# Representation of partially ordered sets over Von Neumann regular algebras.
More prime, non-primitive regular rings

Giuseppe Baccella  Università di L’Aquila, Dipartimento di Ingegneria e
scienze dell’informazione e matematica, Via Vetoio, 67100 L’Aquila, Italy
<EMAIL_ADDRESS>
(Date: January 30, 2024)

###### Abstract.

Given an _infinite_ cardinal $\boldsymbol{\aleph}$, let
${\mathbf{P}\mathbf{o}\mathbf{s}}_{\boldsymbol{\aleph}}$ be the category whose
objects are all partially ordered sets $I$, having a finite cofinal subset and
such that both $I$ and the set of all maximal chains of $I$ have at most
cardinality $\boldsymbol{\aleph}$, while the morphisms are all maps which are
order imbeddings of the domain as an upper subset of the codomain. Given a
field $\mathbb{K}$, for every object $I$ of
${\mathbf{P}\mathbf{o}\mathbf{s}}_{\boldsymbol{\aleph}}$ we find a unital,
locally matricial and hence unit-regular $\mathbb{K}$-algebra
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ such that the lattice of
all its ideals is order isomorphic to the lattice of all lower subsets of $I$.
We show that the Grothendieck group
$\mathbf{K}_{0}(\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I))$, with its
natural partial order, is order isomorphic to the restricted Hahn power of
$\mathbb{Z}$ by $I$; this gives a contribution to solve the _Realization
Problem for Dimension Groups with order-unit_. We also show that the algebra
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ has the following features:
(a) $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ is prime if and only if
$I$ is lower directed; (b) $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$
is primitive if and only if $I$ has a coinitial chain; (c)
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ is semiartinian if and only
if $I$ is artinian, in which the case $I$ is order isomorphic to the primitive
spectra of $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$. Finally we show
that the assignment $I\mapsto\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$
extends to a pair of functors from
${\mathbf{P}\mathbf{o}\mathbf{s}}_{\boldsymbol{\aleph}}$ to the category of
$\mathbb{K}$-algebras, one covariant and the other contravariant.

###### Key words and phrases:

Von Neumann regular ring; partially ordered set; prime ring; primitive ring;
dimension group, Grothendieck group; (restricted) Hahn power.

###### 1991 Mathematics Subject Classification:

Primary 16E50; Secondary 16D60, 16S50, 19A49, 06A06, 06F20

##  0\. Introduction

We recall that if $L$ is a complete lattice, an element $a\in L$ is said to be
_compact_ if whenever $a\leqslant\bigvee M$ for some subset $M\subset L$, then
$a\leqslant\bigvee N$ for some finite subset $N\subset M$. Given a ring $R$
(associative with multiplicative identity), the compact elements of the
complete lattice $\mathbb{L}_{2}(R)$ of ideals of $R$ are the finitely
generated ideals, consequently $\mathbb{L}_{2}(R)$ is an _algebraic_ lattice,
meaning that all its elements are join of (possibly infinitely many) compact
elements. It is well known that if, in addition, $R$ is a von Neumann regular
ring, then $\mathbb{L}_{2}(R)$ is also distributive.

The problem of determining which algebraic and distributive lattices are
isomorphic to $\mathbb{L}_{2}(R)$ for some von Neumann regular ring $R$ is one
among the so called _Realization (or Representation) Problems for Von Neumann
regular rings_. A first answer to this problem was given by Bergman in his
famous 1986 unpublished note (see [12]):

Bergman’s Theorem. _An algebric distributive lattice $D$ in which the greatest
element is compact is isomorphic to $\mathbb{L}_{2}(R)$, for some unital von
Neumann regular ring $R$, if either $D$ has countably many compact elements,
or every element of $D$ is a join of _join-irreducible_ compact elements_.

In [23] Goodearl and Wehrung wrote an extended account about the
representation problems of distributive algebraic lattices and the connection
with the analogous problems for distributive semilattices. The first condition
stated in Bergman’s Theorem raised the obvious question: is it possible to
remove the countability condition on the set of compact elements of $D$? A
negative answer was given by Wehrung in [35], where he proved that Bergman’s
Theorem cannot be extended to those algebraic distributive lattices having
$\boldsymbol{\aleph}_{2}$ or more compact elements. However Wehrung proved in
[37] that Bergman’s Theorem still holds for algebraic distributive lattices
having at most $\boldsymbol{\aleph}_{1}$ compact elements.

Concerning the second condition, we note that the algebric distributive
lattices satisfying it are precisely those isomorphic to the lattice
(partially ordered by inclusion) $\Downarrow\\!\\!I$ of all lower subsets of
some partially ordered set $I$; so, actually, given a partially ordered set
$I$ having a finite cofinal subset, Bergman had built a locally matricial,
unital and hence unit-regular algebra $R$ whose lattice of ideals is
isomorphic to $\Downarrow\\!\\!I$.

Other important realization problems arise when one considers the additive
abelian monoid $\mathcal{V}(R)$ of isomorphism classes of finitely generated
right modules over a ring $R$, with the addition defined as
$[A]+[B]\,\,\colon\\!\\!=[A\oplus B]$, and the corresponding Grothendieck
group $\mathbf{K}_{0}(R)$. If $R$ is a regular ring, then $\mathcal{V}(R)$
enjoys some fundamental and well known properties and the problem of deciding
wether, given an abelian monoid $M$ having the same properties, there exists a
regular ring $R$ such that $\mathcal{V}(R)$ is isomorphic to $M$, was the
first which was named as _Realization Problem for Von Neumann Regular Rings_
(see [6]).

The group $\mathbf{K}_{0}(R)$ carries a natural structure of a pre-ordered and
directed abelian group with order-unit, which is partially ordered in case $R$
is unit-regular (see [19, Ch 15]). This gives rise to the corresponding
_Realization Problem_ : which partially ordered, directed abelian groups with
order unit are isomorphic to $\mathbf{K}_{0}(R)$ for some regular ring $R$ ?
(see [19, Problem 29]).

Partially ordered abelian groups of a special interest are the _dimension
groups_ , namely the directed, unperforated and interpolation groups.
Originally, the name “dimension group” was introduced by Elliott in [16] to
designate a group which is a direct limit of a countable sequence of
simplicial groups, i. e. of groups each order-isomorphic to $\mathbb{Z}^{n}$
for some $n\in\mathbb{N}$, with the partial order induced by the submonoid
$\mathbb{N}^{n}$ as positive cone. Later on it was proved by Effros, Handelman
and Shen (se [15, Theorem 2.2]) that a partially ordered abelian group is a
(arbitrary) direct limit of simplicial groups if and only if it is a directed,
unperforated and interpolation group. It was well known that a countable
dimension group is isomorphic to $\mathbf{K}_{0}(R)$ for an ultramatricial
algebra $R$ over a given field. Goodearl and Handelman proved in [21] that
every dimension group with order unit having cardinality
$\boldsymbol{\aleph}_{1}$ can be realized as $\mathbf{K}_{0}(R)$ for a locally
matricial (and hence unit-regular) algebra $R$ over a given field and, in
addition, every infinite tensor product of such dimension groups can be
realized similarly. By way of contrast, dimension groups which cannot be
realized as $\mathbf{K}_{0}(R)$ for a regular ring $R$ were found by Wehrung
in [34] and [36]. Actually the two realization problems for Von Neumann
regular rings, the first concerning algebric distributive lattices and the
second concerning dimension groups are intimately connected, as illustrated by
P. P. Růžička in [31] and [32].

Recently L.Vas accomplished in [33] the task of extending the concepts of a
simplicial group and of a dimension group to partially ordered abelian groups
having an additional action of a group $\varGamma$; by considering the
$\varGamma$-graded Grothendieck group $\mathbf{K}_{0}^{\varGamma}(R)$ of a
$\varGamma$-graded ring $R$ (as introduced in [24]) she studied the
corresponding realization problems of the above graded dimension groups as
graded Grothendieck groups.

The deep role of countable dimension groups in the theory of topological
dynamical systems is widely displayed in the monograph [14].

If $I$ is a partially ordered set and $G(I)$ denotes the free abelian group
generated by $I$, the _(restricted) Hahn power_ of $\mathbb{Z}$ by $I$ is the
group $G(I)$, equipped with the partial order defined by declaring positive
those elements $x=\sum\\{x_{i}i\mid x_{i}\in\mathbb{Z},i\in I\\}$ of $G(I)$
such that $x_{i}\geqslant 0$ whenever $i$ is a maximal element of the support
of $x$ (see Section 5 for the general definition). In [11] we proved that
$G(I)$ is a dimension group and, if $I$ has a finite cofinal subset, the sum
of all maximal elements of $I$ is a order unit; thus we proved that if $I$ is
an artinian poset with a finite cofinal subset, then $G(I)$ is realizable as
$\mathbf{K}_{0}(R)$, where $R$ is a unit-regular and semiartinian ring whose
primitive spectrum is isomorphic to $I$.

Our main goal in the present work is to show that if $I$ is any partially
ordered set, with the only condition of having a finite cofinal subset, then
there exists a locally matricial unital algebra $R$ such that
$\mathbf{K}_{0}(R)$ is isomorphic to $G(I)$. In order to give some more
detail, for a given _infinite_ cardinal $\boldsymbol{\aleph}$, let us call
_$\boldsymbol{\aleph}$ -poset_ every partially ordered set $I$ such that both
the cardinalities of $I$ and of the set $\mathcal{M}(I)$ of all maximal chains
of $I$ do not exceed $\boldsymbol{\aleph}$. Given a field $\mathbb{K}$, for
every $\boldsymbol{\aleph}$-poset $I$ _having a finite cofinal subset_ we
build a locally matricial, unital $\mathbb{K}$-algebra
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ having the following main
properties: the lattice of all ideals of
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ is isomorphic to the
lattice of all lower subsets of $I$ and the Grothendieck group
$\mathbf{K}_{0}(\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I))$ is order
isomorphic to $G(I)$. Thus, while the first property simply confirms one part
of Bergman’s Theorem, with the second property we enlarge the class of
dimension groups with order unit which are realizable as $\mathbf{K}_{0}$ of
some unital locally matricial algebra, by adding the restricted Hahn powers of
$\mathbb{Z}$ by any partially ordered set $I$ having a finite cofinal subset.

The first two sections have a merely technical nature. We start with an
$\boldsymbol{\aleph}$-poset $I$ and we consider the set
$X(I)\,\,\colon\\!\\!=\boldsymbol{\aleph}^{(I)}\times\mathcal{M}(I)$, where
$\boldsymbol{\aleph}^{(I)}$ denotes the set of all maps from $p\colon
I\to\boldsymbol{\aleph}$ such that $p(i)=\emptyset$ for all but a finite
number of $i\in I$, and we establish the algebra
$Q(I)\,\,\colon\\!\\!=\mathbb{C}\mathbb{F}\mathbb{M}_{\mathbb{K}}(X(I))$ of
all $X(I)\times X(I)$ column-finite matrices over $\mathbb{K}$ as the scenario
for the construction of the algebra
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$. Note that
$Q(I)\simeq\mathbb{C}\mathbb{F}\mathbb{M}_{\mathbb{K}}(\boldsymbol{\aleph})$,
as $|X(I)|=\boldsymbol{\aleph}$. By using suitable partitions of $X(I)$, for
every $i\in I$ we operate two subsequent selections of specially shaped
matrices in $Q(I)$. With the first one we obtain a $\mathbb{K}$-subalgebra
$Q(I,i)$ of $Q(I)$, isomorphic to
$\mathbb{C}\mathbb{F}\mathbb{M}_{\mathbb{K}}(\boldsymbol{\aleph})$, in such a
way that if $i,j\in I$, then $Q(I,i)\subset Q(I,j)$ if and only if $j\leqslant
i$. With the second selection we find for each $i\in I$ a
$\mathbb{K}$-subalgebra $S(I,i)$ of $Q(I,i)$, still isomorphic to
$\mathbb{C}\mathbb{F}\mathbb{M}_{\mathbb{K}}(\boldsymbol{\aleph})$, in such a
way that if $i,j\in I$, then $S(I,i)S(I,j)=0$ if and only if $i,j$ are not
comparable, while $S(I,i)S(I,j)\cup S(I,j)S(I,i)\subset S(I,i)$ if $i\leqslant
j$.

In the third section we define the family $(H(I,i))_{i\in I}$ of “building
blocks” which will enable us to define the algebra
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$. For every $i\in I$ we take
$H(I,i)=\operatorname{Soc}(S(I,i))$ if $i$ is not a maximal element of $I$, so
that $H(I,i)$ is isomorphic to the algebra of all finite ranked linear
tranformations of an $\boldsymbol{\aleph}$-dimensional vector space over
$\mathbb{K}$, hence it is a simple, semisimple non-artinian ring; if $i$ is a
maximal element of $I$ we select $H(I,i)$ as a suitable subalgebra isomorphic
to $\mathbb{K}$. We prove that $(H(I,i))_{i\in I}$ is an independent family of
$\mathbb{K}$-subspaces of $Q(I)$ with the following properties: given $i,j\in
I$, if $i,j$ are not comparable then $H(I,i)H(I,j)=\\{\mathbf{0}\\}$
otherwise, if $i\leqslant j$ and $\mathbf{0}\neq\mathbf{a}\in H(I,j)$, then
both $\mathbf{a}H(I,i)$ and $H(I,i)\mathbf{a}$ are not zero and are contained
in $H(I,i)$. As a consequence, for every subset $J\subset I$ the subspace
$H(I,J)=\bigoplus_{j\in J}H(I,j)$ is a $\mathbb{K}$-subalgebra of $Q(I)$. We
explore the conditions under which $H(I,J)$ has a multiplicative identity; in
particular this turns out to be the case if $J$ is an upper subset of $I$
which is bounded by above by a finite number of maximal elements of $I$.

Starting with the fourth section we assume that the
$\boldsymbol{\aleph}$-poset $I$ _has a finite cofinal subset_ and, finally, we
define
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)\,\,\colon\\!\\!=H(I,I)$,
which is a unital $\mathbb{K}$-algebra by the above. We provide an explicit
description of the ideals of
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$; more generally we prove
that, given any subset $J$ of $I$, the assignment $L\mapsto H(I,L)$ defines an
isomorphism from the lattice of all lower subsets of $J$ to the lattice of
ideals of $H(I,J)$. We also prove that
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ is a locally matricial
$\mathbb{K}$-algebra, so that it is unit-regular. As previously announced,
this represent a confirmation of the part of Bergman’s Theorem which states
that every algebric distributive lattice $D$, in which the greatest element is
compact and every element of $D$ is a join of _join-irreducible_ compact
elements, is isomorphic to the lattice of ideals of some unital von Neumann
regular ring $R$.

The fifth section is entirely devoted to the computation of the Grothendieck
group of $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ and, as we already
mentioned previously, the main result is that
$\mathbf{K}_{0}(\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I))$ is
isomorphic to $G(I)$, the restricted Hahn power of $\mathbb{Z}$ by $I$.

In the sixth section we investigate when the algebra
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ is prime and when it is
primitive. In his great two volumes book “Ring Theory” Rowen wrote: “The
abstract theory of noncommutative regular rings has been a disappointment”
(see [29, p.277]) because some known basic facts about commutative regular
rings, which were hoped to be true in general, turned out to fail. For
instance, the question whether every prime regular ring is primitive has an
obvious positive answer in the commutative case. By 1970 Kaplansky conjectured
an affirmative answer in general (see [26, p.2]) and, again, Rowen commented
in [30, p.308] that “…until recently the major question in the theory of (von
Neumann) regular rings was whether a prime regular ring need be primitive”, a
comment already mentioned in [2]. A first counterexample to Kaplansky
conjecture was found in 1977 by Domanov [13], but only thirty seven years
later a whole class of counterexamples was found by Abrams, Bell and
Rangaswamy; in [2] they found necessary and sufficient conditions for an
oriented graph $E$ in order that the Leavitt algebra $L(E)$ (over any field)
is prime and those for which $L(E)$ is primitive. These conditions involve the
preorder induced in the set $E_{0}$ of vertices of $E$ by the paths; namely,
if $v,w\in E_{0}$, then declare that $v\leqslant w$ if and only if there is a
path from $v$ to $w$ (including vertices as zero length paths); thus they
proved that $L(E)$ is prime if and only if $E_{0}$ is downward directed, while
$L(E)$ is primitive if and only if: (i) $E_{0}$ is downward directed, (ii)
every cycle in $E$ has an exit and (iii) $E_{0}$ has a countable coinitial
subset $S$. Now $L(E)$ is regular if and only if $E$ is acyclic (see [2,
Theorem 1], in which the case the above preorder $\leqslant$ is a partial
order; as a consequence, Abrams, Bell and Rangaswamy result gives rise to
whole classes of prime and non-primitive regular rings.

With our result we get new classes of regular algebras which are prime but not
primitive. In fact we prove that
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ is prime if and only if the
poset $I$ is downward directed, while it is right primitive if and only if $I$
has a coinitial chain, in which the case it is also left primitive; if
$\boldsymbol{\aleph}>\boldsymbol{\aleph}_{0}$ there are plenty of downward
directed $\boldsymbol{\aleph}$-posets without a coinitial chain. Curiously
enough, the set of conditions on an acyclic graph $E$ for primitivity of the
Leavitt algebra $L(E)$ is very close to the condition on a poset $I$ in order
that $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ is primitive; in fact,
it is not difficult to show that if a poset is downward directed and has a
countable coinitial subset, then it has a countable coinitial chain, therefore
Abrams, Bell and Rangaswamy result may be restated as follows: if $E$ is an
acyclic graph, then the regular algebra $L(E)$ is primitive if and only if the
set of vertices $E_{0}$, partially ordered by paths, has a countable coinitial
chain.

By combining our result with the characterization of the ideals of
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$, we obtain that an ideal
$\mathfrak{P}$ of $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ is
primitive if and only if there is a (not necessarily unique) chain $A$ of $I$
such that $\mathfrak{P}=H(I,\\{A\nleqslant\\})$, where $\\{A\nleqslant\\}$ is
the complement set in $I$ of the upper subset $\\{A\leqslant\\}$ generated by
$A$; of course there is a $\mathbb{K}$-algebra isomorphism
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)/\mathfrak{P}\simeq
H(I,\\{A\leqslant\\})$. In particular, for every $i\in I$ we have the
primitive ideal $H(I,\\{\\{i\\}\nleqslant\\})$ of
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$; by denoting with
${\mathbf{P}\mathbf{r}\mathbf{i}\mathbf{m}}_{\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)}$
the primitive spectrum of $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$
(partially ordered by inclusion), the assignment $i\mapsto
H(I,\\{\\{i\\}\nleqslant\\})$ defines an order embedding $\Psi\colon
I\to{\mathbf{P}\mathbf{r}\mathbf{i}\mathbf{m}}_{\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)}$.
We prove that $\Psi$ is an order isomorphism if and only if $I$ is artinian,
if and only if the algebra $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$
is semiartinian. This means that every artinian poset, with a finite cofinal
subset, can be realized as the primitive spectrum of a regular ring.

In the seventh and last section we explore the possible functorial extensions
of the assignments $I\mapsto G(I)$ and
$I\mapsto\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$. We find that if we
consider the category ${\mathbf{P}\mathbf{o}\mathbf{s}}$ whose objects are all
posets, while the morphisms are all maps which are order imbeddings of the
domain as an upper subset of the codomain, then the assignment $I\mapsto G(I)$
extends in a “natural” way to a pair of functors $G(-)$, $G^{*}(-)$, the first
covariant and the second contravariant, from
${\mathbf{P}\mathbf{o}\mathbf{s}}$ to the category whose objects are all
partially ordered abelian groups and isotone group homomorphisms. On the other
hand, given an infinite cardinal $\boldsymbol{\aleph}$ and a field
$\mathbb{K}$, the assignment
$I\mapsto\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ extends to a pair
of functors $\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(-)$,
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}^{*}(-)$, the first covariant
and the second contravariant, from the subcategory
${\mathbf{P}\mathbf{o}\mathbf{s}}_{\boldsymbol{\aleph}}$ of
${\mathbf{P}\mathbf{o}\mathbf{s}}$ of all $\boldsymbol{\aleph}$-posets _having
a finite cofinal subset_ to the category
${\mathbf{A}\mathbf{l}\mathbf{g}}_{\mathbb{K}}$ of _unital_
$\mathbb{K}$-algebras and _not necessarily unital_ $\mathbb{K}$-algebra
homomorphisms. It turns out that these four functors match with the functor
$\mathbf{K}_{0}(-)$, in the sense that there are natural equivalences
$\rho(-)\colon G(-)\approx\mathbf{K}_{0}(-)\cdot\mathfrak{B}(-)$ and
$\rho^{*}(-)\colon G^{*}(-)\approx\mathbf{K}_{0}(-)\cdot\mathfrak{B}^{*}(-)$.

Concerning notations and terminology, we follow the (by now) universally
established practice in rings and modules theory. However, we use the symbol
$A\subset B$ with the meaning “$A$ is a subset of $B$” and write $A\subsetneqq
B$ to indicate that “$A$ is a proper subset of $B$”. Finally, we always
consider the number $0$ as a member of the set $\mathbb{N}$ of natural
numbers.

##  1\. Representing a partially ordered set $I$: the main frame.

Let us start by recalling some basic concept and introduce notations
concerning partially ordered sets that we shall use extensively. If $I$ is a
partially oredred set, we denote by $\mathcal{M}(I)$ the set of all maximal
chains of $I$; the _Hausdorff maximal principle_ (which is equivalent to the
Choice Axiom) states that every chain of $I$ is contained in some maximal one.
If not otherwise stated, we shall consider every subset $J$ of $I$ as a
partially ordered set endowed with the partial order induced by that of $I$.
We denote by $\min(I)$ and $\max(I)$ respectively the smallest and the
greatest element of $I$ (when they exists); also, we denote by $\mathbf{M}(I)$
the set of all maximal elements of $I$.

For every subset $J\subset I$ we define

$\displaystyle\\{\leqslant J\\}\,\,\colon\\!\\!=\\{k\in I\mid k\leqslant
j\text{ for al }j\in J\\},$
$\displaystyle\\{J\leqslant\\}\,\,\colon\\!\\!=\\{k\in I\mid j\leqslant
k\text{ for al }j\in J\\},$ $\displaystyle\\{\nleqslant
J\\}\,\,\colon\\!\\!=I\setminus\\{\leqslant J\\}$
$\displaystyle\\{J\nleqslant\\}\,\,\colon\\!\\!=I\setminus\\{J\leqslant\\};$

if $J=\\{i\\}$ for a single element $i\in I$, we shall simplify notations by
writing $\\{i\leqslant\\}$ or $\\{i\nleqslant\\}$, for instance, in place of
$\\{\\{i\\}\leqslant\\}$ or $\\{\\{i\\}\nleqslant\\}$ respectively. A lower
subset (resp. upper subset) of $I$ is a subset $J\subset I$ such that if $j\in
J$, then $\\{\leqslant j\\}\subset J$ (resp. $\\{j\leqslant\\}\subset J$).
Clearly $\\{\leqslant J\\}$ and $\\{J\leqslant\\}$ are respectively the
smallest lower subset and the smallest upper subset of $I$ which contain $J$.
While the above notations are largely standard, in order to simplify notations
when nested parentheses should occur and provided that it does not create any
conflict in the specific context, we will consistently omit almost everywhere
curled braces; thus we will write

$\leqslant J\,,\quad J\leqslant\,,\quad\leqslant i\,,\quad
i\leqslant\,,\quad\nleqslant J\,,\quad J\nleqslant\,,\quad\nleqslant i\,,\quad
i\nleqslant\,.$

instead of

$\\{\leqslant J\\}\,,\quad\\{J\leqslant\\}\,,\quad\\{\leqslant
i\\}\,,\quad\\{i\leqslant\\}\,,\quad\\{\nleqslant
J\\}\,,\quad\\{J\nleqslant\\}\,,\quad\\{\nleqslant
i\\}\,,\quad\\{i\nleqslant\\}$

respectively. We denote by $\Uparrow\\!\\!I$ (resp. $\Downarrow\\!\\!I$) the
set of all upper subsets (resp. lower subsets) of $I$; both $\Uparrow\\!\\!I$
and $\Downarrow\\!\\!I$ are complete and distributive lattices and the map
$J\mapsto I\setminus J$ is an anti-isomorphism from $\Uparrow\\!\\!I$ to
$\Downarrow\\!\\!I$.

Throughout the present paper $\boldsymbol{\aleph}$ stands for a given infinite
cardinal. For every set $A$, we denote by $\boldsymbol{\aleph}^{(A)}$ the set
of all maps $p$ from $A$ to $\boldsymbol{\aleph}$ such that $p(a)=\emptyset$
for all but a finite number of $a\in A$ 111As it is the standard practice
within ZFC set theory, we consider $\boldsymbol{\aleph}$ as a special ordinal;
so $\boldsymbol{\aleph}$ is the set of all ordinals strictly less than
$\boldsymbol{\aleph}$ and thus, in particular,
$\emptyset\in\boldsymbol{\aleph}$..

######  Proposition 1.1.

If $A$ is a set such that $0<|A|\leqslant\boldsymbol{\aleph}$, then

$\left|\boldsymbol{\aleph}^{(A)}\right|=\boldsymbol{\aleph}.$

###### Proof.

Let $\mathcal{F}$ be the set of all finite subsets of $A$ and, for every
$K\in\mathcal{F}$, let $X_{K}$ be the set of those maps $p\colon
A\to\boldsymbol{\aleph}$ such that $p(A\setminus K)=\\{\emptyset\\}$. Then

$|X_{K}|=\left|\boldsymbol{\aleph}^{K}\right|=|\boldsymbol{\aleph}|$

for every $K\in\mathcal{F}$. Since

$\boldsymbol{\aleph}^{(A)}=\bigcup\\{X_{K}\mid K\in\mathcal{F}\\}$

and $|\mathcal{F}|=|A|$ in case $A$ is infinite, we infer that

$\left|\boldsymbol{\aleph}^{(A)}\right|=|\mathcal{F}|\cdot\boldsymbol{\aleph}=\boldsymbol{\aleph},$

as claimed. ∎

If $A$ is a set with $|A|=\boldsymbol{\aleph}$, we say that a partition
$\mathcal{E}$ of $A$ is an $\boldsymbol{\aleph}$-partition if
$|\mathcal{E}|=\boldsymbol{\aleph}$ and $|B|=\boldsymbol{\aleph}$ for all
$B\in\mathcal{E}$. Given two partitions $\mathcal{E}$, $\mathcal{F}$ of $A$,
we say that $\mathcal{F}$ is $\boldsymbol{\aleph}$-coarser than $\mathcal{E}$
if each element of $\mathcal{F}$ is the union of $\boldsymbol{\aleph}$
elements of $\mathcal{E}$.

We say that a partially ordered set $I$ is an _$\boldsymbol{\aleph}$ -poset_
in case both $I$ and $\mathcal{M}(I)$ have cardinality at most
$\boldsymbol{\aleph}$. As we are going to see, if $I$ is a non-empty
$\boldsymbol{\aleph}$-poset, then $\boldsymbol{\aleph}^{(I)}$ admits a family
$(\mathcal{E}_{i})_{i\in I}$ of partitions with the following features: (1)
$\mathcal{E}_{i}$ is an $\boldsymbol{\aleph}$-partition for every
$i\neq\min(I)$, (2) if $i<j$, then $\mathcal{E}_{j}$ is
$\boldsymbol{\aleph}$-coarser than $\mathcal{E}_{i}$, (3) if $i\nleqslant j$,
then no member of $\mathcal{E}_{i}$ is contained in any member of
$\mathcal{E}_{j}$. First observe that, given $i\in I$, every
$p\in\boldsymbol{\aleph}^{(I)}$ decomposes uniquely as a disjoint union

$p=r\cup u,$

where $r\in\boldsymbol{\aleph}^{(i\nleqslant)}$ and
$u\in\boldsymbol{\aleph}^{(i\leqslant)}$: simply take
$r=p\left|{}_{\\{i\nleqslant\\}}\right.$ and
$u=p\left|{}_{\\{i\leqslant\\}}\right.$. For every
$u\in\boldsymbol{\aleph}^{(i\leqslant)}$, let us consider the subset

(1.1) $E_{u}\,\,\colon\\!\\!=\left\\{r\cup
u\left|\,\,r\in\boldsymbol{\aleph}^{(i\nleqslant)}\right.\right\\}=\left\\{\left.p\in\boldsymbol{\aleph}^{(I)}\right|\,\,p\left|{}_{\\{i\leqslant\\}}\right.=u\right\\}\neq\emptyset$

of $\boldsymbol{\aleph}^{(I)}$; clearly, if
$u,u^{\prime}\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $u\neq u^{\prime}$,
then $E_{u}\cap E_{u^{\prime}}=\emptyset$ and therefore

(1.2)
$\mathcal{E}_{i}\,\,\colon\\!\\!=\left\\{E_{u}\left|\,\,u\in\boldsymbol{\aleph}^{(i\leqslant)}\right.\right\\}$

is a partition of $\boldsymbol{\aleph}^{(I)}$. If $i\neq\min(I)$, then
$\mathcal{E}_{i}$ is an $\boldsymbol{\aleph}$-partitionas by Proposition 1.1;
if $i=\min(I)$, namely $\\{i\nleqslant\\}=\emptyset$, then
$\boldsymbol{\aleph}^{(i\nleqslant)}=\\{\emptyset\\}$ and therefore
$E_{u}=\\{\emptyset\cup u\\}=\\{u\\}$ for every
$u\in\boldsymbol{\aleph}^{(i\leqslant)}=\boldsymbol{\aleph}^{(I)}$, so that
$\mathcal{E}_{i}=\left\\{\\{u\\}\left|\,\,u\in\boldsymbol{\aleph}^{(I)}\right.\right\\}$.

######  Lemma 1.2.

With the above settings and notations, the following properties hold:

  1. (1)

If $i,j\in I$ and $i<j$, then $\mathcal{E}_{j}$ is
$\boldsymbol{\aleph}$-coarser than $\mathcal{E}_{i}$; specifically, for every
$v\in\boldsymbol{\aleph}^{(j\leqslant)}$ we have that

(1.3)
$E_{v}=\bigcup\left\\{E_{u}\left|\,\,u\in\boldsymbol{\aleph}^{(i\leqslant)}\text{
and }u\left|{}_{\\{j\leqslant\\}}\right.=v\right.\right\\}.$

  2. (2)

Given $i\in I$ and a finite subset $J\subset I$, assume that there are

$B\in\mathcal{E}_{i}\quad\text{and}\quad C_{1},\ldots,C_{n}\in\bigcup_{j\in
J}\mathcal{E}_{j}$

such that $B\subset C_{1}\cup\cdots\cup C_{n}$. Then $i\in\\{\leqslant J\\}$.
Consequently, if $j\in I$ and $i\nleqslant j$, then no member of
$\mathcal{E}_{i}$ is contained in some member of $\mathcal{E}_{j}$.

###### Proof.

(1) Assume that $i<j$ and observe that

(1.4)
$\\{\,\\{i\nleqslant\\},\,\,\\{i\leqslant\\}\setminus\\{j\leqslant\\},\,\\{j\leqslant\\}\,\\}$

is a partition of $I$. Accordingly, every element
$p\in\boldsymbol{\aleph}^{(I)}$ decomposes uniquely as a disjoint union

$p=r\cup s\cup v,$

where $r\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
$s\in\boldsymbol{\aleph}^{(\\{i\leqslant\\}\setminus\\{j\leqslant\\})}$ and
$v\in\boldsymbol{\aleph}^{(j\leqslant)}$. Consequently, for every
$v\in\boldsymbol{\aleph}^{(j\leqslant)}$ we have that

$E_{v}=\left\\{r\cup s\cup
v\left|\,\,r\in\boldsymbol{\aleph}^{(i\nleqslant)}\text{ and
}s\in\boldsymbol{\aleph}^{(\\{i\leqslant\\}\setminus\\{j\leqslant\\})}\right.\right\\}$

and hence (1.3) follows.

(2) Under the assumption of (2) suppose that, on the contrary,
$i\not\in\\{\leqslant J\\}$. There are
$u\in\boldsymbol{\aleph}^{(i\leqslant)}$, $j_{1},\ldots,j_{n}\in J$ and
$u_{\alpha}\in\boldsymbol{\aleph}^{[\\{j_{\alpha}\leqslant\\}]}$, for
$\alpha\in\\{1,\ldots,n\\}$, such that

$B=E_{u}\quad\text{and}\quad C_{\alpha}=E_{u_{\alpha}}.$

According to the assumption, if $p\in\boldsymbol{\aleph}^{(I)}$ and
$\displaystyle p\left|{}_{\\{i\leqslant\\}}\right.=u$, then there is some
$\alpha\in\\{1,\ldots,n\\}$ such that $\displaystyle
p\left|{}_{\\{j_{\alpha}\leqslant\\}}\right.=u_{\alpha}$. Set

$D=u_{1}(\\{j_{1}\leqslant\\})\cup\cdots\cup u_{n}(\\{j_{n}\leqslant\\}).$

By the finiteness of $D$, we can choose some $b\in\boldsymbol{\aleph}\setminus
D$ and, since $j_{\alpha}\not\in\\{i\leqslant\\}$ for all
$\alpha\in\\{1,\ldots,n\\}$, we can define the map
$p\in\boldsymbol{\aleph}^{I}$ by setting

$p\left|{}_{\\{i\leqslant\\}}\right.=u,\quad
p(j_{1})=\cdots=p(j_{n})=b\quad\text{and}\quad
p(I\setminus(\\{i\leqslant\\}\cup\\{j_{1},\ldots,j_{n}\\})=\\{\emptyset\\}.$

Then we have that $p\in E_{u}$. However, by the choice of $b$ we have that
$p(j_{\alpha})=b\neq u_{\alpha}(j_{\alpha})$ for every
$\alpha\in\\{1,\ldots,n\\}$. This means that $p\not\in C_{1}\cup\cdots\cup
C_{n}$ and we have a contradiction. ∎

Starting from this point we let $\mathbb{K}$ be a given field. Given an
$\boldsymbol{\aleph}$-poset $I$, we set

$X(I)\,\,\colon\\!\\!=\boldsymbol{\aleph}^{(I)}\times\mathcal{M}(I);$

the main frame within which we will find the regular $\mathbb{K}$-algebra
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$ associated to $I$ is the
$\mathbb{K}$-algebra

$Q(I)\,\,\colon\\!\\!=\mathbb{C}\mathbb{F}\mathbb{M}_{X(I)}(\mathbb{K})$

of all finite-column $X(I)\times X(I)$-matrices with entries in $\mathbb{K}$;
note that $|X(I)|=\boldsymbol{\aleph}$ by Proposition 1.1, therefore
$Q(I)\simeq\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}}(\mathbb{K})$.
If $I=\emptyset$, then $X(I)=\\{(\emptyset,\emptyset)\\}$ and $Q(I)$ is
identified with $\mathbb{K}$.

######  Notations 1.3.

With the setting, we adopt the following notations:

  * •

We denote respectively by $\mathbf{0}$ and $\mathbf{1}$ the zero and the
unital matrices of Q(I).

  * •

If $\mathbf{a}\in Q(I)$ and $x,y\in X(I)$, we use the symbol $\mathbf{a}(x,y)$
to denote the entry at the intersection of the $x$-th row with the $y$-th
column of $\mathbf{a}$ (i. e. the $(x,y)$-entry of $\mathbf{a}$), instead of
the more traditional symbol $\mathbf{a}_{xy}$; since we often use more complex
arrays, other than single letters, in order to designate the position of the
entries of the matrices we deal with, our choice should guarantee a better
readability. If $Y,Z\subset X(I)$, then $\mathbf{a}(Y,Z)$ denotes the
$(Y,Z)$-block of $\mathbf{a}$, that is the submatrix $(\mathbf{a}(x,y))_{y\in
Y,z\in Z}$ of $\mathbf{a}$.

  * •

For every $Y\subset X(I)$, we denote with $\mathbf{e}_{Y}$ the idempotent
diagonal matrix such that $\mathbf{e}_{Y}(x,x)$ is $1$ if $x\in Y$ and is $0$
otherwise. If $x,y\in X(I)$, we write $\mathbf{e}_{x}$ instead of
$\mathbf{e}_{\\{x\\}}$, while $\mathbf{e}_{x,y}$ stands for the usual
$(x,y)$-matrix unit, namely the matrix whose $(x,y)$-entry is $1$ and all
others are zero; so, in particular $\mathbf{e}_{x}=\mathbf{e}_{x,x}$.

  * •

$\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K})$ denotes the subset of $Q(I)$ of all
matrices having only finitely many nonzero rows.

Of course we shall consider $\mathbb{K}$ as a subring of $Q(I)$ by identifying
each element of $\mathbb{K}$ with the corresponding scalar matrix in $Q(I)$.
Without any assumption on the ring $\mathbb{K}$ (apart being unitary), it is
well known that $\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K})$ is a fully invariant
ideal of $Q(I)$ and the following equality holds:

$\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K})=\bigoplus\\{\mathbf{e}_{x}Q(I)\mid
x\in X(I)\\};$

moreover, for every $x\in X(I)$ the following hold:

$\displaystyle\mathbf{e}_{x}Q(I)=\mathbf{e}_{x}\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K}),$
$\displaystyle\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K})=\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K})\mathbf{e}_{x}\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K})=Q(I)\mathbf{e}_{x}Q(I).$

With our assumption that $\mathbb{K}$ is a field, $Q(I)$ is regular and right
selfinjective; moreover
$\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K})=\operatorname{Soc}(Q(I))$ is a
minimal ideal and $\\{\mathbf{e}_{x}\mid x\in X(I)\\}$ is a set of pairwise
orthogonal primitive idempotents; it is _complete_ , in the sense that it
generates $\operatorname{Soc}(Q(I))$ as a right ideal.

If $R$ is any ring and $\varphi\colon Q(I)\to R$ is a ring isomorphism, then
$\mathcal{E}=\\{\varphi(\mathbf{e}_{x})\mid x\in X(I)\\}$ generates
$\operatorname{Soc}(R)$ as a right ideal; we emphasize that this latter does
not depend on the specific isomorphism $\varphi$. We say that a set
$\mathcal{S}$ of idempotents of $R$ is a _complete set of pairwise orthogonal
and primitive idempotents_ if $\mathcal{S}=\varphi(\mathcal{E})$ for some
isomorphism $\varphi$ as above.

The family $(\mathcal{E}_{i})_{i\in I}$ of partitions of
$\boldsymbol{\aleph}^{(I)}$ previously introduced induces the family
$(\mathcal{P}_{i})_{i\in I}$ of partitions of $X(I)$ defined as follows. Given
$i\in I$, for every
$(u,A)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)$ we consider
the subset

$\displaystyle X(I)_{(u,A)}\,\,\colon\\!\\!=E_{u}\times\\{A\\}=\left\\{(r\cup
u,A)\left|\,\,r\in\boldsymbol{\aleph}^{(i\nleqslant)}\right.\right\\}$
$\displaystyle=\left\\{(p,A)\left|\,\,p\in\boldsymbol{\aleph}^{(I)}\text{ and
}p\left|{}_{\\{i\leqslant\\}}\right.\right.=u\right\\}$

(see (1.1)) of $X(I)$ and set

(1.5)
$\mathcal{P}_{i}\,\,\colon\\!\\!=\left\\{X(I)_{(u,A)}\left|\,\,(u,A)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)\right.\right\\}.$

Note that if $i=\min(I)$, then $X(I)_{(u,A)}=\\{(u,A)\\}$ and hence

$\mathcal{P}_{\min(I)}=\left\\{\\{(u,A)\\}\left|\,\,u\in\boldsymbol{\aleph}^{(I)},A\in\mathcal{M}(I)\right.\right\\}=\\{\\{x\\}\mid
x\in X(I)\\}.$

As a consequence of Lemma 1.2 we have the following:

######  Corollary 1.4.

Given $i\in I$, if $i\neq\min(I)$ then $\mathcal{P}_{i}$ is an
$\boldsymbol{\aleph}$-partition of $X(I)$. if $i,j\in I$, then
$\mathcal{P}_{j}$ is $\boldsymbol{\aleph}$-coarser than $\mathcal{P}_{i}$ in
case $i<j$, while no member of $\mathcal{P}_{i}$ is contained in any member of
$\mathcal{P}_{j}$ if $i\nless j$.

As a first step toward our goals, we determine in $Q(I)$ a family
$(Q(I,i))_{i\in I}$ of unital $\mathbb{K}$-subalgebras, each isomorphic to
$Q(I)$, such that $Q(I,i)\subset Q(I,j)$ exactly when $i\leqslant j$.

For every $i\in I$, let $Q(I,i)$ be the subset of $Q(I)$ consisting of those
matrices $\mathbf{a}$ satisfying the following condition: for every
$r,s,t\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$

(1.6) $\displaystyle\mathbf{a}((r\cup u,A),(s\cup v,B))$
$\displaystyle=\delta(r,s)\,\mathbf{a}((t\cup u,A),(t\cup v,B));$

equivalently, for every
$(u,A),(v,B)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)$ there
is a scalar $d\in\mathbb{K}$ such that

$\mathbf{a}((r\cup u,A),(s\cup v,B))=\delta(r,s)d$

for all $r,s\in\boldsymbol{\aleph}^{(i\nleqslant)}$. We observe that if
$(u,A)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)$, then the
assignment $r\mapsto(r\cup u,A)$ defines a bijection from
$\boldsymbol{\aleph}^{(i\nleqslant)}$ to $X(I)_{(u,A)}$; consequently, for
every $(u,A),(v,B)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)$,
we are allowed to consider the block $\mathbf{a}(X(I)_{(u,A)},X(I)_{(v,B)})$
of $\mathbf{a}$ as a
$\boldsymbol{\aleph}^{(i\nleqslant)}\times\boldsymbol{\aleph}^{(i\nleqslant)}$-matrix
and, as such, the above condition (1.6) just impose that this is a scalar
matrix.

If $i=\min(I)$, it is clear that $Q(I,i)=Q(I)$.

######  Theorem 1.5.

With the above settings and notations, the following properties hold:

  1. (1)

For every $i\in I$, the map

$\varphi_{i}\colon\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)}(\mathbb{K})\longrightarrow
Q(I)$

defined by

(1.7) $\varphi_{i}(\mathbf{a})((r\cup u,A),(s\cup
v,B))=\delta(r,s)\,\mathbf{a}((u,A),(v,B))$

for all
$\mathbf{a}\in\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)}(\mathbb{K})$,
$r,s,\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$, is a
unital and injective $\mathbb{K}$-algebra homomorphism such that
$\operatorname{Im}(\varphi_{i})=Q(I,i)$; consequently $Q(I,i)$ is a unital
$\mathbb{K}$-subalgebra of $Q(I)$ isomorphic to
$\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}}(\mathbb{K})$.

  2. (2)

For every $i\in I$, $u\in\boldsymbol{\aleph}^{(i\leqslant)}$ and
$A\in\mathcal{M}(I)$

(1.8) $\varphi_{i}(\mathbf{e}_{(u,A)})=\mathbf{e}_{X(I)_{(u,A)}};$

consequently $\\{\mathbf{e}_{X}\mid X\in\mathcal{P}_{i}\\}$ is a complete set
of pairwise orthogonal and primitive idempotents of $Q(I,i)$.

  3. (3)

For every $i\in I$, a matrix $\mathbf{b}\in Q(I,i)$ belongs to

$\operatorname{Soc}(Q(I,i))=\varphi_{i}\left(\mathbb{F}\mathbb{R}_{\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)}(\mathbb{K})\right)$

if and only if it satisfies the following condition:

$\displaystyle\,\,\,\,(\star)\quad$ there are
$X_{1},\ldots,X_{k}\in\mathcal{P}_{i}$ such that if the $x$-th
$\displaystyle\text{row of $\mathbf{b}$ is not zero, then $x\in
X_{1}\cup\cdots\cup X_{k}$}.$

  4. (4)

If $i,j\in I$, then $Q(I,j)\subset Q(I,i)$ if and only if $i\leqslant j$.

  5. (5)

The set $\\{\operatorname{Soc}(Q(I,i))\mid i\in I\\}$ of
$\mathbb{K}$-subspaces of $Q(I)$ is independent.

###### Proof.

(1) Let $i\in I$. Given $r,s,t\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$, we have
that

$\displaystyle\varphi_{i}(\mathbf{a})((r\cup u,A),(s\cup v,B))$
$\displaystyle=\delta(r,s)\,\mathbf{a}((u,A),(v,B))$
$\displaystyle=\delta(r,s)\,\delta(t,t)\,\mathbf{a}((u,A),(v,B))$
$\displaystyle=\delta(r,s)\,\varphi_{i}(\mathbf{a})((t\cup u,A),(t\cup v,B)),$

therefore $\operatorname{Im}(\varphi_{i})\subset Q(I,i)$ in view of (1.6).
Conversely, given $\mathbf{b}\in Q(I,i)$, let
$\mathbf{a}\in\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)}(\mathbb{K})$
be the matrix defined as follows: take an arbitrary
$r\in\boldsymbol{\aleph}^{(i\nleqslant)}$ and, for every
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$, set

$\mathbf{a}((u,A),(v,B))=\mathbf{b}((r\cup u,A),(r\cup v,B));$

this definition does not depend on the choice of $r$, thanks to the condition
(1.6). Then, for every $r,s,\in\boldsymbol{\aleph}^{(\nleqslant i)}$,
$u,v\in\boldsymbol{\aleph}^{(\leqslant i)}$ and $A,B\in\mathcal{M}(I)$, we
have

$\displaystyle\varphi_{i}(\mathbf{a})((r\cup u,A),(s\cup v,B))$
$\displaystyle=\delta(r,s)\,\mathbf{a}((u,A),(v,B))$
$\displaystyle=\delta(r,s)\,\mathbf{b}((r\cup u,A),(r\cup v,B))$
$\displaystyle=\mathbf{b}((r\cup u,A),(s\cup v,B))\qquad\text{again by
\eqref{magicmatrix}};$

consequently $\mathbf{b}=\varphi_{i}(\mathbf{a})$ and hence
$\operatorname{Im}(\varphi_{i})=Q(I,i)$. It is immediate to check that
$\varphi_{i}(\mathbf{1})=\mathbf{1}$ and that $\varphi_{i}$ is
$\mathbb{K}$-linear. Given
$\mathbf{a},\mathbf{b}\in\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)}(\mathbb{K})$,
for all $r,s,\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$ we have
the following:

$\displaystyle\quad\varphi_{i}(\mathbf{a}\mathbf{b})((r\cup u,A),(s\cup
v,B))=\delta(r,s)\,(\mathbf{a}\mathbf{b})((u,A),(v,B))\qquad\qquad\qquad$
$\displaystyle\begin{aligned}
&=\sum_{(w,C)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)}\delta(r,s)\,\mathbf{a}((u,A),(w,C))\,\mathbf{b}((w,C),(v,B))\\\
&=\sum_{\begin{subarray}{c}(w,C)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)\\\
t\in\boldsymbol{\aleph}^{(i\nleqslant)}\end{subarray}}\delta(r,t)\,\delta(t,s)\,\mathbf{a}((u,A),(w,C))\,\mathbf{b}((w,C),(v,B))\\\
&=\sum_{\begin{subarray}{c}(w,C)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)\\\
t\in\boldsymbol{\aleph}^{(i\nleqslant)}\end{subarray}}\delta(r,t)\,\mathbf{a}((u,A),(w,C))\,\delta(t,s)\,\mathbf{b}((w,C),(v,B))\\\
&=\sum_{\begin{subarray}{c}(w,C)\in\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)\\\
t\in\boldsymbol{\aleph}^{(i\nleqslant)}\end{subarray}}\varphi_{i}(\mathbf{a})((r\cup
u,A),(t\cup w,C))\cdot\varphi_{i}(\mathbf{b})((t\cup w,C),(s\cup v,B))\\\
&=\left(\varphi_{i}(\mathbf{a})\,\varphi_{i}(\mathbf{b})\right)((r\cup
u,A),(s\cup v,B)).\end{aligned}$

Thus $\varphi_{i}$ is a $\mathbb{K}$-algebra homomorphism. Finally, let
$\mathbf{a}\in\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)}(\mathbb{K})$
and assume that $\mathbf{a}((u,A),(v,B))\neq 0$ for some
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$. Then

$\displaystyle(\varphi_{i}(\mathbf{a}))((r\cup u,A),(r\cup v,B))$
$\displaystyle=\delta(r,r)\,\mathbf{a}((u,A),(v,B))=\mathbf{a}((u,A),(v,B))\neq
0$

for all $r\in\boldsymbol{\aleph}^{(i\nleqslant)}$; this shows that
$\varphi_{i}$ is injective. As
$\left|\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(I)\right|=\boldsymbol{\aleph}$,
we infer that
$Q(I,i)\simeq\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}}(\mathbb{K})$.

(2) Equality (1.8) follows by a direct computation.

(3) It follows from (2) that $\operatorname{Soc}(Q(I,i))$, as a right ideal of
$Q(I,i)$, is generated by the set $\\{\mathbf{e}_{X}\mid
X\in\mathcal{P}_{i}\\}$ of pairwise orthogonal idempotents. Thus, given
$\mathbf{b}\in Q(I,i)$, we have that $\mathbf{b}\in\operatorname{Soc}(Q(I,i))$
if and only if there are $X_{1},\ldots,X_{k}\in\mathcal{P}_{i}$ such that

$\mathbf{b}=(\mathbf{e}_{X_{1}}+\cdots+\mathbf{e}_{X_{k}})\mathbf{b},$

hence if and only if $\mathbf{b}$ satisfies ($\star$).

(4) Let $i,j\in I$, suppose that $i<j$ and let $\mathbf{a}\in Q(I,j)$. For
every $x\in X(I)$, let us denote by $r_{x}$, $s_{x}$, $u_{x}$ and $A_{x}$ the
unique elements of $\boldsymbol{\aleph}^{(i\nleqslant)}$,
$\boldsymbol{\aleph}^{(\\{i\leqslant\\}\setminus\\{j\leqslant\\})}$,
$\boldsymbol{\aleph}^{(j\leqslant)}$ and $\mathcal{M}(I)$, respectively, such
that

$x=(r_{x}\cup s_{x}\cup u_{x},A_{x}).$

Then, given $x,y\in X(I)$ and any $t\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
by using (1.6) with $j$ in place of $i$ we have the following equalities:

$\displaystyle\mathbf{a}(x,y)$ $\displaystyle=\delta(r_{x}\cup s_{x},r_{y}\cup
s_{y})\,\mathbf{a}((t\cup s_{x}\cup u_{x},A_{x}),(t\cup s_{x}\cup
u_{y},A_{y}))$
$\displaystyle=\delta(r_{x},r_{y})\,\delta(s_{x},s_{y})\,\mathbf{a}((t\cup
s_{x}\cup u_{x},A_{x}),(t\cup s_{x}\cup u_{y},A_{y}))$
$\displaystyle=\delta(r_{x},r_{y})\,\delta(t\cup s_{x},t\cup
s_{y})\,\mathbf{a}((t\cup s_{x}\cup u_{x},A_{x}),(t\cup s_{x}\cup
u_{y},A_{y}))$ $\displaystyle=\delta(r_{x},r_{y})\,\mathbf{a}((t\cup s_{x}\cup
u_{x},A_{x}),(t\cup s_{y}\cup u_{y},A_{y})).$

Noting that both $s_{x}\cup u_{x}$ and $s_{y}\cup u_{y}$ are in
$\boldsymbol{\aleph}^{(i\leqslant)}$, again from (1.6) we infer that
$\mathbf{a}\in Q(I,i)$.

Conversely, assume that $Q(I,j)\subset Q(I,i)$ and take any
$X\in\mathcal{P}_{j}$. Then $\mathbf{0}\neq\mathbf{e}_{X}\in Q(I,j)\subset
Q(I,i)$ and so, by property (2), there is some $Y\in\mathcal{P}_{i}$ such that
$\mathbf{e}_{Y}\mathbf{e}_{X}\neq\mathbf{0}$, therefore $X\cap
Y\neq\emptyset$. Let $x\in X\cap Y$. As $x=(p,A)$ for some
$p\in\boldsymbol{\aleph}^{(I)}$ and $A\in\mathcal{M}(I)$, we infer that
$X=X(I)_{(u,A)}$ and $Y=X(I)_{(v,A)}$, where $u=p|_{\\{j\leqslant\\}}$ and
$v=p|_{\\{i\leqslant\\}}$. As $\mathbf{e}_{X}(x,x)=1$ and $x\in Y$, since
$\mathbf{e}_{X}\in Q(I,i)$ we infer that $\mathbf{e}_{X}(y,y)=1$ for every
$y\in Y$, proving that $Y\subset X$. Finally it follows from Lemma 1.2,(2)
that $i\leqslant j$, by taking into account that $X=E_{u}\times\\{A\\}$ and
$Y=E_{v}\times\\{A\\}$.

(5) For every subset $J$ of $I$ let us consider the family

$\mathcal{F}_{J}\,\,\colon\\!\\!=\\{\operatorname{Soc}(Q(I,j))\mid j\in J\\}$

and let $J$ be any finite subset of $I$. Since $\mathcal{F}_{J}$ is
independent if $J=\emptyset$, given a positive integer $n$, assume that
$\mathcal{F}_{J}$ is independent whenever $|J|<n$ and suppose that $|J|=n$. We
can choose a maximal element $m$ of $J$, so that if
$J^{\prime}=J\setminus\\{m\\}$, then $\mathcal{F}_{J^{\prime}}$ is independent
by the inductive assumption. We claim that

(1.9) $\operatorname{Soc}(Q(I,m))\cap\bigoplus J^{\prime}=\\{\mathbf{0}\\},$

which proves the independence of $\mathcal{F}_{J}$. Assume that
$\mathbf{0}\neq\mathbf{a}\in\operatorname{Soc}(Q(I,m))$. Then there are
$Y,Z\in\mathcal{P}_{m}$ such that the the block $\mathbf{a}(Y,Z)$ is a nonzero
scalar
$\boldsymbol{\aleph}^{(i\nleqslant)}\times\boldsymbol{\aleph}^{(i\nleqslant)}$-matrix;
in particular the $y$-th row of $\mathbf{a}$ is not zero for all $y\in Y$. On
the other hand, if $\mathbf{0}\neq\mathbf{b}\in\bigoplus J^{\prime}$, it
follows from the already proved property (3) that there are
$Y_{1},\ldots,Y_{n}\in\bigcup\\{\mathcal{P}_{j}\mid j\in J^{\prime}\\}$ such
that the $y$-th row of $\mathbf{b}$ is not zero only if $y\in
Y_{1}\cup\cdots\cup Y_{n}$. Since $m\nleqslant j$ for every $j\in J^{\prime}$,
it follows from (2) of Lemma 1.2 that $Y\not\subset Y_{1}\cup\cdots\cup
Y_{n}$. We conclude that $\mathbf{a}\not\in\bigoplus J^{\prime}$ and hence
(1.9) follows. ∎

##  2\. Representing a partially ordered set $I$: the family $(S(I,i))_{i\in
I}$ of subalgebras of $\mathbb{C}\mathbb{F}\mathbb{M}_{X(I)}(\mathbb{K})$.

As a further step toward the construction of the $\mathbb{K}$-algebra
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$, we start from the family
$(Q(I,i))_{i\in I}$ of $\mathbb{K}$-subalgebras of
$Q(I)=\mathbb{C}\mathbb{F}\mathbb{M}_{X(I)}(\mathbb{K})$ defined in the
previous section and select in each $Q(I,i)$ a $\mathbb{K}$-subalgebra
$S(I,i)$, still isomorphic to $Q(I)$, in such a way that
$S(I,i)S(I,j)=\\{\mathbf{0}\\}$ exactly when $i$ and $j$ are not comparable,
while $S(I,i)S(I,j)\cup S(I,j)S(I,i)\subset S(I,i)$ if $i\leqslant j$. In
order to reach this goal we need an additional setup.

######  Notations and Settings 2.1.

By keeping the same setting and notations as in the previous section, we add
the following data and further notations:

  * •

We let $I$ be a given $\boldsymbol{\aleph}$-poset. For every $i\in I$ we
denote by $\mathcal{M}(\leqslant i\leqslant)$ the set of all maximal chains
which pass through $i$:

$\mathcal{M}(\leqslant i\leqslant)\,\,\colon\\!\\!=\\{A\in\mathcal{M}(I)\mid
i\in A\\}.$

  * •

Given $i\in I$ and $C\in\mathcal{M}(\leqslant i)$, we define

$\displaystyle\mathcal{P}_{i,C}$
$\displaystyle\,\,\colon\\!\\!=\left\\{X(I)_{(u,C\cup
E)}\left|\,\,u\in\boldsymbol{\aleph}^{(i\leqslant)},E\in\mathcal{M}(i\leqslant)\right.\right\\}\subset\mathcal{P}_{i},$
$\displaystyle\mathcal{N}_{C}$
$\displaystyle\,\,\colon\\!\\!=\boldsymbol{\aleph}^{(I)}\times\\{C\cup E\mid
E\in\mathcal{M}(i\leqslant)\\}=\bigcup\mathcal{P}_{i,C}\subset X(I),$
$\displaystyle X(I)_{i}$
$\displaystyle\,\,\colon\\!\\!=\boldsymbol{\aleph}^{(I)}\times\mathcal{M}(\leqslant
i\leqslant)=\bigcup\\{\mathcal{N}_{C}\mid C\in\mathcal{M}(\leqslant
i)\\}\subset X(I).$

Note that $\mathcal{P}_{i,C}$ is an $\boldsymbol{\aleph}$-partition of
$\mathcal{N}_{C}$, while $\\{\mathcal{N}_{C}\mid C\in\mathcal{M}(\leqslant
i)\\}$ is a partition of $X(I)_{i}$.

  * •

Given $C,C^{\prime}\in\mathcal{M}(\leqslant i)$, we consider the map

$f_{C^{\prime}C}\colon\mathcal{N}_{C}\longrightarrow\mathcal{N}_{C^{\prime}}$

defined by

$f_{C^{\prime}C}(p,C\cup E)=(p,C^{\prime}\cup E)$

for every $p\in\boldsymbol{\aleph}^{(I)}$ and $E\in\mathcal{M}(i\leqslant)$.
It is clear that $f_{C^{\prime}C}$ is a bijection; moreover, for every
$C,C^{\prime},C^{\prime\prime}\in\mathcal{M}(\leqslant i)$ we have

(2.1)
$f_{C^{\prime\prime}C^{\prime}}\,f_{C^{\prime}C}=f_{C^{\prime\prime}C}\,\quad\text{and}\quad
f_{CC}=1_{\mathcal{N}_{C}.}$

######  Proposition 2.2.

Two elements $i,j\in I$ are comparable if and only if

$X(I)_{i}\cap X(I)_{j}\neq\emptyset.$

Consequently, if every maximal chain of $I$ has a supremum, then the set

$\\{X(I)_{m}\mid m\in\mathbf{M}(I)\\}$

is a partition of $X(I)$. In particular $X(I)_{\max(X(I))}=X(I)$.

###### Proof.

First observe that $X(I)_{i}\cap X(I)_{j}\neq\emptyset$ if and only if
${\mathcal{M}(\leqslant i\leqslant)}\cap\mathcal{M}({\leqslant
j\leqslant})\neq\emptyset$, if and only if there is some $A\in\mathcal{M}(I)$
such that $i,j\in A$. By the Hausdorff Maximal Principle this latter condition
holds if and only if $i$ and $j$ are comparable. ∎

######  Lemma 2.3.

With the above notations, if $i,j\in I$, then the following properties hold:

  1. (1)

Given $C\in\mathcal{M}(\leqslant i)$ and $D\in\mathcal{M}(\leqslant j)$, if
$\mathcal{N}_{C}\cap\mathcal{N}_{D}\neq\emptyset$, then $i$ and $j$ are
comparable and $\mathcal{N}_{D}\subset\mathcal{N}_{C}$ (resp.
$\mathcal{N}_{C}\subset\mathcal{N}_{D}$) in case $i\leqslant j$ (resp
$j\leqslant i$).

  2. (2)

Given $C\in\mathcal{M}(\leqslant i)$ and $D\in\mathcal{M}(\leqslant j)$, if
$C\subset D$ then $\mathcal{N}_{D}\subset\mathcal{N}_{C}$.

  3. (3)

If $i$ and $j$ are not comparable, then
$\mathcal{N}_{C}\cap\mathcal{N}_{D}=\emptyset$ for every
$C\in\mathcal{M}(\leqslant i)$ and $D\in\mathcal{M}(\leqslant j)$.

  4. (4)

If $i\leqslant j$, for every $C\in\mathcal{M}(\leqslant i)$ there exists some
$D\in\mathcal{M}(\leqslant j)$ such that
$\mathcal{N}_{D}\subset\mathcal{N}_{C}$.

  5. (5)

Assume that $i\leqslant j$, let $C,C^{\prime}\in\mathcal{M}(\leqslant i)$ and
let $D\in\mathcal{M}(\leqslant j)$ be such that
$\mathcal{N}_{D}\subset\mathcal{N}_{C}$. Then
$f_{C^{\prime}C}(\mathcal{N}_{D})=\mathcal{N}_{D^{\prime}}$ for some
$D^{\prime}\in\mathcal{M}(\leqslant j)$ and we have

(2.2) $f_{C^{\prime}C}(x)=f_{D^{\prime}D}(x)\qquad\text{for every
$x\in\mathcal{N}_{D}$.}$

###### Proof.

(1). Assume that $\mathcal{N}_{C}\cap\mathcal{N}_{D}\neq\emptyset$. Then there
is $p\in\boldsymbol{\aleph}^{(I)}$, $E\in\\{i\leqslant\\}$ and
$F\in\\{j\leqslant\\}$ such that $(p,C\cup E)=(p,D\cup F)$ and hence $C\cup
E=D\cup F$. This implies that $i,j$ are comparable. If $i\leqslant j$, then
$D=C\cup(D\cap[i\leqslant j])$ and this implies that
$\mathcal{N}_{D}\subset\mathcal{N}_{C}$. A similar argument applies if
$j\leqslant i$.

(2) is clear from the definition and (3) is an obvious consequence of (1).

(4) Suppose that $i\leqslant j$ and let $C\in\\{\leqslant i\\}$. If
$C^{\prime}$ is any maximal chain of $[i,j]$, by taking $D=C\cup C^{\prime}$
it is clear from (2) that $\mathcal{N}_{D}\subset\mathcal{N}_{C}$.

(5) Suppose that $i<j$ and let $C,C^{\prime},D$ be as in the assumptions. Then
$D=C\cup E$ for some $E\in\mathcal{M}[i,j]$. By taking
$D^{\prime}=C^{\prime}\cup E$ we see that, for every
$p\in\boldsymbol{\aleph}^{(I)}$ and $F\in\mathcal{M}\\{j\leqslant\\}$, the
following equalities hold:

$f_{C^{\prime}C}(p,D\cup F)=f_{C^{\prime}C}(p,A\cup E\cup F)=(p,C^{\prime}\cup
E\cup F)=(p,D^{\prime}\cup F)=f_{D^{\prime}D}(p,D\cup F).$

This proves that $f_{C^{\prime}C}(\mathcal{N}_{D})=\mathcal{N}_{D^{\prime}}$
and (2.2). ∎

######  Remark 2.4.

Let $i<j$ in $I$, let $C\in\mathcal{M}(\leqslant i)$,
$D\in\mathcal{M}(\leqslant j)$ and assume that $C\subset D$; then
$\mathcal{N}_{D}\subset\mathcal{N}_{C}$ according to Lemma 2.3, (2). If
$Y\in\mathcal{P}_{j,D}$, then $Y$ is the union of $\boldsymbol{\aleph}$
members of $\mathcal{P}_{i,C}$. In fact, since $\mathcal{P}_{j}$ is
$\boldsymbol{\aleph}$-coarser than $\mathcal{P}_{i}$ and
$\mathcal{P}_{j,D}\subset\mathcal{P}_{j}$, then $Y$ is the union of
$\boldsymbol{\aleph}$ members of $\mathcal{P}_{i}$. But if
$Z\in\mathcal{P}_{i}$ and $Z\subset Y$, then $Z\subset
Y\subset\mathcal{N}_{D}\subset\mathcal{N}_{C}$ and so $Z\in\mathcal{P}_{i,C}$.

The necessary setup is now ready and we can proceed to find, for each $i\in
I$, a $\mathbb{K}$-subalgebra $S(I,i)$ of $Q(I,i)$ having the properties
announced at the beginning of this section.

Given $i\in I$, by their definitions each $\mathcal{N}_{C}$ (for
$C\in\mathcal{M}(\leqslant i)$) and $X(I)_{i}$ are union of members of
$\mathcal{P}_{i}$, therefore all the idempotents
$\mathbf{e}_{\mathcal{N}_{C}}$ and $\mathbf{e}_{X(I)_{i}}$ belong to $Q(I,i)$.
Let us denote by $S(I,i)$ the subset of $Q(I,i)$ of those matrices
$\mathbf{a}$ satisfying the following two conditions:

  1. ($*_{i}$)

for every $p,q\in\boldsymbol{\aleph}^{(I)}$ and $A,B\in\mathcal{M}(I)$,

$\mathbf{a}((p,A),(q,B))\neq 0\text{ only if }i\in A\cap B\text{ and
}A\cap\\{\leqslant i\\}=B\cap\\{\leqslant i\\},$

  2. ($**_{i}$)

for every $p,q\in\boldsymbol{\aleph}^{(I)}$, $C,D\in\mathcal{M}(\leqslant i)$
and $E,F\in\mathcal{M}(i\leqslant)$,

$\mathbf{a}((p,C\cup E),(q,C\cup F))=\mathbf{a}((p,D\cup E),(q,D\cup F)).$

Note that these conditions may be restated as follows:

  1. ($*_{i}$)

$\mathbf{a}=\mathbf{e}_{X(I)_{i}}\mathbf{a}=\mathbf{a}\mathbf{e}_{X(I)_{i}}$
and

(2.3)
$\mathbf{e}_{\mathcal{N}_{A}}\mathbf{a}=\mathbf{e}_{\mathcal{N}_{A}}\mathbf{a}\mathbf{e}_{\mathcal{N}_{A}}=\mathbf{a}\mathbf{e}_{\mathcal{N}_{A}}\quad\text{for
every $A\in\mathcal{M}(\leqslant i)$},$

  2. ($**_{i}$)

for every $C,C^{\prime}\in\mathcal{M}(\leqslant i)$ and
$x,y\in\mathcal{N}_{C}$,

(2.4) $\mathbf{a}(x,y)=\mathbf{a}(f_{C^{\prime}C}(x),f_{C^{\prime}C}(y)).$

Roughly speaking, $S(I,i)$ consists of those matrices of $Q(I,i)$ which have
zero entries outside the $(\mathcal{N}_{C},\mathcal{N}_{C})$-blocks for
$C\in\mathcal{M}(\leqslant i)$ (which are mutually disjoint) and, if
$C,C^{\prime}\in\mathcal{M}(\leqslant i)$, the
$(\mathcal{N}_{C^{\prime}},\mathcal{N}_{C^{\prime}})$-block coincides with the
$(\mathcal{N}_{C},\mathcal{N}_{C})$-block “ up to the bijection
$f_{C^{\prime}C}$ ”.

For every $i\in I$ let us consider the set

$X(i\leqslant)\,\,\colon\\!\\!=\boldsymbol{\aleph}^{(i\leqslant)}\times\mathcal{M}(i\leqslant).$

As we shall see with the next result, $S(I,i)$ turns out to be a unital
$\mathbb{K}$-subalgebra of $\mathbf{e}_{X(I)_{i}}Q(I,i)\mathbf{e}_{X(I)_{i}}$
isomorphic to $\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})$ and
hence to $\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}}(\mathbb{K})$.

######  Proposition 2.5.

With the above notations and settings, for every ${i\in I}$ let us consider
the map

$\psi_{I,i}\colon\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})\longrightarrow
Q(I)$

defined as follows: for all
$\mathbf{a}\in\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})$,
$r,s\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$,

$\psi_{I,i}(\mathbf{a})((r\cup u,A),(s\cup
v,B))=\begin{cases}\delta(r,s)\,\mathbf{a}((u,A\cap\\{i\leqslant\\}),(v,B\cap\\{i\leqslant\\}),\\\
\text{ if }i\in A\cap B\text{ and }A\cap\\{\leqslant i\\}=B\cap\\{\leqslant
i\\},\\\ 0,\text{ otherwise. }\end{cases}$

Then $\psi_{I,i}$ is an injective (not necessarily unital)
$\mathbb{K}$-algebra homorphism and

(2.5)
$\psi_{I,i}(\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K}))=S(I,i).$

Moreover the following properties hold:

  1. (1)

$S(I,i)S(I,j)=0$ if and only if $i,j$ are not comparable.

  2. (2)

If $i\leqslant j$, then $S(I,i)S(I,j)\cup S(I,j)S(I,i)\subset S(I,i)$.

###### Proof.

Let $\mathbf{a}\in\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})$.
Given $r,s,t\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$ we have
that

$\psi_{I,i}(\mathbf{a})((r\cup u,A),(s\cup
v,B))=\delta(r,s)\psi_{I,i}(\mathbf{a})((t\cup u,A),(t\cup v,B)),$

consequently $\psi_{I,i}(\mathbf{a})\in Q(I,i)$. It is clear that
$\psi_{I,i}(\mathbf{a})$ satisfies ($*_{i}$). Assume that
$C,D\in\mathcal{M}(\leqslant i)$ and $E,F\in\mathcal{M}(i\leqslant)$. Then

$\displaystyle\psi_{I,i}(\mathbf{a})((r\cup u,C\cup E),(s\cup v,C\cup
F))=\delta(r,s)\mathbf{a}((u,E),(v,F))$
$\displaystyle=\psi_{I,i}(\mathbf{a})((r\cup u,D\cup E),(s\cup v,D\cup F))$

therefore $\psi_{I,i}(\mathbf{a})$ satisfies ($**_{i}$) and then
$\psi_{I,i}(\mathbf{a})\in S(I,i)$. Conversely, suppose that $\mathbf{b}\in
S(I,i)$ and let us define the matrix
$\mathbf{a}\in\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})$ as
follows: given $u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and
$E,F\in\mathcal{M}(i\leqslant)$, choose any $D\in\mathcal{M}(\leqslant i)$,
$r\in\boldsymbol{\aleph}^{(i\nleqslant)}$ and set

(2.6) $\mathbf{a}((u,E),(v,F)=\mathbf{b}((r\cup u,D\cup E),(r\cup v,D\cup
F)).$

We note that, since $\mathbf{b}\in Q(I,i)$, the second member of (2.6) does
not depend on the choice of $r$ in $\boldsymbol{\aleph}^{(i\nleqslant)}$; it
does not depend on the choice of $D$ in $\mathcal{M}(\leqslant i)$ either,
because $\mathbf{b}$ satisfies condition ($**_{i}$). A straightforward
computation shows that $\psi_{I,i}(\mathbf{a})=\mathbf{b}$ and we conclude
that (2.5) holds.

The map $\psi_{I,i}$ is $\mathbb{K}$-linear; let
$\mathbf{a}\in\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})$ and
suppose that $\psi_{I,i}(\mathbf{a})=\textbf{0}$. Given
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and
$E,F\in\mathcal{M}({i\leqslant})$, by choosing any $D\in\mathcal{M}({\leqslant
i})$ we have that

$\delta(r,s)\mathbf{a}((u,E),(v,F))=\psi_{I,i}(\mathbf{a})((r\cup u,D\cup
E),(s\cup v,D\cup F))=0$

for every $r,s\in\boldsymbol{\aleph}^{(i\nleqslant)}$, hence
$\mathbf{a}((u,E),(v,F))=0$. We conclude that $\mathbf{a}=\textbf{0}$ and this
shows that $\psi_{I,i}$ is injective.

Next, given
$\mathbf{a},\mathbf{b}\in\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})$,
we must prove that

$\psi_{I,i}(\mathbf{a}\mathbf{b})=\psi_{I,i}(\mathbf{a})\psi_{I,i}(\mathbf{b}),$

that is, given $r,s,\in\boldsymbol{\aleph}^{(i\nleqslant)}$,
$u,v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and $A,B\in\mathcal{M}(I)$, by
writing $x=(r\cup u,A)$ and $y=(s\cup v,B)$ we must show that the equality

(2.7) $\psi_{I,i}(\mathbf{a}\mathbf{b})(x,y)=\sum_{z\in
X(I)}\psi_{I,i}(\mathbf{a})(x,z)\psi_{I,i}(\mathbf{b})(z,y)$

holds. If either $i\not\in A$, or $i\not\in B$, or $i\in A\cap B$ but
$A\cap\\{\leqslant i\\}\neq B\cap\\{\leqslant i\\}$, then it follows from the
definition of $\psi_{I,i}$ that the first member and each summand in the
second member of (2.7) are zero. Otherwise, suppose that $i\in A\cap B$, set
$D=A\cap\\{\leqslant i\\}=B\cap\\{\leqslant i\\}$ and note that
$x,y\in\mathcal{N}_{D}$. Then we have:

$\psi_{I,i}(\mathbf{a}\mathbf{b})(x,y)=\delta(r,s)(\mathbf{a}\mathbf{b})((u,A\cap\\{i\leqslant\\}),(v,B\cap\\{i\leqslant\\})\\\
\begin{aligned} &=\delta(r,s)\sum_{(w,E)\in
X(i\leqslant)}\mathbf{a}((u,A\cap\\{i\leqslant\\}),(w,E))\mathbf{b}((w,E),(v,B\cap\\{i\leqslant\\}))\\\
&=\sum_{\begin{subarray}{c}t\in\boldsymbol{\aleph}^{(i\nleqslant)},w\in\boldsymbol{\aleph}^{(i\leqslant)}\\\
C\in\\{D\cup F\mid
F\in\mathcal{M}(i\leqslant)\\}\end{subarray}}\bigg{(}\delta(r,t)\delta(t,s)\mathbf{a}((u,A\cap\\{i\leqslant\\}),(w,C\cap\\{i\leqslant\\}))\\\
&\phantom{=\sum_{\begin{subarray}{c}t\in\boldsymbol{\aleph}^{(i\nleqslant)},w\in\boldsymbol{\aleph}^{(i\leqslant)}\\\
C\in\\{D\cup F\mid
F\in\mathcal{M}(i\leqslant)\\}\end{subarray}}\bigg{(}\delta(r,t)\delta(t,s)}\cdot\mathbf{b}((w,C\cap\\{i\leqslant\\}),(v,B\cap\\{i\leqslant\\}))\bigg{)}\\\
&=\sum_{\begin{subarray}{c}t\in\boldsymbol{\aleph}^{(i\nleqslant)},w\in\boldsymbol{\aleph}^{(i\leqslant)}\\\
C\in\\{D\cup F\mid
F\in\mathcal{M}(i\leqslant)\\}\end{subarray}}\psi_{I,i}(\mathbf{a})(x,(t\cup
w,C))\psi_{I,i}(\mathbf{b})((t\cup w,C),y)\\\
&=\sum_{z\in\mathcal{N}_{D}}\psi_{I,i}(\mathbf{a})(x,z)\psi_{I,i}(\mathbf{b})(z,y)\\\
&=\sum_{z\in
X(I)}\psi_{I,i}(\mathbf{a})(x,z)\psi_{I,i}(\mathbf{b})(z,y),\end{aligned}$

as wanted, and we conclude that $\psi_{I,i}$ is a $\mathbb{K}$-algebra
homorphism.

(1) Let $i,j\in I$ and assume that $i,j$ are not comparable. Then, given
${C\in\mathcal{M}(\leqslant i)}$ and $D\in\mathcal{M}(\leqslant j)$, we have
$\mathcal{N}_{C}\cap\mathcal{N}_{D}=\emptyset$ according to Lemma 2.3, (3) and
therefore $X(I)_{i}\cap X(I)_{j}=\emptyset$. Consequently, if $\mathbf{a}\in
S(I,i)$ and $\mathbf{b}\in S(I,j)$, then
$\mathbf{a}\mathbf{b}=\mathbf{e}_{X(I)_{i}}\mathbf{a}\mathbf{e}_{X(I)_{i}}\mathbf{e}_{X(I)_{j}}\mathbf{b}\mathbf{e}_{X(I)_{j}}=0$.
If, on the contrary, $i\leqslant j$ and $A$ is any maximal chain of $I$ such
that $i,j\in A$, then $A\in\mathcal{M}(\leqslant
i\leqslant)\cap\mathcal{M}(\leqslant j\leqslant)$ and hence $X(I)_{i}\cap
X(I)_{j}\neq\emptyset$. Consequently
$\mathbf{0}\neq\mathbf{e}_{X(I)_{i}}\,\mathbf{e}_{X(I)_{j}}=\mathbf{e}_{X(I)_{j}}\,\mathbf{e}_{X(I)_{i}}\in
S(I,i)S(I,j)\cap S(I,j)S(I,i)$.

(2) Suppose that $i<j$, let $\mathbf{a}\in S(I,i)$, $\mathbf{b}\in S(I,j)$ and
assume that $0\neq(\mathbf{a}\mathbf{b})(x,y)=\sum_{z\in
X(I)}\mathbf{a}(x,z)\mathbf{b}(z,y)$ for some $x,y\in X(I)$. Then
$\mathbf{a}(x,z)\neq 0\neq\mathbf{b}(z,y)$ for some $z\in X(I)$ and therefore,
by conditions ($*_{i}$) and ($*_{j}$), there are some
${C\in\mathcal{M}(\leqslant i)}$ and $D\in\mathcal{M}(\leqslant j)$ such that
$x,z\in\mathcal{N}_{C}$ and $z,y\in\mathcal{N}_{D}$, so that
$\mathcal{N}_{C}\cap\mathcal{N}_{D}\neq\emptyset$ and hence
$\mathcal{N}_{D}\subset\mathcal{N}_{C}$ in view of Lemma 2.3, (1). We infer
that the matrix $\mathbf{a}\mathbf{b}$ has zero entries outside the
$(\mathcal{N}_{C},\mathcal{N}_{C})$-blocks for $C\in\mathcal{M}(\leqslant i)$.
Suppose that $C,C^{\prime}\in\mathcal{M}(\leqslant i)$ and let us prove that

(2.8)
$(\mathbf{a}\mathbf{b})(x,y)=(\mathbf{a}\mathbf{b})(f_{C^{\prime}C}(x),f_{C^{\prime}C}(y))$

for all $x,y\in\mathcal{N}_{C}$. It follows from Lemma 2.3, (5) that there is
no $D\in\mathcal{M}(\leqslant j)$ such that $y\in\mathcal{N}_{D}$ if and only
if there is no $D^{\prime}\in\mathcal{M}(\leqslant j)$ such that
$f_{C^{\prime}C}(y)\in\mathcal{N}_{D^{\prime}}$; if it is the case, since
$\mathbf{b}\in S(I,j)$, both members of (2.8) are zero. Otherwise there is
$D\in\mathcal{M}(\leqslant j)$ such that $y\in\mathcal{N}_{D}$; again by Lemma
2.3, (5) we have that $\mathcal{N}_{D}\subset\mathcal{N}_{C}$ and, by setting
$D^{\prime}=f_{C^{\prime}C}(B)$, we may compute as follows:

$\displaystyle(\mathbf{a}\mathbf{b})(x,y)$
$\displaystyle=\sum_{z\in\mathcal{N}_{C}}\mathbf{a}(x,z)\,\mathbf{b}(z,y)=\sum_{z\in\mathcal{N}_{D}}\mathbf{a}(x,z)\,\mathbf{b}(z,y)$
$\displaystyle=\sum_{z\in\mathcal{N}_{D}}[\mathbf{a}(f_{C^{\prime}C}(x),f_{C^{\prime}C}(z))]\,[\mathbf{b}(f_{D^{\prime}D}(z),f_{D^{\prime}D}(y))]$
$\displaystyle=\sum_{u\in\mathcal{N}_{D^{\prime}}}[\mathbf{a}(f_{C^{\prime}C}(x),u)]\,[\mathbf{b}(u,f_{D^{\prime}D}(y))]$
$\displaystyle=\sum_{u\in\mathcal{N}_{C^{\prime}}}[\mathbf{a}(f_{C^{\prime}C}(x),u)]\,[\mathbf{b}(u,f_{C^{\prime}C}(y))]$
$\displaystyle=(\mathbf{a}\mathbf{b})(f_{C^{\prime}C}(x),f_{C^{\prime}C}(y)).$

Thus (2.8) holds for all $x,y\in\mathcal{N}_{C}$, showing that
$\mathbf{a}\mathbf{b}\in S(I,i)$. The proof that $\mathbf{b}\mathbf{a}\in
S(I,i)$ is similar. ∎

######  Remarks 2.6.

Clearly we have that $\psi_{I,i}(\mathbf{1})=\mathbf{e}_{X(I)_{i}}$ and
$S(I,i)$ is a unital subalgebra of $Q(I,i)$ if and only if $X(I)_{i}=X(I)$,
namely $I={\\{\leqslant i\\}\cup\\{i\leqslant\\}}$; if it is the case, then
$S(I,i)=Q(I,i)$ if and only if $\\{\leqslant i\\}$ is a chain. As a result, we
have that $S(I,i)=Q(I,i)$ for all $i\in I$ if and only if $I$ is a chain.

As $\psi_{I,i}$ restricts to an isomorphism from
$\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})$ to $S(I,i)$, we
have in $S(I,i)$ the complete system
$\left\\{\psi_{I,i}(\mathbf{e}_{(u,E)})\left|(u,E)\in
X(i\leqslant)\right.\right\\}$ of pairwise orthogonal and primitive
idempotents, each of which is actually a diagonal matrix; to see this and for
subsequent purposes we introduce the following additional

######  Notations 2.7.

xxxx

  1. $\bullet$

For every $u\in\boldsymbol{\aleph}^{(i\leqslant)}$ and
$E\in\mathcal{M}(i\leqslant)$ we define

(2.9) $\displaystyle Y_{(u,E)}\colon$
$\displaystyle=\bigcup\left\\{X(I)_{(u,C\cup
E)}\left|\,\,C\in\mathcal{M}(\leqslant i)\right.\right\\}$
$\displaystyle=\left\\{(p,A)\in
X(I)\left|\,\,p|_{\boldsymbol{\aleph}^{(i\leqslant)}}=u,A\cap\\{i\leqslant\\}=E\right.\right\\}\subset
X(I)_{i}$

and set

(2.10)
$\mathcal{R}_{i}\,\,\colon\\!\\!=\left\\{Y_{(u,E)}\left|\,\,u\in\boldsymbol{\aleph}^{(i\leqslant)},E\in\mathcal{M}(i\leqslant)\right.\right\\}.$

  2. $\bullet$

For every $C\in\mathcal{M}(\leqslant i)$ and $V\in\mathcal{P}_{i,C}$ define

(2.11) $\overline{V}\,\,\colon\\!\\!=\bigcup\\{f_{C^{\prime}C}(V)\mid
C^{\prime}\in\mathcal{M}(\leqslant i)\\}.$

Now direct computations show that, given $i\in I$, for every
$u\in\boldsymbol{\aleph}^{(i\leqslant)}$, $C\in\mathcal{M}(\leqslant i)$ and
$C\in\mathcal{M}(i\leqslant)$ we have

(2.12) $\psi_{I,i}(\mathbf{e}_{(u,E)})=\mathbf{e}_{Y_{(u,E)}}$

and

(2.13) $Y_{(u,E)}=\overline{X(I)_{(u,C\cup E)}};$

consequently

$\mathcal{R}_{i}=\\{\overline{V}\mid V\in\mathcal{P}_{i,C}\\}.$

######  Lemma 2.8.

With the above notations, let $i,j\in I$. Then the following hold:

  1. (1)

If $Y\in\mathcal{R}_{i}$, $Z\in\mathcal{R}_{j}$ and $i,j$ are not comparable,
then $Y\cap Z=\emptyset$, while if $i\leqslant j$ and $Y\cap Z\neq\emptyset$,
then $Y\subset Z$.

  2. (2)

$\mathcal{R}_{i}$ is a partition of $X(I)_{i}$ of cardinality
$\boldsymbol{\aleph}$; it is an $\boldsymbol{\aleph}$-partition if
$i\neq\min(I)$.

  3. (3)

If $Z\in\mathcal{R}_{j}$ and $j$ is not a minimal element of $I$, then $Z$ is
the union of members of $\mathcal{R}_{i}$, for $i$ ranging in $\\{<j\\}$. In
addition, if $i<j$, then $Z$ contains $\boldsymbol{\aleph}$ elements of
$\mathcal{R}_{i}$; precisely, if $Z=Y_{(u,E)}\in\mathcal{R}_{j}$, where
${u\in\boldsymbol{\aleph}^{(j\leqslant)}}$ and $E\in\mathcal{M}(j\leqslant)$,
given $v\in\boldsymbol{\aleph}^{(i\leqslant)}$ and
$F\in\mathcal{M}(i\leqslant)$, we have that $Y_{(v,F)}\subset Z$ if and only
if $v\left|{}_{\\{j\leqslant\\}}=u\right.$ and $D\cap\\{j\leqslant\\}=E$.

  4. (4)

If $i$ is not a maximal element of $I$ and $Y\in\mathcal{R}_{i}$, then there
is some $j>i$ and $Z\in\mathcal{R}_{j}$ such that $Y\subset Z$.

###### Proof.

(1) Let $Y\in\mathcal{R}_{i}$, $Z\in\mathcal{R}_{j}$. If $i,j$ are not
comparable then $X(I)_{i}\cap X(I)_{j}=\emptyset$ by Proposition 2.2. Thus
$Y\cap Z=\emptyset$, because $Y\subset X(I)_{i}$ and $Z\subset X(I)_{j}$.
Assume that $i\leqslant j$ and write $Y=Y_{(u,E)}$, $Z=Y_{(v,F)}$ for some
$u\in\boldsymbol{\aleph}^{(i\leqslant)}$,
$v\in\boldsymbol{\aleph}^{(j\leqslant)}$, $E\in\mathcal{M}(i\leqslant)$ and
$F\in\mathcal{M}(j\leqslant)$. If there is some $(p,A)\in X(I)$ which belongs
to $Y\cap Z$, then necessarily $p\left|{}_{\\{i\leqslant\\}}=u\right.$,
$p\left|{}_{\\{j\leqslant\\}}=v\right.$, $A\cap\\{i\leqslant\\}=E$ and
$A\cap\\{j\leqslant\\}=F$. Inasmuch as
$\\{j\leqslant\\}\subset\\{i\leqslant\\}$, then
$u\left|{}_{\\{j\leqslant\\}}=v\right.$ and $F\subset E$. From the definition
(2.9) we easily infer that $Y\subset Z$.

(2) We already know that $\bigcup\mathcal{R}_{i}\subset X(I)_{i}$. If
$Y,Z\in\mathcal{R}_{i}$ and $Y\neq Z$, then it follows from (1) that $Y\cap
Z=\emptyset$. Given $x\in X(I)_{i}$, that is $x=(p,A\cup B)$ for some
$p\in\boldsymbol{\aleph}^{(I)}$, $C\in\mathcal{M}(\leqslant i)$ and
$E\in\mathcal{M}(i\leqslant)$, by setting $u=p|_{\\{i\leqslant\\}}$ we have
that $x\in Y_{(u,E)}$. Thus $\bigcup\mathcal{R}_{i}=X(I)_{i}$. The remaining
part of the statement comes clear from the definitions.

(3) Assume that $j$ is not minimal and let $Z=Y_{(u,E)}\in\mathcal{R}_{j}$,
where $u\in\boldsymbol{\aleph}^{(j\leqslant)}$ and
$E\in\mathcal{M}(j\leqslant)$. Given $C\in\mathcal{M}(\leqslant j)$, we can
choose some $i\in C$ with $i<j$ and we obtain from Lemma 1.2 that

$X(I)_{(u,C\cup E)}=\bigcup\left\\{X(I)_{(v,C\cup
E)}\left|\,\,v\in\boldsymbol{\aleph}^{(i\leqslant)}\text{ and
}v|_{\\{j\leqslant\\}}=u\right.\right\\}.$

On the other hand, given $v\in\boldsymbol{\aleph}^{(i\leqslant)}$ with
$v|_{\\{j\leqslant\\}}=u$, we have that $X(I)_{(v,C\cup E)}\subset Y_{(v,F)}$,
where $F=(C\cap\\{i\leqslant\\})\cup E\in\mathcal{M}(i\leqslant)$. Thus, by
recalling (2.9) we infer that $Y_{(v,F)}\cap Z\neq\emptyset$ and therefore
$Y_{(v,F)}\subset Z$ by (1). This proves the first statement of (3) and the
second follows since the set
$\left.\left\\{v\in\boldsymbol{\aleph}^{(i\leqslant)}\right|v|_{\\{j\leqslant\\}}=u\right\\}$
has cardinality $\boldsymbol{\aleph}$.

(4) Assume that $i$ is not a maximal element and let
$Y_{(u,E)}\in\mathcal{R}_{i}$, where $u\in\boldsymbol{\aleph}^{(i\leqslant)}$
and $E\in\mathcal{M}(i\leqslant)$. By the assumption there is some $j\in E$
such that $i<j$ and, by taking $v=u|_{\\{j\leqslant\\}}$ and
$F=E\cap\\{j\leqslant\\}$ we see that $Y_{i,(u,E)}\subset Y_{j,(v,F)}$. ∎

##  3\. Representing a partially ordered set $I$: the $\mathbb{K}$-algebra
$H(I,J)$ associated to a subset $J$ of $I$.

By keeping the same data, settings and notations as in the previous sections,
our next objective is to associate to each element $i$ of the
$\boldsymbol{\aleph}$-poset $I$ a (not necessarily unital)
$\mathbb{K}$-subalgebra $H(I,i)$ of $S(I,i)$, in such a way that
$\mathcal{H}=\\{H(I,i)\mid i\in I\\}$ is an independent set of
$\mathbb{K}$-subspaces of $Q(I)$ with the following features: if $i$ is a
maximal element of $I$, then $H(I,i)$ is isomorphic to $\mathbb{K}$; if $i$ is
not maximal, then $H(I,i)$ is isomorphic to
$\mathbb{F}\mathbb{R}_{X(I)}(\mathbb{K})$, therefore it is a simple,
semisimple and non-artinian algebra; moreover $H(I,i)H(I,j)=0$ if $i,j$ are
not comparable, while both $H(I,i)H(I,j)$ and $H(I,j)H(I,i)$ are nonzero and
are contained in $H(I,i)$ if $i\leqslant j$. Thus, for every subset $J\subset
I$, the subspace

$H(I,J)\,\,\colon\\!\\!=\bigoplus_{j\in J}H(I,j)$

turns out to be a $\mathbb{K}$-subalgebra of $Q(I)$, possibly without an
identity.

For every $i\in I$ let us define the $\mathbb{K}$-subalgebra $H(I,i)$ of
$Q(I)$ as follows:

$H(I,i)\,\,\colon\\!\\!=\begin{cases}\psi_{I,i}\left(\mathbb{F}\mathbb{R}_{X(i\leqslant)}(\mathbb{K})\right)=\operatorname{Soc}(S_{i}),\hbox{
if $i$ \text@underline{is not} a maximal element of $I$;}\\\
\psi_{I,i}\left(\mathbf{1}_{\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})}\mathbb{K}\right)=\mathbf{e}_{X(I)_{i}}\mathbb{K},\hbox{
if $i$ \text@underline{is} a maximal element of $I$.}\end{cases}$

Of course $H(I,i)\neq H(I,j)$ if $i\neq j$; also note that $H(I,i)$ is a
unital subring of $Q(I)$ if and only if $i$ is the greatest element of $I$, in
which the case $H(I,i)=\mathbb{K}$.

######  Lemma 3.1.

Assume that $i$ is not a maximal element of $I$. Then

(3.1)
$H(I,i)=\bigoplus\left\\{\mathbf{e}_{Y}H(I,i)\left|Y\in\mathcal{R}_{i}\right.\right\\}$

and

(3.2) $H(I,i)=H(I,i)\,\mathbf{e}_{Y}H(I,i)$

for every $Y\in\mathcal{R}_{i}$. Moreover, given $\mathbf{a}\in S(I,i)$, the
following conditions are equivalent:

  1. (1)

$\mathbf{a}\in H(I,i)$.

  2. (2)

$\mathbf{e}_{\mathcal{N}_{C}}\mathbf{a}\,\mathbf{e}_{\mathcal{N}_{C}}\in\operatorname{Soc}(Q(I,i))$
for all $C\in\mathcal{M}(\leqslant i)$.

  3. (3)

There are $Y_{1},\ldots,Y_{n}\in\mathcal{R}_{i}$ such that the $x$-th row of
$\mathbf{a}$ is not zero only if $x\in Y_{1}\cup\cdots\cup Y_{n}$;
equivalently

$\mathbf{a}=\left(\mathbf{e}_{Y_{1}}+\cdots+\mathbf{e}_{Y_{n}}\right)\mathbf{a}.$

###### Proof.

The first statement follows from the fact that $\\{\mathbf{e}_{Y}\mid
Y\in\mathcal{R}_{i}\\}$ is a complete set of pairwise orthogonal and primitive
idempotents of $S(I,i)$, while the equivalence (1)$\Leftrightarrow$(3) is
clear from (3.1).

(3)$\Rightarrow$(2). Assume (3) and let $C\in\mathcal{M}(\leqslant i)$. There
are $V_{i},\ldots,V_{n}\in\mathcal{P}_{i,C}$ such that
$Y_{r}=\overline{V_{r}}$ for all $r\in\\{1,\ldots,n\\}$ and we infer from the
definition (2.11) that

$(Y_{1}\cup\cdots\cup Y_{n})\cap\mathcal{N}_{i,C}=V_{1}\cup\cdots\cup V_{n}.$

As a result we have that

$\mathbf{e}_{\mathcal{N}_{C}}(\mathbf{e}_{Y_{1}}+\cdots+\mathbf{e}_{Y_{n}})=\mathbf{e}_{V_{1}}+\cdots+\mathbf{e}_{V_{n}}$

and (3) of Theorem 1.5 imply that
$\mathbf{e}_{\mathcal{N}_{C}}\mathbf{a}\,\mathbf{e}_{\mathcal{N}_{C}}\in\operatorname{Soc}(Q(I,i))$,
because $\mathcal{P}_{i,C}\subset\mathcal{P}_{i}$, $\mathbf{a}\in Q(I,i)$ and
$\operatorname{Soc}(Q(I,i))$ is an ideal of $Q(I,i)$.

(2)$\Rightarrow$(3) Suppose (2) and take any $C\in\mathcal{M}(\leqslant i)$.
Then it follows from (3) of Theorem 1.5 that there are
$Z_{1},\ldots,Z_{n}\in\mathcal{P}_{i}$ such that, for every $x\in X(I)$, the
$x$-th row of
$\mathbf{e}_{\mathcal{N}_{C}}\mathbf{a}\,\mathbf{e}_{\mathcal{N}_{C}}$ is not
zero only if $x\in Z_{1}\cup\cdots\cup Z_{n}$. As
$\mathbf{e}_{\mathcal{N}_{C}}\mathbf{a}\,\mathbf{e}_{\mathcal{N}_{C}}$ has
zero entries outside the
$\mathbf{e}_{\mathcal{N}_{C}}\times\mathbf{e}_{\mathcal{N}_{C}}$-block,
necessarily $Z_{1},\ldots,Z_{n}$ are contained in $\mathcal{N}_{C}$ and so
they belongs to $\mathcal{P}_{i,C}$. Given $x,y\in X(I)$, assume that
$\mathbf{a}(x,y)\neq 0$. Then $x,y\in\mathcal{N}_{C^{\prime}}$ for some
$C^{\prime}\in\mathcal{M}(\leqslant i)$ and
$\mathbf{a}(f_{CC^{\prime}}(x),f_{CC^{\prime}}(y))=\mathbf{a}(x,y)\neq 0$,
therefore $f_{CC^{\prime}}(x)\in Z_{1}\cup\cdots\cup Z_{n}$ and hence $x\in
f_{C^{\prime}C}(Z_{1})\cup\cdots\cup
f_{C^{\prime}C}(Z_{n})\subset\overline{Z_{1}}\cup\cdots\cup\overline{Z_{n}}$.
∎

######  Remark 3.2.

In general we have that
$\mathbf{e}_{\mathcal{N}_{C}}H(I,i)\,\mathbf{e}_{\mathcal{N}_{C}}\not\subset
H(I,i)$, unless $\\{\leqslant i\\}$ is a chain. If $i$ is not a maximal
element of $I$, then $H(I,i)\subset\operatorname{Soc}(Q(I,i))$ _if and only if
$\\{\leqslant i\\}$ has finitely many maximal chains_.

######  Remark 3.3.

We feel worth to emphasize that the main features of the algebras $H(I,i)$ we
are going to investigate bear on the particular shape of a matrix
$\mathbf{a}\neq\mathbf{0}$ which belongs to $H(I,i)$. Firstly, the set $X$ of
those $x\in X(I)$ such that the $x$-th row of $\mathbf{a}$ is not zero is a
union of members of $\mathcal{P}_{i}$, because $\mathbf{a}\in Q(I,i)$. Since
$\mathbf{a}\in S(I,i)$, if $x\in X$, by condition ($*_{i}$) we have that
$x\in\mathcal{N}_{C}$ for a necessarily unique $C\in\mathcal{M}(\leqslant i)$.
If $i$ is not a maximal element of $I$, given $C\in\mathcal{M}(\leqslant i)$,
by (3) of Theorem 1.5 and Lemma 3.1 we have that $X\cap\mathcal{N}_{C}$ is the
union of finitely many members of $\mathcal{P}_{i,C}$ and condition ($**_{i}$)
entails that

$X\cap\mathcal{N}_{C^{\prime}}=f_{C^{\prime}C}(X\cap\mathcal{N}_{C})\quad\text{for
every $C,C^{\prime}\in\mathcal{M}(\leqslant i)$}.$

######  Lemma 3.4.

Let $j_{1}<\cdots<j_{n}$ be a finite chain of $I$ with $n>1$, assume that
$\mathbf{0}\neq\mathbf{a}_{1}\in\nolinebreak
H(I,j_{1})$,…,$\mathbf{0}\neq\mathbf{a}_{n}\in H(I,j_{n})$ and let
$C_{n}\in\mathcal{M}(\leqslant j_{n})$ 222 _WARNING_! In a context like the
present one, in which we have a poset $I$, a subset $J\subset I$ and an
element $j\in J$, by the symbol $\\{\leqslant j\\}$ we mean the set $\\{k\in
I\mid k\leqslant j\\}$ and not $\\{k\in J\mid k\leqslant j\\}$! be such that
$j_{1},\ldots,j_{n-1}\in C_{n}$. Then the
$(\mathcal{N}_{C_{n}}\times\mathcal{N}_{C_{n}})$-block of
$\mathbf{a}=\mathbf{a}_{1}+\cdots+\mathbf{a}_{n}$ is not zero; specifically,
there is some $x\in\mathcal{N}_{C_{n}}$ such that the $x$-th row of
$\mathbf{a}$ is not zero and coincides with the $x$-th row of
$\mathbf{a}_{n}$.

###### Proof.

For every $r\in\\{1,\ldots,n\\}$ set $C_{r}=C_{n}\cap\\{\leqslant j_{r}\\}$,
let $Y_{r}$ be the subset of those $x\in\mathcal{N}_{C_{r}}$ such that the
$x$-th row of $\mathbf{a}_{r}$ is not zero and note that $Y_{r}\neq\emptyset$.
If $r<n$, the element $j_{r}$ is not maximal in $I$, therefore $Y_{r}$ is a
(disjoint) union of finitely many elements of
$\mathcal{P}_{j_{r},C_{r}}\subset\mathcal{P}_{j_{r}}$ (see Remark 3.3). Since
$\mathcal{P}_{j_{s}}$ is $\boldsymbol{\aleph}$-coarser than
$\mathcal{P}_{j_{r}}$ when $r<s\leqslant n$, in particular $Y_{n}$ is a
(disjoint) union of $\boldsymbol{\aleph}$ elements of $\mathcal{P}_{j_{n-1}}$;
on the other hand, by the above $Y_{1}\cup\cdots\cup Y_{n-1}$ is contained in
the (disjoint) union of finitely many elements of $\mathcal{P}_{j_{n-1}}$.
Consequently

$Y_{n}\setminus(Y_{1}\cup\cdots\cup Y_{n-1})\neq\emptyset$

and therefore, if $x\in Y_{n}\setminus(Y_{1}\cup\cdots\cup Y_{n-1})$, the
$x$-th row of $\mathbf{a}$ coincides with the $x$-th row of $\mathbf{a}_{n}$,
which is not zero. ∎

######  Lemma 3.5.

Let $J$ be a finite subset of $I$, let $\mathbf{a}=\sum_{j\in
J}\mathbf{a}_{j}$, where each $\mathbf{a}_{j}$ is a non-zero element of $\in
H(I,j)$ for $j\in J$, let $j_{1}<\cdots<j_{n}$ be a maximal chain of $J$ and
let $C_{n}\in\mathcal{M}(\leqslant j_{n})$ be such that
$j_{1},\ldots,j_{n-1}\in C_{n}$. Then the
$(\mathcal{N}_{C_{n}}\times\mathcal{N}_{C_{n}})$-blocks of $\mathbf{a}$ and
$\mathbf{a}_{j_{1}}+\cdots+\mathbf{a}_{j_{n}}$ coincide.

###### Proof.

As in the proof of Lemma 3.4, for every $r\in\\{1,\ldots,n\\}$ set
$C_{r}=C_{n}\cap\\{\leqslant j_{r}\\}$ and note that since
$C_{1}\subsetneqq\cdots\subsetneqq C_{n}$, then
$\mathcal{N}_{C_{n}}\subsetneqq\cdots\subsetneqq\mathcal{N}_{C_{1}}$ by (2) of
Lemma 2.3. Consequently, given $j\in J$, if there is some
$D\in\mathcal{M}(\leqslant j)$ such that
$\mathcal{N}_{C_{n}}\cap\mathcal{N}_{D}\neq\emptyset$, then
$\mathcal{N}_{C_{r}}\cap\mathcal{N}_{D}\neq\emptyset$ for all
$r\in\\{1,\ldots,n\\}$ and therefore it follows from Lemma 2.3 that $j$ is
comparable with every $j_{r}$. As a result $j\in\\{j_{1},\ldots,j_{n}\\}$,
because this latter is a maximal chain of $J$. Since every matrix in $H(I,j)$
has zero entries outside the $(\mathcal{N}_{D}\times\mathcal{N}_{D})$-blocks,
for $D\in\mathcal{M}(\leqslant j)$, we infer from the above that if $j\in
J\setminus\\{j_{1},\ldots,j_{n}\\}$, then the
$(\mathcal{N}_{C_{n}}\times\mathcal{N}_{C_{n}})$-block of every matrix in
$H(I,j)$ is zero and, consequently, the
$(\mathcal{N}_{C_{n}}\times\mathcal{N}_{C_{n}})$-blocks of $\mathbf{a}$ and
$\mathbf{a}_{j_{1}}+\cdots+\mathbf{a}_{j_{n}}$ coincide. ∎

######  Theorem 3.6.

With the above notations and settings, the set

$\mathcal{H}=\\{H(I,i)\mid i\in I\\}$

of $\mathbb{K}$-subspaces of $Q(I)$ is independent and the following
conditions hold:

  1. (1)

Given $i\in I$, the subring $H(I,i)$ of $Q(I)$ has an identity, given by
$\mathbf{e}_{X(I)_{i}}$, if and only if $i$ is a maximal element of $I$; in
particular $H(I,i)=\mathbb{K}$ when $i=\max(I)$.

  2. (2)

$H(I,i)H(I,j)=\\{\mathbf{0}\\}$ if $i,j$ are not comparable;

  3. (3)

Given $i\in I$, if $J\subset\\{i\leqslant\\}$ and
$\mathbf{0}\neq\mathbf{a}\in\bigoplus_{j\in J}H(I,j)$, then

$\\{\mathbf{0}\\}\neq H(I,i)\mathbf{a}\subset
H(I,i)\quad\text{and}\quad\\{\mathbf{0}\\}\neq\mathbf{a}H(I,i)\subset H(I,i);$

in particular there are $Y,Z\in\mathcal{R}_{i}$ such that

$\mathbf{0}\neq\mathbf{e}_{Y}\mathbf{a}\in
H(I,i)\quad\text{and}\quad\mathbf{0}\neq\mathbf{a}\mathbf{e}_{Z}\in H(I,i).$

###### Proof.

Assume that $J$ is a finite subset of $I$, suppose that $\mathbf{a}=\sum_{j\in
J}\mathbf{a}_{j}$, where $\mathbf{0}\neq\mathbf{a}_{j}\in H(I,j)$ for $j\in
J$, let us choose a maximal chain $j_{1}<\cdots<j_{n}$ of $J$ and let
$C_{n}\in\mathcal{M}(\leqslant j_{n})$ be such that $j_{1},\ldots,j_{n-1}\in
C_{n}$. Then by Lemma 3.5 the
$(\mathcal{N}_{C_{n}}\times\mathcal{N}_{C_{n}})$-blocks of $\mathbf{a}$ and
$\mathbf{a}^{\prime}=\mathbf{a}_{j_{1}}+\cdots+\mathbf{a}_{j_{n}}$ coincide
and, on the other hand, the
$(\mathcal{N}_{C_{n}}\times\mathcal{N}_{C_{n}})$-block of
$\mathbf{a}^{\prime}$ is not zero by Lemma 3.4. As a consequence
$\mathbf{a}\neq\mathbf{0}$ and this proves the independence of $\mathcal{H}$.

(1) If $i\in I$ and $i$ is not a maximal element, then
$H(I,i)\simeq\mathbb{F}\mathbb{R}_{X(i\leqslant)}(\mathbb{K})$ as rings,
therefore $H(I,i)$ is a ring without an identity. If $i$ is a maximal element
of $I$, then
$H(I,i)=\psi_{i}\left(\mathbf{1}_{\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})}\mathbb{K}\right)\simeq\mathbb{K}$
and
$\mathbf{e}_{X(I)_{i}}=\psi_{i}\left(\mathbf{1}_{\mathbb{C}\mathbb{F}\mathbb{M}_{X(i\leqslant)}(\mathbb{K})}\right)$
is the multiplicative identity of $H(I,i)$. If $i=\max(I)$, then
$X(I)_{i}=X(I)$ and so $H(I,i)=\mathbf{1}\mathbb{K}$ is a unital subring of
$Q(I)$.

(2) follows from the property (1) of Proposition 2.5, since $H(I,i)\subset
S(I,i)$ for all $i\in I$.

(3) Let us prove first that if we take $i,j\in I$ with $i<j$ and
$\mathbf{a}\in H(I,j)$, $\mathbf{b}\in H(I,i)$, then $\mathbf{a}\mathbf{b}$
and $\mathbf{b}\mathbf{a}$ are both in $H(I,i)$. According to Proposition 2.5
we have that $\mathbf{a}\mathbf{b}\in S(I,i)$ and $\mathbf{b}\mathbf{a}\in
S(I,i)$. Given $C\in\mathcal{M}(\leqslant i)$, we have from (2.3) that

(3.3)
$\mathbf{e}_{\mathcal{N}_{C}}(\mathbf{a}\mathbf{b})\mathbf{e}_{\mathcal{N}_{C}}=\mathbf{e}_{\mathcal{N}_{C}}\mathbf{a}(\mathbf{b}\mathbf{e}_{\mathcal{N}_{C}})=\mathbf{e}_{\mathcal{N}_{C}}\mathbf{a}\mathbf{e}_{\mathcal{N}_{C}}\mathbf{b}\mathbf{e}_{\mathcal{N}_{C}}.$

By Lemma 3.1 we have that
$\mathbf{e}_{\mathcal{N}_{C}}\mathbf{b}\mathbf{e}_{\mathcal{N}_{C}}\in\operatorname{Soc}(Q(I,i))$.
On the other hand $\mathbf{a}\in H(I,j)\subset S(I,j)\subset Q(I,j)\subset
Q(I,i)$. Since $\mathbf{e}_{\mathcal{N}_{C}}\in Q(I,i)$ and
$\operatorname{Soc}(Q(I,i))$ is an ideal of $Q(I,i)$, we infer that the third
member of (3.3), and hence the first one, belongs to
$\operatorname{Soc}(Q(I,i))$. As a result $\mathbf{a}\mathbf{b}\in H(I,i)$,
again by Lemma 3.1. The proof that $\mathbf{b}\mathbf{a}\in H(I,i)$ is
similar.

Now, let $J$ be a finite subset of $\\{i\leqslant\\}$, let
$\mathbf{a}=\sum_{j\in J}\mathbf{a}_{j}$, where
$\mathbf{0}\neq\mathbf{a}_{j}\in H(I,j)$ for $j\in J$, let
$j_{1}<\cdots<j_{n}$ be a maximal chain of $J$, choose
${C_{n}\in\mathcal{M}(\leqslant j_{n})}$ in such a way that
$i,j_{1},\ldots,j_{n}\in C_{n}$ and set $D=C_{n}\cap\\{\leqslant
i\\}\in\mathcal{M}(\leqslant i)$. As seen in the first part of the present
proof, the $(\mathcal{N}_{C_{n}}\times\mathcal{N}_{C_{n}})$-block of
$\mathbf{a}$ is not zero. Let $x,y\in\mathcal{N}_{C_{n}}$ be such that
$\mathbf{a}(x,y)\neq\mathbf{0}$. Since
$\mathcal{N}_{C_{n}}\subset\mathcal{N}_{D}$, there are (necessarily unique)
$V,W\in\mathcal{P}_{i,D}$ such that $x\in V$ and $y\in W$. Now
$Y=\overline{V},Z=\overline{W}$ are both members of $\mathcal{R}_{i}$; we have
that $x\in Y$ and $y\in Z$, thus both $\mathbf{e}_{Y}\mathbf{a}$ and
$\mathbf{a}\mathbf{e}_{Z}$ are nonzero and belong to $H(I,i)$ by the above,
because $\mathbf{e}_{Y},\mathbf{e}_{Z}\in H(I,i)$. ∎

A first consequence of Theorem 3.6, as already announced at the beginning of
this section, is that for every subset $J$ of $I$ the $\mathbb{K}$-subspace
$H(I,J)=\bigoplus_{j\in J}H(I,j)$ is a $\mathbb{K}$-subalgebra of $Q(I)$. Of
course we have that $H(I,\emptyset)=\\{\mathbf{0}\\}$ and
$H(I,\\{i\\})=H(I,i)$ for every $i\in I$. Note that $H(I,J)$ may fail to be a
unitary subring of $Q(I)$ and it may even lack multiplicative identity; we
want to investigate under which conditions on $J$ the ring $H(I,J)$ has a
multiplicative identity. Firstly, we observe that the subset $X(I,J)$ of $X$
defined by

$X(I,J)\,\,\colon\\!\\!=\bigcup\\{X(I)_{i}\mid i\in
J\\}=\bigcup\left.\left\\{\boldsymbol{\aleph}^{(I)}\times\mathcal{M}(\leqslant
i\leqslant)\right|i\in J\right\\}$

is the smallest subset of $X(I)$ such that every matrix in $H(I,J)$ has zero
entries outside the $(X(I,J)\times X(I,J))$-block. On the other hand, if a
matrix $\mathbf{u}\in Q(I)$ acts as a multiplicative identity on $H(I,J)$,
then the following equalities hold:

(3.4)
$\mathbf{u}\mathbf{e}_{X(I,J)}=\mathbf{e}_{X(I,J)}=\mathbf{e}_{X(I,J)}\mathbf{u}.$

In fact, given $x\in X(I,J)$, there are $i\in J$, $C\in\mathcal{M}(\leqslant
i)$ and $V\in\mathcal{P}_{i,C}$ such that $x\in V$ (remember that, for every
$i\in J$, we have $X(I)_{i}=\bigcup\\{\mathcal{N}_{C}\mid
C\in\mathcal{M}(\leqslant i)\\}$ and
$\mathcal{N}_{C}=\bigcup\mathcal{P}_{i,C}$ for every
$C\in\mathcal{M}(\leqslant i)$). Inasmuch as $\mathbf{e}_{\overline{V}}\in
H(I,i)\subset H(I,J)$, we have that
$\mathbf{e}_{\overline{V}}\mathbf{u}=\mathbf{e}_{\overline{V}}=\mathbf{u}\,\mathbf{e}_{\overline{V}}$
and, since $x\in{\overline{V}}$, we infer that
$\mathbf{u}(x,y)=\delta(x,y)=\mathbf{u}(y,x)$ for every $y\in X(I)$, which
proves (3.4). As a consequence, $H(I,J)+\mathbf{e}_{X(I,J)}\,\mathbb{K}$ is
the smallest $\mathbb{K}$-subalgebra of $Q(I)$ which has a multiplicative
identity (given by $\mathbf{e}_{X(I,J)}$) and contains $H(I,J)$ as an ideal;
this means that the ring $H(I,J)$ has a multiplicative identity exactly when
$\mathbf{e}_{X(I,J)}\in H(I,J)$.

Let us denote by $J^{\bigstar}$ the set of those maximal elements of $I$ which
are comparable with some element of $J$:

$J^{\bigstar}\,\,\colon\\!\\!=\mathbf{M}(I)\cap\\{J\leqslant\\}$

(recall that $\mathbf{M}(I)$ denotes the set of all maximal elements of $I$).
Of course it may happen that $J\not\subset\\{\leqslant J^{\bigstar}\\}$, in
particular that $J^{\bigstar}=\emptyset$, while $J\neq\emptyset$. If every
element of $I$ is bounded by a maximal element or, equivalently, all maximal
chains of $I$ have a greatest element, then it is clear that $X(I,J)\subset
X(I,J^{\bigstar})$; this inclusion is an equality if and only if, given $m\in
J^{\bigstar}$, every maximal chain of $I$ which is bounded by $m$ contains an
element of $J$. Obviously this is the case if $J^{\bigstar}\subset J$, in
particular when $J$ is an upper subset of $I$; in this latter case it is clear
that $J^{\bigstar}=\mathbf{M}(J)$.

######  Definition 3.7.

We say that a subset $J$ of $I$ is _finitely sheltered in_ $I$ if the
following three conditions hold:

$J^{\bigstar}\text{ is finite,}\quad J\subset\\{\leqslant
J^{\bigstar}\\}\quad\text{ and }\quad J^{\bigstar}\subset J.$

If $J$ is an upper subset of $I$, then $J$ is finitely sheltered in $I$ if and
only if $J$ has a finite cofinal subset; in particular $I$ is finitely
sheltered in $I$ exactly when $I$ has a finite cofinal subset and, if it is
the case, then every upper subset of $I$ is finitely sheltered in $I$.

######  Proposition 3.8.

Given a nonempty subset $J$ of $I$, the following conditions are equivalent:

  1. (1)

$H(I,J)$ has a multiplicative identity (which is necessarily
$\mathbf{e}_{X(I,J)}$).

  2. (2)

$H(I,J)\cap(\mathbf{e}_{X(I,J)}\mathbb{K})\neq 0$.

  3. (3)

$J$ is finitely sheltered in $I$.

If any (and hence all) of these conditions holds, then

(3.5) $\mathbf{e}_{X(I,J)}=\mathbf{e}_{X(I,J^{\bigstar})}=\sum_{m\in
J^{\bigstar}}\mathbf{e}_{X(I)_{m}}.$

###### Proof.

(1)$\Rightarrow$(2) is obvious.

(2)$\Rightarrow$(3). Suppose that (2) holds. This means that there is a finite
subset $F\subset J$ and nonzero matrices $\mathbf{d}_{i}\in H(I,i)$, for $i\in
F$, such that

(3.6) $\mathbf{d}\,\,\colon\\!\\!=\sum_{i\in
F}\mathbf{d}_{i}\in\mathbf{e}_{X(I,J)}\mathbb{K}$

is not zero. Let us prove first that $X(I,F)=X(I,J)$. Clearly $F\subset J$
implies that $X(I,F)\subset X(I,J)$. On the other hand, given $x\in X(I,J)$,
since the $x$-th row of $\mathbf{d}$ is not zero, then the $x$-th row of
$\mathbf{d}_{i}$ is not zero for some $i\in F$. This means that there exists
$C\in\mathcal{M}(\leqslant i)$ such that $x\in\mathcal{N}_{C}\subset
X(I)_{i}\subset X(I,F)$. Thus $X(I,J)\subset X(I,F)$ and so $X(I,F)=X(I,J)$.

Next, we claim that every maximal element of $F$ is maximal in $I$. Let
$i_{1}<\ldots<i_{n}$ be a maximal chain of $F$. If we consider any
$C\in\mathcal{M}(\leqslant i_{n})$ such that $i_{1},\ldots,i_{n-1}\in C$, then
it follows from Lemma 3.5 that the
$\mathcal{N}_{C}\times\mathcal{N}_{C}$-blocks of $\mathbf{d}$ and
$\mathbf{d}_{i_{1}}+\cdots+\mathbf{d}_{i_{n}}$ coincide. If $i_{n}$ were not
maximal in $I$, it would follow from Lemma 3.1, (2) and Theorem 1.5, (3) that
there are $X_{1},\ldots,X_{r}\in\mathcal{P}_{i_{n}}$ such that, given
$x\in\mathcal{N}_{C}$, the $x$-th row of $\mathbf{d}$ is not zero only if
$x\in X_{1}\cup\cdots\cup X_{r}$. On the other hand, the
$\mathcal{N}_{C}\times\mathcal{N}_{C}$-block of $\mathbf{d}$ is a nonzero
scalar matrix and therefore the $x$-th row of $\mathbf{d}$ is not zero for
every $x\in\mathcal{N}_{C}$. Since $\mathcal{N}_{C}$ is the union of
$\boldsymbol{\aleph}$ elements of the partition $\mathcal{P}_{i}$ (see
Notations and Settings 2.1), we have a contradiction and our claim is proved.

Now, given $j\in J$, let $A$ be any maximal chain of $I$ such that $j\in C$.
Inasmuch as $X(I,F)=X(I,J)$, this means that $C\cap F\neq\emptyset$ and, by
the above proved claim, $C\cap F$ contains a maximal element of $I$. This
proves that $J\subset\\{\leqslant F\\}$.

Let us show that $J^{\bigstar}\subset F$, from which it will follow that
$J^{\bigstar}$ is finite and is contained in $J$, completing the proof that
$J$ is finitely sheltered. Assume, on the contrary, that there is some maximal
element $m$ of $I$ and some $j\in J$ such that $m\not\in F$ but $j<m$, let
$C\in\mathcal{M}(I)$ be such that $j,m\in C$ and put $D=C\cap\\{\leqslant
j\\}\in\mathcal{M}(\leqslant j)$. We have from Lemma 2.3, (2) that

$\emptyset\neq\mathcal{N}_{C}\subset\mathcal{N}_{D}\subset X(I)_{j}\subset
X(I,J),$

therefore the $\mathcal{N}_{C}\times\mathcal{N}_{C}$-block of $\mathbf{d}$ is
not zero. If $i_{1},\ldots,i_{r}$ are the elements of $F$ such that the
$\mathcal{N}_{C}\times\mathcal{N}_{C}$-blocks of
$\mathbf{d}_{i_{1}},\ldots,\mathbf{d}_{i_{r}}$ are not zero, then
$i_{1},\ldots,i_{r}\in\\{<m\\}$. Indeed, given $t\in\\{1,\ldots,r\\}$, there
is some $C\in\mathcal{M}(\leqslant i_{t})$ such that
$\mathcal{N}_{D}\cap\mathcal{N}_{C}\neq\emptyset$, hence $i_{t}$ and $m$ are
related by Lemma 2.3, (1) and therefore $i_{t}<m$. We have now

$\mathbf{0}\neq\mathbf{d}(\mathcal{N}_{C},\mathcal{N}_{C})=\mathbf{d}_{i_{1}}(\mathcal{N}_{C},\mathcal{N}_{C})+\cdots+\mathbf{d}_{i_{r}}(\mathcal{N}_{C},\mathcal{N}_{C})$

Let $x\in\mathcal{N}_{C}$ be such that the $x$-row of $\mathbf{d}$ is not
zero. Then there is some $t\in\\{1,\ldots,r\\}$ such that the $x$-row of
$\mathbf{d}_{i_{t}}$ is not zero and necessarily $x\in\mathcal{N}_{D}$ for
some $D\in\mathcal{M}(\leqslant i_{t})$. Inasmuch as $i_{t}$ is not a maximal
element of $I$ then, according to Lemma 3.1, (2) and Theorem 1.5, (3) we have
that $x$ belongs to the union of finitely many elements of
$\mathcal{P}_{i_{t}}$. Since $i_{t}<m$, the partition $\mathcal{P}_{m}$ is
coarser than $\mathcal{P}_{i_{t}}$ and therefore $x$ belongs to the union of
finitely many elements of $\mathcal{P}_{m}$. As a result we have that there
are $Y_{1},\ldots,Y_{s}\in\mathcal{P}_{m,C}\subset\mathcal{P}_{m}$ such that,
given $x\in\mathcal{N}_{C}$, the $x$-th row of $\mathbf{d}$ is not zero only
if $x\in Y_{1}\cup\cdots\cup Y_{s}$. This leads to a contradiction; in fact,
since $\mathbf{d}\in\mathbf{e}_{X(I,J)}\mathbb{K}$ and $\mathcal{N}_{C}\subset
X(I,J)$, for every $x\in\mathcal{N}_{C}$ the $x$-th row of $\mathbf{d}$ is not
zero and $\mathcal{N}_{C}$ is the union of $\boldsymbol{\aleph}$ members of
$\mathcal{P}_{m}$. This shows that $J^{\bigstar}\subset F$, as wanted.

(3)$\Rightarrow$(1) Assume (3). Then it follows that $\\{X(I)_{m}\mid m\in
J^{\bigstar}\\}$ is a finite partition of $X(I,J)$ (see the proof of
Proposition 2.2). Thus (3.5) holds and, since $\mathbf{e}_{X(I)_{m}}$ is in
$H(I,m)$ for all $m\in J^{\bigstar}$, it follows that $\mathbf{e}_{X(I,J)}\in
H(I,J)$. ∎

######  Corollary 3.9.

The ring $H(I,I)$ has a multiplicative identity if and only if $I$ has a
finite cofinal subset. If it is the case and $J\subset I$, then $H(I,J)$ is a
unitary subring of $H(I,I)$ if and only if all maximal elements of $I$ belong
to $J$.

##  4\. Representing a partially ordered set $I$: the unit-regular
$\mathbb{K}$-algebra $\mathfrak{B}(I)$.

Henceforth, unless otherwise stated, we assume that the
$\boldsymbol{\aleph}$-poset $I$ we consider _has a finite cofinal subset_.
Starting from this section we focuse our interest on the main object of our
work: the $\mathbb{K}$-subalgebra

$\mathfrak{B}(I)\,\,\colon\\!\\!=H(I,I)=\bigoplus_{i\in I}H(I,i)$

of $Q(I)=\mathbb{C}\mathbb{F}\mathbb{M}_{X(I)}(\mathbb{K})$, which is unital
by Proposition 2.2 and Corollary 3.9. Obviously $\mathfrak{B}(I)$ depends on
the cardinal $\boldsymbol{\aleph}$ and the field $\mathbb{K}$; if we need to
emphasize this we use the notation
$\mathfrak{B}_{\boldsymbol{\aleph},\mathbb{K}}(I)$.

Our first result is a complete description of ideals of $H(I,J)$, for any
subset $J\subset I$, and hence of the whole $\mathbb{K}$-algebra
$\mathfrak{B}(I)$.

######  Theorem 4.1.

Given a subset $J$ of $I$, the assignment $L\mapsto H(I,L)$ defines a lattice
isomorphism from the lattice ${\Downarrow\\!\\!J}$ of all lower subsets of $J$
to the lattice $\mathbb{L}_{2}(H(I,J))$ of all ideals of $H(I,J)$. In
particular, the lattice of all ideals of $\mathfrak{B}(I)$ is isomorphic to
$\Downarrow\\!\\!I$.

###### Proof.

At first, let us prove that if $L\subset J\subset I$, then $H(I,L)$ is an
ideal of $H(I,J)$ if and only if $L$ is a lower subset of $J$. Assume that
$H(I,L)$ is an ideal of $H(I,J)$ and take $i\in L$, $j\in J$ with $j\leqslant
i$. If $j\not\in L$, then $H(I,j)\subset H(I,J\setminus L)$, while it follows
from Theorem 3.6 that

$\\{\mathbf{0}\\}\neq H(I,i)H(I,j)\subset H(I,j)\cap H(I,L)\subset
H(I,J\setminus L)\cap H(I,L)=\\{\mathbf{0}\\},$

hence a contradiction. Thus necessarily $j\in L$. Conversely, suppose that $L$
is a lower subset of $J$ and let $i\in L$, $j\in J$. Then exactly one of the
following conditions occurs: a) $j\in L$, b) $j\not\in L$ and $i,j$ are
uncomparable, c) $j\not\in L$ and $j>i$. In all cases it follows from Theorem
3.6 that $H(I,j)H(I,i)\cup H(I,i)H(I,j)\subset H(I,L)$. Since $H(I,L)$ is
already a $\mathbb{K}$-subspace of $H(I,J)$, this is sufficient to conclude
that it is an ideal of $H(I,J)$.

Suppose now that $\mathfrak{I}$ is an ideal of $H(I,J)$. As
$\\{\mathbf{0}\\}=H(I,\emptyset)$, we may assume that
$\mathfrak{I}\neq\\{\mathbf{0}\\}$ and observe that if $\mathfrak{I}\cap
H(I,j)\neq\\{\mathbf{0}\\}$ for some $j\in J$, then necessarily
$H(I,j)=\mathfrak{I}\cap H(I,j)\subset\mathfrak{I}$, because $H(I,j)$ is a
simple ring. Our goal is to prove that $\mathfrak{I}=H(I,L)$, where $L=\\{j\in
J\mid H(I,j)\subset\mathfrak{I}\\}$. Suppose that
$\mathbf{0}\neq\mathbf{a}\in\mathfrak{I}$ and let us consider the unique
decomposition

$\mathbf{a}=\sum\\{\mathbf{a}_{j}\mid j\in F\\}$

for some finite subset $F$ of $J$ and nonzero elements $\mathbf{a}_{j}\in
H(I,j)$, for $j\in F$. We will reach our goal once we show that
$\mathbf{a}_{j}\in\mathfrak{I}$ for every $j\in F$. As the statement is
obvious in case $|F|=1$, let us consider an integer $n\geqslant 2$, assume
inductively that the statement is true whenever $|F|<n$ and suppose that
$|F|=n$. We claim that $\mathfrak{I}\cap H(I,j_{0})\neq\\{\mathbf{0}\\}$ for
at least one $j_{0}\in J$. If $F$ is an antichain, given any $j_{0}\in F$ and
an idempotent $\mathbf{e}$ of $H(I,j_{0})$ such that
$\mathbf{a}_{j_{0}}=\mathbf{e}\mathbf{a}_{j_{0}}$, it follows from (2) of
Theorem 3.6 that

$\mathbf{0}\neq\mathbf{a}_{j_{0}}=\mathbf{e}\mathbf{a}_{j_{0}}=\mathbf{e}\mathbf{a}\in\mathfrak{I}\cap
H(I,j_{0}).$

Otherwise, let us choose a maximal element $k$ of $F$ which is not minimal in
$F$, set $\\{j_{1},\ldots,j_{r}\\}=\\{<k\\}\cap F$ and choose an element
$Z\in\mathcal{R}_{k}$ such that $\mathbf{e}_{Z}\mathbf{a}_{k}\neq\mathbf{0}$.
It follows from Lemma 3.1, (3) that, for every $\alpha\in\\{1,\ldots,r\\}$,
there are $Y_{\alpha 1},\ldots,Y_{\alpha
s_{\alpha}}\in\mathcal{R}_{j_{\alpha}}$ such that, by setting
$Y_{\alpha}=Y_{\alpha 1}\cup\cdots\cup Y_{\alpha s_{\alpha}}$, the $x$-th row
of $\mathbf{a}_{j_{\alpha}}$ is not zero only if $x\in Y_{\alpha}$. Set
$Y=Y_{1}\cup\cdots\cup Y_{r}$ and let us choose any
${j_{0}}\in\\{j_{1},\ldots,j_{r}\\}$. We have that
$\mathbf{a}_{j_{\alpha}}=\mathbf{e}_{Y}\mathbf{a}_{j_{\alpha}}$ for every
$\alpha\in\\{1,\ldots,r\\}$ and, on the other side, it follows from (3) Lemma
2.8 that there is some $T\in\mathcal{R}_{{j_{0}}}$ such that $T\subset Z$ and
$T\cap Y=\emptyset$, thus

$\mathbf{e}_{T}\mathbf{a}_{j_{1}}=\cdots\mathbf{e}_{T}\mathbf{a}_{j_{r}}=\mathbf{0}.$

In addition, if $l\in F\setminus\\{<k\\}$ and $U\in\mathcal{R}_{l}$, then
$U\cap Z=\emptyset$ by Lemma 2.8, (1) and so $U\cap T=\emptyset$.
Consequently, by (2) of Theorem 3.6 we have that
$\mathbf{e}_{T}\mathbf{a}_{l}=\mathbf{0}$ for every $l\in F\setminus\\{<k\\}$.
Since the $x$-th row of $\mathbf{a}_{k}$ is not zero for all $x\in Z$ and
$\emptyset\neq T\subset Z$, we have that
$\mathbf{e}_{T}\mathbf{a}_{{j_{0}}}\neq\mathbf{0}$. By taking (3) of Theorem
3.6 into account, we conclude that

$\mathbf{0}\neq\mathbf{e}_{T}\mathbf{a}_{{j_{0}}}=\mathbf{e}_{T}\mathbf{a}\in\mathfrak{I}\cap
H(I,{j_{0}})$

and our claim is proved. As seen previously, we have that
$H(I,{j_{0}})\subset\mathfrak{I}$ and therefore
$\mathbf{a}_{{j_{0}}}\in\mathfrak{I}$. Consequently
$\mathbf{a}-\mathbf{a}_{{j_{0}}}\in\mathfrak{I}$ and the inductive assumption
gives us that $\mathbf{a}_{j}\in\mathfrak{I}$ for all $j\in
F\setminus\\{j_{0}\\}$. We have so far established the first of the two
inclusions

$\mathfrak{I}\subset H(I,L)=\bigoplus_{l\in L}H(I,l)\subset\mathfrak{I},$

the second being clear; thus we have the equality, as wanted. Finally $L$ is a
lower subset of $J$, as we have seen at the beginning of the present proof,
while the the fact that the assignment $L\mapsto H(I,L)$ defines a lattice
isomorphism from $\Downarrow\\!\\!J$ to $\mathbb{L}_{2}(H(I,J))$ is
straightforward. ∎

######  Remark 4.2.

If $J\subset I$ and $L$ is a lower subset of $J$, then the projection map

$\varphi_{L,J}\colon H(I,J)\longrightarrow H(I,J\setminus L)$

with kernel $H(I,L)$, that is

(4.1)
$\varphi_{L,J}(\mathbf{a}^{\prime}+\mathbf{a}^{\prime\prime})=\mathbf{a}^{\prime}\quad\text{
for all\,\, $\mathbf{a}^{\prime}\in H(I,J\setminus
L),\,\,\mathbf{a}^{\prime\prime}\in H(I,L)$},$

is a surjective $\mathbb{K}$-algebra homomorphism which induces an isomorphism
of $\mathbb{K}$-algebras

$H(I,J)/H(I,L)\simeq H(I,J\setminus L).$

Let us say that an ideal $\mathfrak{I}$ of a $\mathbb{K}$-algebra $R$ is
_splitting_ if there is a $\mathbb{K}$-subalgebra $S$ of $R$ such that
$R=\mathfrak{I}\oplus S$ as $\mathbb{K}$-vector spaces or, equivalently, the
canonical ring epimorphism $R\to R/\mathfrak{I}$ restricts to an isomorphism
from $S$ to $R/\mathfrak{I}$. Then it follows from Theorem 4.1 that every
ideal of $\mathfrak{B}(I)$ is splitting.

Another remarkable feature of the algebra $\mathfrak{B}(I)$ is that it is
locally matricial. We recall that a _matricial $\mathbb{K}$-algebra_ is any
algebra isomorphic to

$\mathbb{M}_{n_{1}}(\mathbb{K})\times\cdots\times\mathbb{M}_{n_{r}}(\mathbb{K})$

for some positive integers $n_{1},\ldots,n_{r}$ (see [19], p.217, for
instance). A $\mathbb{K}$-algebra $R$ is _locally matricial_ if every _finite_
subset of $R$ is contained in some matricial $\mathbb{K}$-subalgebra. Given
$i\in I$, since $H(I,i)$ is isomorphic to $\operatorname{Soc}(Q(I,i))$, by
Litoff Theorem (see [25], p.90) $H(I,i)$ is locally matricial. Note that if
$\mathbf{e}$ is any idempotent of $H(I,i)$, then $\mathbf{e}H(I,i)\mathbf{e}$
is a matricial $\mathbb{K}$-algebra, as it is isomorphic to the endomorphism
ring of a finite direct sum of copies of the simple left $H(I,i)$-module
$\mathbb{K}^{(X(I))}$, whose endomorphism ring is isomorphic to $\mathbb{K}$.

######  Proposition 4.3.

If $J$ is any subset of $I$, then the $\mathbb{K}$-algebra $H(I,J)$ is locally
matricial. In particular $\mathfrak{B}(I)$ is locally matricial.

###### Proof.

Clearly, we may assume that $J$ is finite and proceed by induction on the
number $n=|J|$. If $n=0$ then $H(I,J)=\\{\mathbf{0}\\}$ and there is nothing
to prove. Given $n>0$, assume that $H(I,J)$ is locally matricial whenever
$|J|=n-1$ and let us consider the case in which $|J|=n$. After choosing a
minimal element $m$ of $J$ and by setting $S=H(I,m)$, we have that

$H(I,J)=S\oplus H(I,J\setminus\\{m\\})$

as $\mathbb{K}$-vector spaces. Thus, in order to prove that $H(I,J)$ is
locally matricial, it is sufficient to prove that if $F$ and $G$ are finite
subsets of $S$ and $H(I,J\setminus\\{m\\})$ respectively, then there is a
matricial $\mathbb{K}$-subalgebra of $H(I,J)$ containing $F\cup G$.

By Litoff Theorem and according to its alternative proof given by Faith and
Utumi in [17], there is an idempotent $\mathbf{e}\in S$ such that the
matricial $\mathbb{K}$-algebra $\mathbf{e}S\mathbf{e}$ contains $F$. By the
inductive assumption, there is a matricial $\mathbb{K}$-subalgebra $T$ of
$H(I,J\setminus\\{m\\})$ containing $G$. By setting $\widehat{J}=J\cup
J^{\bigstar}$, it follows from Proposition 3.8 that $H(I,\widehat{J})$ has a
multiplicative identity $\mathbf{u}$; set $U=T+\mathbb{K}\mathbf{u}$ and note
that $U$ is still a matricial $\mathbb{K}$-algebra, because
$T+\mathbb{K}\mathbf{u}=T+\mathbb{K}(\mathbf{u}-\mathbf{f})\simeq
T\times\mathbb{K}$, where $\mathbf{f}$ is the multiplicative identity of $T$
(note that $\mathbf{u}-\mathbf{f}$ is a central idempotent of
$T+\mathbb{K}\mathbf{u}$).

Now $S$ is a $(U,U)$-submodule of $H(I,\widehat{J})$, because it is an ideal
of $H(I,\widehat{J})$ by Theorem 4.1; thus we may consider the
$(U,U)$-submodule $U\mathbf{e}S\mathbf{e}U$ of $S$. As $\mathbf{u}\in U$, we
have that $\mathbf{e}\in\mathbf{e}U\mathbf{e}$, thus
$\mathbf{0}\neq\mathbf{e}\in S\mathbf{e}U\mathbf{e}S$. Since $S$ is a simple
ring, we infer that $S\mathbf{e}U\mathbf{e}S=S$ and therefore

$(U\mathbf{e}S\mathbf{e}U)(U\mathbf{e}S\mathbf{e}U)=U\mathbf{e}S\mathbf{e}U\mathbf{e}S\mathbf{e}U=U\mathbf{e}S\mathbf{e}U,$

so that $V\,\,\colon\\!\\!=U\mathbf{e}S\mathbf{e}U$ is a
$\mathbb{K}$-subalgebra of $S$. If $\mathfrak{I}$ is a nonzero ideal of $V$,
then

$U\mathfrak{I}=UV\mathfrak{I}=V\mathfrak{I}=\mathfrak{I}=\mathfrak{I}V=\mathfrak{I}VU=\mathfrak{I}U$

and, consequently, by using the simplicity of $S$ we get

$\mathfrak{I}=V\mathfrak{I}V=U\mathbf{e}S\mathbf{e}U\mathfrak{I}U\mathbf{e}S\mathbf{e}U=U\mathbf{e}S\mathbf{e}\mathfrak{I}\mathbf{e}S\mathbf{e}U=U\mathbf{e}S\mathbf{e}U=V;$

this shows that $V$ is a simple ring. On the other hand, if $A$ and $B$ are
bases of the $\mathbb{K}$-vector spaces $\mathbf{e}S\mathbf{e}$ and $U$
respectively, from the above we infer that the set

$C=\\{\mathbf{xyz}\mid\mathbf{y}\in A,\mathbf{x},\mathbf{z}\in B\\}$

generates $V$ as a $\mathbb{K}$-vector space. Since $A$ and $B$ are finite,
then so is $C$ and we conclude that $V$ is simple and artinian, so it has a
multiplicative identity $\mathbf{g}$. Note that if $\mathbf{t}\in U$, then
$\mathbf{t}\mathbf{g}$ and $\mathbf{g}\mathbf{t}$ belong to $V$ and this
entails that
$\mathbf{g}\mathbf{t}=\mathbf{g}\mathbf{t}\mathbf{g}=\mathbf{t}\mathbf{g}$;
thus $\mathbf{g}$ centralizes $U$. As a consequence

$U\mathbf{g}S\mathbf{g}U=\mathbf{g}USU\mathbf{g}=\mathbf{g}S\mathbf{g}$

and, by recalling that $T\subset U$, we may consider the
$\mathbb{K}$-subalgebra $W=\mathbf{g}S\mathbf{g}+T$ of $H(I,J)$, in which
$\mathbf{g}$ is a central idempotent. We have that $\mathbf{g}S\mathbf{g}\cap
T=\\{\mathbf{0}\\}$ and $\mathbf{g}W=\mathbf{g}S\mathbf{g}$, thus
$(\mathbf{1}-\mathbf{g})W\simeq T$ as $\mathbb{K}$-algebras and therefore
$W\simeq\mathbf{g}S\mathbf{g}\times T$. Since $\mathbf{g}S\mathbf{g}$ is a
matricial $\mathbb{K}$-algebra, then so is $W$. Finally, since
$\mathbf{e}S\mathbf{e}\subset V\subset\mathbf{g}S\mathbf{g}$ we conclude that
$F\cup G\subset W$, as wanted. ∎

We recall that a ring $R$ is _unit-regular_ if for every $x\in R$ there exists
a unit $y\in R$ such that $x=xyx$. It is well known that a regular ring $R$ is
unit regular if and only if, given three finitely generated projective right
$R$-modules $A,B,C$, the condition $A\oplus C\simeq B\oplus C$ implies
$A\simeq B$. If $R$ is a locally matricial $\mathbb{K}$-algebra, then $R$ is
regular and it is unit-regular in case it has multiplicative identity. Indeed,
given $x\in R$, there is a matricial $\mathbb{K}$-subalgebra $S$ of $R$ such
that $x\in S$. Since $S$ is unit-regular, then there is a unit $y$ of $S$ such
that $x=xyx$. If $e$ denotes the multiplicative identity of $S$, we see
immediately that $z=y+1-e$ is a unit of $R$ and $x=xzx$. Thus, given $J\subset
I$, the algebra $H(I,J)$ is regular and, by Proposition 3.8, it is unit-
regular in case $J$ is finitely sheltered in $I$; in particular
$\mathfrak{B}(I)$ is unit-regular.

If $J\subset I$, let us denote by $J_{1}$ the set of all minimal elements of
$J$. The dual classical Krull filtration of the poset $J$ is the ascending
chain $(J_{\alpha})_{0\leqslant\alpha}$ of subsets of $J$, indexed over the
ordinals, defined recursively as follows (see [3]):

$\displaystyle J_{0}\,\,\colon\\!\\!=\emptyset,$ $\displaystyle
J_{\alpha+1}\,\,\colon\\!\\!=J_{\alpha}\cup\left(J\setminus
J_{\alpha}\right)_{1}\quad\text{ for all }\alpha,$ $\displaystyle
J_{\alpha}\,\,\colon\\!\\!=\bigcup_{\beta<\alpha}J_{\beta}\quad\text{ if
$\alpha$ is a limit ordinal.}$

Note that each $J_{\alpha}$ is a lower subset of $J$. Clearly there exists a
smallest ordinal $\xi$ such that $J_{\xi+1}=J_{\xi}$ and hence
$J_{\alpha}=J_{\xi}$ for every $\alpha>\xi$; moreover $J$ is artinian (i. e.
it satisfies the DCC or, equivalently, every chain of $J$ is well ordered) if
and only if $J=J_{\xi}$ and, in this case, the ordinal $\xi$ is called the
dual classical Krull dimension of $J$ and the family
$\left(J_{\alpha+1}\setminus J_{\alpha}\right)_{\alpha<\xi}$ is a partition of
$J$.

Given $J\subset I$, the next result illustrates the tight connection between
the dual classical Krull filtration of $J$ and the Loewy chain
$(\operatorname{Soc}_{\alpha}(H(I,J)))_{0\leqslant\alpha}$ of ideals of
$H(I,J)$.

We recall that if $R$ is a regular ring (unitary or not), then every
homogeneous component of $\operatorname{Soc}(R_{R})$ is a minimal ideal and
$\operatorname{Soc}(R_{R})=\operatorname{Soc}({{}_{R}R})$, from which one can
easily infer that
$\operatorname{Soc}_{\alpha}(R_{R})=\operatorname{Soc}_{\alpha}({{}_{R}R})$
for every ordinal $\alpha$.

######  Theorem 4.4.

Let $J$ be a subset of $I$. For every ordinal $\alpha$ we have that

(4.2) $\operatorname{Soc}_{\alpha}(H(I,J))=H(I,J_{\alpha}).$

Consequently $H(I,J)$ is semiartinian if and only if $J$ is artinian, in which
the case the Loewy length of $H(I,J)$ equals the dual classical Krull
dimension of $J$. In particular $\mathfrak{B}(I)$ is semiartinian if and only
if $I$ is artinian.

###### Proof.

Obviously (4.2) holds if $\alpha=0$. Now the ideal $H(I,J_{1})$ is semisimple
as a right ideal of $H(I,J)$, because it is a direct sum of the simple and
semisimple rings $H(I,i)$, for $i\in J_{1}$. On the other hand, since $H(I,J)$
is regular, each homogeneous component of $\operatorname{Soc}(H(I,J))$ is a
minimal ideal and so, by Theorem 4.1, it is of the form $H(I,m)$ for some
minimal element $m$ of $J$. This shows that (4.4) holds if $\alpha=1$. Given
an ordinal $\alpha>0$, assume inductively that

$\operatorname{Soc}_{\beta}(H(I,J))=H(I,J_{\beta})$

whenever $\beta<\alpha$. If $\alpha$ is a limit ordinal, since
$J_{\alpha}=\bigcup_{\beta<\alpha}J_{\alpha}$, then we have

$\operatorname{Soc}_{\alpha}(H(I,J))=\bigcup_{\beta<\alpha}\operatorname{Soc}_{\beta}(H(I,J))=\bigcup_{\beta<\alpha}H(I,J_{\beta})=H(I,J_{\alpha}).$

Suppose that $\alpha=\beta+1$ for some $\beta$, set
$J_{\beta}^{\bullet\bullet}=J\setminus J_{\beta}$ and
$J_{\beta+1}^{\bullet}=(J\setminus J_{\beta})_{1}$, so that $J_{\beta+1}$ is
the disjoint union of $J_{\beta}$ and $J_{\beta+1}^{\bullet}$. The projection
map

$\varphi_{J_{\beta},J}\colon H(I,J)\longrightarrow
H(I,J_{\beta}^{\bullet\bullet})$

with kernel $H(I,J_{\beta})$ induces an isomorphism of $\mathbb{K}$-algebras

$H(I,J)/H(I,J_{\beta})\simeq H(I,J_{\beta}^{\bullet\bullet})$

(see Remark 4.2). Since

$\displaystyle\varphi_{J_{\beta},J}(H(I,J_{\beta+1}))=\varphi_{J_{\beta},J}[H(I,J_{\beta})\oplus
H(I,J_{\beta+1}^{\bullet})]=H(I,J_{\beta+1}^{\bullet})$
$\displaystyle=H(I,(J_{\beta}^{\bullet\bullet})_{1})=\operatorname{Soc}[H(I,(J_{\beta}^{\bullet\bullet})],$

we infer that

$H(I,J_{\beta+1})/H(I,J_{\beta})=\operatorname{Soc}[H(I,J)/H(I,J_{\beta})]$

and hence

$H(I,J_{\beta+1})=\operatorname{Soc}_{\alpha}(H(I,J)).$

The last statement follows directly from the definitions, toghether with the
fact that if $K,L\subset I$, then $H(I,K)\subset H(I,L)$ if and only if
$K\subset L$. ∎

######  Remark 4.5.

Given a division ring $D$ and an artinian partially ordered set $I$, in [10]
we presented a construction which produces a semiartinian and unit-regular
ring $D_{I}$, which is isomorphic to a subring of
$\mathbb{C}\mathbb{F}\mathbb{M}_{\boldsymbol{\aleph}}(D)$ containing $D$ (this
latter identified with the subring of all scalar matrices), where
$\boldsymbol{\aleph}$ is the first infinite cardinal such that
$\boldsymbol{\aleph}\geqslant|I|$ and
$\boldsymbol{\aleph}\geqslant|\mathcal{M}(I)|$. The construction, heavily
based on ordinal arithmetic, turns out to be rather tricky and cumbersome with
no possibility to extend it to non-artinian partially ordered sets. The
present construction overcomes this problem and, in addition, if $I$ is an
artinian $\boldsymbol{\aleph}$-poset having a finite cofinal subset, the
algebra $\mathfrak{B}(I)$ satisfies the same properties (1), (2), (3), (4),
(5) and (8) of $D_{I}$, as stated in [10, Theorem 9.5]; _mutatis mutandis_ ,
their proofs are identical.

##  5\. $\mathbf{K}_{0}(\mathfrak{B}(I))$.

We keep going on with the assumption that $I$ is an
$\boldsymbol{\aleph}$-poset having a finite cofinal subset. In this section we
afford the task of investigating the structure of the partially ordered
abelian group $\mathbf{K}_{0}(\mathfrak{B}(I))$. It will turn out that
$\mathbf{K}_{0}(\mathfrak{B}(I))$ is order isomorphic to a group which is a
special instance of a more general construction, well known within the
literature of lattice-ordered groups (see [4], page 6 for instance).
Specifically, given a family $\mathcal{H}=(H_{i})_{i\in I}$ of partially
ordered abelian groups, the _Hahn group_ $V(I,H_{i})$ of $\mathcal{H}$ is the
subgroup of the direct product $\prod_{i\in I}H_{i}$ consisting of all
elements having noetherian support, endowed with the partial order induced by
declaring positive those elements having positive coordinates relative to the
maximal elements of the support. The subgroup $\varSigma(I,H_{i})$ of
$V(I,H_{i})$ of those elements having finite support, endowed with the induced
partial order, is called the _restricted Hahn group_ ; of course, as a group,
$\varSigma(I,H_{i})$ is merely the direct sum $\bigoplus_{i\in I}H_{i}$. When
all $H_{i}$ coincide with a single group $H$, the groups $V(I,H_{i})$ and
$\varSigma(I,H_{i})$ are called respectively the _Hahn power_ and the
_restricted Hahn power of $H$ by $I$_. Our main result is to show that
$\mathbf{K}_{0}(\mathfrak{B}(I))$ is order isomorphic to the restricted Hahn
power of $\mathbb{Z}$ by $I$, whose support is just the free abelian group
with $I$ as a basis; we give here a description of it with some more details,
suited for our needs.

Let us denote by $G(I)$ the free abelian group with $I$ as a basis; we keep
the standard practice by representing each element $x\in G(I)$ as a unique
linear combination of elements of $I$ with coefficients in $\mathbb{Z}$ and,
if $i\in I$, we denote by $x_{i}$ the coefficient of $i$. As usual, the
_support_ of an element $x\in G(I)$ is the set
$\operatorname{Supp}(x)\,\,\colon\\!\\!=\\{i\in I\mid x_{i}\neq 0\\}$. As it
was shown in [11], the set

(5.1) $M(I)\,\,\colon\\!\\!=\\{x\in G(I)\mid x_{i}>0\text{ if $i$ is a maximal
element of $\operatorname{Supp}(x)$}\\}$

is a conical submonoid of $G(I)$, in the sense that if $x,y\in M(I)$ and
$x+y=0$, then $x=0=y$. We will consider $G(I)$ partially ordered with $M(I)$
as positive cone, so that $G(I)$ is the _restricted Hahn power_ of
$\mathbb{Z}$ by $I$. It was shown in [11, Proposition 4.2] that $G(I)$ is a
dimension group in which the sum of all maximal elements of $I$ is an order-
unit and, consequently, $G(I)$ is directed and $M(I)$ is a refinement monoid,
in the sense that if $x,y,z\in M(I)$ and $x\leqslant y+z$, then there exist
$a,b\in M(I)$ such that $a\leqslant x$, $b\leqslant y$ and $x=a+b$ (see [20,
Ch. 3]).

The following two results will not be used in the present work; nonetheless
these propositions illustrate remarkable properties of the group $G(I)$. We
recall that an element $p$ of an abelian monoid $M$ is _prime_ if it is not
invertible and, whenever $p\leqslant x+y$ for some $x,y\in M$, then either
$p\leqslant x$ or $p\leqslant y$ (here “$\leqslant$” is the algebraic
preorder: if $a,b\in M$, then $a\leqslant b$ means that $b=a+c$ for some $c\in
M$). If $M$ is a conical, cancellative refinement monoid (hence the algebraic
preorder is a partial order), it is easy to see that a non-invertible element
$p\in M$ is prime if and only if, given $a,b\in M$, the equality $p=a+b$
implies that either $p=a$ or $p=b$.

######  Proposition 5.1.

The prime elements of the monoid $M(I)$ are precisely the minimal elements of
$I$.

###### Proof.

Let us consider the subset

$\mathcal{A}=\\{i+x\mid i\in I,\operatorname{Supp}(x)\subset\\{<i\\}\text{ and
$x_{j}<0$ for all $j\in\operatorname{Supp}(x)\setminus\\{i\\}$}\\}$

of $M(I)$, let $\langle\mathcal{A}\rangle$ be the submonoid of $M(I)$
generated by $\mathcal{A}$ and let us prove first that
$\langle\mathcal{A}\rangle=M(I)$. Given $x\in M(I)$, if
$|\operatorname{Supp}(x)|\leqslant 1$, then it is clear that
$x\in\langle\mathcal{A}\rangle$. Given a positive integer $n$, assume that
$x\in\langle\mathcal{A}\rangle$ whenever $x\in M(I)$ and
$|\operatorname{Supp}(x)|<n$, let $x\in M(I)$ be such that
$|\operatorname{Supp}(x)|=n$ and let $m$ be a maximal element of
$\operatorname{Supp}(x)$. By setting

$y=x_{m}m+\sum\\{x_{i}i\mid i\in\\{<m\\}\text{ and }x_{i}<0\\},$

we have that $x^{\prime}=x-y\in M(I)$. In fact, let $k$ be a maximal element
of $\operatorname{Supp}(x^{\prime})$. If $k$ is maximal in
$\operatorname{Supp}(x)$ too, since
$\operatorname{Supp}(x^{\prime})=\operatorname{Supp}(x)\setminus\operatorname{Supp}(y)$
we have that $x^{\prime}_{k}=x_{k}>0$. Otherwise there is some
$h\in\operatorname{Supp}(x)$ such that $k<h$. Necessarily
$h\in\operatorname{Supp}(y)\subset\\{\leqslant m\\}$, from which $k\leqslant
m$. Since $k\neq m$ and $k\not\in\operatorname{Supp}(y)$, we conclude that
$x_{k}>0$. As $|\operatorname{Supp}(x^{\prime})|<n$, by the inductive
assumption we have that $x^{\prime}\in\langle\mathcal{A}\rangle$; since
$y\in\langle\mathcal{A}\rangle$, it follows that
$x=x^{\prime}+y\in\langle\mathcal{A}\rangle$, as wanted.

Let $p$ be a prime element of $M(I)$. We have that $p=x_{(1)}+\cdots+x_{(n)}$
for some $x_{(1)},\ldots,x_{(n)}\in\mathcal{A}$. Since $M(I)$ is a conical,
cancellative refinement monoid, then $p=x_{(\alpha)}$ for some
$\alpha\in\\{1,\ldots,n\\}$, that is there are $i\in I$ and $y\in G(I)$, with
$\operatorname{Supp}(y)\subset\\{<i\\}$ and $y_{j}<0$ for
$j\in\operatorname{Supp}(y)$, in such a way that $p=i+y$. Thus $p=(i+2y)+(-y)$
and, since both $i+2y$ and $-y$ are in $M(I)$ and $p\neq-y$, primality of $p$
implies that $p=i+2y$. Consequently $y=0$ and so $p=i$. If $i$ were not
minimal, that is $i>j$ for some $j\in I$, then $p=i=(i-j)+j$. Again, since
both $i-j$ and $j$ are in $M(I)$, primality of $p$ implies that either $i=i-j$
or $i=j$, which is impossible.

Conversely, let $i$ be a minimal element of $I$, let $x,y\in M(I)$, assume
that $i=x+y$ and let $j$ be a maximal element of
$\operatorname{Supp}(x)\cup\operatorname{Supp}(y)$ such that $j\neq i$. Then
$x_{j}+y_{j}=0$ and so either $x_{j}<0$, or $y_{j}<0$: a contradiction with
the assumption that $x$ and $y$ are nonzero elements of $M(I)$. This means
that $\operatorname{Supp}(x)\cup\operatorname{Supp}(y)=\\{i\\}$ and, since
$M(I)$ is cancellative, we conclude that either $x=i$ or $y=i$. ∎

If $G$ is a partially ordered abelian group, we denote by $G^{+}$ the positive
cone of $G$, namely $G^{+}=\\{x\in G\mid 0\leqslant x\\}$. A subgroup $H$ of
$G$ is an _ideal_ of $G$ if $H$ is a directed and convex subgroup (see [20],
page 8). We recall that $H$ is directed if and only if for every $x\in H$
there are $x^{\prime},x^{\prime\prime}\in H^{+}=G^{+}\cap H$ such that
$x=x^{\prime}-x^{\prime\prime}$ (see [19, Proposition 15.16]).

######  Proposition 5.2.

The assignment $L\mapsto G(L)$ defines an isomorphism from the lattice of all
lower subsets of $I$ to the lattice of all ideals of $G(I)$.

###### Proof.

Let $L$ be a lower subset of $I$ and let us prove first that $G(L)$ is convex.
Given $x\in G(I)$ and $y\in G(L)$, assume that $0<x\leqslant y$ but $x\not\in
G(L)$. Then there exists some $i\in I\setminus L$ such that $x_{i}\neq 0$ and,
since $\operatorname{Supp}(x)$ is finite, there is a maximal element $m$ of
$\operatorname{Supp}(x)$ such that $i\leqslant m$ and $x_{m}>0$, because $x\in
M(I)$; note that $m\not\in L$ otherwise, since $L$ is a lower subset of $I$,
from $i\leqslant m$ we would get $i\in L$. By setting $z=y-x$, we observe that
$m$ must be maximal in
$\operatorname{Supp}(z)\subset\operatorname{Supp}(x)\cup\operatorname{Supp}(y)$,
otherwise there would be some $l\in\operatorname{Supp}(y)\subset L$ such that
$i\leqslant m\leqslant l$, from which $i\in L$. Since $z\in M(I)$, then
$z_{m}>0$, in contradiction with the fact that, $z_{m}=-x_{m}<0$, because
$y_{m}=0$. Thus necessarily $x\in G(L)$ and so $G(L)$ is convex.

Next, let $0\neq x\in G(L)$. Then $x=x^{\prime}-x^{\prime\prime}$, where

$x^{\prime}=\sum\\{x_{i}\cdot i\mid
i\in\operatorname{Supp}(x),x_{i}>0\\}\text{ and
}x^{\prime\prime}=\sum\\{-x_{i}\cdot i\mid
i\in\operatorname{Supp}(x),x_{i}<0\\}.$

Since $x^{\prime},x^{\prime\prime}\in M(L)=G(L)\cap M(I)$, we infer that
$G(L)$ is directed and we conclude that $G(L)$ is an ideal of $G(I)$.

Conversely, let $H$ be an ideal of $G(I)$. Obviously, if $H=\\{0\\}$ then
$H=G(\emptyset)$. Assume that $H\neq\\{0\\}$, set
$L=\bigcup\\{\operatorname{Supp}(x)\mid x\in H^{+}\\}$ and let us prove that
$H=G(L)$. To this purpose, inasmuch as $H$ is directed it is sufficient to
show that $L\subset H$. Given $x\in H^{+}\setminus\\{0\\}$, if $x=x_{i}\cdot
i$ for some $i\in I$ and $x_{i}\in\mathbb{N}^{*}$, then obviously $i\in H$ by
the convexity of $H$. Given an integer $n>1$, assume inductively that
$\operatorname{Supp}(x)\subset H$ whenever $x\in H^{+}$ and
$|\operatorname{Supp}(x)|<n$. Given $x\in H^{+}$ with
$|\operatorname{Supp}(x)|=n$, set $K=\\{i\in I\mid x_{i}<0\\}$ and let us
consider the elements

$y=\sum_{i\in\operatorname{Supp}(x)\setminus K}x_{i}\cdot
i\quad\text{and}\quad z=\sum_{i\in K}(-x_{i})\cdot i,$

so that $x=y-z$. Note that $y,z$ and $y-2z$ belong to $M(I)$; since
$x=z+y-2z$, we infer that $0\leqslant z<x$ and hence $z\in H^{+}$, because $H$
is convex. Consequently from $y=x+z$ it follows that $y\in H^{+}$ as well. If
$K\neq\emptyset$, then $0<|\operatorname{Supp}(x)\setminus K|<n$ and $|K|<n$;
since $\operatorname{Supp}(x)\setminus K=\operatorname{Supp}(y)$ and
$K=\operatorname{Supp}(z)$, from the inductive assumption it follows that
$\operatorname{Supp}(x)\subset H$. If $K=\emptyset$, then $x=y$ and from the
convexity of $H$ it is easy to infer that $\operatorname{Supp}(x)\subset H$
and this completes the proof that $H=G(L)$.

Now, if $i,j\in I$ are such that $i\leqslant j$ and $j\in L$, then $0\leqslant
i\leqslant j$ in $H$. By the convexity of $H$ we infer that $i\in H$ and,
since $i\in H^{+}$, we conclude that $i\in L$, showing that $L$ is a lower
subset of $I$.

Finally, if $L$ is any subset of $I$ such that $G(L)$ is an ideal of $G(I)$,
by the above we have first that $G(L)=G(L^{\prime})$, where
$L^{\prime}=\bigcup\\{\operatorname{Supp}(x)\mid x\in G(L)^{+}=M(L)\\}$; this
implies that $L=L^{\prime}$ and therefore $L$ is a lower subset of $I$.

∎

Given a ring $R$, we denote by $\mathbf{F}\mathbf{P}_{R}$ the class of
finitely generated and projective right $R$-modules. Concerning basic notions
and notations about the Grothendiek group $\mathbf{K}_{0}(R)$ we follow [19,
Ch 15]; thus, in particular, for a single $A\in\mathbf{F}\mathbf{P}_{R}$ we
denote by $[A]$ the stable isomorphism class of $A$ and, for a subset
$\mathbf{A}$ of $\mathbf{F}\mathbf{P}_{R}$, we set

$[\mathbf{A}]\,\,\colon\\!\\!=\\{[A]\mid A\in\mathbf{A}\\}.$

Recall that if $R$ is unit-regular, then $[A]$ is merely the isomorphism class
of $A$, for every $A\in\mathbf{F}\mathbf{P}_{R}$.

Given $i\in I$, every minimal right ideal of $H(I,i)$ is generated by a
primitive idempotent and any two minimal right ideals of $H(I,i)$ are
isomorphic. If $\mathbf{u}$ is a primitive idempotent of $H(I,i)$, then the
element $[\mathbf{u}\mathfrak{B}(I)]$ of $\mathbf{K}_{0}(\mathfrak{B}(I))$
does not depend on the choice of $\mathbf{u}$. Indeed, if $\mathbf{v}$ is
another primitive idempotent of $H(I,i)$, since $\mathbf{u}H(I,i)$ and
$\mathbf{v}H(I,i)$ are isomorphic minimal right ideals of $H(I,i)$, then
$\mathbf{u}\mathfrak{B}(I)\simeq\mathbf{v}\mathfrak{B}(I)$, as a consequence
of the following very basic lemma (the proof is an exercise).

######  Lemma 5.3.

Let $R$ be a not necessarily unital ring and let $e,f$ be idempotents of $R$.
Then $eR_{R}$ and $fR_{R}$ are isomorphic if and only if there are $a,b\in R$
such that

$a=fae,\qquad b=ebf,\qquad ba=e,\qquad ab=f.$

Consequently, if $S$ is a not necessarily unital ring and $\eta\colon R\to S$
is any ring homomorphism, in particular if $R$ is a subring of $S$, then every
isomorphism from $eR_{R}$ to $fR_{R}$ gives rise to an isomorphism from
$\eta(e)S_{S}$ to $\eta(f)S_{S}$.

For every $i\in I$ let us choose a primitive idempotent $\mathbf{u}_{i}$ of
$H(I,i)$ and let us consider the subset

$\mathbf{U}\,\,\colon\\!\\!=\\{\mathbf{u}_{i}\mathfrak{B}(I)\mid i\in I\\}$

of $\mathbf{F}\mathbf{P}_{\mathfrak{B}(I)}$. Then, by the above, the subset
$[\mathbf{U}]$ of $\mathbf{K}_{0}(\mathfrak{B}(I))$ does not depend on the
choice of the $\mathbf{u}_{i}$’s; as we shall see, the structure of
$\mathbf{K}_{0}(\mathfrak{B}(I))$ is completely determined by $[\mathbf{U}]$.
In several instances we shall use the fact that, for every $i\in I$ and
$Y\in\mathcal{R}_{i}$, we have that
$\mathbf{u}_{i}H(I,i)\simeq\mathbf{e}_{Y}H(I,i)$ (see Lemma 3.1) and hence
$\mathbf{u}_{i}\mathfrak{B}(I)\simeq\mathbf{e}_{Y}\mathfrak{B}(I)$.

######  Lemma 5.4.

Let $e,f$ be two idempotents of a regular ring $R$. If $eR\lesssim fR$, then
$ReR\subset RfR$.

###### Proof.

If $eR\lesssim fR$ then, by applying [19, Lemma 2.7] we see that there are
orthogonal idempotents $f_{1},f_{2}\in R$ such that $fR=f_{1}R\oplus f_{2}R$
and $f_{1}R\simeq eR$. As a result
$ReR=\operatorname{Tr}_{R}(eR)=\operatorname{Tr}_{R}(f_{1}R)=Rf_{1}R\subset\operatorname{Tr}_{R}(fR)=RfR$.
∎

######  Proposition 5.5.

With the above notations, by considering $[\mathbf{U}]$ equipped with the
partial order induced by that of $\mathbf{K}_{0}(\mathfrak{B}(I))$, the
assignment $i\mapsto[\mathbf{u}_{i}\mathfrak{B}(I)]$ defines an order
isomorphism from $I$ to $[\mathbf{U}]$.

###### Proof.

Given $i,j\in I$, assume that $i\leqslant j$ and take any
$Z\in\mathcal{R}_{j}$, so that
$\mathbf{u}_{j}\mathfrak{B}(I)\simeq\mathbf{e}_{Z}\mathfrak{B}(I)$. According
to Lemma 2.8, (3), we can choose an element $Y\in\mathcal{R}_{i}$ such that
$Y\subset Z$. As a consequence we have that

$\mathbf{u}_{i}\mathfrak{B}(I)\simeq\mathbf{e}_{Y}\mathfrak{B}(I)\subset\mathbf{e}_{Z}\mathfrak{B}(I)\simeq\mathbf{u}_{j}\mathfrak{B}(I)$

and so
$[\mathbf{u}_{i}\mathfrak{B}(I)]\leqslant[\mathbf{u}_{j}\mathfrak{B}(I)]$.
Conversely, assume that this latter condition holds, meaning that
$\mathbf{u}_{i}\mathfrak{B}(I)\lesssim\mathbf{u}_{j}\mathfrak{B}(I)$. Then it
follows from Lemma 5.4 that

$H(I,\\{\leqslant
i\\})=\mathfrak{B}(I)\mathbf{u}_{i}\mathfrak{B}(I)\subset\mathfrak{B}(I)\mathbf{u}_{j}\mathfrak{B}(I)=H(I,\\{\leqslant
j\\}),$

therefore $i\leqslant j$. ∎

For the purposes of our investigation on the group
$\mathbf{K}_{0}(\mathfrak{B}(I))$, it is necessary to recall some basic facts
about regular and semiartinian rings; thus, in what follows we let $R$ be such
a ring and $\xi+1$ be its Loewy length. For every right $R$-module $M$ there
is a smallest ordinal $h(M)\leqslant\xi+1$ such that
$M=M\cdot\operatorname{Soc}_{h(M)}(R)$ and $h(M)$ is a successor ordinal in
case $M$ is finitely generated. In particular, if $e=e^{2}\in R$ and $\alpha$
is the unique ordinal such that
$e\in\operatorname{Soc}_{\alpha+1}(R)\setminus\operatorname{Soc}_{\alpha}(R)$,
then $h(eR)=\alpha+1$. If $A$ is a finitely generated and projective righ
$R$-module, then
$A\cdot\operatorname{Soc}_{\alpha}(R)=\operatorname{Soc}_{\alpha}(A)$ for
every $\alpha\leqslant\xi+1$ (see [9, Proposition 1.5]), therefore $A/A\cdot
Soc_{h(A)-1}(R)$ is always a semisimple $R/Soc_{h(A)-1}(R)$-module; we say
that $A$ is _eventually simple_ in case $A/A\cdot Soc_{h(A)-1}(R)$ is simple
and we consider the class

$\mathbf{E}\mathbf{S}\mathbf{F}\mathbf{P}_{R}\,\,\colon\\!\\!=\\{A\in\mathbf{F}\mathbf{P}_{R}\mid\text{$A$
is eventually simple}\\}.$

We say that a subset
$\mathbf{A}\subset\mathbf{E}\mathbf{S}\mathbf{F}\mathbf{P}_{R}$ is _complete
and irredundant_ if for every simple right $R$-module $U$ there is a unique
$A\in\mathbf{A}$ such that $A/A\cdot Soc_{h(A)-1}(R)\simeq U$. It can be shown
that $R$ admits a set $E$ of idempotents such that $\\{eR\mid e\in E\\}$ is a
complete and irredundant subset of
$\mathbf{E}\mathbf{S}\mathbf{F}\mathbf{P}_{R}$ (see [8, Page 70]).

Let us denote by ${\mathbf{S}\mathbf{i}\mathbf{m}\mathbf{p}}_{R}$ a set of