# Differential Projective Modules over Differential Rings, II

Lourdes Juan  Department of Mathematics
Texas Tech University
Lubbock TX
<EMAIL_ADDRESS>and  Andy Magid  Department of Mathematics
University of Oklahoma
Norman OK 73019
<EMAIL_ADDRESS>
(Date: October 5, 2020)

###### Abstract.

Differential modules over a commutative differential ring $R$ which are
finitely generated projective as ring modules, with differential
homomorphisms, form an additive category, so their isomorphism classes form a
monoid. We study the quotient monoid by the submonoid of isomorphism classes
of free modules with component wise derivation. This quotient monoid has the
reduced $K_{0}$ of $R$ (ignoring the derivation) as an image and contains the
reduced $K_{0}$ of the constants of $R$ as its subgroup of units. This monoid
provides a description of the isomorphism classes of differential projective
$R$ modules up to an equivalence.

###### 1991 Mathematics Subject Classification:

12H05

##  1\. Introduction

Let $R$ be a commutative differential ring with derivation $D_{R}$, or just
$D$ if the context is clear. A _differential projective module_ over $R$ is a
finitely generated projective $R$ module $M$ and an additive endomorphism
$D_{M}$, or just $D$ if the context is clear, such that
$D_{M}(rm)=D_{R}(r)m+rD_{M}(m)$ for all $r\in R$ and $m\in M$. For example, if
$A\in M_{n}(R)$ defining $D$ on the free $R$ module $R^{n}$ of column
$n$-tuples of elements of $R$ by
$D((x_{1},\dots,x_{n})^{T})=(x_{1}^{\prime},\dots,x_{n}^{\prime})^{T}+A(x_{1},\dots,x_{n})^{T}$
makes the free module into a differential projective module denoted
$(R^{n},A)$ whose differential structure is denoted $D_{A}$. (Note: in [4] we
used a similar construction for differential structures on free modules
$R^{(n)}$ of row $n$-tuples.) When $A$ is a zero matrix, denoted $0$
regardless of size, we call the differential module, and those differentially
isomorphic to it, _differentially trivial_. Thus $(R,0)^{(n)}$, which is
differentially isomorphic to $(R^{n},0)$ is differentially trivial.

In [4] we investigated differential projective modules and looked at their
classification. It is our goal in this work to advance this classification. In
[4], we took a $K$ theoretic approach. We review briefly that approach and its
difficulties.

The isomorphism classes of differential projective modules, with an addition
operation induced from direct sum, form a commutative monoid with identity the
class of the zero module; the most general group to which this monoid maps is
denoted $K_{0}^{\text{diff}}(R)$ [4, §4]. We use brackets to denote the image
of a differential projective module in $K_{0}^{\text{diff}}(R)$. For
differential projective modules $M$ and $N$, $[M]=[N]$ provided there is a
differential projective module $P$ such that $M\oplus P$ is differentially
isomorphic to $N\oplus P$. There is a differential projective module $Q$ such
that $P\oplus Q$ is differentially isomorphic to $(R^{n},C)$ for suitable $n$
and $C$ [4, Corollary 3, p. 4342] so we conclude that $[M]=[N]$ provided there
is a $C$ with $M\oplus(R^{n},C)$ differentially isomorphic to
$N\oplus(R^{n},C)$.

Forgetting the differential structure defines a homomorphism from this
differential $K$ group to the usual $K$ group, $K_{0}^{\text{diff}}(R)\to
K_{0}(R)$. This is an epimorphism by [4, Theorem 2, p. 4341], and every
element of the kernel has the form $[(R^{n},C)]-[(R^{n},D)]$ by [4, §4].
Informally, these results say that, setting aside the non–differential
$K$-theory of $R$, the differential $K$ theory of $R$ is controlled by the
classes of differential projective modules whose underlying $R$ module is
free. Thus the classification problem, using $K$ theory, comes down to the
relation of declaring $(R^{n},A)$ and $(R^{n},B)$ equivalent provided there is
a $C$ with $(R^{n},A)\oplus(R^{m},C)$ differentially isomorphic to
$(R^{n},B)\oplus(R^{m},C)$.

Differential projective modules $(R^{n},A)$ and $(R^{n},B)$ are differentially
isomorphic if there is an invertible matrix $T$ with $T^{\prime}=AT-TB$. Thus
$(R^{n},A)\oplus(R^{m},C)$ is differentially isomorphic to
$(R^{n},B)\oplus(R^{m},C)$ provided there is an invertible matrix $S$ with
$S^{\prime}=(A\oplus C)S-S(B\oplus C)$. Deciding whether $(R^{n},A)$ and
$(R^{n},B)$ are equivalent thus requires considering all possible $C$’s and
$S$’s for which this equation holds.

To simplify this differential matrix algebra problem, we propose a finer
classification, where we restrict the $C$’s to be zero matrices. (Example 1
below shows that this classification can actually be finer.) This yields a
quotient of the monoid of isomorphism classes which is a monoid, but need not
be a group.

Starting with the monoid of isomorphism classes of differential projective
modules we look at the quotient by the submonoid of isomorphism classes of
modules of the form $(R^{n},0)$. While this quotient differential monoid is
not always a group, it maps surjectively to the similar object defined for $R$
ignoring the derivation, which is a group, and which we show is isomorphic to
$K_{0}(R)/<[R]>$. When $R$ is connected, this quotient is the kernel of the
rank function. Moreover, we show that the group of units of the quotient
monoid is isomorphic to $K_{0}(R^{D})/<[R^{D}]>$. Also, we will see that the
part of the quotient monoid not described by the $K$ theory of $R$ and $R^{D}$
is captured by the quotient of the submonoid of differential isomophism
classes of the modules of the form $(R^{n},A)$. (This approach is similar to
the one adopted for differential Azumaya algebras in [6], an approach taken
because of the difficulties similar to those noted above that appear in [3]).

##  2\. The Monoids

A monoid is a set endowed with an associative binary operation and an identity
element. Every monoid considered here will commutative. We use addition for
the operation, and $0$ for the identity. Except for the notation for the
operation, the following observations are identical with [6]: Throughout this
section we will be concerned with monoids of isomorphism classes of
differential modules and various quotients thereof. If $N\subseteq M$ is a
submonoid, by $M/N$ we mean the set of equivalence classes on $M$ under the
equivalence relation $m_{1}\sim m_{2}$ if there are $n_{1},n_{2}\in N$ such
that $m_{1}+n_{1}=m_{2}+n_{2}$. Sums of equivalent elements are equivalent,
which defines a commutative operation on $M/N$ by adding representatives of
equivalence classes, and the equivalence class of the identity is an identity.
Thus $M/N$ is a monoid. The projection $p:M\to M/N$ ($p(m)$ is the class of
$m$) is an epimorphism, $p(N)=0$, and any monoid homomorphism $q:M\to Q$ with
$q(N)=0$ factors uniquely through $p$. Note, however, that we can have $M/N=0$
but $N\neq M$, for example $\mathbb{Q}^{*}/\mathbb{Z}^{\times}$, the
multiplicative group of non-zero rationals modulo the multiplicative monoid of
non-zero integers.. Thus $p^{-1}(0)$ can be strictly larger than $N$.

If $N$ is a subgroup of the monoid $M$, $M/N$ is the set of $N$ orbits in $M$
where $N$ acts by translation.

The set of invertible elements $U(M)$ of $M$ is its maximal subgroup. The set
of elements $a\in M$ whose equivalence classes in $M/N$ are invertible is

$\\{a\in M|\exists b\in M,n\in N\text{ such that }a+b+n\in N\\}.$

The monoid $M/U(M)$ has no invertible elements except the identity.

We will be considering the following monoids. In the definitions we sometimes
consider $R$ as a commutative ring and sometimes as a differential ring, and
similarly for differential projective $R$ modules.

######  Definition 1.

$P(R)$ denotes the monoid of isomorphism classes of finitely generated
projective $R$ modules with the operation induced by direct sum and the
identity the isomorphism class of $0$.

$P^{\text{diff}}(R)$ denotes denotes the monoid of isomorphism classes of
differential finitely generated projective $R$ modules with the operation
induced by direct sum and the identity the isomorphism class of $0$.

$F(R)$ denotes the submonoid of $P(R)$ whose elements are isomorphism classes
of free $R$ modules.

$F^{\text{diff}}(R)$ denotes the submonoid of $P^{\text{diff}}(R)$ whose
elements are isomorphism classes of differential projective modules of the
form $(R^{n},Z)$.

$F_{0}^{\text{diff}}(R)$ denotes the submonoid of $P^{\text{diff}}(R)$ whose
elements are isomorphism classes of differential projective modules of the
form $(R^{n},0)$.

$PC(R)$ denotes $P(R)/F(R)$

$PC^{\text{diff}}(R)$ denotes $P^{\text{diff}}(R)/F_{0}^{\text{diff}}(R)$.

$FC^{\text{diff}}(R)$ denotes $F^{\text{diff}}(R)/F_{0}^{\text{diff}}(R)$.

$PC(R)$ is called the _projective class monoid_ of $R$.

$PC^{\text{diff}}(R)$ is called the _differential projective class monoid_ of
$R$.

$FC^{\text{diff}}(R)$ is called the _diffferential free module class monoid_
of $R$.

All of the monoids in Definition 1 are functors.

Using Definiton 1 and the definition of quotient monoids we see that if $P$
and $Q$ are finitely generated projective $R$ modules (respectively
differential finitely generated projective $R$ modules) then their classes in
$PC(R)$ (respectively $PC^{\text{diff}}(R)$) will be equal provided for some
$m$ and $n$ $P\oplus R^{n}$ is isomorphic to $Q\oplus R^{m}$ (respectively
$P\oplus(R^{n},0)$ is differentially isomorphic to $Q\oplus(R^{m},0)$). In
particular, $P$ is zero in $PC(R)$ (respectively $PC^{\text{diff}}(R)$)
provided for some $m$ and $n$ $P\oplus R^{n}$ is isomorphic to $R^{m}$
(respectively $P\oplus(R^{n},0)$ is differentially isomorphic to $(R^{m},0)$).
In this case $P$ is called stably free (respectively _stably differentially
trivial_).

######  Proposition 1.

$PC(R)$ is a group isomorphic to $K_{0}(R)/<[R]>$.

###### Proof.

Let $P$ be a finitely generated projective $R$ module. The existence of a $Q$
such that $P\oplus Q$ is a free module shows that the class of $Q$ is an
inverse to the class of $P$ in $P(R)/F(R)$. Thus $PC(R)$ is a group. Hence the
map $P(R)\to PC(R)$ factors through $K_{0}(R)$, and since $[R]\mapsto 0$ it
factors through $K_{0}(R)/<[R]>$. On the other hand, the monoid map $P(R)\to
K_{0}(R)$ maps $F(R)$ into $<[R]>$ so that $P(R)\to K_{0}(R)/<[R]>$ factors
through $P(R)/F(R)$. The composites in either order are the identity, so
$PC(R)$ is isomorphic to $K_{0}(R)/<[R]>$. ∎

Suppose $R$ is connected, so that the rank homomorphism
$rk:K_{0}(R)\to\mathbb{Z}$ is defined. By [2, Proposition 3.2 p. 460]
$K_{0}(R)/<[R[>$ is isomorphic to $\text{Ker}(rk)$.

The kernel of $rk$ is denoted $Rk_{0}(R)$ in [2, p. 459]. Sometimes the kernel
of $rk$ is called the _reduced_ $K$ group and denoted $\tilde{K}_{0}(R)$.

The homomorphism $P^{\text{diff}}(R)\to P(R)$ obtained by forgetting the
derivation is surjective [4, Theorem 2, p. 4341], so the induced map
$PC^{\text{diff}}(R)\to PC(R)$ is an epimorphism. This map carries
$FC^{\text{diff}}(R)$ to $0$. In fact we have the following:

######  Proposition 2.

The map $PC^{\text{diff}}(R)\to PC(R)$ is an epimorphism. The map
$p:PC^{\text{diff}}(R)/FC^{\text{diff}}(R)\to PC(R)$ is an isomorphism

###### Proof.

As we have remarked, [4, Theorem 2, p. 4341] implies the first assertion and
so $p$ is an epimorphism. To see that it is an isomorphism, we first show that
it is a homomorphism of groups and then show it has trivial kernel. To see
that $PC^{\text{diff}}(R)/FC^{\text{diff}}(R)$ is a group, let $P$ be a
differential finitely generated projective $R$ module. By [4, Corollary 3 p.
4342] there is a differential projective module $Q$ such that $P\oplus Q$ is
of the form $(R^{n},A)$ for some $n$ and some $A$. That is, the class of $P$
plus the class of $Q$ in $P^{\text{diff}}(R)$ lies in $F^{\text{diff}}(R)$.
Thus $P^{\text{diff}}(R)/F^{\text{diff}}(R)$ is a group, hence so is its image
$PC^{\text{diff}}(R)/FC^{\text{diff}}(R)$. Now suppose again that $P$ is a
differential finitely generated projective module such that its class in
$PC^{\text{diff}}(R)/FC^{\text{diff}}(R)$ is sent to $0$ by $p$. Thus as
projective $R$ modules $P$ and $0$ have the same class in $PC(R)$, so there
are $m$ and $n$ such that $P\oplus R^{n}$ is isomorphic to $R^{m}$. Thus,
using the given differential structure on $P$, $P\oplus(R^{n},0)$ is a
differential module, free of rank $m$ as an $R$ module, and hence of the form
$(R^{m},Z)$. It follows that the class of $P$ in
$PC^{\text{diff}}(R)/FC^{\text{diff}}(R)$ is trivial. Thus $p$ has trivial
kernel, so is injective, and hence an isomorphism. ∎

Informally, Proposition 2 can be understood as asserting that the
classification of differential projective $R$ modules comes down to the
classification of those which are free as $R$ modules, since when the latter
are nullified the differential structures on projective modules depend only on
the underlying $R$ module. More formally:

######  Corollary 1.

If $FC^{\text{diff}}(R)=0$ then $PC^{\text{diff}}(R)=PC(R)$.

######  Corollary 2.

If $PC(R)=0$ then $FC^{\text{diff}}(R)=PC^{\text{diff}}(R)$.

###### Proof.

(Of Corollary 2.) As remarked above, it is possible for the quotient of a
monoid by a proper submonoid to be trivial. We argue as in the proof of
Proposition 2. Since $PC(R)=0$, for any projective $R$ module $P$ there are
$m$,$n$ such that $P\oplus R^{n}$ is isomorphic to $R^{m}$. Now suppose $P$
has a differential structure. Then so does $P\oplus(R^{n},0)$, which is then
isomorphic to $(R^{m},A)$ for some $A$. Since the class of $(R^{m},A)$ is in
$FC^{\text{diff}}(R)$, so is the class of $P\oplus(R^{n},0)$, and this is the
same as the class of $P$. ∎

Proposition 2 is also, in a way, a statement about the functor from
differential modules to modules. A counterpart to that is the functor from
modules over the constants to differential modules, which results in the
following:

######  Proposition 3.

The functor $P_{0}\mapsto R\otimes_{R^{D}}P_{0}$ induces a group isomorphism
$q:PC(R^{D})\to U(PC^{\text{diff}}(R))$.

###### Proof.

The functor induces a map $P(R^{D})\to P^{\text{diff}}(R)$. If $P_{0}$ is a
free $R^{D}$ module of finite rank, then $R\otimes_{R^{D}}P_{0}$ belongs to
$F_{0}^{\text{diff}}(R)$, so the map passes to a homomorphism $PC(R^{D})\to
PC^{\text{diff}}(R)$. If $P_{0}$ is a finitely generated projective $R^{D}$
module, and $Q_{0}$ is a finitely generated projective $R^{D}$ module such
that $P_{0}\oplus Q_{0}$ is free, then in $PC^{\text{diff}}(R)$ the classes
$R\otimes_{R^{D}}P_{0}$ and $R\otimes_{R^{D}}Q_{0}$ add up to the class of
$0$, and hence are inverses of each other. Thus the homomorphism $PC(R^{D})\to
PC^{\text{diff}}(R)$ has image in the units of the range. Suppose $P$ is a
differential finitely generated projective module whose class in
$PC^{\text{diff}}(R)$ is invertible. Then there exist $m$, $n$, and a
differential finitely generated projective module $Q$ such that $P\oplus
Q\oplus(R^{n},0)$ is isomorphic to $(R^{m},0)$. In particular, $P$ is a
differential direct summand of $(R^{m},0)$. By [4, Theorem 1, p. 4340], this
means that $P$ is of the form $R\otimes_{R^{D}}P_{0}$ where $P_{0}$ is a
finitely generated projective $R^{D}$ module. This shows that $q$ is group
epimorphism. Suppose that the class of $P_{0}$ is in the kernel of $q$. Then
the classes of $R\otimes_{R^{D}}P_{0}$ and $0$ are the same in
$PC^{\text{diff}}(R)$ so there are $m$, $n$ such that
$(R\otimes_{R^{D}}P_{0})\oplus(R^{n},0)$ is differentially isomorphic to
$(R^{m},0)$. Now $(R\otimes_{R^{D}}P_{0})\oplus(R^{n},0)$ is differentially
isomorphic to $R\otimes_{R^{D}}(P_{0}\oplus(R^{D})^{n})$ while $(R^{m},0)$ is
differentially isomorphic to $R\otimes_{R^{D}}(R^{D})^{m}$. Taking constants
and applying [4, Lemma 1 p. 4340] we see that $P_{0}\oplus(R^{D})^{n}$ is
$R^{D}$ isomorphic to $(R^{D})^{m}$. That is, $P_{0}$ and $0$ represent the
same class in $PC(R^{D})$. Thus $q$ has trivial kernel and hence is a group
isomorphism. ∎

The functor $P_{0}\mapsto R\otimes_{R^{D}}P_{0}$ also induces a group
homomorphism $PC(R^{D})\to PC(R)$. This need not be either injective or
surjective, in general. However Proposition 3 enables the identification of
its kernel.

######  Lemma 1.

The kernel of

$\phi:PC(R^{D})\to PC(R)$

is identified with $U(FC^{\text{diff}}(R))$ via the isomorphism $q$ of
Proposition 3,

###### Proof.

The inclusion $R^{D}\to R$ induces the group homomorphism

$\phi:PC(R^{D})\to PC(R).$

The kernel of $\phi$ consists of classes of finitely generated projective
$R^{D}$ modules $P_{0}$ such that $P=R\otimes_{R^{D}}P_{0}$ is a stably free
$R$ module: there are $m$, $n$ such that $P\oplus
R^{n}=R\otimes_{R^{D}}(P_{0}\oplus(R^{D})^{n})$ is isomorphic to $R^{m}$.
Since $P_{0}$ and $P_{0}\oplus(R^{D})^{n}$ give rise to the same class in
$PC(R^{D})$, we can replace the former by the latter and assume that $P$ is
actually $R$ free. So the class of $P$ as a differential module lies in
$FC^{\text{diff}}(R)$. Since it comes from $PC(R^{D})$ it also lies in
$U(PC^{\text{diff}}(R))$ and hence in $I=FC^{\text{diff}}(R)\cap
U(PC^{\text{diff}}(R))$. Thus $q(\text{Ker}(\phi))\subseteq I$; since
$FC^{\text{diff}}(R)$ is the kernel of the morphism $PC^{\text{diff}}(R)\to
PC(R)$ (see Proposition 2), we have that $q(\text{Ker}(\phi))\cong I$. We
claim that $I=U(FC^{\text{diff}}(R))$, which will complete the proof. That
$U(FC^{\text{diff}}(R))$ is contained in $I$ is clear. On the other hand $I$,
being the kernel of a group homomorphism, is a (sub)group in
$FC^{\text{diff}}(R)$ and hence contained in its group of units. ∎

Using Proposition 3 we have the following consequences of Corollaries 1 and 2:

######  Corollary 3.

If $FC^{\text{diff}}(R)=0$ then $PC(R^{D})\to PC(R)$ is an isomorphism.

###### Proof.

Suppose $FC^{\text{diff}}(R)=0$. By Corollary 1, $PC^{\text{diff}}(R)=PC(R)$.
By Proposition 1, $PC(R)$ is a group, hence
$PC^{\text{diff}}(R)=U(PC^{\text{diff}}(R))$, so by Proposition 3 $PC(R^{D})$
maps isomorphically to $PC^{\text{diff}}(R)$ and hence $PC(R^{D})\to PC(R)$ is
an isomorphism. ∎

######  Corollary 4.

If $PC(R^{D})=0$ then $PC^{\text{diff}}(R)$ has no invertible elements.

###### Proof.

Combine Corollary 2 and Proposition 3. ∎

Proposition 2 implies that if $PC^{\text{diff}}(R)=0$ then $PC(R)=0$. Of
course if $PC^{\text{diff}}(R)=0$ then its submonoid $FC^{\text{diff}}(R)=0$.
In this case, Corollary 3 further implies that $PC(R^{D})=PC(R)=0$. We record
this fact, along with the similar assertion under the stronger assumption that
all differential projective $R$ modules are trivial:

######  Corollary 5.

If $PC^{\text{diff}}(R)=0$ then $PC(R^{D})=PC(R)=0$. If all finitely generated
differential projective $R$ modules are differentially trivial, then all
finitely generated projective $R$ modules and all finitely generated
projective $R^{D}$ modules are free.

###### Proof.

For the second assertion, let $P$ be a finitely generated projective $R$
module. Let $D_{P}$ be a differential structure on $P$. Since $P$ with $D_{P}$
is differentially trivial, its underlying module $P$ is free. Let $Q$ be a
finitely generated projective $R^{D}$ module. Since $R\otimes_{R^{D}}Q$ is
differentially trivial, it is differentially isomorphic to $(R^{n},0)$ for
some $n$. It follows that
$Q=(R\otimes_{R^{D}}Q)^{D}=(R^{n},0)^{D}=(R^{D})^{n}$. ∎

In Proposition 5 below we will prove a partial converse of Corollary 5

##  3\. Trivial Differential Projective Class Monoid

We consider the condition that $PC^{\text{diff}}(R)=0$, or that all projective
differential modules are stably differentially trivial. We will show that this
condition is much more restrictive than the (non-differential) condition that
all projective $R$-modules be stably free.

If $PC^{\text{diff}}(R)=0$, then the submonoid $FC^{\text{diff}}(R)$ is also
trivial. A partial converse also holds:

######  Proposition 4.

If $FC^{\text{diff}}(R)=0$ and $PC(R^{D})=0$ then $PC^{\text{diff}}(R)=0$.

###### Proof.

Suppose $FC^{\text{diff}}(R)=0$ and let $P$ be a finitely generated
differential projective module. Then there is a differential projective $Q$
such that $P\oplus Q$ is free as an $R$ module so, as an element of
$FC^{\text{diff}}(R)$, $P\oplus Q$ must be $0$. Thus the class of $P$ is
invertible in $PC^{\text{diff}}(R)$. Proposition 3 and the assumption that
$PC(R^{D})=0$ imply that the class of $P$ is $0$ in $PC^{\text{diff}}(R)$. ∎

Proposition 4 says that if all differential $R$ modules which are free as $R$
modules are stably differentially trivial, and if all projective $R^{D}$
modules are stably free, then all differential projective $R$ modules are
stably differentially trivial. This is comparable to the corresponding
non–differential condition that all projective modules are stably free. The
stronger condition, namely that all projective modules are free, has often
been investigated; see for example [5]. The differential analogue would be the
condition that all differential protectives modules are differentially
trivial. As it happens, the differential and non–differential versions of this
condition are related:

######  Proposition 5.

If all finitely generated projective $R^{D}$ modules are free, then every
stably differentially trivial finitely generated differential projective $R$
module $P$ is differentially trivial.

###### Proof.

For a differential $R$ module $P$ to be stably differentially trivial means
that there are $m,n$ such that $P\oplus(R^{m},0)$ is isomorphic to
$(R^{n},0)$. In particular, $P$ is a differential direct summand of a trivial
differential module. By [4, Theorem 1 p. 4340] this implies that $P$ is of the
form $R\otimes_{R^{D}}P_{0}$ for some finitely generated projective $R^{D}$
module $P_{0}$. Since by assumption $P_{0}$ is free, say for rank $k$ (where
of course $k=n-m$), $P$ is differentially isomorphic to $(R^{k},0)$ and hence
differentially trivial. ∎

If $\eta:R\to S$ is a differential ring homomorphism, then there is an induced
monoid homomorphism $h:PC^{\text{diff}}(R)\to PC^{\text{diff}}(S)$ which
carries the submonoid $FC^{\text{diff}}(R)$ of $PC^{\text{diff}}(R)$ to the
submonoid $FC^{\text{diff}}(S)$ of $PC^{\text{diff}}(S)$. Even if $\eta$ is
surjective, $h$ need not be surjective, but its restriction to
$FC^{\text{diff}}(R)$ does surject onto $FC^{\text{diff}}(S)$. If it happens
that $PC(S^{D})=0$, then we can use this to detect whether
$PC^{\text{diff}}(R)\neq 0$.

######  Corollary 6.

Let $\eta:R\to S$ be a surjective differential homomorphism of differential
rings. Assume $PC(S^{D})=0$. If $PC^{\text{diff}}(R)=0$ then
$PC^{\text{diff}}(S)=0$.

###### Proof.

If $PC^{\text{diff}}(R)=0$, then $FC^{\text{diff}}(R)=0$. Since $\eta$ is
surjective, this implies that $FC^{\text{diff}}(S)=0$. Then by Proposition 4
again, $PC^{\text{diff}}(S)=0$. ∎

If $S$ is a simple differential ring, then $S^{D}$ is a field [11, Lemma
1.3.6, p. 11], so all projective $S$ modules are free, and in particular
$PC(S^{D})=0$. So if $R\to S$ is a surjection with $S$ simple, and if
$PC^{\text{diff}}(R)=0$, then $PC^{\text{diff}}(S)=0$.

######  Corollary 7.

Let $R$ be a differential ring with $PC^{\text{diff}}(R)=0$. Then for every
maximal differential ideal $I$ of $R$, $PC^{\text{diff}}(R/I)=0$.

Note that in general $R/I$, for $I$ a maximal differential ideal, may have
differential projective modules which are not obtained from differential
projective $R$ modules by reduction modulo $I$. Indeed every differential $R$
module, projective or not, reduces to a differential projective $R/I$ module
since every finitely generated differential module over the simple
differential ring $R/I$ is a projective $R/I$ module [1, Theorem 2.2.1. p.
455].

We consider as examples a class of differential rings whose differential
projective class monoids are _never_ trivial.

We will be using some facts about Picard–Vessiot extensions, which we now
briefly recall ([9, Chapter 1]). Let $F$ be a characteristic zero differential
field with algebraically closed field of constants $C$. Let $P=(F^{n},A)$ be a
finitely generated differential projective $F$ module. A differential field
extension $E\supseteq F$ is called a _Picard–Vessiot extension_ of $F$ for $P$
provided: (1) $E^{D}=C$; (2) $E\otimes_{F}P$ is differentially trivial; and
(3) No proper differential subfield of $E$ containing $F$ satisfies (1) and
(2). Picard–Vessiot extensions exist and are unique up to $F$-isomorphism. Now
suppose $P_{1}$ and $P_{2}$ are finitely generated projective differential $F$
modules which represent the same class in $PC^{\text{diff}}(F)$. That is,
there are $m,n$ such that $P_{1}\oplus(F^{m},0)$ and $P_{2}\oplus(F^{n},0)$
are differentially isomorphic. Let $E_{i}$ be a Picard–Vessiot extension for
$P_{i}$. Since $E_{1}\otimes_{F}P_{1}$ has a basis of constants, so does
$E_{1}\otimes_{F}(P_{1}\oplus(F^{m},0))$. Because of the isomorphism,
$E_{1}\otimes_{F}(P_{2}\oplus(F^{n},0))$ also has a basis of constants. In
other words, $E_{1}\otimes_{F}(P_{2}\oplus(F^{n},0))$ is differentially
trivial. Thus $E_{1}\otimes_{F}P_{2}$ is stably differentially trivial, and
then by Proposition 5 $E_{1}\otimes_{F}P_{2}$ is differentially trivial, i.e.
has a basis of constants. Since also $E_{1}^{D}=C$, $E_{1}$ contains a
Picard–Vessiot extension for $P_{2}$. Similarly, $E_{2}$ contains a
Picard–Vessiot extension of $P_{1}$. We conclude that $E_{1}$ and $E_{2}$ are
isomorphic differential $F$ algebras. Suppose we have chosen all
Picard–Vessiot extensions of $F$ inside the same Picard–Vessiot closure of
$F$. That is, inside a differential field extension of $F$ with the same
constants as $F$ which is a union of Picard–Vessiot extensions of $F$ and
which contains an isomorphic copy of every Picard–Vessiot extension of $F$.
(See [8] for the proof that Picard–Vessiot closures exist.) Then we have that
$E_{1}=E_{2}$. Conversely, if the Picard–Vessiot extensions for $P_{1}$ and
$P_{2}$ are not equal, then $P_{1}$ and $P_{2}$ have different classes in
$PC^{\text{diff}}(F)$. This same reasoning applies to sets of differential $F$
modules with more than two members, indeed to such sets of any cardinality:

######  Proposition 6.

Let $F$ be a characteristic zero differential field with algebraically closed
field of constants $C$. Let $\\{P_{i}|i\in I\\}$ be a set of finitely
generated differential projective $F$ modules and for each $i\in I$ let
$E_{i}$ be a Picard–Vessiot extension of $F$ for $P_{i}$, all inside a chosen
Picard–Vessiot closure of $F$. Suppose all the $E_{i}$’s are distinct. Then
the classes of the $P_{i}$’s are distinct elements of $PCM^{\text{diff}}(F)$.

Next we look at differential algebras which are finitely generated as $F$
algebras over differential fields $F$ that admit Picard–Vessiot extensions of
arbitrarily large transcendence degree (this is a relatively mild condition
satisfied for example by $F=\mathbb{C}$ with $0$ derivation and
$F=\mathbb{C}(x)$ with derivation $d/dx$) :

######  Proposition 7.

Let $F$ be a characteristic zero differential field with algebraically closed
field of constants $C$. Suppose that $F$ admits Picard–Vessiot extensions of
arbitrarily large transcendence degree over $F$. Let $R$ be a differential $F$
algebra which is finitely generated as an $F$ algebra. Then
$PC^{\text{diff}}(R)\neq 0$.

###### Proof.

By Corollary 7 it is enough to show that for some maximal differential ideal
$I$ of $R$ $PC^{\text{diff}}(R/I)\neq 0$. Thus we can replace $R$ by $R/I$ and
assume that $R$ is differentially simple. It follows that the quotient field
$E$ of $R$ is an extension of $F$ whose field of constants is also $C$ [7,
Corollary 1.18, p. 11]. Note that $E$ has finite transcendence degree over
$F$, so there are finitely generated differential modules over $F$ which do
not admit Picard–Vessiot extensions in $E$. In particular, this means there is
some matrix $A$ over $F$ such that $(F^{n},A)$ does not admit a basis of
constants over $E$, and hence does not have one over $R$. Thus $(F^{n},A)$ is
not a trivial differential $R$ module. Since $R^{D}$ is the field $C$, all
projective $R^{D}$ modules are free. Proposition 5 then implies that
$(F^{n},A)$ is not even stably differentially trivial, and hence its class in
$PCM^{\text{diff}}(R)$ is non–trivial. ∎

Proposition 7 applies to differential affine $\mathbb{C}$ algebras, none of
which, including polynomial rings or Laurent polynomial rings, can have
trivial differential projective class monoids.

##  4\. Cancellation

Throughout this section we assume that $R$ is connected, so that every
projective $R$ module is of constant rank; in fact, we only need this for
finitely generated differential projective $R$ modules. The assumption
guarantees that the number of non-zero summands in a direct sum decomposition
of a module of rank $r$ is at most $r$.

######  Definition 2.

$R$ is said to have _free cancellation_ if, for finitely generated
differential projective $R$ modules $P$ and $Q$, if $P\oplus(R,0)$ is
differentially isomorphic to $Q\oplus(R,0)$ then $P$ is differentially
isomorphic to $Q$.

From the definition and induction, using the identification of $(R^{n},0)$
with $(R,0)^{(n)}$, we see that for a ring with free cancellation cancellation
of $(R,0)$ implies the cancellation of $(R^{n},0)$.

######  Proposition 8.

Let $R$ have free cancellation and let $P$ and $Q$ be finitely generated
differential projective $R$ modules with $P\oplus(R^{n},0)$ differentially
isomorphic to $Q\oplus(R^{n},0)$. Then $P$ is differentially isomorphic to
$Q$.

It is a corollary of Proposition 8 that differentially trivial modules of
different ranks may be canceled:

######  Corollary 8.

Let $R$ have free cancellation and let $P$ and $Q$ be finitely generated
differential projective $R$ modules with $P\oplus(R^{n},0)$ differentially
isomorphic to $Q\oplus(R^{m},0)$ where $n\geq m$. Then $P$ is differentially
isomorphic to $Q\oplus(R^{n-m},0)$.

If $n=m$ in Corollary 8 and below then by $(R^{0},0$) is meant $0$.

######  Definition 3.

Let $P$ be a finitely generated differential projective $R$ module. If $P$ has
no differential direct summand isomorphic to $(R,0)$ then $P$ is said to be
_trivial–free_. A differential projective submodule $C$ of $P$ is called a
_core_ provided

  * 1.

$C$ is trivial–free; and

  * 2.

$C\oplus(R^{n},0)$ is differentially isomorphic to $P$ for some $n$ (possibly
$n=0$).

By condition $2.$, $C$ and $P$ represent the same class in
$PC^{\text{diff}}(R)$.

The limit on the number of summands of a finitely generated projective module
implies that every finitely generated differential projective module has a
core. For example, the zero module is a core of $(R^{n},0)$.

In general, a module may have non-isomorphic cores. For example, in the zero
derivation case, a non-free stably free module will lead to an example with
non-isomorphic cores: see Example 1 below. In the free cancellation case, this
doesn’t happen:

######  Proposition 9.

Assume $R$ has free cancellation and let $P$ be a finitely generated
differential projective module $R$ module. Let $C$ and $D$ be cores of $P$.
Then $C$ and $D$ are differentially isomorphic.

###### Proof.

For some $n,m$ we have both $C\oplus(R^{n},0)$ and $D\oplus(R^{m},0)$
isomorphic to $P$ and hence to each other. Without loss of generality we may
assume $n\geq m$. By Corollary 8 $C$ is differentially isomorphic to
$D\oplus(R^{n-m},0)$. Since $C$ has no summand isomorphic to $(R,0)$ we
conclude that $n=m$ and hence that by Proposition 8 $C$ and $D$ are
differentially isomorphic. ∎

Using Proposition 9 we introduce, in the case where $R$ has free cancellation,
the following notation:

######  Notation.

Let $R$ have free cancellation and let $P$ be a finitely generated
differential projective $R$ modules. Then $H(P)$ denotes a core of $P$; $H(P)$
is unique up to isomorphism.

If $P$ is trivial–free then $H(P)=P$.

Note that $H((R^{n},0))=\\{0\\}$.

Suppose that $R$ has free cancellation and let $P$ be a finitely generated
differential projective module. Then $P$ and $H(P)$ belong to the same class
in $PC^{\text{diff}}(R)$; in fact $H(P)$ is a module of minimal rank in the
class of $P$. This sets up a bijection between the isomorphism classes of
trivial–free finitely generated projective differential modules and
$PC^{\text{diff}}(R)$. This bijection means a monoid structure can be imposed
on the set of isomorphism classes of trivial–free finitely generated
projective differential modules, namely if $H(P)=P$ and $H(Q)=Q$ then
$[P]+[Q]=[H(P\oplus Q)]$. This coincides with the operation on isomorphism
classes induced from direct sums provided the direct sum of trivial–free
modules is trivial–free.

We turn now to a class of differential rings which has free cancellation and
also this property for direct sums.

######  Proposition 10.

Suppose $R^{D}$ is a field. Then $R$ has free cancellation.

###### Proof.

Let $C$ denote the field $R^{D}$. Let $M$ be a finitely generated differential
projective $R$ module, and suppose $P=M\oplus(R,0)$ can be written as an
internal direct sum of differential submodules $N_{1}$ and $R_{1}$ where
$R_{1}$ is differentially isomorphic to $(R,0)$. To prove the proposition, we
must show that $M$ is differentially isomorphic to $N_{1}$.

We write $P=M\oplus(R,0)$ as $\\{(m,r)|m\in P,r\in R\\}$, and regard $N_{1}$
and $R_{1}$ as differential submodules of the latter with $N_{1}+R_{1}=P$ and
$N_{1}\cap R_{1}=0$. By assumption, $(R,0)$ is differentially isomorphic to
$R_{1}$; let $x=(y,z)$ be the $R$ generator of $R_{1}$ corresponding to $1$.
Since $D(1)=0$, $D(x)=(D(y),D(z))=(0,0)$. Thus $D(y)=0$ and $D(z)=0$ so $z\in
C$.

We consider first the case that $z\neq 0$.

Then $z$ is a unit of $R$, and $R_{1}=Rx=Rz^{-1}x$ where $z^{-1}x=(w,1)$. Note
that $D(w)=D(z^{-1}y)=0$. Define $T:P\to P$ by $(m,r)\mapsto(m+rw,r)$. $T$ is
a differential endomorphism, and in fact an automorphism with inverse $S$
given by $(m,r)\mapsto(m-rw,r)$. Then $T((0,1))=(w,1)$ which implies that
$T(0\oplus(R,0))=R_{1}$ and $S(R_{1})=0\oplus(R,0)$. Also $T(M\oplus
0)=M\oplus 0$; in fact $T$ is the identity on $M\oplus 0$. Let $N=S(N_{1})$.
Then $N+(0\oplus(R,0))=S(N_{1})+S(R_{1})=S(N_{1}+R_{1})=S(P)=P$. Let $(m,r)$
be any element of $P$. Since $P=N+(0\oplus(R,0))$, $(m,r)=n+s(0,1)$ for some
$n\in N$ and $s\in R$. Thus $n=(m,r-s)$. In particular, every element of $M$
appears as the first entry of an element of $N$. Also
$N\cap(0\oplus(R,0))=S(N_{1})\cap S(R_{1})=S(N_{1}\cap R_{1})=0$. The latter
implies that if $(0,r)\in N$ then $r=0$. So if $(m,s)\in N$ and $(m,t)\in N$
then $(m,s)-(m,t)=(0,s-t)\in N$ so $s=t$. That is, the second entry of an
element of $N$ is determined by the first entry. Thus we have a function
$f:M\to R$ such that $N=\\{(m,f(m))|m\in M\\}$. Since $N$ is a differential
submodule of $P$, $f$ is a differential homomorphism. It follows that $M\to N$
by $m\mapsto(m,f(m))$ is a differential homorphism, in fact a differential
isomorphism whose inverse $N\to M$ is given by $(m,f(m))\mapsto m$. This
proves the result in the case $z\neq 0$.

Now consider the case that $z=0$.

Then $R_{1}=R(y,0)$ where $D(y)=0$. This implies $R_{1}\subset M\oplus 0$.Thus

$(N_{1}\cap(M\oplus 0))+R_{1}\subseteq M\oplus 0.$

Let $m\in M$. Since $N_{1}+R_{1}=P$, there are $n_{1}\in N_{1}$ and
$r_{1}=r(y,0)\in R_{1}$ such that $(m,0)=n_{1}+r_{1}$. That is
$n_{1}=(m-ry,0)\in M\oplus 0$. Thus $n_{1}\in N_{1}\cap(M\oplus 0)$. Thus
$M\oplus 0=(N_{1}\cap(M\oplus 0))+R_{1}$. Let $N_{2}=N_{1}\cap(M\oplus 0)$.
Then $M\oplus 0$ is the internal direct sum of $N_{2}$ and $R_{1}$.
($N_{2}\cap R_{1}=\\{0\\}$ since $N_{1}\cap R_{1}=\\{0\\}$.) Reverting to
direct sum notation, we have that $M$ is the direct sum $N_{2}\oplus R_{1}$
which means that $P$ is the direct sum $N_{2}\oplus R_{1}\oplus(R,0)$, while
$P$ is also the direct sum to $N_{1}\oplus R_{1}$. Thus on the one hand we
have $P/R_{1}$ differentially isomorphic to $N_{2}\oplus(R,0)$ and on the
other that $P/R_{1}$ is differentially isomorphic to $N_{1}$. We conclude that
$N_{1}$ is differentially isomorphic to $N_{2}\oplus(R,0)$. Since the latter
is differentially isomorphic to $N_{2}\oplus R_{1}$, and this in turn is
differentially isomorphic to $M$, we have that $M$ and $N_{1}$ are
differentially isomorphic, completing the proof in the case $z=0$ and hence of
the proposition. ∎

Next, we see that the same condition on $R$ – that $R^{D}$ is a field – which
Proposition 10 shows implies free cancellation also implies that the direct
sum of trivial–free modules is trivial–free.

######  Proposition 11.

Suppose $R^{D}$ is a field. Then if $P_{1}$ and $P_{2}$ are finitely generated
differential projective modules such that $P_{1}\oplus P_{2}$ has a
differential direct summand differentially isomorphic to $(R,0)$ then either
$P_{1}$ or $P_{2}$ has such a summand.

###### Proof.

Suppose there is a direct summand as in the proposition Let $g:(R,0)\to
P_{1}\oplus P_{2}$ and $f:P_{1}\oplus P_{2}\to(R,0)$ be differential
homomorphisms with $fg$ the identity. Let $f_{i}:P_{i}\to(R,0)$ be $f$
preceded by inclusion on summand $i$ and let $g_{i}:(R,0)\to P_{i}$ be $g$
followed by projection on summand $i$. For each $i$, $f_{i}g_{i}$ is a
differential endomorphism of $(R,0)$ and hence given by multiplication by a
constant $a_{i}$. For $r\in(R,0)$, $r=fg(r)=f_{1}(g_{1}(r))+f_{2}(g_{2}(r))$.
Thus at least one of the $f_{i}g_{i}$ is non-zero, say for $i=1$. Thus the
constant $a_{1}$ is non-zero, and since $R^{D}$ is a field $a_{1}$ is a unit.
It follows that $(a_{1}^{-1}f_{1})g_{1}$ is the identity, showing that
$g_{1}((R,0))$ is a differential direct summand of $P_{1}$ isomorphic to
$(R,0)$. ∎

######  Corollary 9.

Suppose $R^{D}$ is a field. Let $P$ and $Q$ be finitely generated differential
projective $R$ modules. Then $H(P)\oplus H(Q)$ is differentially isomorphic to
$H(P\oplus Q)$.

###### Proof.

By Proposition 10, $R$ has cancellation, so both $P$ and $Q$ have cores unique
up to differential isomorphism, so the symbols $H(P)$ and $H(Q)$ make sense.
For some $m,n$ we have $P$ isomorphic to $H(P)\oplus(R^{m},0)$ and $Q$
isomorphic to $H(Q)\oplus(R^{n},0)$. Thus $P\oplus Q$ is differentially
isomorphic to $H(P)\oplus H(Q)\oplus(R^{m+n},0)$. By Proposition 11, or rather
its contrapositive, $H(P)\oplus H(Q)$ is trivial–free, and the previous
differential isomorphism shows that it is a core of $P\oplus Q$. ∎

Corollary 9 implies that, in the case that $R^{D}$ is a field, (1) the
isomorphism classes of trivial–free finitely generated projective differential
modules form a monoid with the operation induced from direct sum; and (2) that
this monoid is isomorphic to $PC^{\text{diff}}(R)$:

######  Theorem 1.

Suppose $R^{D}$ is a field. Then

  * 1.

The differential isomorphism classes of trivial–free finitely generated
projective differential $R$ modules form a monoid $M$ with the operation
induced from direct sum.

  * 2.

$P\mapsto H(P)$ induces an isomorphism $PC^{\text{diff}}(R)\to M$.

###### Proof.

For assertion $1.$, suppose $P$ and $Q$ are trivial–free finitely generated
projective differential $R$ modules. Then $H(P)$ is differentially isomorphic
to $P$ and $H(Q)$ is differentially isomorphic to $Q$. Since $H(P\oplus Q)$ is
differentially isomorphic to $H(P)\oplus H(Q)$ by Corollary 9, $H(P\oplus Q)$
is differentially isomorphic to $P\oplus Q$, which means $P\oplus Q$ is
trivial free. Thus the monoid $M$ is well-defined. For assertion $2.$, we
begin by noting that by Proposition 9 and Proposition 10 $P\mapsto H(P)$
defines an endomorphism of the monoid $P^{\text{diff}}(R)$ of isomorphism
classes of finitely generated projective differential $R$ modules with the
operation induced from direct sum as in Definition 1. If $P$ and $Q$ are
finitely generated projective differential $R$ modules and there are $m,n$
with $P\oplus(R^{n},0)$ differentially isomorphic to $Q\oplus(R^{m},0)$ then
by Corollary 9 $H(P\oplus(R^{n},0))$ is isomorphic to $H(P)\oplus
H((R^{n},0))$ and since the second summand is $\\{0\\}$ we conclude that
$H(P\oplus(R^{n},0))$ is differentially isomorphic to $H(P)$. Similarly,
$H(Q\oplus(R^{m},0))$ is differentially isomorphic to $H(Q)$. Hence $H(P)$ is
differentially isomorphic to $H(Q)$. Thus the monoid homomorphism
$P^{\text{diff}}(R)\to M$ induced by $P\mapsto H(P)$ factors through
$PC^{\text{diff}}(R)$ and defines a monoid homomorphism
$h:PC^{\text{diff}}(R)\to M$. To see that $h$ is onto, let $P$ be a
trivial–free finitely generated differential projective $R$ module. Then since
$H(P)$ is differentially isomorphic to $P$, $h$ applied to the class of $P$ in
$PC^{\text{diff}}$ is the class of $P$ in $M$. So $h$ is onto. Suppose $P$ and
$Q$ are sent to the same element by $h$. That means $P$ and $Q$ have
differentially isomorphic cores. Let $C$, $D$ be cores of $P$, $Q$
respectively. Then by Definition 3 $2.$ there are $m,n$ such that $P$ is
differentially isomorphic to $C\oplus(R^{n},0)$ and $Q$ is differentially
isomorphic to $D\oplus(R^{m},0)$. Since $C$ and $C\oplus(R^{n},0)$
(respectively $D$ and $D\oplus(R^{m},0)$) represent the same element of
$PC^{\text{diff}}(R)$, and so do $C$ and $D$, we have that $P$ and $Q$
represent the same element of $PC^{\text{diff}}(R)$. This shows that $h$ is
injective and hence an isomorphism. ∎

It is a consequence of Theorem 1 that if $P$ and $Q$ are non-isomorphic
trivial–free finitely generated projective differential $R$ modules, for
instance if they have different ranks, then since they represent different
classes in $M$ they represent different classes in $PC^{\text{diff}}(R)$.

We can think of $P\mapsto H(P)$ as inducing an endomorphism of
$P^{\text{diff}}(R)$ which is the identity on $M$ and whose image is $M$.
Theorem 1 says that $M$ is isomorphic to $PC^{\text{diff}}(R)$. Restated:

######  Corollary 10.

Suppose $R^{D}$ is a field. Then the monoid epimorphism

$P^{\text{diff}}(R)\to PC^{\text{diff}}(R)$

has a right inverse.

The monoid epimorphism $P^{\text{diff}}(R)\to PC^{\text{diff}}(R)$ is
injective on $M$ and hence on submonoids of $M$. As an example of some
calculations using this observation, we consider the case $R=\mathbb{C}[x]$
with $D=d/dx$. We have $R^{D}=\mathbb{C}$ is a field, so Theorem 1 applies. We
consider modules of the form $(R,f)$ for $f\in R$, where we have identified
$1\times 1$ matrices with their only entry. A differential homomorphism
$(R,f)\to(R,g)$ is given by multiplication by $a\in R$ satisfying
$a^{\prime}=fa-ag$. Thus $a^{\prime}=a(f-g)$ which, considering polynomial
degrees, is only possible if $a=0$ or, if $a\neq 0$, when both
$a\in\mathbb{C}$ and $f=g$. Only the second possibility holds if
$(R,f)\to(R,g)$ is an isomorphism. In particular, if $f\neq 0$ then $(R,f)$ is
not isomorphic to $(R,0)$, implying that $(R,f)$ is trivial free. Thus if
$f_{1},\dots,f_{n}\in R$ are all non-zero, by Corollary 9
$P=(R,f_{1})\oplus\dots\oplus(R,f_{n})$ is trivial–free. Similarly, if
$g_{1},\dots,g_{m}$ are non-zero elements of $R$ then
$Q=(R,g_{1})\oplus\dots\oplus(R,g_{m})$ is trivial–free. Thus $P$ and $Q$ will
represent different classes in $PC^{\text{diff}}(R)$ unless $P$ and $Q$ are
isomophic. This requires that $m=n$, and then we can represent a differential
isomorphism $T:P\to Q$ by an $n\times n$ matrix whose entries are differential
homomorphisms $(R,f_{i})\to(R,g_{j})$. As we have seen, these are all given by
multiplication by elements of $\mathbb{C}$, where the elements are zero unless
$f_{i}=g_{j}$. Thus if no $f_{i}$ equals any $g_{j}$ then $P$ and $Q$ are not
differentially isomorphic. We can use these observations to produce
uncountably many distinct elements of the monoid of isomorphism classes of
trivial–free finitely generated differential projective $R$ modules and hence
of $PC^{\text{diff}}(R)$. Of course in this case both $PC(R^{D})=0$ and
$PC(R)=0$ as well.

##  5\. Zero Derivation Rings

Any commutative ring can be made the underlying ring of a differential ring by
giving it the zero derivation. A derivation on a module over a zero derivation
ring is just a linear endomorphism, so any module over a zero derivation ring
can be made into a differential module by choosing such an endomorphism. Thus
examples of projective modules, such as non–free projective modules, or
non–free stably free projective modules, can be used to give examples of
differential projective modules whose underlying projective modules have these
properties. In particular, the endomorphism chosen could be the zero map
(multiplication by $0$), the identity map (multiplication by $1$) or
multiplication by any other scalar, as we do in the following result.

######  Proposition 12.

Let $A,B$, and $C$ be modules over the commutative ring $R$. Give $R$ the zero
derivation and let $A$ and $B$ be given differential structures by using their
identity endomorphisms. Let $C$ be given differential structure by the zero
endomorphism. Suppose $A\oplus C$ is differentially isomorphic to $B\oplus C$.
Then $A$ is differentially isomorphic to $B$.

###### Proof.

Let $f:A\oplus C\to B\oplus C$ be a differential isomorphism. Since $(A\oplus
C)^{D}$ is a differential submodule of $A\oplus C$, $(B\oplus C)^{D}$ is a
differential submodule of $B\oplus C$, and $f((A\oplus C)^{D})=(B\oplus
C)^{D}$, $f$ induces a differential isomorphism $(A\oplus C)/(A\oplus
C)^{D}\to(B\oplus C)/(B\oplus C)^{D}$. Since $(A\oplus C)^{D}=\\{0\\}\oplus C$
and $(B\oplus C)^{D}=\\{0\\}\oplus C$ this induced isomorphism is a
differential isomorphism between $A$ and $B$. ∎

In Proposition 12 if $C$ is the free module $R^{n}$ then $C$ with the zero
derivative is $(R^{n},0)$. Thus if $A$ and $B$ have the identity derivation,
and if $A\oplus(R^{n},0)$ is differentially isomorphic to $B\oplus(R^{n},0)$,
then $A$ is differentially isomorphic to $B$.

######  Example 1.

Let $R$ be a connected commutative ring and let $A$ be a non-free stably free
$R$ module; for example, [10, Example 1, p. 269]. That is, there are $m,n$
such that there is an $R$ isomorphism $f:A\oplus R^{m}\to R^{n}$. If follows
that $A$ is projective of rank $p=n-m$. Moreover $f$ can be regarded as an
isomorphism from $A\oplus R^{m}$ to $R^{p}\oplus R^{m}$. However $A$ is not
isomorphic to $R^{p}$. Now give $R$ the zero derivation structure and $A$,
$R^{p}$, and $R^{m}$ the differential structure coming from the identity. Then
$f$ is a differential homomorphism. If we let $B$ denote $R^{p}$ with the
identity derivation and $P$ denote $R^{m}$ with the identity derivation then
$A\oplus P$ is differentially isomorphic to $B\oplus P$, but, since $A$ and
$B$ are not isomorphic, Proposition 12 shows that for no $q$ is
$A\oplus(R^{q},0)$ isomorphic to $B\oplus(R^{q},0)$.

In the notation of Example 1, we see that the differential projective $R$
modules $A$ and $B$ give rise to distinct elements of $PC^{\text{diff}}(R)$
but the same element of $K_{0}^{\text{diff}}(R)$.

We have recalled the fact [4, Theorem 2, p. 4341] that for any differential
ring $R$ every finitely generated projective $R$ module $M$ can be endowed
with a differential structure. This makes the forgetful functor $\mathcal{U}$
from the additive category of finitely generated differential projective $R$
modules to the additive category of finitely generated projective $R$ modules
surjective on objects. In the case that $R$ has the zero derivation, as we
noted above, we can make a projective module $M$ a differential module using
the zero endomorphism of $M$. And if $M$ and $N$ are both finitely generated
projective $R$ modules and $f:M\to N$ an $R$ module homomorphism, then $f$ is
a differential homomorphism when both $M$ and $N$ are given differential
structure using their zero endomorphisms. That is, using the zero endomorphism
differential structures produces an additive functor $\mathcal{F}$ from the
category of finitely generated projective $R$ modules to the category of
finitely generated differential projective $R$ modules with
$\mathcal{U}\mathcal{F}$ the identity. Note that $\mathcal{F}(R)=(R,0)$. Could
such a functor exist in general for any differential commutative ring $R$?

More precisely: Let $R$ be a differential ring and let $\mathcal{U}$ denote
the forgetful functor from the additive category of finitely generated
differential projective $R$ modules to the additive category of finitely
generated projective $R$ modules. We ask: can there be an additive functor
$\mathcal{F}$ from the additive category of finitely generated projective $R$
modules to the additive category of finitely generated differential projective
$R$ modules with

  * 1.

$\mathcal{U}\mathcal{F}$ naturally equivalent to the identity; and

  * 2.

$\mathcal{F}(R)=(R,0)$.

The additive functors $\mathcal{U}$ and $\mathcal{F}$ will induce monoid
homomorphisms

$u:P^{\text{diff}}(R)\to P(R)$ $f:P(R)\to P^{\text{diff}}(R)$

which, by condition $1.$, are such that $uf$ is the identity. Since
$\mathcal{U}((R^{n},0))=R^{n}$, $u$ induces a monoid homomorphism
$\overline{u}:PC^{\text{diff}}(R)\to PC(R)$. An additive functor $\mathcal{F}$
satisfying condition $2.$ will have $\mathcal{F}(R^{n})=(R,0)^{(n)}$ which is
differentially isomorphic to $(R^{n},0)$. Thus $f$ carries $F(R)$ to
$F_{0}^{\text{diff}}(R)$ and hence induces a monoid homomorpism
$\overline{f}:PC(R)\to PC^{\text{diff}}(R)$ with $\overline{u}\overline{f}$
equaling the identity. Since (by Proposition 1) $PC(R)$ is a group, so is its
isomorphic image $\overline{f}(PC(R))$. Now suppose that $R$ is such that
$PC(R^{D})=0$ and $PC(R)\neq 0$. Then by Corollary 4 $PC^{\text{diff}}(R)$ has
no invertible elements, so its subgroup $\overline{f}(PC(R))$ is trivial. This
contradiction means that for such a ring $R$ there is no such functor
$\mathcal{F}$. For a concrete example of such an $R$ we can take the
Picard–Vessiot ring $R$ of a Picard–Vessiot extension of a differential field
$F$ with differential Galois group $G$ such that the coordinate ring $F[G]$
has non stably free projective modules. Then $R^{D}=F$ so $PC(R^{D})=0$ while
$PC(R)=PC(F[G])\neq 0$.

This paper has been published in _Communications in Algebra_

https://doi.org/10.1080/00927872.2022.2045493.

## References

  * [1] André, Y. _Solution algebras of differential equations and quasi–homogeneous varieties_ , Ann. Sci. École Norm. Sup. 47, 449-467
  * [2] Bass, H. _Algebraic K Theory_ , W. A. Benjamin, New York, 1968
  * [3] Juan, L., and Magid, A. _Differential central simple algebras and Picard–Vessiot representations_ , Proc. Amer. Math. Soc. 136 (2008), 1911-1918
  * [4] Juan, L., and Magid, A. _Differential projective modules over differential rings_ , Communications in Algebra 47 (2019), 4336-4346
  * [5] Lam, T. Y. _Serre’s Problem_ , Springer, Berlin, 2006
  * [6] Magid, A. _Differential Brauer monoids_ , ArXiv 123456
  * [7] Magid, A. _Lectures on Differential Galois Theory_ , University Lecture Series 7, American Mathematical Society, Providence 1994 (second printing with corrections, 1997)
  * [8] Magid, A. _The Picard-Vessiot closure in differential Galois theory. Differential Galois theory_ (Bedlewo, 2001), 151-164, Banach Center Publ., 58, Polish Acad. Sci. Inst. Math., Warsaw, 2002.
  * [9] Singer, M., and van der Put, M. _Galois Theory of Linear Differential Equations_ , G. der math. Wiss. 328, Springer, Berlin, 2003
  * [10] Swan, R. S., _Vector Bundles and Projective Modules_ , Trans. Amer. Math Soc. 105 (1962), 264-277
  * [11] Wibmer, M. _Differential Galois theory, an introduction to the Galois theory of linear differential equations_ , retrieved from https://sites.google.com/view/wibmer