system of supports $\mathcal{S}$ and $B\cap{\mathbb{N}}_{0}^{s}=A$. We will
show that $B_{\min}(A)\subseteq B$. Let $\mathbf{c}\in B_{\mathrm{min}}(A)$,
and choose $H\in\mathcal{S}_{1}$ and $\mathbf{x}\in A_{H}^{\mathcal{S}_{1}}$
with $\varepsilon_{H}^{\mathcal{S}_{1}}(\mathbf{x})=\mathbf{c}$. (See Remark
4.14.) If $H=\emptyset$, then $\mathbf{c}=\mathbf{x}\in B$, so we assume that
$H\neq\emptyset$. Then, $H\supseteq\mbox{\rm supp}\,(\mathbf{a})$ for some
non-zero $\mathbf{a}\in A$, and we put $\mathbf{b}=\infty a$. Then
$\mathbf{b}\in B\setminus A$ and $\infty-\mbox{\rm
supp}\,(\mathbf{c})=H\supseteq\infty-\mbox{\rm supp}\,(\mathbf{b})$. Now part
(3) of Proposition 7.2 yields $\mathbf{c}\in B$.

$(3)$. As in the proof of $(2)$, $B_{\max}\cap{\mathbb{N}}_{0}^{s}=A$. Suppose
$B$ is given by a system of supports $\mathcal{S}$ and
$B\cap{\mathbb{N}}_{0}^{s}=A$. If $\mathbf{c}\in B$, then
$\mathbf{c}=\varepsilon_{H}^{\mathcal{S}}(\mathbf{x})$, with $\mathbf{x}\in
A_{H}^{\mathcal{S}}$ and $H\in\mathcal{S}\subseteq\\{1,\dots,s\\}$. Then $H\in
S_{2}$, $\mathbf{x}\in A_{H}^{\mathcal{S}_{2}}$ and
$\varepsilon_{H}^{\mathcal{S}}(\mathbf{x})=\varepsilon_{H}^{\mathcal{S}_{\max}}(\mathbf{x})\in
B_{\max}$.

Keep the notation in Proposition 7.4. By Proposition 4.21, $A+\infty\cdot
A\subseteq B_{\mathrm{min}}(A)$. We will see examples in which the containment
is proper. It seems to be an interesting problem to determine for which
monoids $A$ this is true. The following lemma gives a characterization of when
this happens.

######  Lemma 7.5.

Let $B$ be a submonoid of $({\mathbb{N}}_{0}^{*})^{s}$ with order-unit
$(n_{1},\dots,n_{s})\in\mathbb{N}^{s}$, given by an almost-free system of
supports $\mathcal{S}(n_{1},\dots,n_{s})$. Then $B=A_{\emptyset}+\infty\cdot
A_{\emptyset}$ if and only if

  * (i)

$\mathcal{S}$ is the set of supports of the elements of $A_{\emptyset}$; and

  * (ii)

if $H$ is a minimal element of $\mathcal{S}\setminus\\{\emptyset\\}$, then
$\pi_{H}(A_{\emptyset})={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}(=A_{H})$
where
$\pi_{H}\colon{\mathbb{N}}_{0}^{s}\to{\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus
H}$ denotes the canonical projection.

If this happens, then
$\pi_{H}(A_{\emptyset})={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$ for
every $H\in\mathcal{S}\setminus\\{\emptyset\\}$.

Proof. If $B=A_{\emptyset}+\infty\cdot A_{\emptyset}$ then the system of
supports that defines $B$ must satisfy (i) by Proposition 4.21 (i). To prove
(ii) and the last statement, let $H\in\mathcal{S}\setminus\\{\emptyset\\}$. By
Proposition 4.21 (i), $\pi_{H}(A_{\emptyset})=A_{H}$ and, since
$\mathcal{S}(n_{1},\dots,n_{s})$ is almost-free,
$A_{H}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$.

Conversely, assume that $B$ is a monoid defined by an almost-free system of
supports satisfying (i) and (ii). We claim that
$\pi_{H}(A_{\emptyset})={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}=A_{H}$
for each non-empty $H\in\mathcal{S}$. To see this, choose a minimal element
$K$ of $\mathcal{S}\setminus\\{\emptyset\\}$ with $K\subseteq H$, and let
$\pi\colon{\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus
K}\to{\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$ be the canonical
projection. Then
$\pi_{H}(A_{\emptyset})=\pi\circ\pi_{K}(A_{\emptyset})=\pi({\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus
K})={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}=A_{H}$ (the last equality
holds because the system of supports is almost-free). Since, by (i) of
Definition 4.13, $(n_{1},\dots,n_{s})\in A_{\emptyset}$, we can apply
Proposition 4.21 (i) with $A=A_{\emptyset}$ to see that
$A_{\emptyset}+\infty\cdot A_{\emptyset}$ is given by the same family of
supports as is $B$. Hence $A_{\emptyset}+\infty\cdot A_{\emptyset}=B$.

Notice that in Lemma 7.5, condition (ii) is intrinsic to the submonoid
$A_{\emptyset}$, and it gives a characterization of when
$A_{\emptyset}+\infty\cdot A_{\emptyset}$ is given by an almost-free system of
supports.

######  Lemma 7.6.

Let $A$ be a full submonoid of ${\mathbb{N}}_{0}^{s}$ with order-unit. Let
$\mathcal{S}$ be the set of supports of elements of $A$.

Then $B:=A+\infty\cdot A$ is given by an almost-free system of supports if and
only if, for each minimal element $H$ of
$\mathcal{S}\setminus\\{\emptyset\\}$, one has
$\pi_{H}(A)={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$, where
$\pi_{H}\colon{\mathbb{N}}_{0}^{s}\to{\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus
H}$ denotes the canonical projection.

Moreover, when this happens, one has
$\pi_{H}(A)={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$ for every
$H\in\mathcal{S}\setminus\\{\emptyset\\}$.

Proof. Fix an order-unit $(n_{1},\dots,n_{s})\in A\cap\mathbb{N}^{s}$. Recall,
from Proposition 4.21 that $A+\infty\cdot A$ is always given by the full
system of supports $\mathcal{S}(n_{1},\dots,n_{s})$, where $\mathcal{S}$ is
indeed the set of supports of elements of $A$, and where $A_{H}=\pi_{H}(A)$
for each $H\in\mathcal{S}$. In particular, $A_{\emptyset}=A$.

If $\mathcal{S}(n_{1},\dots,n_{s})$ is almost-free, then Lemma 6.4 delivers
the result we want. For the converse, we see, as in the proof of Lemma 7.5,
that $A_{H}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$ for every
$H\in\mathcal{S}\setminus\\{\emptyset\\}$, and hence
$\mathcal{S}(n_{1},\dots,n_{s})$ is almost-free.

The direct sum decompositions of monoids studied in Section 6, easily give
examples of monoids that fail to satisfy the conditions in Lemma 7.6. In the
proof we will refer to the algebraic partial order on $A$; since $A$ is a full
submonoid of ${\mathbb{N}}_{0}^{s}$, it is easy to see that this is just the
coordinate-wise partial order on ${\mathbb{N}}_{0}^{s}$.

######  Proposition 7.7.

Let $A$ be a full submonoid of ${\mathbb{N}}_{0}^{s}$ with order-unit, such
that it has a non-trivial direct sum decomposition $A=A_{1}\oplus A_{2}$. Then
$A+\infty\cdot A$ is given by an almost-free system of supports if and only if
the following hold for $i=1,2$ (with the notation of Proposition 6.3):

  * (i)

$B_{i}={\mathbb{N}}_{0}^{I_{i}}$, and

  * (ii)

$\mbox{\rm supp}\,(f_{i}(b))=I_{3}$ for every $b\in B_{i}\setminus\\{0\\}$.

When this happens, $A$ is a free commutative monoid.

Proof. By Remark 4.17 and Proposition 4.21, the unique system of supports
giving the monoid $A+\infty\cdot A$ has $\mathcal{S}=\\{\text{supports of
elements of}\ A\\}$, and $A_{H}=p_{H}(A)$ for each $H\in\mathcal{S}$. (As in
Proposition 4.21,
$p_{H}:{\mathbb{N}}_{0}^{s}\to{\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$
is the canonical projection.) Thus we have to show this system of supports is
almost free if and only if (i) and (ii) are satisfied. We will use,
repeatedly, the following formula from Proposition 6.3 (iii):

$a=\varepsilon_{i}(\pi_{i}(a))+\varepsilon_{3}(f_{i}(a_{i})),\qquad\text{for}\
a\in A_{i},\quad i=1,2\,.$ (3)

Assume the system of supports $\mathcal{S}$ is almost-free. We prove first
that (i) is satisfied. Choose $a_{2}\in A_{2}$ with $\mbox{\rm
supp}\,(a_{2})=\mbox{\rm supp}\,(A_{2})=:H$. Then $\\{1,\dots,s\\}\setminus
H=I_{1}$, and
$B_{1}=\pi_{1}(A_{1})=\pi_{1}(A)=p_{H}(A)={\mathbb{N}}_{0}^{I_{1}}$ by Lemma
7.6. Symmetrically, $B_{2}={\mathbb{N}}_{0}^{I_{2}}$.

In order to prove (ii), it will suffice, by symmetry, to show that $\mbox{\rm
supp}\,(f_{1}(b))=I_{3}$ for every $b\in B_{1}\setminus\\{0\\}$. Now $b$ is
divisible by a minimal element $e_{1}$ of $B_{1}\setminus\\{0\\}$; then
$f_{1}(e_{1})\mid f_{1}(b)$, and hence $\mbox{\rm
supp}\,(f_{1}(e_{1}))\subseteq\mbox{\rm supp}\,(f_{1}(b))$. Therefore, it
suffices to show that $\mbox{\rm supp}\,(f_{1}(e_{1}))=I_{3}$. By (i), we
have, after a harmless permutation of $\\{1,\dots,s\\}$, that
$e_{1}=(1,0,\dots,0)$. Suppose, by way of contradiction, that there is an
index $j\in I_{3}\setminus\mbox{\rm supp}\,(f_{1}(e_{1}))$. Put
$e=(0,\dots,0,1,0\dots 0)\in{\mathbb{N}}_{0}^{s}$, with $\mbox{\rm
supp}\,(e)=\\{j\\}$.

Write $e_{1}=\pi_{1}(a_{1})$, where $a_{1}\in A_{1}$, and put
$g=\infty\cdot(\varepsilon_{1}(e_{1})+\varepsilon_{3}(f_{1}(e_{1}))+e$. From
equation (3), we get
$a_{1}=\varepsilon_{1}(e_{1})+\varepsilon_{3}(f_{1}(e_{1}))$ and hence
$g=\infty\cdot a_{1}+e$. We claim that $g$ belongs to $A+\infty\cdot A$.
Obviously $\infty\cdot a_{1}\in A+\infty\cdot A$. Let $H=\mbox{\rm
supp}\,(a_{1})=\\{1\\}\cup\mbox{\rm
supp}\,\big{(}\varepsilon_{3}(f_{1}(e_{1}))\big{)}$, and let $d\in
A_{H}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$ be the element that has
$1$ in the $j^{\text{th}}$ spot and $0$ elsewhere. Then $\varepsilon_{H}(d)$
is in $A+\infty\cdot A$ (the monoid given by the system of supports
$\mathcal{S}$), whence $\infty\cdot a_{1}+\varepsilon_{H}(d)\in A+\infty\cdot
A$. But clearly $\infty\cdot a_{1}+\varepsilon_{H}(d)=\infty\cdot a_{1}+e=g$,
and the claim is verified.

Write $g=a+\infty\cdot a^{\prime}$, with $a,a^{\prime}\in A$. Then

$(\mbox{\rm supp}\,(a+a^{\prime}))\cap(I_{1}\cup I_{2})=(\mbox{\rm
supp}\,(g))\cap(I_{1}\cup I_{2})=\\{1\\}\,.$

Put $a+a^{\prime}=c\in A$, write $c=c_{1}+c_{2}$, with $c_{i}\in A_{i}$, and
let $b_{i}=\pi_{i}(c_{i})\in B_{i}$. By equation (3),
$c_{i}=\varepsilon_{i}(b_{i})+\varepsilon_{3}(f_{i}(b_{i}))$, for $i=1,2$. Now
$(\mbox{\rm supp}\,(c_{i}))\cap(I_{1}\cup I_{2})\subseteq\\{1\\}$, and it
follows that $b_{1}\in{\mathbb{N}}_{0}e_{1}$ and $b_{2}=0$. Writing
$b_{1}=ne_{1}$, with $n\in{\mathbb{N}}_{0}$,
$a+a^{\prime}=c=n(\varepsilon_{1}(e_{1})+\varepsilon_{3}(f_{1}(e_{1}))$. In
particular, $j\notin(\mbox{\rm supp}\,(a))\cup(\mbox{\rm
supp}\,(a^{\prime}))$, but $j\in\mbox{\rm supp}\,(e)\subset\mbox{\rm
supp}\,(g)$, contradicting the equality $g=a+a^{\prime}\cdot\infty$. This
completes the proof of (ii).

Suppose, conversely, that conditions (i), (ii) of Proposition 7.7 hold. We
will show that the system of supports described above is almost-free.

We claim that $\mathcal{S}\setminus\\{\emptyset\\}$, the set of supports of
non-zero elements of $A$, is exactly the set of subsets of $\\{1,\dots,s\\}$
that contain $I_{3}$ strictly. To see this, suppose $a\in A\setminus\\{0\\}$,
and write $a=a_{1}+a_{2}$, with $a_{i}\in A_{i}$, say, $a_{1}\neq 0$. Then
equation (3) and condition (ii) show that $\mbox{\rm supp}\,(a)\supsetneqq
I_{3}$. On the other hand, if $I_{3}\subsetneqq H\subseteq\\{1,\dots,s\\}$, we
seek an element $a\in A$ such that $\mbox{\rm supp}\,(a)=H$. We may assume,
harmlessly, that $H\cap I_{1}\neq\emptyset$. Choose, using condition (i), an
element $b_{1}\in B_{1}$ such that $(\mbox{\rm supp}\,(b_{1}))\cap I_{1}=H\cap
I_{1}$. By condition (ii) and equation (3), the element $a_{1}\in A_{1}$ with
$\pi_{1}(a_{1})=b_{1}$ has $(\mbox{\rm supp}\,(a_{1}))\cap(I_{1}\cup
I_{3})=H\cap(I_{1}\cup I_{3})$. If $H\cap I_{2}=\emptyset$, we set $a=a_{1}$.
Otherwise, we choose $a_{2}\in A_{1}$ with $(\mbox{\rm
supp}\,(a_{2}))\cap(I_{2}\cup I_{3})=H\cap(I_{2}\cup I_{3})$, and set
$a=a_{1}+a_{2}$. This proves the claim.

Let $A^{\prime}\subseteq({\mathbb{N}}_{0}^{*})^{s}$ be the set consisting of
$0$ and all elements in $({\mathbb{N}}_{0}^{*})^{s}$ whose infinite support
strictly contains $I_{3}$. By the claim, the almost-free system of supports
based on the set $\mathcal{S}$ gives the monoid $A+A^{\prime}$. Therefore, by
Remark 4.17, it will suffice to show that $A+A^{\prime}=A+\infty\cdot A$. The
inclusion $A+\infty\cdot A\subseteq A+A^{\prime}$ is clear. For the opposite
inclusion, let $0\neq a\in A^{\prime}$, and for $I\subseteq\\{1,\dots,s\\}$
let $v_{I}=(v_{1},\dots,v_{s})$, where $v_{j}=\infty$ if $j\in I$ and
$v_{j}=0$ if $j\notin I$. Extending the notation of Proposition 6.3, we let
$\pi_{i}:({\mathbb{N}}_{0}^{*})^{s}\to({\mathbb{N}}_{0}^{*})^{I_{i}}$ and
$\varepsilon_{i}:({\mathbb{N}}_{0}^{*})^{I_{i}}\to({\mathbb{N}}_{0}^{*})^{s}$
be the canonical projections and injections. To see that $a\in A+\infty\cdot
A$, we note that

$a=(\varepsilon_{1}\pi_{1}(a)+v_{I})+(\varepsilon_{2}\pi_{2}(a)+v_{I})\,,$ (4)

where $I=\infty-\mbox{\rm supp}\,(a)$. We will show that the right-hand side
of (4) belongs to $A+\infty\cdot A$.

Let
$\pi:({\mathbb{N}}_{0}^{*})^{s}\to({\mathbb{N}}_{0}^{*})^{\\{1,\dots,s\\}\setminus
I}$ be the canonical projection, and choose $c\in{\mathbb{N}}_{0}^{s}$ with
$\pi(c)=\pi(a)$. Using condition (i), choose, for $i=1,2$, elements $d_{i}\in
A_{i}$ such that $\pi_{i}(d_{i})=\pi_{i}(c)$, and let $d=d_{1}+d_{2}\in A$.
Choose an element $b\in A$ with $\mbox{\rm supp}\,(b)=I$; then $\infty\cdot
b=v_{I}$. For $i=1,2$,
$\varepsilon_{i}\pi_{i}(a)+v_{I}=d_{i}+v_{I}=d_{i}+\infty\cdot b\in
A+\infty\cdot A$. From (4), we get $a\in A+\infty\cdot A$.

Now we characterize the single linear equations that define an almost-free
system of supports.

######  Proposition 7.8.

Fix $s\geq 2$. Let $\mathbf{a}=(a_{1},\dots,a_{s})$ and
$\mathbf{b}=(b_{1},\dots,b_{s})$ be nonzero elements of ${\mathbb{N}}_{0}^{s}$
such that $\mathbf{a}\neq\mathbf{b}$. Let $B$ be the monoid of solutions in
$({\mathbb{N}}_{0}^{*})^{s}$ of the equation
$\mathbf{a}\mathbf{T}=\mathbf{b}\mathbf{T}$, where
$\mathbf{T}=(t_{1},\dots,t_{s})^{\text{tr}}$. Then

  * (i)

$B\cap\mathbb{N}^{s}\neq\emptyset$ if and only if $\mathbf{a}-\mathbf{b}$ has
a positive entry and a negative entry.

  * (ii)

Assume $B\cap\mathbb{N}^{s}\neq\emptyset$. Then $B$ is given by an almost-free
system of supports if and only if
$\mathrm{supp}(\mathbf{a})\bigcup\mathrm{supp}(\mathbf{b})=\\{1,\dots,s\\}$.

  * (iii)

Assume that $B\cap\mathbb{N}^{s}\neq\emptyset$ and that
$\mathrm{GCD}(a_{1},\dots,a_{s},b_{1},\dots,b_{s})=1$. Then $B$ is given by an
almost-free system of supports satisfying $B=A_{\emptyset}+\infty\cdot
A_{\emptyset}$ if and only if

    * (iii.1)

$\mathbf{a}+\mathbf{b}\in\mathbb{N}^{s}$;

    * (iii.2)

$\mathrm{supp}(\mathbf{a})\cap\mathrm{supp}(\mathbf{b})=\emptyset$; and

    * (iii.3)

$\mathrm{GCD}(a_{i},b_{j})=1$ for each $i\in\mathrm{supp}(\mathbf{a})$ and
each $j\in\mathrm{supp}(\mathbf{b})$.

Proof. Throughout the proof, for any $i\in\\{1,\dots,s\\}$, let
$e_{i}=(0,\dots,1^{i)},\dots,0)$.

(i) If the equation $(\mathbf{a}-\mathbf{b})\mathbf{T}=0$ has a solution in
$\mathbb{N}^{s}$ then not all the entries of $\mathbf{a}-\mathbf{b}$ can have
the same sign. Conversely, if there are $i\neq j\in\\{1,\dots,s\\}$ such that
$a_{i}>b_{i}$ and $a_{j}<b_{j}$ then it is easy to find a solution in
$\mathbb{N}^{s}$ of the equation $(\mathbf{a}-\mathbf{b})\mathbf{T}=0$.

(ii) Assume that $B$ is given by an almost-free system of supports, and that
$i\in\\{1,\dots,s\\}$ is such that $a_{i}=0$. If also $b_{i}=0$, then
$\infty\cdot e_{i}$ is in $B$. Thus, $H=\\{i\\}$ is an infinite support of
$B$, and hence $A_{H}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus\\{i\\}}$. By
Proposition 5.4, all elements of
$A_{H}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus\\{i\\}}$ satisfy the
equation obtained by deleting the variable $t_{i}$. This implies that
${\mathbb{N}}_{0}^{s}\subseteq B$ and hence that $\mathbf{a}=\mathbf{b}$, a
contradiction. Therefore $b_{i}\neq 0$.

To prove the converse assume that $\mathbf{a}$ and $\mathbf{b}$ have the
required properties. We already know that the monoid of solutions of a single
linear equation is a submonoid of $({\mathbb{N}}_{0}^{*})^{s}$ given by a full
system of supports, with $\mathcal{S}$ the set of infinite supports of
elements of $B$. (See Proposition 4.16.) By Lemma 7.1, to check that the
system of supports is almost-free, we need to check that the monoids
associated to the minimal non-empty infinite supports of elements in $B$ are
free.

We prove first that, in the case of monoids defined by a single equation, the
minimal nonempty infinite supports have either one element or two elements.
Let $i\in\\{1,\dots,s\\}$; then $a_{i}=0=b_{i}$ or $a_{i}\neq 0\neq b_{i}$ if
and only if $\infty\cdot e_{i}=(0,\dots,\infty^{i)},\dots,0)\in B$. In this
case, $\\{i\\}$ is a minimal non-empty infinite support of $B$.

Let $H$ be a minimal element of $\mathcal{S}\setminus\\{\emptyset\\}$, and
assume that $H$ has at least two elements. Choose $i_{0}\in H$. By the
previous argument, we may assume that $a_{i_{0}}\neq 0$ and $b_{i_{0}}=0$.
Choose $c\in B$ such that $\infty-\mbox{\rm supp}\,(c)=H$. Then $c$ is a
solution of the equation $\mathbf{a}\mathbf{T}=\mathbf{b}\mathbf{T}$, and one
can check that $\sum_{i\in H}\infty\cdot e_{i}$, the element obtained from $c$
by replacing the entries outside $H$ by $0$, is also a solution. This implies
that there exists $j_{0}\in H\setminus\\{i_{0}\\}$ such that $b_{j_{0}}\neq
0$. Hence $\infty\cdot e_{i_{0}}+\infty\cdot e_{j_{0}}\in B$. The minimality
of $H$ implies that $H=\\{i_{0},j_{0}\\}$.

Fix $i\in\\{1,\dots,s\\}$. Assume that $H=\\{i\\}$ is an infinite support. By
our previous arguments, and because we are assuming that
$\mathrm{supp}(\mathbf{a})\bigcup\mathrm{supp}(\mathbf{b})=\\{1,\dots,s\\}$,
this implies that $a_{i}\cdot b_{i}\neq 0$. Hence, all elements
$\mathbf{x}=(x_{1},\dots,x_{s})\in({\mathbb{N}}_{0}^{*})^{s}$ with
$x_{i}=\infty$ are solutions of the equation
$\mathbf{a}\mathbf{T}=\mathbf{b}\mathbf{T}$. By Proposition 4.16,
$A_{\\{i\\}}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus\\{i\\}}$ as wanted.

Assume now that $\\{i\\}$ is not an infinite support. This means that either
$a_{i}\neq 0$ and $b_{i}=0$ or $a_{i}=0$ and $b_{i}\neq 0$. By symmetry, we
may assume $a_{i}\neq 0$. Then, by our previous arguments, the minimal
infinite supports that also contain $i$ are of the form $H_{j}=\\{i,j\\}$ with
$j\neq i$ and such that $b_{j}\neq 0$. Hence, all elements
$\mathbf{x}=(x_{1},\dots,x_{s})\in({\mathbb{N}}_{0}^{*})^{s}$ with
$x_{i}=\infty=x_{j}$ are solutions of the equation
$\mathbf{a}\mathbf{T}=\mathbf{b}\mathbf{T}$. By Proposition 4.16,
$A_{H_{j}}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H_{j}}$ as desired.

(iii). Assume that $B$ is given by an almost-free system of supports and
satisfies $B=A_{\emptyset}+\infty\cdot A_{\emptyset}$.

We claim that $B$ does not have infinite supports with only one element.
Indeed, if the equation $\mathbf{a}\mathbf{T}=\mathbf{b}\mathbf{T}$ has a
solution $(t_{1},\dots,t_{s})\in({\mathbb{N}}_{0}^{*})^{s}$ with
$t_{i}=\infty$ and $t_{j}<\infty$ for $j\neq i$, then $\infty\cdot
e_{i}=(0,\dots,\infty^{i)},\dots 0)$ is also a solution. Now
$\\{i\\}\in\mathcal{S}$ by (3) of Definition 4.13. As
$B=A_{\emptyset}+\infty\cdot A_{\emptyset}$, there exists $r\in\mathbb{N}$
such that $re_{i}\in A_{\emptyset}$. This implies $a_{i}=b_{i}$. Since
$A_{\\{i\\}}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus\\{i\\}}$, for any
$j\neq i$ there exists $r_{j}\in{\mathbb{N}}_{0}$ such that
$e_{j}+r_{j}e_{i}\in A_{\emptyset}$ is a solution of
$\mathbf{a}\mathbf{T}=\mathbf{b}\mathbf{T}$. This implies that $a_{j}=b_{j}$.
Hence, we have proved that $\mathbf{a}-\mathbf{b}=0$, a contradiction.

In view of the arguments in (ii), the non-existence of minimal supports with
just one single element implies that $a_{i}\cdot b_{i}=0$ for every
$i\in\\{1,\dots,s\\}$. Therefore
$\mathrm{supp}(\mathbf{a})\cap\mathrm{supp}(\mathbf{b})=\emptyset$. By (ii),
we know that
$\mathrm{supp}(\mathbf{a})\bigcup\mathrm{supp}(\mathbf{b})=\\{1,\dots,s\\}$.
Hence, we deduce that $(iii.1)$ and $(iii.2)$ hold.

Now we prove that $(iii.3)$ is also satisfied. Let
$i\in\mathrm{supp}(\mathbf{a})$ and $j\in\mathrm{supp}(\mathbf{b})$. Then
$H:=\\{i,j\\}$ is a minimal infinite support of $B$, and
$A_{H}={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$. Since
$B=A_{\emptyset}+\infty\cdot A_{\emptyset}$, the equation
$a_{i}t_{i}+a_{k}=b_{j}t_{j}$ must have a solution in ${\mathbb{N}}_{0}$ for
every $k\in\mathrm{supp}(\mathbf{a})\setminus\\{i\\}$, and this happens if and
only if $\mathrm{GCD}(a_{i},b_{j})$ divides $a_{k}$. Similarly, one deduces
that $\mathrm{GCD}(a_{i},b_{j})$ divides $b_{k}$ for every
$k\in\mathrm{supp}(\mathbf{b})\setminus\\{j\\}$.

Since $\mathrm{GCD}(a_{1},\dots,a_{s},b_{1},\dots,b_{s})=1$, we deduce that
condition $(iii.3)$ holds.

Assume now that the equation $\mathbf{a}\mathbf{T}=\mathbf{b}\mathbf{T}$
satisfies $(iii.1)$, $(iii.2)$ and $(iii.3)$. Set
$H_{1}=\mathrm{supp}(\mathbf{a})$ and $H_{2}=\mathrm{supp}(\mathbf{b})$.

By $(iii.1)$, the condition in (ii) is satisfied, and hence $B$ is given by an
almost-free system of supports. By (iii) of Proposition 4.16, $\mathcal{S}$ is
the set of infinite supports of elements in $B$. By $(iii.1)$ and $(iii.2)$,
$H\in\mathcal{S}\setminus\\{\emptyset\\}$ if and only if $H\cap H_{1}$ and
$H\cap H_{2}$ are nonempty. It is easy to see that such $H$ is the support of
an element of $A_{\emptyset}$, so condition (i) of Lemma 7.5 holds.

Let $H$ be a minimal element of $\mathcal{S}\setminus\\{\emptyset\\}$. Then
$H=\\{i,j\\}$ for some $i\in H_{1}$ and $j\in H_{2}$. By $(iii.3)$, for any
$k\in\\{1,\dots,s\\}\setminus H$, there exists an element
$\mathbf{c}=(c_{1},\dots,c_{s})\in A_{\emptyset}$ such that
$\mathrm{supp}(\mathbf{c})\subseteq\\{i,j,k\\}$ and $c_{k}=1$. This implies
that $p_{H}(A_{\emptyset})={\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus H}$,
where
$p_{H}\colon{\mathbb{N}}_{0}^{s}\to{\mathbb{N}}_{0}^{\\{1,\dots,s\\}\setminus
H}$ denotes the canonical projection. Since this happens for any non-empty
minimal support of $B$, condition (ii) of Lemma 7.5 holds, and we deduce that
$B=A_{\emptyset}+\infty\cdot A_{\emptyset}$.

######  Remark 7.9.

1) It is straightforward to check that if $M_{1}$ and $M_{2}$ are submonoids
of $({\mathbb{N}}_{0}^{*})^{s}$ given by almost-free systems of supports
containing a common element of $\mathbb{N}^{s}$, then so is $M_{1}\cap M_{2}$.
Therefore, any monoid defined as the solution of a system of equations, each
one satisfying the condition in Proposition 7.8 (ii), is given by an almost-
free system of supports.

Therefore, the set of solutions in $({\mathbb{N}}_{0}^{*})^{s}$ of a system of
the form $E_{1}\mathbf{T}=E_{2}\mathbf{T}$, where $E_{i}\in M_{\ell\times
s}({\mathbb{N}}_{0})$ for $i=1,2$, and such that $E_{1}+E_{2}\in M_{\ell\times
s}(\mathbb{N})$, is given by almost-free system of supports provided
$\mathbb{N}^{s}$ contains a solution the system.

2) Let $A$ be a full submonoid of ${\mathbb{N}}_{0}^{s}$ with order-unit,
defined by a system of equations $E_{1}\mathbf{T}=E_{2}\mathbf{T}$ where, for
$i=1,2$, $E_{i}\in M_{\ell\times s}({\mathbb{N}}_{0})$. Let $F\in
M_{\ell\times s}({\mathbb{N}}_{0})$ be the matrix with all its entries equal
$1$. By 1), the solutions of the system
$(E_{1}+F)\mathbf{T}=(E_{2}+F)\mathbf{T}$ is an almost-free system of supports
with $A_{\emptyset}=A$ and such that any element in
$\mathcal{P}(\\{1,\dots,s\\})\setminus\emptyset$ is an infinite support.

3) Again, let $A$ be a submonoid of ${\mathbb{N}}_{0}^{s}$, with order-unit,
defined by a system of equations $E\mathbf{T}=0$ where $E\in M_{\ell\times
s}(\mathbb{Z})$. Fix $H\in M_{\ell\times s}(\mathbb{N})$ to be a matrix such
that all the entries of $E+H$ are strictly positive. Consider the system

$(E+H)\mathbf{T}=H\mathbf{T}.$

Let $\mathbf{a}\in{\mathbb{N}}_{0}^{s}$. Then $\mathbf{a}\in A$ if and only if
$\mathbf{a}$ is a solution of the above system.

To compute the solutions in $({\mathbb{N}}_{0}^{*})^{s}$ observe that, again,
any element in $2^{\\{1,\dots,s\\}}\setminus\\{\emptyset\\}$ is an infinite
support and that the monoid of the solutions in $({\mathbb{N}}_{0}^{*})^{s}$
is given by an almost-free system of supports. Therefore, the set of solutions
of this system is the monoid $B_{\mathrm{max}}(A)$, cf. Proposition 7.4.

Now we study further examples showing that there are full submonoids $A$ of
${\mathbb{N}}_{0}^{s}$ with the following property: If $\mathcal{S}$ is an
almost-free system of supports with $A_{\emptyset}\cong A$, then
$A_{\emptyset}+\infty\cdot A_{\emptyset}\subsetneq M$. In view of Corollary
4.26, this implies that whenever $X$ is a torsion-free finitely generated
module over a commutative noetherian domain of Krull dimension $1$ such that
$V(X)\cong A$, then $\mathrm{Add}(X)$ contains objects that are not direct
sums of finitely generated modules.

######  Example 7.10.

Let $t_{1},t_{2}\in\mathbb{N}$. Let $B_{1}$ be any full submonoid of
${\mathbb{N}}_{0}^{t_{1}}$ with order-unit, and assume that $B_{1}$ is not
free. Let $B_{2}={\mathbb{N}}_{0}^{t_{2}}$. Let $s\geq t_{1}+t_{2}$, and let
$A$ be any full submonoid of $\mathbb{N}_{0}^{s}$ with order-unit and
isomorphic to $B_{1}\times B_{2}$. Then $A+\infty\cdot A$ is not a submonoid
of $(\mathbb{N}_{0}^{*})^{s}$ given by an almost-free system of supports.

Proof. Let $g\colon B_{1}\times B_{2}\to A$ be the isomorphism of monoids.
Then $A=g(B_{1})\times g(B_{2})$. Since, by hypothesis, $g(B_{1})$ is not a
free monoid, the conclusion follows from Lemma 7.5 and Proposition 7.7.

##  8\. An interesting particular case: the stable category

Given a ring $R$ and an additive category $\mathcal{C}$ of $R$-modules, we
denote by $\underline{\mathcal{C}}$ the _stable_ category, which has the same
objects as $\mathcal{C}$ and with morphisms $\mbox{\rm
Hom}_{\underline{\mathcal{C}}}(M,N)=\mbox{\rm Hom}_{R}(M,N)/L$, where a
homomorphism $f\in\mbox{\rm Hom}_{R}(M,N)$ belongs to $L$ if and only if it
factors through a free $R$-module. Let $M$ be a finitely generated right
$R$-module. Since modules in $\mathrm{add}(M)$ are also in
$\mathrm{Add}(R\oplus M)$, there is a full embedding of categories
$\Psi:\mathrm{add}(M)\to\mathrm{Add}(R\oplus M)$.

The next proposition specializes the type of equivalence described in Theorem
2.11. Recall that these equivalences restrict well to countably generated
modules, but the behavior with countably generated modules is more subtle.
This is explained in statements $(c)$ and $(d)$.

######  Proposition 8.1.

Let $M$ be a finitely generated right $R$-module, with
$S=\mathrm{End}_{R}(M)$. With the notation introduced above:

  1. (a)

The embedding $\Psi:\mathrm{add}(M)\to\mathrm{Add}(R\oplus M)$ induces a full,
faithful functor
$\overline{\Psi}\colon\underline{\mathrm{add}}(M)\to\mathrm{Add}(R\oplus
M)/\mathcal{J}_{R}.$

  2. (b)

$\mathrm{Add}(R\oplus M)/\mathcal{J}_{R}$ is equivalent to the category
$\mathrm{Add}\left(S/\mathcal{I}_{R}(M,M)\right)$.

  3. (c)

$\mathrm{Add}_{\aleph_{0}}(R\oplus M)/\mathcal{J}_{R}$ is equivalent to the
category $\mathrm{Add}_{\aleph_{0}}\left(S/\mathcal{I}_{R}(M,M)\right)$.

  4. (d)

The category $\underline{\mathrm{add}}(M)$ is equivalent to a subcategory of
$\mathrm{add}\left(S/\mathcal{I}_{R}(M,M)\right)$.

  5. (e)

Let $X$ be an object in $\mathrm{add}(M)$, and suppose $\overline{P_{1}}$ and
$\overline{P_{2}}$ are objects in the category
$\mathrm{add}\left(S/\mathcal{I}_{R}(M,M)\right)$ such that

$\overline{P_{1}}\oplus\overline{P_{2}}\cong\mathrm{Hom}_{R}(M,X)/\mathrm{Hom}_{R}(M,X)\mathcal{I}_{R}(M,M)\,.$

Then there exist objects $Y_{1}$ and $Y_{2}$ in
$\mathrm{Add}_{\aleph_{0}}(R\oplus M)$ such that

$\mathrm{Hom}_{R}(R\oplus M,Y_{i})/\mathrm{Hom}_{R}(R\oplus
M,Y_{i})\mathcal{I}_{R}(M,M)\cong\overline{P_{i}}$

for $i=1,2$, and $Y_{1}\oplus Y_{2}\cong X\oplus R^{(\omega)}$ as right
$R$-modules.

Proof. $(a)$ Let $K$ and $L$ be modules in $\mathrm{add}(M)$. Since $K$ is
finitely generated, $\mathcal{J}_{R}(K,L)=\mathcal{I}_{R}(K,L)$. Therefore,
$\Psi$ induces the claimed full and faithful functor $\overline{\Psi}$.

To prove statement $(b)$, we will do some computations that will show that the
statement is a direct consequence of the equivalence described in Theorem
2.11(ii), applied to the module $R\oplus M$ in place of $M$, and with $X=R$.

First, we need a suitable description of $T:=\mathrm{End}_{R}(R\oplus M)$:

$\mathrm{End}_{R}(R\oplus
M)\cong\left(\begin{array}[]{cc}\mathrm{Hom}_{R}(R,R)&\mathrm{Hom}_{R}(M,R)\\\
\mathrm{Hom}_{R}(R,M)&\mathrm{End}_{R}(M)\end{array}\right)\cong\left(\begin{array}[]{cc}R&\mathrm{Hom}_{R}(M,R)\\\
M&S\end{array}\right)$

Moreover, we need a description of the trace ideal of the projective right
$T$\- module $\mathrm{Hom}_{R}(R\oplus M,R)$. To this aim, notice that

$T_{T}=\left(\begin{array}[]{cc}R&\mathrm{Hom}_{R}(M,R)\\\
0&0\end{array}\right)\oplus\left(\begin{array}[]{cc}0&0\\\
M&S\end{array}\right)=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}T\oplus\begin{pmatrix}0&0\\\ 0&1\end{pmatrix}T,$

where $\mathrm{Hom}_{R}(R\oplus M,R)\cong\begin{pmatrix}1&0\\\
0&0\end{pmatrix}T$ and $\mathrm{Hom}_{R}(R\oplus
M,M)\cong\begin{pmatrix}0&0\\\ 0&1\end{pmatrix}T$. Hence the trace ideal in
$T$ of $\mathrm{Hom}_{R}(R\oplus M,R)$ is given by the following:

$I=T\begin{pmatrix}1&0\\\
0&0\end{pmatrix}T=\left(\begin{array}[]{cc}R&\mathrm{Hom}_{R}(M,R)\\\
M&\mathrm{Hom}_{R}(R,M)\mathrm{Hom}_{R}(M,R)\end{array}\right)=\left(\begin{array}[]{cc}R&\mathrm{Hom}_{R}(M,R)\\\
M&\mathcal{I}_{R}(M,M)\end{array}\right)$

Comparing the expressions for $T$ and $I$,

$T/I=\begin{pmatrix}0&0\\\ 0&S/\mathcal{I}_{R}(M,M)\end{pmatrix}\cong
S/\mathcal{I}_{R}(M,M)\,.$

Now (b) follows immediately from Theorem 2.11(ii) if we replace $M$ with
$M\oplus R$, $S$ with $T$, and $X$ with $R$.

By Theorem 2.11(iii), the equivalence described in $(b)$ restricts well to
countably generated modules, and this proves $(c)$.

Now we prove statement $(d)$. From $(a)$ and $(b)$, we already know that
$\underline{\mathrm{add}}(M)$ is equivalent to a suitable subcategory of
$\mathrm{Add}\left(S/\mathcal{I}_{R}(M,M)\right)$. If $Y$ is an object of
$\mathrm{add}(M)$, then $\mathrm{Hom}_{R}(M,Y)$ is a finitely generated
projective $S$-module, so isomorphic to a direct summand of $S^{n}$ for some
finite $n$. Therefore, the image of $Y$ in the equivalence given by $(b)$ is a
direct summand of $\left(S/\mathcal{I}_{R}(M,M)\right)^{n}$, and hence an
object in $\mathrm{add}\left(S/\mathcal{I}_{R}(M,M)\right)$.

(e) As noticed above, $T/I\simeq S/\mathcal{I}_{R}(M,M)$; therefore
$\mathrm{add}(S/\mathcal{I}_{R}(M,M))$ is equivalent to $\mathrm{add}(T/I)$.
So we have a decomposition

$Q_{1}^{\prime}\oplus Q_{2}^{\prime}\simeq\mathrm{Hom}_{R}(R\oplus
M,X)/\mathrm{Hom}_{R}(R\oplus M,X)I\,,$

where
$Q_{1}^{\prime}\otimes_{T/I}S/\mathcal{I}_{R}(M,M)\simeq\overline{P_{1}}$ and
$Q_{2}^{\prime}\otimes_{T/I}S/\mathcal{I}_{R}(M,M)\simeq\overline{P_{2}}$.
Apply Proposition 2.9 to see that there are countably generated projective
$T$-modules $Q_{1},Q_{2}$ such that $Q_{1}/Q_{1}I\simeq Q_{1}^{\prime}$ and
$Q_{2}/Q_{2}I\simeq Q_{2}^{\prime}$. Let $Z_{1},Z_{2}$ be objects of
$\mathrm{Add}_{\aleph_{0}}\left(R\oplus M\right)$ such that
$Q_{1}\simeq\mathrm{Hom}_{R}(R\oplus M,Z_{1})$ and
$Q_{2}\simeq\mathrm{Hom}_{R}(R\oplus M,Z_{2})$, and define $Y_{1}=Z_{1}\oplus
R^{(\omega)}$, $Y_{2}=Z_{2}\oplus R^{(\omega)}$. By Proposition 3.4(i),
$Y_{1}\oplus Y_{2}\simeq X\oplus R^{(\omega)}$.

######  Remark 8.2.

Notice that, by the Eilenberg Swindle (Remark 2.5) and Lemma 2.7, the modules
$Y_{1}$ and $Y_{2}$ in Proposition 8.1(e) can be chosen so that $Y_{i}\oplus
R^{(\omega)}\cong Y_{i}$ for $i=1,2$.

Using Proposition 8.1 we will see that, if $M$ is indecomposable, the
idempotents of the endomorphism ring of $M$ in the stable category, namely
$S/\mathcal{I}_{R}(M,M)$, correspond to suitable countably generated direct
summands of $R^{(\omega)}\oplus M$. The following examples will illustrate
this phenomenon.

######  Example 8.3.

Let $R$ be a commutative local noetherian domain of Krull dimension 1 with
maximal ideal $\mathfrak{m}$ and residue field $k$. Let $M=\overline{R}$ be
the integral closure of $R$ in its field of fractions, and assume that
$\overline{R}$ is finitely generated over $R$. Let $c$ denote the conductor
ideal. Then

$T:=\mathrm{End}_{R}(R\oplus\overline{R})=\left(\begin{array}[]{cc}R&c\\\
\overline{R}&\overline{R}\end{array}\right)$

is a semilocal ring with $J(T)=\left(\begin{array}[]{cc}\mathfrak{m}&c\\\
\overline{R}&J(\overline{R})\end{array}\right)$ and with $T/J(T)\cong
k\times\overline{R}/J(\overline{R}).$ Assume that
$\overline{R}/J(\overline{R})\cong F_{1}\times\cdots\times F_{s}$, where
$F_{1},\dots,F_{s}$ are fields, and that $s\geq 2$. By the results in §4 there
is a monoid morphism $\mathbf{D}_{R\oplus\overline{R}}\colon
V^{*}(R\oplus\overline{R})\to{\mathbb{N}}_{0}^{*}\times({\mathbb{N}}_{0}^{*})^{s}$.
(Recall Notation 3.6 and Notation 4.3.)

Notice that $\mathbf{D}_{R\oplus\overline{R}}(R)=(1,0,\dots,0)$ and
$\mathbf{D}_{R\oplus\overline{R}}(\overline{R})=(0,1,\dots,1)$. If
$K\in\mathrm{add}(R\oplus\overline{R})$ then
$\mathbf{D}_{R\oplus\overline{R}}(K)=(m,n_{1},\dots,n_{s})\in{\mathbb{N}}_{0}\times({\mathbb{N}}_{0})^{s}$.
Since $(m,n_{1},\dots,n_{s})=m(1,0,\dots,0)+(0,n_{1},\dots,n_{s})$, it follows
that $K\cong R^{m}\oplus Q$ where $Q\in\mathrm{add}(_{R}\overline{R})$. Since
$\mathrm{End}_{R}(\overline{R})\cong\overline{R}$ is a semilocal commutative
domain, it follows that all projective modules over $\overline{R}$ are free
[16]. In particular, $Q\cong\overline{R}^{n}$ for some $n\in{\mathbb{N}}_{0}$.
Hence, $n_{1}=\cdots=n_{s}=n$. This shows that
$\mathrm{add}(R\oplus\overline{R})$ has only trivial objects and that
$\mathbf{D}_{R\oplus\overline{R}}(V(R\oplus\overline{R}))=(1,0,\dots,0){\mathbb{N}}_{0}+(0,1,\dots,1){\mathbb{N}}_{0}$.

Consider the finitely generated projective right $T$-module
$P=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}T\cong\mathrm{Hom}_{R}(R\oplus\overline{R},R)$. Its trace
ideal is $I=T\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}T=\begin{pmatrix}R&c\\\
\overline{R}&c\end{pmatrix}$. Then $T/I\cong\overline{R}/c$, which is an
artinian ring, and its reduction modulo its Jacobson radical is isomorphic to
$F_{1}\times\cdots\times F_{s}$. Hence, $\overline{R}/c$ is the product of $s$
artinian local rings.

By Proposition 8.1 (or Theorem 3.8), $\overline{R}/c$ is the endomorphism ring
of $M=\overline{R}$ in the stable category. By Proposition 8.1 (or Corollary
2.12), the indecomposable projective modules over $\overline{R}/c$ correspond
to direct summands of $R^{(\omega)}\oplus\overline{R}$. Therefore,
$\mathbf{D}_{R\oplus\overline{R}}(V^{*}(R\oplus\overline{R}))$ contains
$x_{1}:=(\infty,1,0,\dots,0),\dots,x_{s}:=(\infty,0,\dots,1)$. We claim that
$\mathbf{D}_{R\oplus\overline{R}}(V^{*}(R\oplus\overline{R}))=(1,0,\dots,0){\mathbb{N}}_{0}^{*}+(0,1,\dots,1){\mathbb{N}}_{0}^{*}+x_{1}{\mathbb{N}}_{0}^{*}+\cdots+x_{s}{\mathbb{N}}_{0}^{*}$.

Assume that
$x=(m,n_{1},\dots,n_{s})\in\mathbf{D}_{R\oplus\overline{R}}(V^{*}(R\oplus\overline{R}))\setminus\mathbf{D}_{R\oplus\overline{R}}(V(R\oplus\overline{R}))$.
If $m=\infty$, then

$x=(1,0,\dots,0)\cdot\infty+x_{1}n_{1}+\cdots+x_{s}n_{s}.$

If $m\in{\mathbb{N}}_{0}$ then $x=m(1,0,\dots,0)+(0,n_{1},\dots,n_{s})$, and
Lemma 4.2 implies that
$(0,n_{1},\dots,n_{s})\in\mathbf{D}_{R\oplus\overline{R}}(V^{*}(R\oplus\overline{R}))$.
Hence, there exists a countably generated but not finitely generated module
$Q\in\mathrm{Add}(\overline{R})$ such that
$\mathbf{D}_{R\oplus\overline{R}}([Q])=(0,n_{1},\dots,n_{s})$. But, as we
showed before, all objects in $\mathrm{Add}(\overline{R})$ are isomorphic to
$\overline{R}^{(C)}$ for some set $C$. Therefore
$(0,n_{1},\dots,n_{s})=(0,1,\dots,1)\cdot\infty$. This finishes the proof of
our claim.

It is not difficult to see that
$\mathbf{D}_{R\oplus\overline{R}}(V^{*}(R\oplus\overline{R}))$ is the set of
solutions $(x,y_{1},\dots,y_{s})$ in
${\mathbb{N}}_{0}^{*}\times({\mathbb{N}}_{0}^{*})^{s}$ of the system of $s-1$
equations: $x+y_{1}=x+y_{2},x+y_{2}=x+y_{3},\dots,x+y_{s-1}=x+y_{s}$.

Following the notation of Proposition 8.1 with $M=\overline{R}$, notice that,
in this case, $S/\mathcal{I}_{R}(M,M)\cong\overline{R}/c$. So
$\underline{\mathrm{add}}(M)$ is equivalent to a suitable subcategory of
$\mathrm{add}\left(S/\mathcal{I}_{R}(M,M)\right)$. Moreover, by Proposition
8.1, the non-free objects in the category
$\mathrm{add}\left(S/\mathcal{I}_{R}(M,M)\right)$ correspond to objects in
$\mathrm{add}(R^{(\omega)}\oplus\overline{R})$ that are not direct sums of
finitely generated modules.

If $X_{1},\dots,X_{s}$ are modules in
$\mathrm{add}(R^{(\omega)}\oplus\overline{R})$ such that
$\mathbf{D}_{R\oplus\overline{R}}([X_{i}])=x_{i}$ for $i=1,\dots,s$, then
$\mathbf{D}_{R\oplus\overline{R}}([R^{(\omega)}\oplus\overline{R}])=\sum_{i=1}^{s}x_{i}$.
This implies that $R^{(\omega)}\oplus\overline{R}\cong X_{1}\oplus\dots\oplus
X_{s}$. Notice also that, for each $i=1,\dots,s$, $X_{i}\cong X_{i}\oplus
R^{(\omega)}$.

######  Example 8.4.

Let $(R,\mathfrak{m})$ be a local Bass domain, that is, a one-dimensional
noetherian local domain whose integral closure $\overline{R}$ is two-generated
as an $R$-module. Then $R$ has finite Cohen-Macaulay type [29, Theorem 2.11],
that is, only finitely many isomorphism classes of indecomposable lattices
(finitely generated torsion-free $R$-modules). Moreover, each indecomposable
lattice is isomorphic to an ideal of $R$. (For the main properties of Bass
domains, we refer the reader to [21] and Bass’s “ubiquity” paper [4].) We will
assume that the maximal ideal $\mathfrak{m}$ splits in $\overline{R}$, and we
let $\overline{R}/J(R)=F_{1}\times F_{2}$, where the $F_{i}$ are fields.

Item (iii) of [21, Theorem 2.1] tells us that every non-zero ideal $I$ of $R$
is a principal ideal of its endomorphism ring $E(I):=\\{z\in\overline{R}\mid
zI\subseteq I\\}$. Therefore, if $I$ and $J$ are non-isomorphic ideals, the
rings $E(I)$ and $E(J)$ are not isomorphic as $R$-modules. Moreover, if $E(I)$
is properly contained in $\overline{R}$, then $E(I)$ is local (see [15,
Proposition 2.5]). In this way we get a finite list $R=R_{0},\dots,R_{\ell}$
of local intermediate rings, all properly contained in $\overline{R}$, such
that the $R$-modules $R_{0},\dots,R_{\ell}$ and $\overline{R}$ represent a
complete list of isomorphism classes of indecomposable lattices. The category
of $R$-lattices is then
$\mathrm{add}\big{(}(\oplus_{i=0}^{\ell}R_{i})\oplus\overline{R}\big{)}$. Set
$M=\big{(}\oplus_{i=0}^{\ell}R_{i}\big{)}\oplus\overline{R}$ and
$T=\mathrm{End}_{R}(M)$. Then $T$ is a semilocal ring, and $T/J(T)\cong
K_{0}\times\dots,\times K_{\ell}\times F_{1}\times F_{2}$, where
$K_{i}=R_{i}/J(R_{i})$ for $i=0,\dots,\ell$.

We have the usual injective monoid morphism $\mathbf{D}_{M}\colon
V^{*}(M)\to({\mathbb{N}}_{0}^{*})^{\ell+1}\times({\mathbb{N}}_{0}^{*})^{2}$
defined by $\mathbf{D}_{M}([N])=(n_{0},\dots,n_{\ell},m_{1},m_{2})$, where $N$
is a countably generated module in $\mathrm{Add}(M)$ such that
$\mathrm{Hom}_{R}(M,N)/\mathrm{Hom}_{R}(M,N)J(T)\cong\left(\oplus_{i=0}^{\ell}K_{i}^{(I_{i})}\right)\oplus(F_{1}^{(J_{1})}\times
F_{2}^{(J_{2})})$, and where $I_{i}=\omega_{0}$ if $n_{i}=\infty$ and
$I_{i}=n_{i}$ otherwise, and $J_{i}=\omega_{0}$ if $m_{i}=\infty$ and
$J_{i}=m_{i}$ otherwise.

Notice that $\mathbf{D}_{M}(V(M))$ contains the elements
$e_{i}:=(0,\dots,1^{(i},\dots,0,0,0)=\mathbf{dim}_{R}([R_{i}])$, for
$i=0,\dots,\ell$, and also contains
$v:=(0,\dots,0,1,1)=\mathbf{dim}_{R}([\overline{R}])$. As in Example 8.3, it
is easy to deduce that $\mathbf{D}_{M}(V(M))$ is the submonoid of
${\mathbb{N}}_{0}^{\ell+1}\times{\mathbb{N}}_{0}^{2}$ generated by
$e_{0},\dots,e_{\ell},v$.

It follows from Example 8.3 that, for $i=0,\dots,\ell$,
$\mathrm{Add}(R_{i}\oplus\overline{R})$ contains countably generated modules
$X_{i}$ and $Y_{i}$ such that

$\displaystyle\mathrm{Hom}_{R}(M,X_{i})/\mathrm{Hom}_{R}(M,X_{i})J(T)$
$\displaystyle\cong K_{i}^{(\omega_{0})}\oplus F_{1}\quad\text{and}$
$\displaystyle\mathrm{Hom}_{R}(M,Y_{i})/\mathrm{Hom}_{R}(M,Y_{i})J(T)$
$\displaystyle\cong K_{i}^{(\omega_{0})}\oplus F_{2}\,.$

Therefore $\mathbf{D}_{M}(V^{*}(M))$ contains
$a_{i}:=(0,\dots,\infty^{i)},\dots,0,1,0)=\mathbf{dim}_{R}([X_{i}])$ and
$b_{i}:=(0,\dots,\infty^{i)},\dots,0,0,1)=\mathbf{dim}_{R}([Y_{i}])$, for
$i=0,\dots,\ell$ for $i=1,\dots,\ell$.

Arguing as in Example 8.3, one can show that $\mathbf{D}_{M}(V^{*}(M))$ is the
submonoid of $({\mathbb{N}}_{0}^{*})^{\ell+1}\times({\mathbb{N}}_{0}^{*})^{2}$
generated by $e_{0},\dots,e_{\ell},v,a_{0},b_{0},\dots,a_{\ell},b_{\ell}$.
Notice that this submonoid is the set of solutions
$(x_{0},\dots,x_{\ell},y_{1},y_{2})$ in ${\mathbb{N}}_{0}^{*}$ of the equation
$x_{0}+\cdots+x_{\ell}+y_{1}=x_{0}+\cdots+x_{\ell}+y_{2}$.

Finally, we specialize to a very well-known example.

######  Example 8.5.

The ring $R=(\mathbb{C}[x,y])_{(x,y)}/(x^{2}-y^{3}-y^{2})$ is a particular
example of Examples 8.4 and 8.3. In this case $\overline{R}=R[\frac{x}{y}]$,
which has two maximal ideals. Observe that the maximal ideal of $R$ coincides
with the conductor ideal $c$.

The class of finitely generated torsion-free modules is
$\mathrm{add}\,(R\oplus\overline{R})$. Moreover,
$X\in\mathrm{add}\,(R\oplus\overline{R})$ if and only if $X\cong
R^{m}\oplus\overline{R}^{n}$ for some $m,n\ \in{\mathbb{N}}_{0}$.

However, and in view of Example 8.4,
$R^{(\omega)}\oplus\overline{R}=X_{1}\oplus X_{2}$ where neither $X_{1}$ nor
$X_{2}$ is a direct sum of finitely generated modules. A concrete construction
of $X_{1}$ and $X_{2}$ can be traced back using the results in [23] or in
[14]. In this case, $V^{*}(R\oplus\overline{R})$ is the submonoid of
$({\mathbb{N}}_{0}^{*})^{3}$ given by the set of solutions of the equation
$x_{0}+y_{1}=x_{0}+y_{2}$; then
$\mathbf{D}_{R\oplus\overline{R}}([X_{1}])=(\infty,1,0)$ and
$\mathbf{D}_{R\oplus\overline{R}}([X_{2}])=(\infty,0,1)$.

##  9\. Completions and pure-projective modules

In this section we show how nicely the completion fits into our setting by
proving that, over a commutative noetherian local ring, pure-projective
modules are isomorphic if and only if their extensions to the completion are
isomorphic.

We begin with a dreary paragraph where we review basic definitions and
connections among them. Let $R$ be a ring, not necessarily commutative. A
short exact sequence

$0\to A\overset{f}{\to}B\overset{g}{\to}C\to 0$ (5)

of right $R$-modules is pure provided $f\otimes 1_{X}:A\otimes_{R}X\to
B\otimes_{R}X$ is injective for every left $R$-module $X$. We also refer to an
injection $A\overset{f}{\hookrightarrow}B$ as a pure injection provided the
short exact sequence (5), with $C=\textrm{Coker}(f)$ is pure. Since every
module is a direct limit of finitely presented modules, it suffices to check
the criterion for every finitely presented module $X$. A consequence is that
the inclusion $R\hookrightarrow\widehat{R}$, where $\widehat{R}$ is the
$\mathfrak{m}$-adic completion of a Noetherian local ring $(R,\mathfrak{m})$,
is a pure ring extension (that is, the inclusion $R\hookrightarrow\widehat{R}$
is a pure injection as $R$-modules). An equivalent condition for the exact
sequence (5) to be pure is that $\mbox{\rm Hom}_{R}(E,B)\to\mbox{\rm
Hom}_{R}(E,C)$ be surjective for every finitely presented right $R$-module
$E$. (Actually this criterion is given as the definition of purity in [5,
§1.4], and [5, Theorem 1.27] shows that the criterion is equivalent to the
usual definition.) A right $R$-module $F$ is pure-projective provided
$\mbox{\rm Hom}_{R}(F,B)\to\mbox{\rm Hom}_{R}(F,C)$ is surjective for every
pure exact sequence (5). Since every module is a pure epimorphic image of a
direct sum of finitely presented modules, the usual argument characterizing
projective modules as direct summands of free modules shows that the pure
projectives are the direct summands of direct sums of finitely presented
modules.

By Kaplansky’s Theorem [17, Theorem 1], every pure-projective module is a
direct sum of countably generated modules. One of the main goals of our
investigation is to find examples of pure-projective modules, over commutative
Noetherian local rings, that are not direct sums of finitely presented modules
and, as indicated, in the title of the paper, we will do it by _comparing with
the completion_.

For further quoting we recall that over a complete noetherian ring all-pure
projective modules are direct sums of finitely generated modules with local
endomorphism ring, so that the structure of direct summands of copies of a
finitely generated modules is completely determined.

######  Theorem 9.1.

Let $\widehat{R}$ be a complete local noetherian ring. Then:

  * (i)

All finitely generated $\widehat{R}$-modules decompose as a direct sum of
(indecomposable) modules with local endomorphism ring. Moreover, such
decomposition is unique up to isomorphism.

  * (ii)

Let $\widehat{M}$ be a finitely generated $\widehat{R}$-module. Let
$M=L_{1}^{n_{1}}\oplus\cdots\oplus L_{s}^{n_{s}}$ where $L_{1},\dots,L_{s}$
are pairwise non-isomorphic indecomposable modules (hence, they have local
endomorphism ring) and $n_{i}\geq 1$ for $i\in\\{1,\dots,s\\}$. Then a module
$X$ is in $\mathrm{Add}(\widehat{M})$ if and only if $M\cong
L_{1}^{(I_{1})}\oplus\cdots\oplus L_{s}^{(I_{s})}$ where $I_{1},\dots,I_{s}$
are suitable sets. In this situation, the cardinality of $I_{1},\dots,I_{s}$
is uniquely determined.

  * (iii)

Any pure-projective $\widehat{R}$-module decomposes as a direct sum of
finitely generated modules with local endomorphism ring. Moreover, such
decomposition is unique up to isomorphism.

Proof. [5, Theorem 2.8 and Corollary 2.53]

Our first step (Proposition 9.6) is to prove descent of pure-projectivity for
pure extensions of commutative rings. The proof is modeled after the _proof_
by Raynaud and Gruson [26] of descent of projectivity.

######  Remark 9.2.

Raynaud and Gruson stated the descent of projectivity via pure morphisms.
However, as noted by Gruson in the paper [7], statement [26, Proposition
II.2.5.2] is wrong. The descent of projectivity via pure monomorphisms is
stated in [26, Examples II.3.1.4] that are presented as a consequence of the
wrong statement, and no correction for that is given in [7]. However, to
conclude the descent of projectivity via pure monomorphisms only [26,
Proposition II.2.5.1, Théorème II.3.1.3] are needed, and these results are
perfectly correct in the original paper. To clarify the matter, the reader may
want to have a look at [11].

In view of Remark 9.2, we have decided to include the complete proof of the
descent of pure-projectivity via pure morphisms. A crucial ingredient is the
notion of Mittag-Leffler module. Recall that a right module $M$ over an
arbitrary ring $R$ is Mittag-Leffler provided the canonical map

$M\otimes_{R}\prod_{i\in I}Y_{i}\to\prod_{i\in I}(M\otimes_{R}Y_{i})$ (6)

is injective for every family $\\{Y_{i}\\}_{i\in I}$ of left $R$-modules. One
of the properties of Mittag-Leffler modules that is crucial to us is that
countably generated Mittag-Leffler modules are pure-projective. (See [26,
Corollaire 2.2.2 p. 74]).

######  Remark 9.3.

The map in (6) is obviously bijective if $M$ is a finitely generated free
module, and then an easy diagram chase shows that it is bijective if $M$ is
finitely presented. Thus, finitely presented modules are Mittag-Leffler. Since
the class of Mittag-Leffler modules is closed under direct summands and
arbitrary direct sums, all pure-projective modules are Mittag-Leffler modules.

A straightforward diagram chase shows that if $M_{R}$ is Mittag-Leffler and
$R\to S$ is any ring homomorphism, then $M\otimes_{R}S$ is Mittag-Leffler as
an $S$-module. The converse holds if $R\hookrightarrow S$ is a pure extension
of commutative rings:

######  Lemma 9.4.

[26, Proposition 2.5.1 p. 78] Let $R\subseteq S$ be a pure extension of
commutative rings. Let $M$ be an $R$-module. Then $M_{R}$ is Mittag-Leffler as
$R$-module if and only if $M\otimes_{R}S$ is Mittag-Leffler as $S$-module.

######  Lemma 9.5.

[26, Théorème 2.2.1 p. 73] Any countably generated submodule $X$ of a Mittag-
Leffler module $Y$ is contained in a countably generated pure submodule
$Y^{\prime}$ of $Y$. In addition, as $Y^{\prime}$ is Mittag-Leffler it is also
pure-projective.

######  Proposition 9.6.

_(A variation of[26, Théorème 3.1.3 p. 78])_ Let $R\subseteq S$ be a pure
extension of commutative rings. Let $M$ be an $R$-module. Then $M_{R}$ is pure
projective if and only if $M\otimes_{R}S$ is pure projective as an $S$-module.

Proof. If $M_{R}$ is pure projective, then clearly $M\otimes_{R}S$ is pure
projective as an $S$-module. For the converse, assume that $M\otimes_{R}S$ is
a pure projective $S$-module. By Remark 9.3 $M\otimes_{R}S$ is a Mittag-
Leffler $S$-module, and then Lemma 9.4 implies that $M$ is a Mittag-Leffler
module $R$-module. We need to prove that, in addition, $M_{R}$ is pure-
projective.

Write $M\otimes_{R}S=\oplus_{i\in I}Q_{i}$ with $\mathcal{F}=\\{Q_{i}\\}_{i\in
I}$, where each $Q_{i}$ is a countably generated $S$-module. Let
$\mathcal{F}=\\{Q_{i}\\}_{i\in I}$. A submodule $X$ of $M$ is said to be
_adapted_ (to $\mathcal{F}$) if it is pure, and the canonical image of
$X\otimes_{R}S$ in $M\otimes_{R}S$ is a direct sum of modules in
$\mathcal{F}$. If $X$ is an adapted submodule of $M$, then the sequence

$0\to X\otimes_{R}S\to M\otimes_{R}S\to(M/X)\otimes_{R}S\to 0$

is split exact. Therefore $X\otimes_{R}S$ and $(M/X)\otimes_{R}S$, being
isomorphic to direct summands of $M\otimes_{R}S$, are pure-projective as
$S$-modules.

Step 1. _Every countably generated pure submodule of $M$ is contained in a
countably generated adapted submodule of $M$. _

Let $X$ be a countably generated pure submodule of $M$. As $M$ is Mittag-
Leffler and the modules $Q_{i}$ are countably generated, we can construct a
sequence $(X_{n},I_{n})_{n\in{\mathbb{N}}_{0}}$ such that

  * (1)

$X_{0}=X$ and, for every $n\geq 0$, $X_{n}$ is a countably generated pure
submodule of $M$ and $X_{n}\subseteq X_{n+1}$;

  * (2)

for any $n\geq 0$, $I_{n}$ is a countable subset of $I$ and it consists of the
elements $i\in I$ such that the canonical projection of $X_{n}\otimes_{R}S$ in
$Q_{i}$ is different from zero;

  * (3)

for any $n\geq 0$, the image of $X_{n+1}\otimes_{R}S$ contains $\oplus_{i\in
I_{n}}Q_{i}$.

To be more specific, suppose $X_{n},I_{n}$ were defined. Since each $Q_{i}$ is
countably generated, there exists a countable set $G_{n}\subseteq M$ such that
the canonical image of $G_{n}R\otimes_{R}S$ in $M\otimes_{R}S$ contains
$\oplus_{i\in I_{n}}Q_{i}$. By Lemma 9.5, there exists a countably generated
$X_{n+1}$ which is a pure submodule of $M$ containing $X_{n}+G_{n}R$. Then
$I_{n+1}$ is chosen as described in (2).

Set $Y=\bigcup_{n\in{\mathbb{N}}_{0}}X_{n}$. By construction, $Y$ is an
adapted submodule of $M$.

Step 2. _Let $X$ be an arbitrary adapted submodule of $M$ such that $X\neq M$.
Then there exists an adapted submodule $X^{\prime}$ of $M$ such that $X\subset
X^{\prime}$ and $X^{\prime}/X$ is a countably generated adapted submodule of
$M/X$. Hence, $X^{\prime}/X$ is pure-projective; therefore the pure exact
sequence_

$0\to X\to X^{\prime}\to X^{\prime}/X\to 0$

_splits._

By definition, if $X$ is an adapted submodule of $M$ then
$(M/X)\otimes_{R}S\cong\oplus_{i\in I^{\prime}}Q_{i}$ for a certain
$I^{\prime}\subseteq I$. Hence, it makes sense to talk about adapted
submodules of $M/X$ with respect to the decomposition induced by
$\mathcal{F}^{\prime}=\\{Q_{i}\\}_{i\in I^{\prime}}$. By Step 1, there exists
a submodule $X^{\prime}$ of $M$ containing $X$ and such that $0\neq
X^{\prime}/X$ is a countably generated adapted submodule of $M/X$. Therefore,
$X^{\prime}$ is also an adapted submodule of $M$. Since $X$ is an adapted
submodule of $X^{\prime}$ the rest of the statement is clear.

Finally, combining the first and the second steps, we deduce that there exists
an ordinal $\kappa$ and a continuous chain $\\{X_{\alpha}\\}_{\alpha<\kappa}$
of adapted submodules of $M$ such that

  * (i)

$X_{0}=0$, and

  * (ii)

for any $\alpha+1<\kappa$, $X_{\alpha+1}/X_{\alpha}$ is pure projective and a
direct summand of $X_{\alpha+1}$.

In this situation $M\cong\oplus(X_{\alpha+1}/X_{\alpha})$ (cf. [26, Lemme
3.1.2, p. 81]). Therefore, $M$ is pure projective.

Let $R$ be a commutative local noetherian ring with maximal ideal
$\mathfrak{m}$. Let $\widehat{R}$ denote its $\mathfrak{m}$-adic completion;
then the maximal ideal of $\widehat{R}$ is $\widehat{\mathfrak{m}}$. If $M,N$
are finitely generated $R$-modules and $f\colon M\to N$ is an
$R$-homomorphism, we define $\widehat{M}:=M\otimes_{R}\widehat{R}$,
$\widehat{N}:=N\otimes_{R}\widehat{R}$ and
$\widehat{f}:=f\otimes_{R}\widehat{R}\colon\widehat{M}\to\widehat{N}$.

Since the extension $R\to\widehat{R}$ is pure, Proposition 9.6 yields the
following:

######  Corollary 9.7.

Let $R$ be a commutative local noetherian ring, and let $P_{R}$ be an
$R$-module. Then $P$ is pure-projective if and only if
$P\otimes_{R}\widehat{R}$ is pure-projective.

Now we turn to the problem of descending isomorphism from the completion to
the ring. The next lemma is crucial for that.

######  Lemma 9.8.

Let $R$ be a commutative local noetherian ring with maximal ideal
$\mathfrak{m}$. Let $M$ and $N$ be finitely generated $R$-modules. Let
$h\colon\widehat{M}\to\widehat{N}$ be an $\widehat{R}$-homomorphism. Then
there exists a homomorphism of $R$-modules $f\colon M\to N$ such that
$(h-\widehat{f})\widehat{M}\subseteq\widehat{N}\widehat{\mathfrak{m}}$.

Proof. A way to derive this result is to use the lifting number of the pair of
modules $M$ and $N$ (cf. [19, Chap. 1, §3]). Let $e=e(M,N)$ be such a number.
Then $h$ induces a homomorphism

$\overline{h}\colon\widehat{M}/\widehat{M}\widehat{\mathfrak{m}}^{e+1}=M/M\mathfrak{m}^{e+1}\to
N/N\mathfrak{m}^{e+1}=\widehat{N}/\widehat{N}\widehat{{\mathfrak{m}}}^{e+1}.$

By the definition of lifting number, there exists $f\colon M\to N$ such that
the induced morphism $\overline{f}\colon M/M\mathfrak{m}\to N/N\mathfrak{m}$
satisfies that
$\overline{f}(m+M\mathfrak{m})=\overline{h}(m+M\mathfrak{m}^{e+1})+N\mathfrak{m}$.
Hence, $\widehat{f}$ satisfies the desired property.

######  Lemma 9.9.

Let $R$ be a commutative local noetherian ring with maximal ideal
$\mathfrak{m}$. Suppose that $P$ and $Q$ are pure-projective modules over $R$
such that $P\otimes_{R}\widehat{R}\cong Q\otimes_{R}\widehat{R}$. Then there
are homomorphisms $f\colon P\to Q$ and $g\colon Q\to P$ such that
$(1_{P}-gf)\otimes 1_{R/\mathfrak{m}}=0$ and $(1_{Q}-fg)\otimes
1_{R/\mathfrak{m}}=0$.

Proof. Suppose that $\overline{h}\colon\oplus_{i\in
I}\widehat{M_{i}}\to\oplus_{j\in J}\widehat{N_{j}}$ is a homomorphism of
$\widehat{R}$-modules, where $M_{i},N_{j}$ are finitely generated $R$-modules.
We claim that there exists an $R$-homomorphism $h\colon\oplus_{i\in
I}M_{i}\to\oplus_{j\in J}N_{j}$ such that $((h\otimes
1_{\widehat{R}})-\overline{h})\otimes 1_{R/\mathfrak{m}}=0$.

In order to prove the claim, display $\overline{h}$ as a column-finite matrix
$(\overline{h}_{j,i})_{j\in J,i\in I}$ where
$\overline{h}_{j,i}\colon\widehat{M_{i}}\to\widehat{N_{j}}$. By Lemma 9.8, for
every $j\in J,i\in I$ there exists $h_{j,i}\colon M_{i}\to N_{j}$ such that
the image of $\widehat{h}_{j,i}-\overline{h}_{j,i}$ is contained in
$\widehat{N}_{j}\mathfrak{m}$; moreover, if $\overline{h}_{j,i}=0$ we take
$h_{j,i}=0$. Then the matrix $(h_{j,i})_{j\in J,i\in I}$ is column-finite, and
it defines an $R$-homomorphism $h\colon\oplus_{i\in I}M_{i}\to\oplus_{j\in
J}N_{j}$. Using the fact that the tensor product commutes with direct sums, we
see that ${\rm Im}\ ((h\otimes
1_{\widehat{R}})-\overline{h})\subseteq\left(\oplus_{i\in
J}\widehat{N}_{j}\right)\mathfrak{m}$. Therefore $((h\otimes
1_{\widehat{R}})-\overline{h})\otimes 1_{R/\mathfrak{m}}=0$. This completes
the proof of the claim.

Now suppose that there are finitely generated $R$-modules $M_{i},i\in I$ and
$N_{j},j\in J$ such that $P$ is a direct summand of $\oplus_{i\in I}M_{i}$ and
$Q$ is a direct summand of $\oplus_{j\in J}N_{j}$. Let $\overline{f}\colon
P\otimes_{R}\widehat{R}\to Q\otimes_{R}\widehat{R}$ and $\overline{g}\colon
Q\otimes_{R}\widehat{R}\to P\otimes_{R}\widehat{R}$ be mutually inverse
isomorphisms. Let $\iota_{P}\colon P\to\oplus_{i\in I}M_{i}$,
$\pi_{P}\colon\oplus_{i\in I}M_{i}\to P$, $\iota_{Q}\colon Q\to\oplus_{j\in
J}N_{j}$ and $\pi_{Q}\colon\oplus_{j\in J}N_{j}\to Q$ be such that
$\pi_{P}\iota_{P}=1_{P}$ and $\pi_{Q}\iota_{Q}=1_{Q}$. Using the claim above,
we get homomorphisms $f_{0}\colon\oplus_{i\in I}M_{i}\to\oplus_{j\in J}N_{j}$
and $g_{0}\colon\oplus_{j\in J}N_{j}\to\oplus_{i\in I}M_{i}$ such that

$\displaystyle f_{0}\otimes 1_{\widehat{R}}\otimes 1_{R/\mathfrak{m}}$
$\displaystyle=({\iota}_{Q}\otimes
1_{\widehat{R}})\overline{f}({\pi}_{P}\otimes 1_{\widehat{R}})\otimes
1_{R/\mathfrak{m}}\quad\text{and}$ $\displaystyle g_{0}\otimes
1_{\widehat{R}}\otimes 1_{R/\mathfrak{m}}$ $\displaystyle=({\iota}_{P}\otimes
1_{\widehat{R}})\overline{g}({\pi}_{Q}\otimes 1_{\widehat{R}})\otimes
1_{R/\mathfrak{m}}\,.$

Put $f=\pi_{Q}f_{0}\iota_{P}$ and $g=\pi_{P}g_{0}\iota_{Q}$. Then, writing
$\tilde{\alpha}$ or $\alpha^{\sim}$ for the reduction of an $R$-homomorphism
$\alpha$ modulo $\mathfrak{m}$,

$\displaystyle(1_{P}-gf)^{\sim}=\tilde{1_{P}}-\tilde{\pi_{P}}\ \tilde{g_{0}}\
\tilde{\iota_{Q}}\ \tilde{\pi_{Q}}\ \tilde{f_{0}}\ \tilde{i_{P}}$
$\displaystyle=$ $\displaystyle\tilde{1_{P}}\ -\tilde{\pi_{P}}\
\tilde{\iota_{P}}\ \tilde{\overline{g}}\ \tilde{\pi_{Q}}\ \tilde{\iota_{Q}}\
\tilde{\pi_{Q}}\ \tilde{\iota_{Q}}\tilde{\overline{f}}\ \tilde{\pi_{P}}\
\tilde{\iota_{P}}$ $\displaystyle=\tilde{1_{P}}-\tilde{\overline{g}}\
\tilde{\overline{f}}=0\,.$

Thus $(1_{P}-gf)\otimes 1_{R/\mathfrak{m}}=0$ and, by symmetry,
$(1_{Q}-fg)\otimes 1_{R/\mathfrak{m}}=0$.

######  Lemma 9.10.

Let $R$ be a commutative local noetherian ring with maximal ideal
${\mathfrak{m}}$. Let $P,Q$ be pure-projective $R$-modules, and let $f\colon
P\to Q$ and $g\colon Q\to P$ be homomorphisms such that
$\mathrm{Im}(1_{P}-gf)\subseteq P{\mathfrak{m}}$,
$\mathrm{Im}(1_{Q}-fg)\subseteq Q{\mathfrak{m}}$. Then, for every finite set
$X\subseteq P$, there exists a homomorphism $g_{1}\colon Q\to P$ such that
$\mathrm{Im}(1_{P}-g_{1}f)\subseteq P{\mathfrak{m}}$,
$\mathrm{Im}(1_{Q}-fg_{1})\subseteq Q{\mathfrak{m}}$ and $g_{1}f(x)=x$ for
every $x\in X$.

Proof. By Kaplansky’s theorem [17, Theorem 1], we can assume that $P$ and $Q$
are countably generated modules. Then there are finitely generated $R$-modules
$M_{1},M_{2},\dots$ and $N_{1},N_{2},\dots$ such that $P$ is a direct summand
of $\oplus_{i=1}^{\infty}M_{i}$ and $Q$ is a direct summand of
$\oplus_{i=1}^{\infty}N_{i}$, say

$P\oplus P^{\prime}=\bigoplus_{i=1}^{\infty}M_{i}\quad\text{and}\quad Q\oplus
Q^{\prime}=\bigoplus_{i=1}^{\infty}N_{i}\,.$ (7)

If we put $K=\oplus_{i=1}^{\infty}(M_{i}\oplus N_{i})$, a variant of the
Eilenberg Swindle (Remark 2.5(1)) gives $P^{\prime}\oplus K^{(\omega)}\simeq
K^{(\omega)}$ and $Q^{\prime}\oplus K^{(\omega)}\simeq K^{(\omega)}$. Now
$P\oplus K^{(\omega)}\cong(\oplus_{i=1}^{\infty}M_{i})\oplus K^{(\omega)}$.
The right-hand side is still a countable direct sum of finitely generated
modules, which we simply rearrange and rename as $\oplus_{i=1}^{\infty}M_{i}$.
After another similar change of notation,

$P\oplus K^{(\omega)}\cong\bigoplus_{i=1}^{\infty}M_{i}\quad\text{and}\quad
Q\oplus K^{(\omega)}\cong\bigoplus_{i=1}^{\infty}N_{i}\,.$

Therefore, in the decompositions (7), we can assume that $P^{\prime}\cong
Q^{\prime}$. Choose mutually inverse isomorphisms $f^{\prime}\colon
P^{\prime}\to Q^{\prime}$ and $g^{\prime}\colon Q^{\prime}\to P^{\prime}$. Set

$\varphi:=\begin{pmatrix}f&0\\\
0&f^{\prime}\end{pmatrix}\colon\oplus_{i=1}^{\infty}M_{i}\to\oplus_{i=1}^{\infty}N_{i}\quad\text{and}\quad\psi:=\begin{pmatrix}g&0\\\
0&g^{\prime}\end{pmatrix}\colon\oplus_{i=1}^{\infty}N_{i}\to\oplus_{i=1}^{\infty}M_{i}\,.$

Observe that $\mathrm{Im}\ (1_{P\oplus P^{\prime}}-\psi\varphi)\subseteq
P{\mathfrak{m}}$ and $\mathrm{Im}(1_{Q\oplus Q^{\prime}}-\varphi\psi)\subseteq
Q{\mathfrak{m}}$. Given a finite set $X\subseteq P$, there exists
$m\in\mathbb{N}$ such that $X$ and $gf(X)$ are subsets of
$\oplus_{i=1}^{m}M_{i}$. Let
$\pi\colon\oplus_{i=1}^{\infty}M_{i}\to\oplus_{i=1}^{m}M_{i}$ be the canonical
projection, and let
$\iota\colon\oplus_{i=1}^{m}M_{i}\to\oplus_{i=1}^{\infty}M_{i}$ be the
canonical embedding. Put $h=\pi(1_{P\oplus
P^{\prime}}-\psi\varphi)\iota:\oplus_{i=1}^{m}M_{i}\to\oplus_{i=1}^{m}M_{i}$.
Observe that $\mathrm{Im}\
h\subseteq\left(\oplus_{i=1}^{m}M_{i}\right){\mathfrak{m}}$ and, by Nakayama’s
Lemma, $1-h$ is an automorphism of $\oplus_{i=1}^{m}M_{i}$. Moreover, it is
easy to see that $(1-h)^{-1}$ is of the form $1-h^{\prime}$, where
$h^{\prime}:\oplus_{i=1}^{m}M_{i}\to\oplus_{i=1}^{m}M_{i}$ and $\mathrm{Im}\
h^{\prime}\subseteq\left(\oplus_{i=1}^{m}M_{i}\right){\mathfrak{m}}$. Put

$\tau:=1_{P\oplus P^{\prime}}-\left(\begin{array}[]{cl}h^{\prime}&0\\\
0&0_{\oplus_{i=m+1}^{\infty}M_{i}}\end{array}\right)\colon\oplus_{i=1}^{\infty}M_{i}\to\oplus_{i=1}^{\infty}M_{i}\,.$

We claim that $g_{1}=\pi_{P}\tau\iota_{P}g$ satisfies the properties claimed
in the statement, where $\pi_{P}:\oplus_{i=1}^{\infty}M_{i}\to P$ and
$\iota_{P}:P\to\oplus_{i=1}^{\infty}M_{i}$ are the canonical projection and
injection relative to the first decomposition in (7). The properties
$\mathrm{Im}(1_{P}-g_{1}f)\subseteq P{\mathfrak{m}}$ and
$\mathrm{Im}(1_{Q}-fg_{1})\subseteq Q{\mathfrak{m}}$ follow from the equality
$\tau\otimes 1_{R/\mathfrak{m}}=1_{P\oplus
P^{\prime}}\otimes_{R}1_{R/\mathfrak{m}}$. Let $x\in X$. Then
$g_{1}f(x)=\pi_{P}\tau\iota_{P}gf(x)=\pi_{P}\tau\psi\varphi\iota_{P}(x)=\pi_{P}\tau\iota(1-h)\pi\iota_{P}(x)=\pi_{P}\iota\pi\iota_{P}(x)$.
Now, as $x\in\oplus_{i=1}^{m}M_{i}$, we have
$\iota\pi\iota_{P}(x)=\iota_{P}(x)$, and hence
$g_{1}f(x)=\pi_{P}\iota_{P}(x)=x$.

######  Theorem 9.11.

Let $P,Q$ be pure-projective modules over a commutative local noetherian ring
$R$. Then, $P\simeq Q$ if and only if $P\otimes_{R}\widehat{R}\simeq
Q\otimes_{R}\widehat{R}$.

Proof. We proceed as in [22]. Since pure-projective modules are direct sums of
countably generated modules, we only need to prove the statement when $P$ and
$Q$ are countably generated.

Suppose that $X_{1}^{\prime},X_{2}^{\prime},\cdots\subseteq P$ are finite sets
such that $P$ is generated by $\bigcup_{i\in\mathbb{N}}X_{i}^{\prime}$ and
that $Y_{1}^{\prime},Y_{2}^{\prime},\cdots\subseteq Q$ are finite sets such
that $Q$ is generated by $\bigcup_{i\in\mathbb{N}}Y_{i}^{\prime}$. By
induction, we construct chains of submodules $X_{1}\subseteq
X_{2}\subseteq\cdots\subseteq P$ and $Y_{1}\subseteq
Y_{2}\subseteq\cdots\subseteq Q$, along with homomorphisms
$f_{1},f_{2},\dots\colon P\to Q$ and $g_{1},g_{2},\dots\colon Q\to P$ such
that

  1. (i)

each $X_{i}$ is finitely generated, and $P=\cup_{i\in\mathbb{N}}X_{i}$

  2. (ii)

each $Y_{i}$ is finitely generated, and $Q=\cup_{i\in\mathbb{N}}Y_{i}$

  3. (iii)

$g_{i}f_{i}(x)=x$ for every $x\in X_{i}$, and $f_{i+1}g_{i}(y)=y$ for every
$y\in Y_{i}$

  4. (iv)

$(1_{P}-g_{i}f_{i})\otimes 1_{R/\mathfrak{m}}$, $(1_{Q}-f_{i}g_{i})\otimes
1_{R/\mathfrak{m}}$, $(1_{P}-g_{i}f_{i+1})\otimes 1_{R/\mathfrak{m}}$, and
$(1_{Q}-f_{i+1}g_{i})\otimes 1_{R/\mathfrak{m}}$ are all equal to zero.

First, we set $X_{1}$ to be the submodule of $P$ generated by
$X_{1}^{\prime}$. Combining Lemma 9.9 and Lemma 9.10, we obtain $f_{1}\colon
P\to Q$ and $g_{1}\colon Q\to P$ such that
$(1_{P}-g_{1}f_{1})\otimes_{R}R/\mathfrak{m}=0$,
$(1_{Q}-f_{1}g_{1})\otimes_{R}R/\mathfrak{m}=0$ and $g_{1}f_{1}(x)=x$ for
every $x\in X_{1}.$

Suppose that $X_{1},X_{2},\dots,X_{k}$, $Y_{1},Y_{2},\dots,Y_{k-1}$ and
$f_{1},f_{2},\dots,f_{k},g_{1},g_{2},\dots,g_{k}$ have been defined. Set
$Y_{k}$ to be the submodule of $Q$ generated by $f_{k}(X_{k})\cup
Y_{k}^{\prime}$. Using Lemma 9.9 and Lemma 9.10, we find $f_{k+1}$ such that
$(1_{P}-g_{k}f_{k+1})\otimes_{R}R/\mathfrak{m}=0$,
$(1_{Q}-f_{k+1}g_{k})\otimes_{R}R/\mathfrak{m}=0$ and $f_{k+1}g_{k}(y)=y$ for
every $y\in Y_{k}$.

Suppose that $X_{1},X_{2},\dots,X_{k}$, $Y_{1},Y_{2},\dots,Y_{k}$ and
$f_{1},f_{2},\dots,f_{k},f_{k+1},g_{1},g_{2},\dots,g_{k}$ have been defined.
We define $X_{k+1}$ to be the submodule of $P$ generated by
$X_{k+1}^{\prime}\cup g_{k}(Y_{k})$. Using Lemma 9.9 and Lemma 9.10, we find
$g_{k+1}\colon Q\to P$ such that
$(1_{P}-g_{k}f_{k+1})\otimes_{R}R/\mathfrak{m}=0$,
$(1_{Q}-f_{k+1}g_{k})\otimes_{R}R/\mathfrak{m}=0$, and $g_{k+1}f_{k+1}(x)=x$
for every $x\in X_{k+1}$

Observe that conditions (i)-(iv) are satisfied. Moreover,
$f_{k}(X_{k})\subseteq Y_{k}$ and $g_{k}(Y_{k})\subseteq X_{k+1}$. Therefore,
if $l>1$ and $x\in X_{l-1}$, then
$f_{l}(x)=f_{l}(g_{l-1}f_{l-1}(x))=f_{l-1}(x)$. Similarly, if $l>1$ and $y\in
Y_{l-1}$ then $g_{l}(y)=g_{l}(f_{l}g_{l-1}(y))=g_{l-1}(y)$. So we can define
$f\colon P\to Q$ and $g\colon Q\to P$ by the rule $f(x)=f_{l}(x)$ if $x\in
X_{l}$ and $g(y)=g_{l}(y)$ if $y\in Y_{l}$. By construction, $f$ and $g$ are
mutually inverse isomorphisms.

######  Corollary 9.12.

[26, Examples II.3.1.4] Let $R$ be a local commutative noetherian ring and $M$
an $R$-module. If $M\otimes_{R}\widehat{R}$ is a projective
$\widehat{R}$-module, then $M$ is projective over $R$.

Proof. Apply Proposition 9.6 to see that $M$ is pure projective. Since
$M\otimes_{R}\widehat{R}\simeq R^{(\kappa)}\otimes_{R}\widehat{R}$ for some
$\kappa$, apply Theorem 9.11 for pure projective $R$-modules $M$ and
$R^{(\kappa)}$ to conclude $M\simeq R^{(\kappa)}\,.$

If $P$ and $Q$ are finitely generated modules over a commutative local
noetherian ring, then $P$ is isomorphic to a direct summand of $Q$ if and only
if $\widehat{P}$ is isomorphic to direct summand of $\widehat{Q}$. (See [19,
Corollary 1.16].) The next example shows that this relationship can fail for
pure-projective modules.

######  Example 9.13.

Let $R$ be the ring constructed in [30, (2.3)], with $s=2$ and with
$\mathbf{F}$ any infinite field. This is an analytically unramified local
noetherian domain of Krull dimension 1 whose completion $\widehat{R}$ has
exactly two minimal prime ideals $P_{1}$ and $P_{2}$. Moreover,
$\widehat{R}/P_{1}$ and $\widehat{R}/P_{2}$ both have infinite Cohen-Macaulay
type. (This follows from [30, (2.4)] or from the fact that each ring
$\widehat{R}$ has multiplicity $4$ [29]. Now a result in Silvia Saccon’s Ph.D.
thesis [27, Theorem 3.4.1] (cf. [2, Theorem 2.1]) guarantees that for each
pair $(r_{1},r_{2})$ of non-negative integers, not both zero, there are
infinitely many pairwise non-isomorphic indecomposable torsion-free
$\widehat{R}$-modules $M$ such that
$M\otimes_{\widehat{R}}\widehat{R}_{P_{i}}\cong(\widehat{R}_{P_{i}})^{(r_{i})}$
for $i=1,2$. In particular, $\widehat{R}$ has two infinite families
$\\{A_{i}\\}_{i\geq 1}$ and $\\{B_{i}\\}_{i\geq 1}$ of pairwise non-isomorphic
indecomposable finitely generated torsion-free modules such that
$A_{i}\otimes_{\widehat{R}}(\widehat{R}_{P_{1}}\times\widehat{R}_{P_{2}})\cong\widehat{R}_{P_{1}}$
and
$B_{i}\otimes_{\widehat{R}}(\widehat{R}_{P_{1}}\times\widehat{R}_{P_{2}})\cong\widehat{R}_{P_{2}}$.

By the Levy-Odenthal criteria [19, Corollary 2.8], it follows that none of the
modules $A_{i}$ or $B_{i}$ is extended from a finitely generated torsion-free
$R$-module, while the direct sums $A_{i}\oplus B_{j}$ are. Hence, there exist
$R$-modules $P$ and $Q$ that are direct sums of finitely generated torsion-
free modules, such that $\widehat{P}\cong\bigoplus_{i\geq 1}(A_{i}\oplus
B_{i+1})$ and $\widehat{Q}\cong\bigoplus_{i\geq 1}(A_{i}\oplus B_{i})$. Hence
$\widehat{P}\oplus B_{1}\cong\widehat{Q}$.

However, if $P\oplus L\cong Q$ then, by the Krull-Remak-Schmidt-Azumaya
Theorem [5, Theorem 2.12], $\widehat{L}\cong A_{1}$. Hence, by Lemma 10.2,
$A_{1}$ would be extended from a finitely generated torsion-free $R$-module,
but as we have remarked before this is not possible. Hence, $P$ is not a
direct summand of $Q$.

##  10\. Extended idempotent ideals

In this section, we study the behavior of idempotent ideals of
$S:=\mathrm{End}_{R}(M)$ with respect to localization and completion. If $R\to
T$ is a flat ring homomorphism and $I$ is an ideal of $S$ then the canonical
map $I\otimes_{R}T\to S\otimes_{R}T$ is a monomorphism and its image is an
ideal of $S\otimes_{R}T$. Hence, we consider $I\otimes_{R}T$ as an ideal of
$S\otimes_{R}T$.

######  Lemma 10.1.

Let $R$ be a commutative ring. Let $S$ be an $R$-algebra, and let $R\to T$ be
a flat ring homomorphism

  1. (i)

Let $P_{S}$ be a countably generated projective right $S$-module with trace
ideal $I$. Then $P\otimes_{R}T$ is a countably generated projective right
module over $S\otimes_{R}T$ with trace ideal $I\otimes_{R}T$.

  2. (ii)

Assume that $T$ is a commutative ring. Let $M_{S}$ be a finitely presented
right $S$-module. Then the trace ideal of $M\otimes_{R}T$ is
$\mathrm{Tr}_{S}(M)\otimes_{R}T$.

Proof. (i). By [14, Lemma 1.3], there exists a sequence $\\{m_{k}\\}_{k\geq
1}$ in $\mathbb{N}$ such that $P_{S}$ is the direct limit of a direct system
of right $S$-module homomorphisms of the form

$S^{m_{1}}\stackrel{{\scriptstyle
f_{1}}}{{\to}}S^{m_{2}}\stackrel{{\scriptstyle f_{2}}}{{\to}}\cdots
S^{m_{k}}\stackrel{{\scriptstyle f_{k}}}{{\to}}S^{m_{k+1}}\cdots$

and for each $k>1$ there exists a module homomorphism $g_{k}\colon
S^{m_{k+1}}\to S^{m_{k}}$ such that $g_{k+1}f_{k+1}f_{k}=f_{k}$. In this
situation, if $f_{k}$ is given by left multiplication by the matrix
$X_{k}=(x_{ij}^{k})$ then $I=\sum_{k,i,j}Sx_{ij}^{k}S$.

Hence $P\otimes_{R}T\cong(\varinjlim
S^{m_{k}})\otimes_{R}T\cong\varinjlim(S\otimes_{R}T)^{m_{k}}$. Therefore,
$P\otimes_{R}T$ is also a direct limit of finitely generated free
$S\otimes_{R}T$-modules with transition maps $f_{k}\otimes_{R}T$. Moreover,
for any $k>1$,
$(g_{k+1}\otimes_{R}T)(f_{k+1}\otimes_{R}T)(f_{k}\otimes_{R}T)=f_{k}\otimes_{R}T$.
We can apply [14, Lemma 1.3], and the fact that $R\to T$ is a flat ring
homomorphism, to deduce that the trace ideal of the projective right
$S\otimes_{R}T$-module $P\otimes_{R}T$ is generated, as a two-sided ideal, by
the entries of the matrix $X_{k}\otimes_{R}T$ and that this implies that the
trace ideal of $P\otimes_{R}T$ is $I\otimes_{R}T$.

(ii). Let $f\in\mathrm{Hom}_{S}(M,S)$ and $t\in T$. Then $f\otimes t$ induces
a homomorphism in $\mathrm{Hom}_{S\otimes_{R}T}(M\otimes_{R}T,S\otimes_{R}T)$
satisfying $m\otimes x\mapsto f(m)\otimes xt$ for any $m\in M$ and any $x\in
T$. Such assignment induces the isomorphism
$\varphi\colon\mathrm{Hom}_{S}(M,S)\otimes_{R}T\to\mathrm{Hom}_{S\otimes_{R}T}(M\otimes_{R}T,S\otimes_{R}T)$.
Therefore, if $g\in\mathrm{Hom}_{S\otimes_{R}T}(M\otimes_{R}T,S\otimes_{R}T)$,
there exist $f_{1},\dots,f_{n}\in\mathrm{Hom}_{S}(M,S)$ and
$t_{1},\dots,t_{n}\in T$ such that $g=\varphi(\sum_{i=1}^{n}f_{i}\otimes
t_{i})$. Hence,

$g(M\otimes_{R}T)=\varphi(\sum_{i=1}^{n}f_{i}\otimes
t_{i})(M\otimes_{R}T)\subseteq\mathrm{Tr}_{S}(M)\otimes_{R}T.$

This shows that
$\mathrm{Tr}_{S\otimes_{R}T}(M\otimes_{R}T)\subseteq\mathrm{Tr}_{S}(M)\otimes_{R}T$.

Let $s\in\mathrm{Tr}_{S}(M)$, and $t\in T$. Then
$s=f_{1}(m_{1})+\cdots+f_{n}(m_{n})$ for suitable $m_{1},\dots,m_{n}\in M$ and
$f_{1},\dots,f_{n}\in\mathrm{Hom}_{S}(M,S)$. Therefore, $x\otimes
t\in\sum_{i=1}^{n}\varphi(f_{i}\otimes
t)(M\otimes_{R}T)\subseteq\mathrm{Tr}_{S\otimes_{R}T}(M\otimes_{R}T)$. This
concludes the proof of the statement.

######  Lemma 10.2.

Let $R$ be a commutative ring. Let $S$ be an $R$-algebra, and let $R\to T$ be
a ring homomorphism such that $T$ is faithfully flat as an $R$-module. Let
$M_{S}$ be a right $S$-module. Then $M_{S}$ is finitely, respectively,
countably generated if and only if $M\otimes_{R}T$ is a finitely,
respectively, countably generated $S\otimes_{R}T$-module.

Proof. The “only if” direction is clear. For the converse, choose a finite,
respectively, countable set $X\subseteq M$, such that $\\{x\otimes 1\mid x\in
X\\}$ generates $M\otimes_{R}T$ as an $S\otimes_{R}T$-module. Let
$N_{S}=\sum_{x\in X}xS$. Then $N\otimes_{R}T=M\otimes_{R}T$, and faithful
flatness implies that $N=M$. (The inclusion $N\hookrightarrow$ induces a
surjection “upstairs” so must already be a surjection.)

######  Lemma 10.3.

Let $S$ be a ring.

  1. (i)

If $I$ is an idempotent ideal of $S$, then $I$ is equal to the trace of the
right $S$-module $I$.

  2. (ii)

If $M$ is a right $S$-module with trace $I$, then $I$ is equal to the trace of
the right $S$-module $I$.

  3. (iii)

Let $I$ and $J$ be idempotent ideals of $S$. Then $I\cong J$ as right
$S$-modules if and only if $I=J$.

Proof. (i) Clearly $I\subseteq\mathrm{Tr}_{S}(I)$. Also, if $f\in\mbox{\rm
Hom}_{S}(I,S)$, then $f(I)=f(I^{2})=f(I)I\subseteq I$; hence
$\mathrm{Tr}_{S}(I)\subseteq I$.

(ii) Let $f\in\mbox{\rm Hom}_{S}(I,S)$. We want to show that $f(I)\subseteq
I$. Since $I$ is generated by elements of the form $g(m)$, where
$g\in\mbox{\rm Hom}_{S}(M,S)$ and $m\in M$, it suffices to show that
$f(g(m))\in I$. But the map $h:M\to S$ defined by $h(x)=f(g(x)$ is in
$\mbox{\rm Hom}_{S}(M,S)$, and $f(g(m)=h(m)\in I$.

(iii) Let $f\colon I\to J$ be a surjective $S$-homomorphism. Then
$J=f(I)=f(I^{2})=f(I)I\subseteq I$. If $f$ is invertible, this yields $I=J$.

######  Lemma 10.4.

Let $S$ denote an algebra over a commutative ring $R$.

  1. (i)

Let $I$ and $J$ be two ideals of $S$. Then $I=J$ if and only if
$I\otimes_{R}R_{\mathfrak{m}}=J\otimes_{R}R_{\mathfrak{m}}$ for every maximal
ideal $\mathfrak{m}$ of $R$.

  2. (ii)

Let $I$ be an ideal of $S$. Then $I$ is idempotent if and only if
$I\otimes_{R}R_{\mathfrak{m}}$ is an idempotent ideal of $S_{\mathfrak{m}}$
for each maximal ideal $\mathfrak{m}$ of $R$.

Proof. (i) $I=J$ if and only if the $R$-modules $I/(I\cap J)$ and $J/(I\cap
J)$ are both zero, and this holds if and only if they are both zero locally.
For (ii), apply (i) to the ideals $I$ and $I^{2}$.

######  Proposition 10.5.

Let $S$ denote a noetherian ring, which is an algebra over a commutative ring
$R$. For each maximal ideal $\mathfrak{m}$ of R, let $I(\mathfrak{m})$ be an
idempotent ideal of the ring $S_{\mathfrak{m}}$. Then there exists an
idempotent ideal $J$ of $S$ such that $J\otimes_{R}R_{\mathfrak{m}}\cong
I(\mathfrak{m})$ for each maximal ideal $\mathfrak{m}$ of $R$, if and only if
there exists a finitely generated right $S$-module $M$ such that
$M\otimes_{R}R_{\mathfrak{m}}\cong I(\mathfrak{m})$ for each maximal ideal
$\mathfrak{m}$ of $R$.

Proof. “only if”: Since $S$ is noetherian, we can take $M=J$.

“if”: Put $J=\mathrm{Tr}_{S}(M)$. For each maximal ideal $\mathfrak{m}$ of
$R$,
$J\otimes_{R}R_{\mathfrak{m}}=(\mathrm{Tr}_{S}(M))\otimes_{R}{R_{\mathfrak{m}}}=\mathrm{Tr}_{S_{\mathfrak{m}}}(M\otimes_{R}R_{\mathfrak{m}})=\mathrm{Tr}_{S_{\mathfrak{m}}}(I(\mathfrak{m}))=I(\mathfrak{m})$,
by Lemma 10.3 (i). Since $I(\mathfrak{m})$ is idempotent for each
$\mathfrak{m}$, so is $J\otimes_{R}R_{\mathfrak{m}}$, and now Lemma 10.4(ii)
implies that $J$ is idempotent.

######  Lemma 10.6.

Let $S$ be a module finite algebra over a local noetherian ring $R$ with
maximal ideal $\mathfrak{m}$. Let $\widehat{R}$ be the $\mathfrak{m}$-adic
completion of $R$, and set $\widehat{S}=S\otimes_{R}\widehat{R}$.

  1. (i)

Let $I$ and $J$ be two idempotent ideals of $S$. If
$I\otimes_{R}\widehat{R}\cong J\otimes_{R}\widehat{R}$ then $I=J$.

  2. (ii)

Let $I$ be an ideal of $S$. Then $I\otimes_{R}\widehat{R}$ is idempotent if
and only if $I$ is idempotent.

Proof. Since the map $S\to\widehat{S}$ is faithfully flat, right ideals $A$
and $B$ of $S$ are equal if and only if
$A\otimes_{R}\widehat{R}=B\otimes_{R}\widehat{R}$. Now (ii) follows. As for
(i), we see from Lemma 10.3(iii) that
$I\otimes_{R}\widehat{R}=J\otimes_{R}\widehat{R}$, and hence $I=J$.

######  Proposition 10.7.

Let $S$ be a module-finite algebra over a local noetherian ring $R$ with
maximal ideal $\mathfrak{m}$. Let $\widehat{R}$ be the $\mathfrak{m}$-adic
completion of $R$, and set $\widehat{S}=S\otimes_{R}\widehat{R}$. Let $K$ be
an idempotent ideal of $\widehat{S}$. Then $K$ is extended from a right
$S$-module if and only if there exists an idempotent ideal $J$ of $S$ such
that $J\otimes_{R}\widehat{R}=K$.

Moreover, if $K\cong N\otimes_{R}\widehat{R}$ for some $S$-module $N$, then
$J=\mathrm{Tr}_{S}(N)$.

Proof. Assume there exists a right $S$-module $M$ such that
$M\otimes_{R}\widehat{R}\cong K$. By Lemma 10.2, $M_{S}$ is finitely
generated. By Lemma 10.3(i), $K=\mathrm{Tr}_{\widehat{S}}(K)$, and now
$K=\mathrm{Tr}_{\widehat{S}}(M\otimes_{R}\widehat{R})=\mathrm{Tr}_{S}(M)\otimes_{R}\widehat{R}$
by Lemma 10.1. Put $J=\mathrm{Tr}_{S}(M)$. Then $J\otimes_{R}\widehat{R}=K$,
and Lemma 10.6(ii) shows that $J$ is idempotent.

Suppose now that $K\cong N\otimes_{R}\widehat{R}$, where $N$ is an $S$-module.
The arguments above show that
$(\mathrm{Tr}_{S}(N))\otimes_{R}\widehat{R}=K=J\otimes_{R}\widehat{R}$, and
now $\mathrm{Tr}_{S}(N)=J$ by faithful flatness of $S\to\widehat{S}$.

##  11\. Levy-Odenthal’s criteria for infinite direct sums

Let $R$ be a commutative local noetherian ring with maximal ideal
$\mathfrak{m}$ and $\mathfrak{m}$-adic completion $\widehat{R}$. Let $M_{R}$
be a finitely generated $R$-module with endomorphism ring $S$. Write
$\widehat{M}=M\otimes_{R}\widehat{R}=L_{1}^{n_{1}}\oplus\cdots\oplus
L_{s}^{n_{s}}$ where $L_{1},\dots,L_{s}$ are pairwise non-isomorphic
indecomposable $\widehat{R}$-modules and $n_{i}\geq 1$ for $i=1,\dots,s$.
Since $\widehat{R}$ is complete, each $L_{i}$ has a local endomorphism ring.

As we have recalled in Theorem 9.1, the modules in
$\mathrm{Add}\,(\widehat{M})$ can be written in a unique way as direct sums of
$L_{1},\dots,L_{s}$. We want to show that this prescribed structure allows us
to describe the monoid morphisms $\mathbf{D}_{M}\colon
V^{*}(M)\to({\mathbb{N}}_{0}^{*})^{s}$ and $\mathbf{dim}_{S}\colon
V^{*}(S)\to({\mathbb{N}}_{0}^{*})^{s}$ from §4 in terms of the direct sum
decomposition of the completion of the modules in $\mathrm{Add}\,(M)$.

There is a canonical ring homomorphism

$\varphi\colon
S\to\widehat{S}=S\otimes_{R}\widehat{R}\cong\mathrm{End}_{\widehat{R}}(\widehat{M}).$

Consider also the projection

$\pi\colon\mathrm{End}_{\widehat{R}}(\widehat{M})\to\mathrm{End}_{\widehat{R}}(\widehat{M})/J(\mathrm{End}_{\widehat{R}}(\widehat{M}))\cong
M_{n_{1}}(D_{1})\times\cdots\times M_{n_{s}}(D_{s})\qquad(*)$

where, for $i=1,\dots,s$,
$D_{i}=\mathrm{End}_{\widehat{R}}(L_{i})/J(\mathrm{End}_{\widehat{R}}(L_{i}))$.

######  Remark 11.1.

To claim the isomorphism in $(*)$ we are using that the elements of the
Jacobson radical of the endomorphism ring of a module
$X=X_{1}\oplus\dots\oplus X_{n}$, where $X_{i}$ has local endomorphism ring
for $i=1,\dots,n$, are those endomorphism $f\colon X\to X$ such that for any
$i,j\in\\{1,\dots,n\\}$ the morphism

$X_{i}\stackrel{{\scriptstyle\iota_{i}}}{{\to}}X\stackrel{{\scriptstyle
f}}{{\to}}X\stackrel{{\scriptstyle\pi_{j}}}{{\to}}X_{j}$

were $\iota_{i}$ and $\pi_{j}$ are the canonical morphisms. The reader can
check [8, Lemma 7.1.1] for an interesting reference where this result is
proved.

######  Lemma 11.2.

With the notation above, $\pi\circ\varphi$ is a surjective ring homomorphism
with kernel $J(S)$. Hence, $\pi\circ\varphi$ induces an isomorphism

$S/J(S)\to\mathrm{End}_{\widehat{R}}(\widehat{M})/J(\mathrm{End}_{\widehat{R}}(\widehat{M})).$

In particular, $J(S)\otimes_{R}\widehat{R}=J(S\otimes_{R}\widehat{R})$.

Proof. Note that $S/J(S)$ is a finite-dimensional algebra over
$R/\mathfrak{m}$. In particular, $S/J(S)$ has finite length as an $R$-module,
and it follows that the canonical homomorphism
$S/J(S)\to(S/J(S))\otimes_{R}{\widehat{R}}\cong\mathrm{End}_{\widehat{R}}(\widehat{M})/J(\mathrm{End}_{\widehat{R}}(\widehat{M}))$
is an isomorphism. The rest is clear.

Set $\widehat{S}=\mathrm{End}_{\widehat{R}}(\widehat{M})$. In view of Lemma
11.2, there is a commutative diagram with exact rows

$\begin{array}[]{clcl}0\longrightarrow
J(S)\longrightarrow&S&\longrightarrow&S/J(S)\longrightarrow 0\\\
\downarrow&\downarrow&&\phantom{\cong}\downarrow^{\cong}\\\ 0\longrightarrow
J(\widehat{S})\longrightarrow&\widehat{S}&\longrightarrow&\widehat{S}/J(\widehat{S})\longrightarrow
0\end{array}$ (8)

Let $P$ be a projective right $S$-module. Applying the functor $P\otimes_{S}-$
to diagram (8) we deduce that $P/PJ(S)\cong\widehat{P}/J(\widehat{P})$ where
$\widehat{P}=P\otimes_{S}\widehat{S}\cong P\otimes_{R}\widehat{R}$ is also a
projective right $\widehat{S}$-module.

Let $V_{1},\dots,V_{s}$ be a set of representatives of non-isomorphic simple
right $S$-modules such that, for $i=1,\dots,s$, $\mathrm{End}_{S}(V_{i})\cong
D_{i}$. By Lemma 11.2, they also give a set of representatives of the
isomorphism classes of simple right $\widehat{S}$-modules. Therefore, if $P$
is a projective right $S$-module, then $P/J(P)\cong
V_{1}^{(I_{1})}\oplus\cdots\oplus
V_{s}^{(I_{s})}\cong\widehat{P}/J(\widehat{P})$. In particular,

$\mathbf{dim}_{S}([P])=\mathbf{dim}_{\widehat{S}}([\widehat{P}]).$ (9)

Notice that $\mathbf{dim}_{\widehat{S}}\colon
V^{*}(\widehat{S})\to({\mathbb{N}}_{0}^{*})^{s}$ is an isomorphism, because
$\widehat{M}$ is a direct sum of modules with local endomorphism rings by
Theorem 9.1, so $\widehat{S}$ is a semiperfect ring. All together, we have
completed the commutative diagram (2) in Section 4 in the following way

$\textstyle{\bigsqcup_{I\in\mathcal{T}(S)}V(S/I)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Phi}$$\scriptstyle{\alpha}$$\textstyle{({\mathbb{N}}_{0}^{*})^{s}}$$\textstyle{V^{*}(S)=V(S)\bigsqcup
B(S)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathbf{dim}_{S}}$$\scriptstyle{\beta}$$\textstyle{V^{*}(\widehat{S})=V(\widehat{S})+\infty\cdot
V(\widehat{S})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\,,}$$\scriptstyle{\mathbf{dim}_{\widehat{S}}}$
(10)

where $\beta([P])=[\widehat{P}]=[P\otimes_{R}\widehat{R}]$ for any countably
generated projective right $S$-module $P$. Recall that $\alpha$ and
$\mathrm{dim}_{\widehat{S}}$ are isomorphisms.

By Proposition 2.1, there is an equivalence between the categories $\mbox{\rm
Add}(M_{R})$ and $\mbox{\rm Add}(S_{S})$ (as well as between $\mbox{\rm
Add}(\widehat{M}_{\widehat{R}})$ and $\mbox{\rm
Add}(\widehat{S}_{\widehat{S}})$). The following remark uses such
equivalences.

######  Remark 11.3.

Let $N\in\mbox{\rm Add}(\widehat{M})$, and let
$Q_{\widehat{S}}=\mathrm{Hom}_{\widehat{R}}(\widehat{M},N)$. By construction,
$N\cong L_{1}^{(I_{1})}\oplus\cdots\oplus L_{s}^{(I_{s})}$ if and only if
$Q/J(Q)\cong V_{1}^{(I_{1})}\oplus\cdots\oplus V_{s}^{(I_{s})}$. Therefore,
for $X\in\mbox{\rm Add}(M)$, $X\otimes_{R}\widehat{R}\cong
L_{1}^{(I_{1})}\oplus\cdots\oplus L_{s}^{(I_{s})}$ if and only if

$\mathrm{Hom}_{R}(M,X)/J(\mathrm{Hom}_{R}(M,X))\cong
V_{1}^{(I_{1})}\oplus\cdots\oplus V_{s}^{(I_{s})}.$

Therefore, in this situation,

$\mathbf{D}_{M}([X])=\mathbf{D}_{\widehat{M}}([X\otimes_{R}\widehat{R}])=\mathbf{D}_{\widehat{M}}([\widehat{X}])$

for any countably generated module $X\in\mathrm{Add}\,(M)$ and its value can
be _read_ directly in the decomposition into indecomposable direct summands of
$\widehat{X}$.

Combining the reinterpretation of the dimension functions in the completion
with the results on extensions of idempotent ideals in section 10, we can now
complete Proposition 4.11 in the following way.

######  Proposition 11.4.

Let $R$ be a commutative noetherian local ring with maximal ideal
$\mathfrak{m}$. Denote by $\widehat{R}$ the $\mathfrak{m}$-adic completion of
$R$. Let $M$ be a nonzero finitely generated $R$-module, and write
$\widehat{M}=M\otimes_{R}\widehat{R}=L_{1}^{n_{1}}\oplus\cdots\oplus
L_{s}^{n_{s}}$ where $L_{1},\dots,L_{s}$ are pairwise non-isomorphic
indecomposable $\widehat{R}$-modules and $n_{i}\geq 1$ for $i=1,\dots,s$.

Let $S=\mathrm{End}_{R}(M)$ and $\widehat{S}=S\otimes_{R}\widehat{R}$. Let
$\mathbf{x}\in({\mathbb{N}}_{0}^{*})^{s}$, set
$\Lambda=\infty-\mathrm{supp}(\mathbf{x})$. Let
$\pi_{\Lambda}\colon({\mathbb{N}}_{0}^{*})^{s}\to({\mathbb{N}}_{0}^{*})^{\\{1,\dots,s\\}\setminus\Lambda}$
denote the canonical projection.

Then the following statements are equivalent:

  1. (i)

There exists a countably generated $R$-module $X$ in $\mathrm{Add}(M)$ such
that $\mathbf{D}_{M}([X])=\mathbf{x}$.

  2. (ii)

The ideal
$\mathrm{Tr}_{\widehat{S}}(\mathrm{Hom}_{\widehat{R}}(\widehat{M},\oplus_{i\in\Lambda}L_{i}))$
is extended from an $S$-module $Y$. Moreover, if $I=\mathrm{Tr}_{S}(Y)$ then
there exists a finitely generated projective right $S/I$-module $\overline{Q}$
such that $\mathbf{dim}_{S/I}([\overline{Q}])=\pi_{\Lambda}(\mathbf{x})$.

If, in addition, $R$ has Krull dimension $1$ and $M$ is torsion-free and
$\Lambda\neq\emptyset$, the above statements are also equivalent to:

  * (iii)

The ideal
$\mathrm{Tr}_{\widehat{S}}(\mathrm{Hom}_{\widehat{R}}(\widehat{M},\oplus_{i\in\Lambda}L_{i}))$
is extended from an $S$-module.

  * (iv)

The element $\mathbf{y}\in({\mathbb{N}}_{0}^{*})^{s}$ such that
$\mathrm{supp}\,(\mathbf{y})=\infty-\mathrm{supp}(\mathbf{y})=\Lambda$ is in
$\mathbf{D}_{M}(V^{*}(M))$.

Proof. $(i)\Rightarrow(ii)$ In view of Proposition 4.11, if $(i)$ holds, there
is an idempotent ideal $I^{\prime}$ of $S$ such that
$\mathrm{supp}(\mathbf{dim}_{S}([I^{\prime}]))=\Lambda$. By Lemma 4.8,
$I^{\prime}/I^{\prime}J(S)\cong\oplus_{i\in\Lambda}V^{n_{i}}_{i}\cong\widehat{I^{\prime}}/\widehat{I^{\prime}}J(\widehat{S})$.

Let
$\widehat{I}=\mathrm{Tr}_{\widehat{S}}(\mathrm{Hom}_{\widehat{R}}(\widehat{M},\oplus_{i\in\Lambda}L_{i}))$.
Since $\widehat{I}$ is the trace ideal of a (finitely generated) projective
right $\widehat{S}$-module, it is an idempotent ideal of $\widehat{S}$ and, by
Lemma 4.8, $\mathrm{supp}(\mathbf{dim}_{\widehat{S}}([\widehat{I}]))=\Lambda$.
Therefore
$\widehat{I}/\widehat{I}J(\widehat{S})\cong\widehat{I^{\prime}}/\widehat{I^{\prime}}J(\widehat{S})$,
and this implies that $\widehat{I}=\widehat{I^{\prime}}$ [14, Corollary 2.9].
This proves that $\widehat{I}$ is extended from the $S$-module $Y=I^{\prime}$.

Notice that, since $I^{\prime}$ is idempotent,
$\mathrm{Tr}_{S}(I^{\prime})=I^{\prime}$. The rest of the statement follows
from Proposition 4.11.

$(ii)\Rightarrow(i)$ We show that our hypothesis imply that statement $(ii)$
in Proposition 4.11 holds. Let
$K=\mathrm{Tr}_{\widehat{S}}(\mathrm{Hom}_{\widehat{R}}(\widehat{M},\oplus_{i\in\Lambda}L_{i}))$.
By hypothesis, $K$ is extended from an $S$-module then, by Proposition 10.7,
$K$ is extended from an idempotent ideal $I$ of $S$. By Lemma 4.8,

$\mathrm{supp}(\mathbf{dim}_{\widehat{S}}([K]))=\Lambda=\mathrm{supp}(\mathbf{dim}_{S}([I]))$

and this proves the first part of Proposition 4.11 (ii). The second part of
that statements are part of our hypotheses. This finishes this part of the
proof.

It is clear that $(ii)$ implies $(iii)$. Assume that $R$ is a local domain of
Krull dimension $1$ and that $M$ is torsion-free. We want to prove
$(iii)\Rightarrow(i)$.

Repeating the argument done in $(ii)\Rightarrow(i)$, the hypotheses imply that
there exists an idempotent ideal $I$ in $S$ such that
$\Lambda=\mathrm{supp}(\mathbf{dim}_{S}([I]))$. So Proposition 4.11 (iii)
holds and this implies $(i)$.

The equivalence of $(iii)$ and $(iv)$ is already proved in Proposition 4.11.

Now we are ready to prove a version of the Levy-Odenthal criteria to
infinitely generated modules. In the following result we recall such criteria
in its more general version from [9].

If $R$ is a commutative ring, we denote by $K(R)$ its localization at the
complementary of the union of the minimal prime ideals of $R$.

######  Theorem 11.5.

_([9, Theorem 4.1])_ Let $R$ and $T$ be local noetherian commutative rings
with maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$, respectively. Let
$\varphi\colon R\to T$ be a local morphism satisfying that

  * (1)

$\varphi(\mathfrak{m})T=\mathfrak{n}$,

  * (2)

$\varphi$ induces and isomorphism of residue fields.

Assume also that $R$ has Krull dimension $1$.

Let $S$ be a module finite $R$-algebra. Let $N$ be a finitely generated
$S\otimes_{R}T$-module. Then $N$ is extended from $S$ if and only if
$N\otimes_{S}K(S)$ is extended from an $S\otimes_{R}K(R)$ module. ∎

######  Remark 11.6.

Assume that $R$ is a one dimensional noetherian domain with reduced completion
$\widehat{R}$. Then $K(R)$ is the field of fractions of $R$, and
$K(\widehat{R})=\widehat{R}_{P_{1}}\times\cdots\times\widehat{R}_{P_{\ell}}$,
where $\widehat{R}_{P_{j}}$ is a field for each $j=1,\dots,\ell$ and where
$P_{1},\dots,P_{\ell}$ is the set of minimal primes of $\widehat{R}$.

If $X$ is a finitely generated $R$-module, then $X\otimes_{R}K(R)\cong
K(R)^{n}$ is a finite dimensional $K(R)$-vector space. Therefore
$\widehat{X}\otimes_{\widehat{R}}K(\widehat{R})\cong
K(\widehat{R})^{n}=\widehat{R}_{P_{1}}^{n}\times\cdots\times\widehat{R}_{P_{\ell}}^{n}$.

Hence, in this situation, Theorem 11.5 claims that a finitely generated
$\widehat{R}$-module $N$ is extended from a finitely generated $R$-module if
and only if $N\otimes_{\widehat{R}}K(\widehat{R})\cong K(\widehat{R})^{n}$ for
some $n\geq 0$. Now we want to show that this extends to countably generated
modules in $\mathrm{Add}\,(M)$.

######  Lemma 11.7.

Let $R$ be a local noetherian domain of Krull dimension $1$ with quotient
field $K$ and with reduced completion $\widehat{R}$. Let $K(\widehat{R})$
denote the total quotient ring of $\widehat{R}$. Let $M_{R}$ be a finitely
generated torsion-free $R$-module, and let $N$ be an $\widehat{R}$-module in
$\mbox{\rm Add}(\widehat{M}_{\widehat{R}})$.

If $N$ is not finitely generated as $\widehat{R}$-module then neither is
$N\otimes_{\widehat{R}}K(\widehat{R})$ as a $K(\widehat{R})$-module.

Proof. $N$ is a non finitely generated module in $\mathrm{Add}\,(\widehat{M})$
if and only if there is an infinite set $\Lambda$ and a (split) embedding
$\varepsilon\colon N\to\widehat{M}^{(\Lambda)}$ such that
$\pi_{\lambda}\circ\varepsilon\neq 0$ for any $\lambda\in\Lambda$, where
$\pi_{\lambda}\colon\widehat{M}^{(\Lambda)}\to\widehat{M}$ denotes the
projection onto the component $\lambda$.

Since $M$ is torsion free and the going-down Theorem holds for the embedding
$R\to\widehat{R}$, the annihilator of any non-zero submodule of $\widehat{M}$
is contained in a minimal prime of $\widehat{R}$. This implies that, for any
$\lambda\in\Lambda$, the induced map
$\pi_{\lambda}\otimes_{\widehat{R}}K(\widehat{R})\neq 0$. Therefore,
$N\otimes_{\widehat{R}}K(\widehat{R})$ cannot be a finitely generated
$K(\widehat{R})$-module.

######  Theorem 11.8.

Let $R$ be a local noetherian domain of Krull dimension $1$ with quotient
field $K$ and with reduced completion $\widehat{R}$. Let $K(\widehat{R})$
denote the total quotient ring of $\widehat{R}$. Let $M_{R}$ be a finitely
generated torsion-free $R$-module. Then the following statements are
equivalent, for a countably generated $\widehat{R}$-module $N\in\mbox{\rm
Add}(\widehat{M}_{\widehat{R}})$:

  1. (i)

$N$ is extended from an $R$-module;

  2. (ii)

there exists a countable set $\Lambda$ such that
$N\otimes_{\widehat{R}}K(\widehat{R})\cong K(\widehat{R})^{(\Lambda)}$;

  3. (iii)

$N$ is extended from a countably generated module in $\mbox{\rm Add}(M_{R})$.

Proof. We may assume that $N\neq\\{0\\}$.

$(i)\Rightarrow(ii).$ Assume that there exists an $R$-module $X$ such that
$N\cong X\otimes_{R}\widehat{R}$. Since $K$ is a field, $X\otimes_{R}K\cong
K^{(\Lambda)}$ for some index set $\Lambda$, which must be countable, by Lemma
10.2. Therefore

$N\otimes_{\widehat{R}}K(\widehat{R})\cong
X\otimes_{R}K\otimes_{K}K(\widehat{R})\cong
K^{(\Lambda)}\otimes_{K}K(\widehat{R})\cong K(\widehat{R})^{(\Lambda)}.$

$(ii)\Rightarrow(iii).$ Set $\widehat{M}=L_{1}^{n_{1}}\oplus\cdots\oplus
L_{s}^{n_{s}}$, where $L_{1},\dots,L_{s}$ are pairwise non-isomorphic
indecomposable torsion-free $\widehat{R}$-modules and $n_{i}\in\mathbb{N}$ for
$i=1,\dots,s$. By Theorem 9.1, $N\cong L_{1}^{(\Lambda_{1})}\oplus\cdots\oplus
L_{s}^{(\Lambda_{s})}$ for suitable countable sets
$\Lambda_{1},\dots,\Lambda_{s}$. Reordering the $L_{i}$’s, if needed, we may
assume that $\Lambda_{1},\dots,\Lambda_{t}$ are infinite and that
$\Lambda_{t+1},\dots,\Lambda_{s}$ are finite. If $t=0$ the statement follows
from the usual Levy-Odenthal criteria, see Theorem 11.5. So we may assume that
$t>0$ and, therefore, $N$ is a countably generated $\widehat{R}$-module that
is not finitely generated.

By Remark 11.3, $\mathbf{D}_{\widehat{M}}(N)=(\infty,\stackrel{{\scriptstyle
t)}}{{\dots}},\infty,|\Lambda_{t+1}|,\dots,|\Lambda_{s}|)$. By Proposition
11.4 (iv), $N$ is extended if and only if $(\infty,\stackrel{{\scriptstyle
t)}}{{\dots}},\infty,0,\dots,0)\in\mathbf{D}_{M}(V^{*}(M))$. Equivalently, cf.
Proposition 11.4 (iii), $N$ is extended from an $R$-module if and only if the
idempotent ideal of $\widehat{S}$,
$I=\mathrm{Tr}_{\widehat{S}}(\mathrm{Hom}_{\widehat{R}}(\widehat{M},\oplus_{i=1}^{t}L_{i}))$
is extended from an $S$-module, where
$\widehat{S}=\mathrm{End}_{\widehat{R}}(\widehat{M})\cong\mathrm{End}_{R}(M)\otimes_{R}\widehat{R}$.
To prove $I$ is extended we shall use again the Levy-Odenthal criteria over
the finitely generated $R$-algebra $S$, cf. Theorem 11.8. First, we need some
preparation.

Let $P_{1},\dots,P_{\ell}$ be the set of minimal primes of $\widehat{R}$.
Since $\widehat{R}$ is reduced,
$K(\widehat{R})\cong\widehat{R}_{P_{1}}\times\cdots\times\widehat{R}_{P_{\ell}}$,
where $\widehat{R}_{P_{j}}$ is a field for each $j=1,\dots,\ell$, cf. Remark
11.6. By hypothesis,

$\prod_{j=1}^{\ell}\widehat{R}_{P_{j}}^{(\Lambda)}\cong
N\otimes_{\widehat{R}}K(\widehat{R})\cong\prod_{j=1}^{\ell}\left(\sum_{i=1}^{s}(L_{i}\otimes_{\widehat{R}}\widehat{R}_{P_{j}})^{(\Lambda_{i})}\right).$

By Lemma 11.7, since $M$ is torsion-free and $N$ is not finitely generated,
neither is $N\otimes_{\widehat{R}}K(\widehat{R})$. Hence, $\Lambda$ is
infinite and, therefore, $(L_{1}\oplus\cdots\oplus
L_{t})\otimes_{\widehat{R}}\widehat{R}_{P_{j}}\neq\\{0\\}$ for each
$j=1,\dots,\ell$.

For $i=1,\dots,t$, set $Q_{i}=\mathrm{Hom}_{\widehat{R}}(\widehat{M},L_{i})$,
then $I=\mathrm{Tr}_{\widehat{S}}(Q_{1}\oplus\cdots\oplus Q_{t})$. By Lemma
10.1, for each prime $P_{j}$,
$I\otimes_{\widehat{R}}\widehat{R}_{P_{j}}\cong\mathrm{Tr}_{\widehat{S}\otimes_{\widehat{R}}\widehat{R}_{P_{j}}}((Q_{1}\oplus\cdots\oplus
Q_{t})\otimes_{\widehat{R}}\widehat{R}_{P_{j}})$ which is different from zero
because $(L_{1}\oplus\cdots\oplus
L_{t})\otimes_{\widehat{R}}\widehat{R}_{P_{j}}\neq 0$. Since
$\widehat{R}_{P_{j}}$ is a field for $j=1,\dots,\ell$, it follows that
$\widehat{S}\otimes_{\widehat{R}}\widehat{R}_{P_{j}}\cong\mathrm{End}_{\widehat{R}_{P_{j}}}(\widehat{M}\otimes_{\widehat{R}}\widehat{R}_{P_{j}})$
is a simple artinian ring. Therefore, for $j=1,\dots,\ell$,
$I\otimes_{\widehat{R}}\widehat{R}_{P_{j}}$ is a nonzero idempotent ideal of
the simple artinian ring
$\widehat{S}\otimes_{\widehat{R}}\widehat{R}_{P_{j}}$, so
$I\otimes_{\widehat{R}}\widehat{R}_{P_{j}}=S\otimes_{\widehat{R}}\widehat{R}_{P_{j}}$.
Therefore,
$I\otimes_{\widehat{R}}K(\widehat{R})=\widehat{S}\otimes_{\widehat{R}}K(\widehat{R})$
which is extended from $S\otimes_{R}K(R)$. By Theorem 11.5, we deduce that $I$
is extended from a finitely generated $S$-module. By Proposition 11.4 (iii),
this completes the proof that (ii) $\implies$ (iii). Since (iii) trivially
implies (i), we are done.

The statement and the proof of Theorem 11.8 give also a system of linear
equations that describes the elements of $\mathbf{D}_{M}(V^{*}(M))$.

######  Corollary 11.9.

We continue with the same hypothesis and notation as in Theorem 11.8 and its
proof. Let $\mathbf{x}=(x_{1},\dots,x_{s})\in({\mathbb{N}}_{0}^{*})^{s}$, and
let
$a_{ji}:=\mathrm{dim}_{\widehat{R}_{P_{j}}}L_{i}\otimes_{\widehat{R}}{\widehat{R}}_{P_{j}}$.
Then $\mathbf{x}\in\mathbf{D}_{M}\,(V^{*}(M))$ if and only if

$\sum_{i=1}^{s}a_{ji}x_{i}=\sum_{i=1}^{s}a_{ki}x_{i}$

for any $j$, $k\in\\{1,\dots,\ell\\}$.

Proof. To easy the notation, let $L^{(\infty)}$ denote the countable direct
sum of copies of the module $L$.

Let $N=L_{1}^{(x_{1})}\oplus\dots\oplus L_{s}^{(x_{s})}$. Since, by
hypothesis, $\widehat{M}=L_{1}^{n_{1}}\oplus\dots\oplus L_{s}^{n_{s}}$ for
some $(n_{1},\dots,n_{s})\in\mathbb{N}^{s}$,
$\mathrm{Add}\,(\widehat{M})=\mathrm{Add}\,(L_{1}\oplus\cdots\oplus L_{s})$.
Hence, $N$ is a countably generated module in $\mathrm{Add}\,(\widehat{M})$.
By Theorem 11.8, $N$ is extended from a countably generated $R$-module if and
only if
$\mathrm{dim}_{\widehat{R}_{P_{j}}}N\otimes_{\widehat{R}}\widehat{R}_{P_{j}}=\mathrm{dim}_{\widehat{R}_{P_{k}}}N\otimes_{\widehat{R}}\widehat{R}_{P_{k}}$
for any $j$, $k\in\\{1,\dots,\ell\\}$. This is equivalent to say that
$(x_{1},\dots,x_{s})$ satisfies the equations in the statement.

If there is a countably generated $R$-module $X\in\mathrm{Add}\,(M)$ such that
$X\otimes_{R}\widehat{R}\cong N$ then, using the commutative diagram (10) we
deduce that

$\mathbf{D}_{M}([X])=\mathbf{dim}_{S}\,\mathrm{Hom}_{R}(M,X)=\mathbf{dim}_{\widehat{S}}\,\mathrm{Hom}_{\widehat{R}}(\widehat{M},N)=\mathbf{x}.$

Now we are ready to show that the monoids appearing in Proposition 7.8 that
contain an order-unit can be realized as $V^{*}(M)$ for $M$ a finitely
generated torsion-free module over a one-dimensional noetherian domain with
reduced completion.

######  Example 11.10.

Let $\mathbf{a}=(a_{1},\dots,a_{s})$ and $\mathbf{b}=(b_{1},\dots,b_{s})$ be
elements in ${\mathbb{N}}_{0}^{s}$ such that $\mathbf{a}-\mathbf{b}\neq 0$
and, for each $i\in\\{1,\dots,s\\}$, $a_{i}+b_{i}>0$. Assume also that there
is $(n_{1},\dots,n_{s})\in\mathbb{N}^{s}$ such that
$\sum_{i=1}^{s}a_{i}n_{i}=\sum_{i=1}^{s}b_{i}n_{i}$.

Then there exists a local noetherian domain $R$ of Krull dimension one with
reduced completion $\widehat{R}$, and a finitely generated torsion-free
$R$-module $M$ such that
$M\otimes_{R}\widehat{R}=L_{1}^{n_{1}}\oplus\cdots\oplus L_{s}^{n_{s}}$, where
$L_{1},\dots,L_{s}$ are non-isomorphic indecomposable $\widehat{R}$-modules,
and such that $\mathbf{dim}_{R}(V^{*}(M))$ is exactly the set of solution in
$({\mathbb{N}}_{0}^{*})^{s}$ of the equation
$\sum_{i=1}^{s}a_{i}t_{i}=\sum_{i=1}^{s}b_{i}t_{i}$.

Proof. By using [19, Construction 2.9, Theorem 2.10], we know that there
exists a domain $R$ as in the statement such that its completion $\widehat{R}$
is reduced and has exactly two minimal prime ideals $\mathcal{P}$ and
$\mathcal{Q}$. Moreover, there are finitely generated, non-isomorphic torsion-
free $\widehat{R}$-modules $L_{1},\dots,L_{s}$ such that, for any
$i=1,\dots,s$

$\mathrm{rank}(L_{i})=(\mathrm{dim}_{\widehat{R}_{\mathcal{P}}}(L_{i}\otimes_{\widehat{R}}\widehat{R}_{\mathcal{P}}),\mathrm{dim}_{\widehat{R}_{\mathcal{Q}}}(L_{i}\otimes_{\widehat{R}}\widehat{R}_{\mathcal{Q}}))=(a_{i},b_{i}).$

Notice that this is possible because $(a_{i},b_{i})\neq(0,0)$. By the usual
Levy-Odenthal’s criteria (cf. Theorem 11.5), $L_{1}^{n_{1}}\oplus\cdots\oplus
L_{s}^{n_{s}}$ is extended from a finitely generated torsion-free $R$-module
$M$. By Corollary 11.9, the countably generated modules in $\mbox{\rm Add}(M)$
are precisely the modules of the form $L_{1}^{(\Lambda_{1})}\oplus\cdots\oplus
L_{s}^{(\Lambda_{s})}$ where $(|\Lambda_{1}|,\cdots,|\Lambda_{s}|)$ is a
solution in $({\mathbb{N}}_{0}^{*})^{s}$ of the equation in the statement.

In the next example we discuss the summands that are not direct sums of
finitely generated ones in the realization technique of full monoids defined
by diophantine equations given [30], see either [19, Theorem 2.12].

######  Example 11.11.

Let $A$ be a full submonoid of ${\mathbb{N}}_{0}^{s}$, with order-unit
$(n_{1},\dots,n_{s})$, which is the set of solutions of a system of equations
$E\mathbf{T}=0$ where $E\in M_{\ell\times s}(\mathbb{Z})$. Then there exists a
local noetherian domain $R$ of Krull dimension one with reduced completion
$\widehat{R}$, and a finitely generated torsion-free $R$-module $M$ with
endomorphism ring $S$, such that $\mathbf{D}_{M}\,(V(M))=A$, and such that
$\mathbf{D}_{M}\,(V^{*}(M))$ is the monoid $B_{\mathrm{max}}(V(M))$ (cf.
Remark 7.9 (3) and Proposition 7.4).

In particular, any element of $({\mathbb{N}}_{0}^{*})^{s}$ with non-empty
infinite support must be in $\mathbf{D}_{M}\,(V^{*}(M))$. Equivalently, if
$M\otimes_{R}\widehat{R}\cong L_{1}^{n_{1}}\oplus\cdots\oplus L_{s}^{n_{s}}$,
where $L_{1},\dots,L_{s}$ are indecomposable pairwise non-isomorphic modules,
then any countably generated module of the form
$L_{1}^{(I_{1})}\oplus\dots\oplus L_{s}^{(I_{s})}$, in which some $I_{i}$ is
infinite, must be extended from a countably generated module in $\mbox{\rm
Add}(M)$.

Proof. Take $R$ and $M$ given by the construction in [19, Theorem 2.12]. That
is, $R$ is a one dimensional domain with reduced completion $\widehat{R}$ such
that $\widehat{R}$ has $\ell+1$ minimal prime ideals $P_{1},\dots,P_{\ell+1}$.
The domain $R$ can be chosen such that there exist $L_{1},\dots,L_{s}$
torsion-free $\widehat{R}$-modules and different
$h_{1},\dots,h_{s}\in\mathbb{N}$ such that, for $i\in\\{1,\dots,s\\}$,

$\mathrm{dim}_{\widehat{R}_{P_{j}}}\,(L_{i}\otimes_{\widehat{R}}\widehat{R}_{P_{j}})=e_{ji}+h_{i}>0$

where $j\in\\{1,\dots,\ell\\}$ and $e_{ji}$ denotes the corresponding entry of
the matrix $E$. Moreover,

$\mathrm{dim}_{\widehat{R}_{P_{\ell+1}}}\,(L_{i}\otimes_{\widehat{R}}\widehat{R}_{P_{\ell+1}})=h_{i}>0,$

Since all $h_{i}$’s are different, the $\widehat{R}$-modules $L_{i}$’s are
pair-wise non-isomorphic.

Let $H\in M_{\ell\times s}(\mathbb{Z})$, be the matrix such that, for
$i\in\\{1,\dots,s\\}$, all entries in its $i$-th column are equal to $h_{i}$.
Then the monoid $A$ is also the set of solutions in ${\mathbb{N}}_{0}^{s}$ of
the system $(E+H)\mathbf{T}=H\mathbf{T}$. Notice that if
$\mathbf{x}=(x_{1},\dots,x_{s})\in({\mathbb{N}}_{0}^{*})^{s}$ is a solution of
the system, then the module $N=L_{1}^{(x_{1})}\oplus\dots\oplus
L_{s}^{(x_{s})}$ satisfies that, for any $j=1,\dots,\ell$,

$\mathrm{dim}_{\widehat{R}_{P_{j}}}\,(N\otimes_{\widehat{R}}\widehat{R}_{P_{j}})=\mathrm{dim}_{\widehat{R}_{P_{\ell+1}}}\,(N\otimes_{\widehat{R}}\widehat{R}_{P_{\ell+1}}).$

By Theorem 11.5 and because $(n_{1},\dots,n_{s})\in A$,
$L_{1}^{n_{1}}\oplus\cdots\oplus L_{s}^{n_{s}}$ is extended from a torsion-
free $R$-module $M$. By the usual Levy-Odenthal criteria,
$\mathbf{D}_{M}(V(M))=A$. By Corollary 11.9, $\mathbf{D}_{M}(V^{*}(M))$ is the
set of solutions in $({\mathbb{N}}_{0}^{*})^{s}$ of the system
$(E+H)\mathbf{T}=H\mathbf{T}$.

By Remark 7.9 (3) and because all the entries of $E+H$ are strictly positive,
the monoid of solutions of the system is $B_{\mathrm{max}}(V(M))$. Then any
infinite direct sum of the $L_{i}$’s is extended from a countably generated
module in $\mbox{\rm Add}(M)$.

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