}_{1}(R,F)=0$, and we can conclude that
$\operatorname{Tor}^{\widetilde{R}}_{1}(R,N)=0$.

Part (b): one has $\operatorname{Hom}_{R}(R,N)=0$ for any null-module $N$ over
an idempotent ring $R$, as
$f(r^{\prime}r^{\prime\prime})=r^{\prime}f(r^{\prime\prime})=0$ in $N$ for any
$r^{\prime}$, $r^{\prime\prime}\in R$ and $f\in\operatorname{Hom}_{R}(R,N)$.
To prove the $\operatorname{Ext}^{1}$ vanishing, consider a short exact
sequence of abelian groups $0\longrightarrow N\longrightarrow J\longrightarrow
K\longrightarrow 0$, where $J$ is a product of copies of
$\mathbb{Q}/\mathbb{Z}$, and view it as a short exact sequence of left null-
modules over $R$. We have already seen that
$\operatorname{Hom}_{\widetilde{R}}(R,K)=0$. By Remark 1.6, we have
$\operatorname{Tor}^{\widetilde{R}}_{1}(\mathbb{Z},R)=\operatorname{Tor}^{\widetilde{R}}_{1}(\widetilde{R}/R,R)=0$,
since $R$ is a t-unital left $R$-module. Hence, viewing
$\mathbb{Q}/\mathbb{Z}$ as an abelian group _and_ a left null-module
${}_{\widetilde{R}}\mathbb{Q}/\mathbb{Z}$, and viewing $\mathbb{Z}$ as a right
null-module $\mathbb{Z}_{\widetilde{R}}$, we have
$\operatorname{Ext}^{1}_{\widetilde{R}}(R,{}_{\widetilde{R}}\mathbb{Q}/\mathbb{Z})=\operatorname{Ext}^{1}_{\widetilde{R}}(R,\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_{\widetilde{R}},\mathbb{Q}/\mathbb{Z}))=\operatorname{Hom}_{\mathbb{Z}}(\operatorname{Tor}_{1}^{\widetilde{R}}(\mathbb{Z}_{\widetilde{R}},R),\mathbb{Q}/\mathbb{Z})=0$.
Therefore, $\operatorname{Ext}^{1}_{\widetilde{R}}(R,J)=0$, and we can
conclude that $\operatorname{Ext}^{1}_{\widetilde{R}}(R,N)=0$. ∎

######  Proposition 5.6.

Let $R$ be a t-unital ring and $f\colon L\longrightarrow M$ be a left
$R$-module morphism whose kernel and cokernel are null-modules. Then

(a) the left $R$-module morphism $R\otimes_{R}f\colon
R\otimes_{R}L\longrightarrow R\otimes_{R}M$ is an isomorphism;

(b) the left $R$-module morphism
$\operatorname{Hom}_{R}(R,f)\colon\operatorname{Hom}_{R}(R,L)\longrightarrow\operatorname{Hom}_{R}(R,M)$
is an isomorphism.

###### Proof.

A more general version of part (a) can be found in [22, Proposition 4.3]. Our
argument is based on Lemma 5.5. In both parts (a) and (b), it suffices to
consider two cases: either $f$ is a surjective map with a null kernel, or $f$
is an injective map with a null cokernel. In the former case, we have a short
exact sequence $0\longrightarrow N\longrightarrow L\longrightarrow
M\longrightarrow 0$ with a null-module $N$. Then the vanishing of the tensor
product $R\otimes_{R}N$ implies an isomorphism $R\otimes_{R}L\simeq
R\otimes_{R}M$, and Lemma 5.5(b) implies an isomorphism
$\operatorname{Hom}_{R}(R,L)\simeq\operatorname{Hom}_{R}(R,M)$. In the latter
case, we have a short exact sequence $0\longrightarrow L\longrightarrow
M\longrightarrow N\longrightarrow 0$ with a null-module $N$. Then Lemma 5.5(a)
implies an isomorphism $R\otimes_{R}L\simeq R\otimes_{R}M$, and the vanishing
of the Hom group $\operatorname{Hom}_{R}(R,N)$ implies an isomorphism
$\operatorname{Hom}_{R}(R,L)\simeq\operatorname{Hom}_{R}(R,M)$. ∎

The following corollary is due to Quillen. Part (a) is contained in [22,
Proposition 2.6], part (b) in [22, Proposition 4.1], and part (c) in [22,
Proposition 5.3].

######  Corollary 5.7 ([22]).

Let $R$ be a t-unital ring and $f\colon L\longrightarrow M$ be a left
$R$-module morphism whose kernel and cokernel are null-modules. Then

(a) for any t-unital right $R$-module $D$, the induced map of abelian groups
$D\otimes_{R}L\longrightarrow D\otimes_{R}M$ is an isomorphism;

(b) for any t-unital left $R$-module $E$, the induced map of abelian groups
$\operatorname{Hom}_{R}(E,L)\longrightarrow\operatorname{Hom}_{R}(E,M)$ is an
isomorphism;

(c) for any c-unital left $R$-module $P$, the induced map of abelian groups
$\operatorname{Hom}_{R}(M,P)\longrightarrow\operatorname{Hom}_{R}(L,P)$ is an
isomorphism.

###### Proof.

Part (a) follows from Proposition 5.6(a) and the natural isomorphisms
$D\otimes_{R}\nobreak K\simeq(D\otimes_{R}R)\otimes_{R}K\simeq
D\otimes_{R}(R\otimes_{R}K)$ for all left $R$-modules $K$ (in particular,
$K=L$ or $K=M$). Part (b) follows from Proposition 5.6(b) and the natural
isomorphisms
$\operatorname{Hom}_{R}(E,K)\simeq\operatorname{Hom}_{R}(R\otimes_{R}E,\>K)\simeq\operatorname{Hom}_{R}(E,\operatorname{Hom}_{R}(R,K))$
for all left $R$-modules $K$. Part (c) follows from Proposition 5.6(a) and the
natural isomorphisms
$\operatorname{Hom}_{R}(K,P)\simeq\operatorname{Hom}_{R}(K,\operatorname{Hom}_{R}(R,P))\simeq\operatorname{Hom}_{R}(R\otimes_{R}K,\>P)$
for all left $R$-modules $K$. ∎

The following theorem is a particular case of Quillen’s [22, Theorems 4.5 and
5.6].

######  Theorem 5.8 ([22]).

Let $R$ be a t-unital ring. In this setting:

(a) The additive category $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is abelian,
and the functor $R\otimes_{R}{-}\,\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is exact. This functor factorizes
through the localization functor
$R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$,
inducing a functor
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$, which is an equivalence of abelian
categories
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{t}Mod}}}$. The composition of the inclusion and
localization functors $R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
provides the inverse equivalence.

(b) The additive category $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is abelian,
and the functor $\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\allowbreak\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is exact. This functor factorizes
through the localization functor
$R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$,
inducing a functor
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$, which is an equivalence of abelian
categories
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{c}Mod}}}$. The composition of the inclusion and
localization functors $R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
provides the inverse equivalence.

######  Remark 5.9.

Notice that the functor $R\otimes_{R}{-}\,\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is _not_ exact in general. The functor
$R\otimes_{R}{-}$ is only exact when it is viewed as _taking values in
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$_. The inclusion functor
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is also _not_ exact in general, as
Example 4.7 illustrates. It is only the composition
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
that is exact.

Similarly, the functor $\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is _not_ exact in general. The functor
$\operatorname{Hom}_{R}(R,{-})$ is only exact when it is viewed as _taking
values in $R{\operatorname{\mathsf{--{}^{c}Mod}}}$_. The inclusion functor
$R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is also _not_ exact in general, as
Examples 4.2 and 4.8 illustrate. It is only the composition
$R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
that is exact.

The full subcategory $R{\operatorname{\mathsf{--{}^{c}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is a _Giraud subcategory_ in the sense
of [23, Section X.1]. Dually, the full subcategory
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is a _co-Giraud subcategory_ [4,
Section 1]. This terminology means that the inclusion functor
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is left adjoint to the localization
functor $R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$,
while the inclusion functor
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is right adjoint to the localization
functor.

In the proof of the theorem, the following observation is used. Let
$\mathsf{C}$ and $\mathsf{D}$ be two categories, and let
$\mathsf{S}\subset\mathsf{C}$ and $\mathsf{T}\subset\mathsf{D}$ be two
multiplicative subsets of morphisms. Consider the localizations
$\mathsf{C}[\mathsf{S}^{-1}]$ and $\mathsf{D}[\mathsf{T}^{-1}]$ of the
categories $\mathsf{C}$ and $\mathsf{D}$. Let
$F\colon\mathsf{C}\longrightarrow\mathsf{D}$ and
$G\colon\mathsf{D}\longrightarrow\mathsf{C}$ be a pair of adjoint functors,
with the functor $F$ left adjoint to $G$. Assume that the functors $F$ and $G$
descend to well-defined functors
$\overline{F}\colon\mathsf{C}[\mathsf{S}^{-1}]\longrightarrow\mathsf{D}[\mathsf{T}^{-1}]$
and
$\overline{G}\colon\mathsf{D}[\mathsf{T}^{-1}]\longrightarrow\mathsf{C}[\mathsf{S}^{-1}]$.
Then the functor $\overline{F}$ is left adjoint to the functor $\overline{G}$.
The adjunction unit and counit for the pair $(\overline{F},\overline{G})$ are
induced by the adjunction unit and counit for the pair $(F,G)$.

###### Proof of Theorem 5.8.

Part (a): it is clear from Proposition 5.6(a) that the functor
$R\otimes_{R}{-}\,\colon R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$ factorizes (uniquely) through the
localization functor $R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$,
inducing a well-defined functor from the quotient category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$. Then it follows from Lemma 5.2 that
the resulting functor is right adjoint to the composition
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$.
It remains to show that the adjunction morphisms in
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ and
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
are isomorphisms. The adjunction unit $M\longrightarrow R\otimes_{R}M$ (which
is only defined for $M\in R{\operatorname{\mathsf{--{}^{t}Mod}}}$) is an
isomorphism for every $M\in R{\operatorname{\mathsf{--{}^{t}Mod}}}$ by
construction. The adjunction counit $R\otimes_{R}L\longrightarrow L$ is an
isomorphism in the quotient category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
for every left $R$-module $L$ by Lemma 5.4(a). Finally, the functor
$R\otimes_{R}{-}\,\colon R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is exact as the composition of two
exact functors $R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$ (notice that any equivalence of
abelian categories is an exact functor).

Part (b): it is clear from Proposition 5.6(b) that the functor
$\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\allowbreak\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ factorizes (uniquely) through the
localization functor $R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$,
inducing a well-defined functor from the quotient category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$. Then it follows from Lemma 5.3 that
the resulting functor is left adjoint to the composition
$R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$.
It remains to show that the adjunction morphisms in
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ and
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
are isomorphisms. The adjunction counit
$\operatorname{Hom}_{R}(R,P)\longrightarrow P$ (which is only defined for
$P\in R{\operatorname{\mathsf{--{}^{c}Mod}}}$) is an isomorphism for every
$P\in R{\operatorname{\mathsf{--{}^{c}Mod}}}$ by construction. The adjunction
unit $Q\longrightarrow\operatorname{Hom}_{R}(R,Q)$ is an isomorphism in the
quotient category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
for every left $R$-module $Q$ by Lemma 5.4(b). Finally, the functor
$\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is exact as the composition of two
exact functors $R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$. ∎

######  Corollary 5.10.

Let $R$ be a t-unital ring. Then there is a natural equivalence of three
abelian categories

(16) $R{\operatorname{\mathsf{--{}^{t}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{c}Mod}}}.$

In particular, the functors $R\otimes_{R}{-}$ and
$\operatorname{Hom}_{R}(R,{-})$ restrict to mutually inverse equivalences

(17) $\textstyle{{\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{t}Mod}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{R{\operatorname{\mathsf{--{}^{c}Mod}}}\,:\\!R\otimes_{R}{-}\,.}}$

###### Proof.

The first assertion is obtained by combining the two parts of Theorem 5.8, and
the second assertion follows by looking into the constructions of two category
equivalences in Theorem 5.8. This is sufficient to prove the corollary.

It is instructive, however, to consider the adjunction constructions for the
two mutually inverse equivalences (17). Here one observes an explicit
_ambidextrous adjunction_ [3, 13], meaning a pair of functors that are adjoint
to each other on both sides. This ambidextrous adjunction is easier to
establish than the full assertions of the corollary.

On the one hand, there is a natural adjunction where the functor
$\operatorname{Hom}_{R}(R,{-})$ is the right adjoint, and the functor
$R\otimes_{R}{-}$ is the left adjoint:

(18) $\textstyle{{\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{t}Mod}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{R{\operatorname{\mathsf{--{}^{c}Mod}}}\,:\\!R\otimes_{R}{-}\,.}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

Indeed, consider the pair of adjoint endofunctors on the category of unital
left $\widetilde{R}$-modules induced by the
$\widetilde{R}$-$\widetilde{R}$-bimodule $R$,

(19)
$\textstyle{{\operatorname{Hom}_{\widetilde{R}}(R,{-})\colon\widetilde{R}{\operatorname{\mathsf{--Mod}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\widetilde{R}{\operatorname{\mathsf{--Mod}}}\,:\\!R\otimes_{\widetilde{R}}{-}\,.}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

This is a particular case of the adjunction expressed by the natural
isomorphism (5) from the introduction or (14) from Section 3. The adjunction
(18) is obtained by restricting the adjunction (19) to the full subcategories
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ and
$R{\operatorname{\mathsf{--{}^{c}Mod}}}\subset
R{\operatorname{\mathsf{--{}^{n}Mod}}}\simeq\widetilde{R}{\operatorname{\mathsf{--Mod}}}$.

On the other hand, there is a natural adjunction where the functor
$\operatorname{Hom}_{R}(R,{-})$ is the left adjoint, and the functor
$R\otimes_{R}{-}$ is the right adjoint:

(20) $\textstyle{{\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{t}Mod}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{R{\operatorname{\mathsf{--{}^{c}Mod}}}\,:\\!R\otimes_{R}{-}\,.}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

The adjunction (20) is the composition of the two adjunctions

$\textstyle{{R{\operatorname{\mathsf{--{}^{t}Mod}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{inclusion}}$$\textstyle{{R{\operatorname{\mathsf{--{}^{n}Mod}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{R\otimes_{R}{-}}$$\scriptstyle{\operatorname{Hom}_{R}(R,{-})}$$\textstyle{{R{\operatorname{\mathsf{--{}^{c}Mod}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{inclusion}}$

provided by Lemmas 5.2 and 5.3. ∎

######  Remark 5.11.

It may be also instructive to consider the unit and counit morphisms of the
two adjunctions (18) and (20). For any left $R$-module $M$, the evaluation map
$R\otimes_{R}\operatorname{Hom}_{R}(R,M)\longrightarrow M$ is the counit of
the adjunction (19). In the particular case when $M$ is a t-unital $R$-module,
the same evaluation map is the counit of the adjunction (18). On the other
hand, applying the functor $R\otimes_{R}{-}$ to the natural map
$M\longrightarrow\operatorname{Hom}_{R}(R,M)$, we obtain a natural map
$R\otimes_{R}M\longrightarrow R\otimes_{R}\operatorname{Hom}_{R}(R,M)$, which
in the case of a t-unital $R$-module $M$ provides a natural map
$M\longrightarrow R\otimes_{R}\operatorname{Hom}_{R}(R,M)$. The latter map is
the unit of the adjunction (20). One can easily check that the composition
$M\longrightarrow R\otimes_{R}\operatorname{Hom}_{R}(R,M)\longrightarrow M$ is
the identity map.

Similarly, for any left $R$-module $P$, the natural map
$P\longrightarrow\operatorname{Hom}_{R}(R,\>R\otimes_{R}P)$ is the unit of the
adjunction (19). In the particular case when $P$ is a c-unital $R$-module, the
same map is the unit of the adjunction (18). On the other hand, applying the
functor $\operatorname{Hom}_{R}(R,{-})$ to the multiplication map
$R\otimes_{R}P\longrightarrow P$, we obtain a natural map
$\operatorname{Hom}_{R}(R,\>R\otimes_{R}P)\longrightarrow\operatorname{Hom}_{R}(R,P)$,
which in the case of a c-unital $R$-module $P$ provides a natural map
$\operatorname{Hom}_{R}(R,\>R\otimes_{R}P)\longrightarrow P$. The latter map
is the counit of the adjunction (20). One can easily check that the
composition
$P\longrightarrow\operatorname{Hom}_{R}(R,\>R\otimes_{R}P)\longrightarrow P$
is the identity map.

##  6\. Rings Arising from Small Preadditive Categories

Let $R$ be a (nonunital) ring. A _family of orthogonal idempotents_ $(e_{x}\in
R)_{x\in X}$ in $R$ is a family of elements such that $e_{x}^{2}=e_{x}$ for
all $x\in X$ and $e_{x}e_{y}=0$ for all $x\neq y\in X$. A family of orthogonal
idempotents $(e_{x})_{x\in X}$ in $R$ is said to be _complete_ if
$R=\bigoplus_{x\in X}Re_{x}=\bigoplus_{x\in X}e_{x}R$, or equivalently,
$R=\bigoplus_{x,y\in X}e_{y}Re_{x}$. Following the terminology of [16] (going
back to [6]), we will say that $R$ is _a ring with enough idempotents_ if
there is a chosen complete family of orthogonal idempotents in $R$.

The following lemma is easy and well-known [16].

######  Lemma 6.1.

Any ring with enough idempotents in s-unital (and consequently, t-unital).

###### Proof.

The point is that for any element $r\in R$ one can choose a finite subset
$Z\subset X$ such that the (idempotent) element $e=\sum_{z\in Z}e_{z}\in R$
satisfies $er=r=re$. The assertion in parentheses follows by Corollary 2.7. ∎

######  Lemma 6.2.

Let $R$ be a ring with enough idempotents and $\widetilde{R}=\mathbb{Z}\oplus
R$ be its unitalization. Then $R$ is an (infinitely generated) projective
unital left $\widetilde{R}$-module and projective unital right
$\widetilde{R}$-module. Specifically, one has

(21) $R=\bigoplus\nolimits_{x\in X}\widetilde{R}e_{x}=\bigoplus\nolimits_{x\in
X}e_{x}\widetilde{R},$

where, for every $x\in X$, the left $\widetilde{R}$-module
$\widetilde{R}e_{x}=Re_{x}$ is projective with a single generator $e_{x}$, and
the right $\widetilde{R}$-module $e_{x}\widetilde{R}=e_{x}R$ is projective
with a single generator $e_{x}$.

###### Proof.

Notice that one has $\widetilde{R}r=Rr$ and $r\widetilde{R}=rR$ for any
s-unital ring $R$ and element $r\in R$. With this simple observation, all the
assertions of the lemma follow immediately from the definition of a ring with
enough idempotents. ∎

######  Corollary 6.3.

(a) Let $R$ be a t-unital ring. Then the categories of t-unital and c-unital
left $R$-modules $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ and
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ are abelian.

(b) Let $R$ be a left s-unital ring (for example, $R$ is a ring with enough
idempotents). Then the full subcategory of t-unital left $R$-modules
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is closed under submodules,
quotients, extensions, and direct sums in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. Hence, in particular, the fully
faithful inclusion functor of abelian categories
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is exact.

(c) Let $R$ be a ring with enough idempotents. Then the full subcategory of
c-unital $R$-modules $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is closed under
kernels, cokernels, extensions, and direct products in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. Hence, in particular, the fully
faithful inclusion functor of abelian categories
$R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ is exact.

###### Proof.

Part (a) is a part of Theorem 5.8(a–b). Part (b) is Proposition 2.2 together
with Corollary 2.9(a). Part (c) follows from Lemma 3.5 together with Remark
3.6 and the first assertion of Lemma 6.2. ∎

Let $\mathsf{E}$ be a small preadditive category, i. e., a small category
enriched in abelian groups. This means that, for every pair of objects $x$,
$y\in\mathsf{E}$, there is the abelian group of morphisms
$\operatorname{Hom}_{\mathsf{E}}(x,y)$, and the composition maps

$\operatorname{Hom}_{\mathsf{E}}(y,z)\otimes_{\mathbb{Z}}\operatorname{Hom}_{\mathsf{E}}(x,y)\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip
1.5mu\operatorname{Hom}_{\mathsf{E}}(x,z)$

are defined for all objects $x$, $y$, $z\in\mathsf{E}$ in such a way that the
usual associativity and unitality axioms (for identity elements
${\mathrm{id}}_{x}\in\operatorname{Hom}_{\mathsf{E}}(x,x)$) are satisfied.

Then the abelian group
$R_{\mathsf{E}}=\bigoplus_{x,y\in\mathsf{E}}\operatorname{Hom}_{\mathsf{E}}(x,y)$
has a natural structure of (nonunital) ring with the multiplication map given
by the composition of morphisms in $\mathsf{E}$. The following lemma is
certainly standard and well-known.

######  Lemma 6.4.

A ring $R$ has enough idempotents if and only if it has the form
$R=R_{\mathsf{E}}$ for some small preadditive category $\mathsf{E}$.

###### Proof.

Let $R$ be a ring with enough idempotents, and let $(e_{x}\in R)_{x\in X}$ be
its complete family of orthogonal idempotents. Then $\mathsf{E}$ is the
category with the set of objects $X$ and the groups of morphisms
$\operatorname{Hom}_{\mathsf{E}}(x,y)=e_{y}Re_{x}$. Conversely, if
$R=R_{\mathsf{E}}$ for a small preadditive category $\mathsf{E}$, then the
identity elements ${\mathrm{id}}_{x}\in\operatorname{Hom}_{\mathsf{E}}(x,x)$,
viewed as elements of $R_{\mathsf{E}}$, form a complete family of orthogonal
idempotents in $R$. ∎

Let $\mathsf{E}$ be a small preadditive category. By a _left
$\mathsf{E}$-module_ we mean a covariant additive functor
$\mathsf{E}\longrightarrow\mathsf{Ab}$ from $\mathsf{E}$ to the category of
abelian groups. In other words, a left $\mathsf{E}$-module $\mathsf{M}$ is a
rule assigning to every object $x\in\mathsf{E}$ an abelian group
$\mathsf{M}(x)$, and to every pair of objects $x$, $y\in\mathsf{E}$ the
_action map_

$\operatorname{Hom}_{\mathsf{E}}(x,y)\otimes_{\mathbb{Z}}\mathsf{M}(x)\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip 1.5mu\mathsf{M}(y)$

in such a way that the usual associativity and unitality axioms are satisfied.

Similarly, a _right $\mathsf{E}$-module_ $\mathsf{N}$ is a contravariant
additive functor $\mathsf{E}^{\mathsf{op}}\longrightarrow\mathsf{Ab}$. In
other words, it is a rule assigning to every object $x\in\mathsf{E}$ an
abelian group $\mathsf{N}(x)$, and to every pair of objects $x$,
$y\in\mathsf{E}$ the action map

$\operatorname{Hom}_{\mathsf{E}}(x,y)\otimes_{\mathbb{Z}}\mathsf{N}(y)\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip 1.5mu\mathsf{N}(x)$

satisfying the usual associativity and unitality axioms. We denote the abelian
categories of left and right $\mathsf{E}$-modules by
$\mathsf{E}{\operatorname{\mathsf{--Mod}}}$ and
${\operatorname{\mathsf{Mod--}}}\mathsf{E}$, respectively.

The next proposition is also well-known. One relevant reference is [14,
Section 7].

######  Proposition 6.5.

Let $\mathsf{E}$ be a small preadditive category and $R=R_{\mathsf{E}}$ be the
related ring with enough idempotents. In this context:

(a) The category of t-unital left $R$-modules is naturally equivalent to the
category of left $\mathsf{E}$-modules,

$R_{\mathsf{E}}{\operatorname{\mathsf{--{}^{t}Mod}}}\simeq\mathsf{E}{\operatorname{\mathsf{--Mod}}}.$

The equivalence assigns to a left $\mathsf{E}$-module $\mathsf{M}$ the
t-unital left $R$-module $M=\bigoplus_{x\in\mathsf{E}}\mathsf{M}(x)$, with the
action of $R$ in $M$ induced by the action of $\mathsf{E}$ in $\mathsf{M}$.

(b) The category of t-unital right $R$-modules is naturally equivalent to the
category of right $\mathsf{E}$-modules,

${\operatorname{\mathsf{Mod^{t}--}}}R_{\mathsf{E}}\simeq{\operatorname{\mathsf{Mod--}}}\mathsf{E}.$

The equivalence assigns to a right $\mathsf{E}$-module $\mathsf{N}$ the
t-unital right $R$-module $N=\bigoplus_{x\in\mathsf{E}}\mathsf{N}(x)$, with
the action of $R$ in $N$ induced by the action of $\mathsf{E}$ in
$\mathsf{N}$.

###### Proof.

Let us explain part (a). By Lemmas 6.2 and 6.4, we have an isomorphism of
right $R$-modules $R\simeq\bigoplus_{x\in X}e_{x}R$, where $X$ is the set of
objects of $\mathsf{E}$ and $e_{x}={\mathrm{id}}_{x}$. Consequently, for any
left $R$-module $M$ we have a natural direct sum decomposition

$R\otimes_{R}M\simeq\bigoplus\nolimits_{x\in
X}e_{x}R\otimes_{R}M=\bigoplus\nolimits_{x\in X}e_{x}M.$

The second equality holds because $e_{x}R=e_{x}\widetilde{R}$. When the
$R$-module $M$ is t-unital, we obtain a direct sum decomposition
$M=\bigoplus_{x\in X}e_{x}M$, allowing to recover the corresponding
$\mathsf{E}$-module $\mathsf{M}$ by the rule $\mathsf{M}(x)=e_{x}M$ for all
$x\in\mathsf{E}$. ∎

######  Proposition 6.6.

Let $\mathsf{E}$ be a small preadditive category and $R=R_{\mathsf{E}}$ be the
related ring with enough idempotents. Then the category of c-unital left
$R$-modules is naturally equivalent to the category of left
$\mathsf{E}$-modules,

$R_{\mathsf{E}}{\operatorname{\mathsf{--{}^{c}Mod}}}\simeq\mathsf{E}{\operatorname{\mathsf{--Mod}}}.$

The equivalence assigns to a left $\mathsf{E}$-module $\mathsf{P}$ the
c-unital left $R$-module $P=\prod_{x\in\mathsf{E}}\mathsf{P}(x)$, with the
action of $R$ in $P$ induced by the action of $\mathsf{E}$ in $\mathsf{P}$.

###### Proof.

By Lemmas 6.2 and 6.4, we have an isomorphism of left $R$-modules
$R\simeq\bigoplus_{x\in X}Re_{x}$, where $X$ is the set of objects of
$\mathsf{E}$ and $e_{x}={\mathrm{id}}_{x}$. Consequently, for any left
$R$-module $P$ we have a natural direct product decomposition

$\operatorname{Hom}_{R}(R,P)\simeq\prod\nolimits_{x\in
X}\operatorname{Hom}_{R}(Re_{x},P)=\prod\nolimits_{x\in X}e_{x}P.$

The second equality holds because $Re_{x}=\widetilde{R}e_{x}$. When the
$R$-module $P$ is c-unital, we obtain a direct product decomposition
$P=\prod_{x\in X}e_{x}P$, allowing to recover the corresponding
$\mathsf{E}$-module $\mathsf{P}$ by the rule $\mathsf{P}(x)=e_{x}P$ for all
$x\in\mathsf{E}$. ∎

######  Proposition 6.7.

Let $\mathsf{E}$ be a small preadditive category and $R=R_{\mathsf{E}}$ be the
related ring with enough idempotents. Then there is a commutative square
diagram of category equivalences provided by Propositions 6.5(a) and 6.6
together with Corollary 5.10,

(22) $\begin{gathered}\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
23.16289pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
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0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern 77.16289pt\raise
0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{{R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\hbox{\kern
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0.32094pt\raise-0.9471pt\hbox{\lx@xy<EMAIL_ADDRESS>0.9471pt\hbox{\lx@xy@drawline@}}}}\ignorespaces{\hbox{\hbox{\kern
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0.32011pt\raise
0.94737pt\hbox{\lx@xy@drawline@}}\hbox{\kern-0.32011pt\raise-0.94737pt\hbox{\lx@xy@drawline@}}}}\ignorespaces{\hbox{\hbox{\kern
0.32011pt\raise
0.94737pt\hbox{\lx@xy@drawline@}}\hbox{\kern-0.32011pt\raise-0.94737pt\hbox{\lx@xy<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{R_{\mathsf{E}}{\operatorname{\mathsf{--{}^{t}Mod}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\hbox{\kern
0.30504pt\raise
0.95233pt\hbox{\lx@xy@drawline@}}\hbox{\kern-0.30504pt\raise-0.95233pt\hbox{\lx@xy@drawline@}}}}\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\hbox{\kern
0.30504pt\raise
0.95233pt\hbox{\lx@xy@drawline@}}\hbox{\kern-0.30504pt\raise-0.95233pt\hbox{\lx@xy@drawline@}}}}\ignorespaces{\hbox{\hbox{\kern
0.30504pt\raise
0.95233pt\hbox{\lx@xy@drawline@}}\hbox{\kern-0.30504pt\raise-0.95233pt\hbox{\lx@xy@drawline@}}}}{\hbox{\kern
47.16289pt\raise-40.97pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern
118.0664pt\raise-40.97pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern
188.96992pt\raise-40.97pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern
218.96992pt\raise-40.97pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{R_{\mathsf{E}}{\operatorname{\mathsf{--{}^{c}Mod}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\hbox{\kern
0.30504pt\raise-0.95233pt\hbox{\lx@xy<EMAIL_ADDRESS>0.95233pt\hbox{\lx@xy@drawline@}}}}\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\hbox{\kern
0.30504pt\raise-0.95233pt\hbox{\lx@xy<EMAIL_ADDRESS>0.95233pt\hbox{\lx@xy@drawline@}}}}\ignorespaces{\hbox{\hbox{\kern
0.30504pt\raise-0.95233pt\hbox{\lx@xy<EMAIL_ADDRESS>0.95233pt\hbox{\lx@xy<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern
47.16289pt\raise-79.76pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern
101.33026pt\raise-79.76pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathsf{E}{\operatorname{\mathsf{--Mod}}}}$}}}}}}}\ignorespaces}}}}\ignorespaces\end{gathered}$

###### Proof.

It suffices to check that the composition

$R_{\mathsf{E}}{\operatorname{\mathsf{--{}^{t}Mod}}}\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip
1.5mu\mathsf{E}{\operatorname{\mathsf{--Mod}}}\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip
1.5muR_{\mathsf{E}}{\operatorname{\mathsf{--{}^{c}Mod}}}$

of the category equivalences from Propositions 6.5(a) and 6.6 is isomorphic to
the Hom functor from (17), and/or that the composition of category
equivalences

$R_{\mathsf{E}}{\operatorname{\mathsf{--{}^{c}Mod}}}\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip
1.5mu\mathsf{E}{\operatorname{\mathsf{--Mod}}}\mskip
1.5mu\relbar\joinrel\relbar\joinrel\rightarrow\mskip
1.5muR_{\mathsf{E}}{\operatorname{\mathsf{--{}^{t}Mod}}}$

is isomorphic to the tensor product functor from (17). For this purpose, one
observes that, for any left $R$-module $L$, one has

$e_{x}R\otimes_{R}L\simeq e_{x}L\simeq\operatorname{Hom}_{R}(Re_{x},L).$

∎

##  7\. More on t-Unital and c-Unital Modules

Let $R$ be a t-unital ring. According to Examples 4.2, 4.7, and 4.8, the full
subcategory of t-unital modules $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ _need
not_ be closed under kernels in the abelian category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$, and the full subcategory of c-unital
modules $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ _need not_ be closed under
cokernels in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$. Nevertheless, Corollary
6.3(a) tells that the categories $R{\operatorname{\mathsf{--{}^{t}Mod}}}$ and
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ are abelian; so all kernels and
cokernels exist in them. The following two lemmas and proposition explain
more.

######  Lemma 7.1.

Let $R$ be a t-unital ring. Then all limits and colimits exist in the category
of t-unital modules $R{\operatorname{\mathsf{--{}^{t}Mod}}}$.

###### Proof.

The abelian category
$R{\operatorname{\mathsf{--{}^{n}Mod}}}\simeq\widetilde{R}{\operatorname{\mathsf{--Mod}}}$
obviously has all limits and colimits. By Lemma 1.5, the full subcategory
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is closed under colimits in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. So all colimits exist in
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ and can be computed as the colimits
in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$. On the other hand, Lemma 5.2
tells that the functor $R\otimes_{R}{-}\,\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is right adjoint to the inclusion
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$. Consequently, the functor
$R\otimes_{R}{-}\,\colon R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}$ preserves limits. Now, in order to
compute the limit of a diagram $(M_{\xi})_{\xi\in\Xi}$ in
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$, it suffices to apply the functor
$R\otimes_{R}{-}$ to the limit of the same diagram in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. ∎

######  Lemma 7.2.

Let $R$ be a t-unital ring. Then all limits and colimits exist in the category
of c-unital modules $R{\operatorname{\mathsf{--{}^{c}Mod}}}$.

###### Proof.

By Lemma 3.5, the full subcategory $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is
closed under limits in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$. So all limits
exist in $R{\operatorname{\mathsf{--{}^{c}Mod}}}$ and can be computed as the
limits in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$. On the other hand, Lemma
5.3 tells that the functor $\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is left adjoint to the inclusion
$R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$. Consequently, the functor
$\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ preserves colimits. Now, in order to
compute the colimit of a diagram $(P_{\xi})_{\xi\in\Xi}$ in
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$, it suffices to apply the functor
$\operatorname{Hom}_{R}(R,{-})$ to the colimit of the same diagram in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. ∎

A study of abelian categories of the form
$R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
(for idempotent rings $R$) can be found in the note of Roos [20] (cf. [22,
Section 6.3]). The following result is a particular case of [20, Théorème 1].

######  Proposition 7.3 ([20]).

Let $R$ be a t-unital ring. Then the abelian category
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is a Grothendieck abelian category
with exact functors of infinite product.

###### Proof.

The equivalences of abelian categories
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ hold by Corollary 5.10. Furthermore,
the Serre subcategory $R{\operatorname{\mathsf{--{}^{0}Mod}}}$ is closed under
direct sums and direct products in $R{\operatorname{\mathsf{--{}^{n}Mod}}}$,
as mentioned in the beginning of Section 5; and the localization functor
$R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}$
has adjoints on both sides, as per Remark 5.9.

It is well-known that for any Grothendieck abelian category $\mathsf{A}$ with
a Serre subcategory $\mathsf{S}\subset\mathsf{A}$ such that the localization
functor $\mathsf{A}\longrightarrow\mathsf{A}/\mathsf{S}$ admits a right
adjoint, the quotient category $\mathsf{A}/\mathsf{S}$ is also Grothendieck
[8, Proposition 8 in Section III.4], [23, Proposition X.1.3]. In particular,
the direct limit functors in $\mathsf{A}/\mathsf{S}$ are exact, since they are
exact in $\mathsf{A}$ and the localization functor
$\mathsf{A}\longrightarrow\mathsf{A}/\mathsf{S}$, being a left adjoint,
preserves direct limits. Similarly, if the functor
$\mathsf{A}\longrightarrow\mathsf{A}/\mathsf{S}$ has a left adjoint, then it
preserves direct products; and if the direct products are exact in
$\mathsf{A}$, then it follows that they are also exact in
$\mathsf{A}/\mathsf{S}$.

In fact, it is not difficult to prove directly that for any abelian category
$\mathsf{A}$ with exact functors of coproduct and a Serre subcategory
$\mathsf{S}\subset\mathsf{A}$ closed under coproducts, the localization
functor $\mathsf{A}\longrightarrow\mathsf{A}/\mathsf{S}$ preserves coproducts
(and takes any set of generators of $\mathsf{A}$ to a set of generators of
$\mathsf{A}/\mathsf{S}$). Dually, if an abelian category $\mathsf{A}$ has
exact products and a Serre subcategory $\mathsf{S}\subset\mathsf{A}$ is closed
under products, then the localization functor
$\mathsf{A}\longrightarrow\mathsf{A}/\mathsf{S}$ preserves products. See, e.
g., [2, Lemma 1.5] or [15, Lemma 3.2.10] for a similar argument in the
triangulated category realm. (Cf. the discussion in [22, Section 6.4 and
Proposition 6.5] for module categories.) ∎

######  Corollary 7.4.

Let $R$ be a t-unital ring and $\widetilde{R}=\mathbb{Z}\oplus R$ be its
unitalization. Then the full subcategory
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is closed under kernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$ _if and only if_ the unital right
$\widetilde{R}$-module $R$ is flat.

###### Proof.

The “if” assertion was explained in Remark 1.6; it does not depend on the
assumption that $R$ is t-unital. The “only if” claim is a corollary of Theorem
5.8(a). Assume that the full subcategory
$R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is closed under kernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$; then the inclusion functor
$R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ preserves kernels. The localization
functor $R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{t}Mod}}}$ is exact, so it also preserves
kernels. Thus the composition
$R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{t}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ preserves kernels. But this
composition is the tensor product functor $R\otimes_{R}{-}\,\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$, that is, in other notation, the
tensor product functor
$R\otimes_{\widetilde{R}}{-}\,\colon\widetilde{R}{\operatorname{\mathsf{--Mod}}}\longrightarrow\widetilde{R}{\operatorname{\mathsf{--Mod}}}$.
Alternatively, one could refer to Remark 8.14(3) and Corollary 8.20 below. ∎

######  Corollary 7.5.

Let $R$ be a left s-unital ring and $\widetilde{R}=\mathbb{Z}\oplus R$ be its
unitalization. Then the unital right $\widetilde{R}$-module $R$ is flat.

###### Proof.

By Corollary 2.7, the ring $R$ is t-unital. By Corollary 2.9(a), a left
$R$-module is t-unital if and only if it is s-unital,
$R{\operatorname{\mathsf{--{}^{t}Mod}}}=R{\operatorname{\mathsf{--{}^{s}Mod}}}$.
Proposition 2.2 tells that the full subcategory
$R{\operatorname{\mathsf{--{}^{s}Mod}}}$ of s-unital left $R$-modules is
closed under submodules (hence also under kernels) in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. Finally, it remains to apply the
“only if” assertion of Corollary 7.4. ∎

######  Corollary 7.6.

Let $R$ be a t-unital ring and $\widetilde{R}=\mathbb{Z}\oplus R$ be its
unitalization. Then the full subcategory
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is closed under cokernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$ _if and only if_ the unital left
$\widetilde{R}$-module $R$ is projective.

###### Proof.

The “if” assertion was explained in Remark 3.6; it does not depend on the
assumption that $R$ is t-unital. The “only if” claim is a corollary of Theorem
5.8(b). Assume that the full subcategory
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is closed under cokernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$; then the inclusion functor
$R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ preserves cokernels. The localization
functor $R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}/R{\operatorname{\mathsf{--{}^{0}Mod}}}\simeq
R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is exact, so it also preserves
cokernels. Thus the composition
$R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$ preserves cokernels. But this
composition is the functor $\operatorname{Hom}_{R}(R,{-})\colon
R{\operatorname{\mathsf{--{}^{n}Mod}}}\longrightarrow
R{\operatorname{\mathsf{--{}^{n}Mod}}}$, that is, in other notation, the
functor
$\operatorname{Hom}_{\widetilde{R}}(R,{-})\colon\widetilde{R}{\operatorname{\mathsf{--Mod}}}\longrightarrow\widetilde{R}{\operatorname{\mathsf{--Mod}}}$.
Alternatively, one could refer to Remark 8.2(3) and Corollary 8.6 below. ∎

######  Remark 7.7.

The theory developed in Sections 1, 3, 5, and 7 admits a far-reaching
generalization of the following kind.

Let $\widetilde{R}$ be a unital ring and $R\subset\widetilde{R}$ be a two-
sided ideal. Let us say that a left $\widetilde{R}$-module $M$ is _t-
$R$-unital_ if the natural map $R\otimes_{\widetilde{R}}M\longrightarrow M$ is
an isomorphism, a right $\widetilde{R}$-module $N$ is t-$R$-unital if the
natural map $N\otimes_{\widetilde{R}}R\longrightarrow N$ is an isomorphism,
and a left $\widetilde{R}$-module $P$ is _c- $R$-unital_ if the natural map
$P\longrightarrow\operatorname{Hom}_{\widetilde{R}}(R,P)$ is an isomorphism.
Let us say that the ideal $R\subset\widetilde{R}$ itself is _t- $R$-unital_ if
it is t-$R$-unital as a left/right $\widetilde{R}$-module, i. e., the natural
map $R\otimes_{\widetilde{R}}R\longrightarrow R$ is an isomorphism.

All the results of Sections 1, 3, 5, and 7, as well as of the next Section 8,
can be extended to this more general context. We chose to restrict ourselves
to the more narrow setting of nonunital rings in these five sections in order
to keep our exposition more accessible and facilitate the comparison with
Sections 2 and 6.

A detailed discussion in the context of idempotent two-sided ideals in unital
rings can be found in the manuscript [22]. Moreover, it is explained in [22,
§9] that the resulting theory does not depend on (in our notation) the chosen
embedding of a given nonunital ring $R$ into a unital ring $\widetilde{R}$
where $R$ is an ideal.

##  8\. t-Flat, c-Projective, and t-Injective Modules

We start with a discussion of c-projectivity, then pass to t-injectivity, and
in the end come to t-flatness.

######  Definition 8.1.

Let $R$ be a t-unital ring. We will say that a left $R$-module $Q$ is _c-
projective_ if the covariant functor $\operatorname{Hom}_{R}(Q,{-})\colon
R{\operatorname{\mathsf{--{}^{c}Mod}}}\longrightarrow\mathsf{Ab}$ preserves
cokernels (i. e., takes cokernels in the category
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ to cokernels in the category of
abelian groups $\mathsf{Ab}$).

######  Remarks 8.2.

(0) Notice that, by the definition, a c-projective $R$-module _need not_ be
c-unital.

(1) Furthermore, a projective nonunital $R$-module (i. e., a projective unital
$\widetilde{R}$-module) _need not_ be c-projective. For example, let $T$ be
the commutative ring from Example 4.6 and $0\longrightarrow
F\longrightarrow\widetilde{F}\longrightarrow S\longrightarrow 0$ be the short
exact sequence of $T$-modules from Example 4.8. So $F$ and $\widetilde{F}$ are
c-unital $T$-modules, but $S$ is not. One can easily compute that
$\operatorname{Hom}_{T}(T,S)=\operatorname{Hom}_{S}(S,S)=\widetilde{S}$, where
the basis elements $x^{p}y^{q}\in T$ act by zero in $\widetilde{S}$ when
$p>0$. According to the proof of Lemma 7.2, $\widetilde{S}$ is the cokernel of
the morphism $F\longrightarrow\widetilde{F}$ in
$T{\operatorname{\mathsf{--{}^{c}Mod}}}$. Since the map
$\widetilde{F}\longrightarrow\widetilde{S}$ is not surjective (i. e., not an
epimorphism in $T{\operatorname{\mathsf{--{}^{n}Mod}}}$ or in $\mathsf{Ab}$),
it follows that the free $\widetilde{T}$-module $\widetilde{T}$ is _not_
c-projective as a $T$-module.

(2) Quite generally, the free left $\widetilde{R}$-module $\widetilde{R}$ is
c-projective as an $R$-module _if and only if_ the full subcategory
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is closed under cokernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$. If this is the case (e. g., this
holds for all rings $R$ with enough idempotents, see Corollary 6.3(c)), then
all projective nonunital $R$-modules are c-projective.

(3) The left $R$-module $R$ is c-projective if and only if the left $R$-module
$\widetilde{R}$ is (cf. Proposition 8.4 below). Consequently, the left
$R$-module $R$ is c-projective if and only if the full subcategory
$R{\operatorname{\mathsf{--{}^{c}Mod}}}$ is closed under cokernels in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$.

######  Lemma 8.3.

Let $R$ be a t-unital ring and $Q$ be a t-unital left $R$-module. Then the
$R$-module $Q$ is c-projective if and only if, for every right exact sequence
$L\longrightarrow M\longrightarrow E\longrightarrow 0$ in
$R{\operatorname{\mathsf{--{}^{n}Mod}}}$ with $L$, $M\in
R{\operatorname{\mathsf{--{}^{c}Mod}}}$, the sequence of abelian groups
$\operatorname{Hom}_{R}(Q,L)\longrightarrow\operatorname{Hom}_{R}(Q,M)\longrightarrow\operatorname{Hom}_{R}(Q,E)\longrightarrow
0$ is right exact.

###### Proof.

Following the proof of Lemma 7.2, the c-unital $R$-module
$\operatorname{Hom}_{R}(R,E)$ is the cokernel of the morphism