Realization of monoids with countable sum
Zahra Nazemian




  
For every infinite cardinal number κ, κ-monoids and their realization have recently been introduced and studied in <cit.>. 
As stated in <cit.>, a κ-monoid H has a realization to a ring R 
if there exists an element x ∈ H such that H is ℵ_1 ^--braided over add(ℵ_0 x), and add(ℵ_0 x),  as ℵ_0-monoid, 
 has a realization to R. Furthermore, H has a realization to hereditary rings if there exists an element x ∈ H such that H is 
 braided over add(x). These prompt an investigation into when ℵ_0-monoids have realizations. 
In this paper, we discuss the realization of ℵ_0-monoids and provide a complete characterization for the realization of two-generated ones in hereditary Von Neumann regular rings.






A self-consistent model for dust settling and the vertical shear instability in protoplanetary disks
    Satoshi Okuzumi1
    
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§ INTRODUCTION


If R is a  ring, then V (R), the class isomorphism of finitely genereted projective right modules,  
 is a reduced commutative monoid with order-unit. In their seminal papers Bergman <cit.> and 
Bergman and Dicks <cit.> show that: for every reduced commutative monoid with order-unit and every
 field k, there exists a hereditary k-algebra R with  V(R) ≅ H.
In the past two decades, the investigation of projective modules that are not necessarily finitely generated have
 been a focal point among researchers, as evidenced by works such as those by 
 Herbera, Přihoda,
  and Wiegand <cit.>. 
For every cardinal number κ, κ-monoids, a class of commutative monoids where the sum of κ many elements is allowed, have been recently introduced and studied 
in   <cit.>.
When R is a ring, isomorphism classes of projective (right) modules generated by at most κ many elements, 
denoted by V ^κ (R),  
is an example of a κ-monoid. 
A κ-monoid H is said to have realization if it is isomorphic as a κ-monoid to V^κ(R), for some ring R, see <cit.>. 
Although κ-monoids are reduced monoids,  they do not generally have realizations.

Besides the study of properties of κ-monoids, the realization problem for them has been intensively studied in <cit.>. 
As it is recalled in the following, it turned out that the focus should be on the study of the realization problem for ℵ_0-monoids. 
Let us recall some definitions before delving deep into the problem, and prior to that, we adopt the conventions from <cit.>. Namely,
 we use infinite ordinals and cardinals, but to avoid set-theoretical difficulties, we always consider such cardinals up to a fixed bound κ.
As underlying axiom system we have in mind the usual ZFC (with informal classes).
As usual in this context, we use the von Neumann definition of ordinals, and a cardinal is the smallest ordinal of a given cardinality (but cardinals are by default just considered as sets, forgetting the well-order).
We make liberal use of the axiom of choice, to ensure that the basic properties of cardinal arithmetic are well-behaved.
See <cit.> for background.



If x  (x_i)_i ∈κ∈ H^κ is a family indexed by elements of κ over some set H, and Σ H^κ→ H is a map, we use the notation

    ∑_i ∈κ x_i Σ( (x_i)_i ∈κ),

mimicking the typical summation notation. 


 
  Let κ be an infinite cardinal.
  A κ-monoid is a set H together with an element 0 ∈ H and a map Σ H^κ→ H such that the following conditions are satisfied.
  
  
  *  If x=(x_i)_i ∈κ∈ H^κ with x_i=0 for all i  0, then ∑_i ∈κ x_i=x_0.
  
  * 
    If (x_i,j)_i,j ∈κ∈ H^κ×κ and πκ×κ→κ is a bijection, then
    
    ∑_i ∈κ∑_j ∈κ x_i,j = ∑_k ∈κ x_π^-1(k).

  



Naturally, as it is explained in details in <cit.>,  when  H is a  κ-monoid, we can define a binary operation on H  which makes H a commutative  (reduced) monoid. 
Indeed,  sum of two elements  a, b∈ H is defined to be 
 ∑_i ∈κ x_i, where x_0 = a, x_1 = b and all other x_i = 0. 
 
 Similarly,
when I is a subset of κ and (y_i)_i ∈ I is a family of elements in H, we can define 
∑ _i ∈ I y_i :=  ∑_i  ∈κ x_i, where x_i = y_i for every i ∈ I and y _i = 0 other places. Using this, 
   one can see that  when 
  α≤κ are infinite  cardinal  numbers, then a  κ-monoid is an α-monoid. In particular, every 
  κ-monoid is ℵ_0-monoid, where ℵ_0:= {0, 1, ⋯}. 
  Moreover, 
   by  α a, where a ∈ H and α≤κ is any cardinal number, we mean sum of α-many copies of x, that is ∑_i ∈α a. 
Given a monoid H and  element x ∈ H, we write

    (x) {  y ∈ H : there exist  z ∈ H,  n  ≥ 1  with  y+z=nx  }

for the set of all summands of multiples of x. 
An element  u ∈ H is called an order-unit if  H = (u). 
It is not difficult to see that κ-monoids are  reduced (that is, any equation of the form a+b=0 implies a=b=0), see <cit.>. 


Let 𝒫_fin be a class of finitely generated projective  (right) modules over some (unital, associative) ring R. 


Then the set of isomorphism classes together with the operation induced by the direct sum form a monoid, denoted by (R). 
Every  element in 
 (R) is in form [P], the class isomorphism of a  finitely generated projective module P. 
Obviously,  (R)  is a commutative reduced monoid, where
[R] is an  order-unit. A ring R is called right (resp., left) hereditary if every right (resp., left) ideal is a projective R-module. 
A ring which is both right and left hereditary is called  a hereditary ring.
By Bergman and Dicks' result for every reduced commutative monoid H  with order-unit and every
 field k, there exists a hereditary k-algebra R with  V(R) ≅ H.
 Similarly, 
 ^κ(R) is the κ-monoid of isomorphism classes of projective (right) modules generated by at most κ-many elements. 
 Although κ-monoids are reduced, they may not have a  realization even if they have an order-unit (see <cit.>). 
 For a κ-monoid H and x ∈ H, add(ℵ_0 x) is an ℵ_0-monoid, which plays an important role in the realization problem:


 (<cit.>)
 
 A κ-monoid H has realization  (that it is isomorphic to ^κ(R) for a ring R)  iff there exits x ∈ H such that:
 
 
  * (ℵ_0 x) has a realization as ℵ_0-monoid. 
 
  * 

 
  * For every h ∈ H, there exist a family of elements (x_i) _i ∈κ in (ℵ_0 x)  such that h =  ∑_i  ∈κ x_i.
 
  * When ever  ∑_i  ∈κ x_i  =  ∑_i  ∈κ y_i, there exist 
  indexed partitions (I_μ)_μ∈κ and (J_μ)_μ∈κ of κ with I_μ, J_μ≤ℵ_0, and families (u_μ)_μ∈κ and (v_μ)_μ∈κ in (ℵ_0 x) such that v_μ=0 whenever μ is a limit element, and 
    
    ∑_i ∈ I_μ x_i = v_μ + u_μ,
              and    ∑_j ∈ J_μ y_j = v_μ+1 + u_μ    for all μ∈κ.

 
 
 
 



Therefore, the realization problem for κ-monoids can be restricted to ℵ_0-monoids. In this manuscript, we discuss the realization of ℵ_0-monoids, or, 
in other words, monoids with countable sums. The structure of the paper is as follows: Section 1 delves into the realization problem concerning monoids featuring countable sums.
We show that if an ℵ_0-monoid has a realization, the x given in Theorem <ref> can be better understood, and we refer to such x as a braiding element.
We observe that for an ℵ_0-monoid having a realization to a  hereditary ring is equivalent to having a realization to a ring whose projective 
right modules are direct sums of finitely generated ones (we say such a ring is PDFG).

In Section 3, we focus on two-generated  ℵ_0-monoids. 
Recall that a set X of elements of H is called a generating set if the smallest ℵ_0-submonoid 
of H containing X is equal to H. In this 
case we write 
H =   ⟨ X   ⟩ _ℵ _0. For example, in the case where H is an ℵ_0-monoid,  Condition (2)(i) in Theorem 1.2  asserts that H = ⟨(ℵ_0 x) ⟩_ℵ_0.
 We call an ℵ_0-monoid H cyclic if there exists an element x ∈ H such that H = ⟨ x ⟩_ℵ_0, 
meaning every element in H is of the form α x, where 0 ≤α≤ℵ_0. As we see in Lemma <ref>, a cyclic  ℵ_0-monoid H  has 
realization iff 
ℵ_0 x ≠ nx for every n < ℵ_0 iff
 H has realization to a hereditary ring. 
A  κ-monoid H is called two-generated if it has a generating set with two elements, say H =  ⟨{x_1, x_2}⟩ _ℵ _0. 
That means 
every y ∈ H can be represented as ∑_k ∈ℵ_0 y_k with y_k equal to x_1,  x_2 or 0. Since the order of the elements in the
 family (y_k)_k ∈ℵ_0 does not matter, it is convenient to represent such a family as α X_1 + β X_2, where 0 ≤α≤ℵ_0 is the number of elements in the family that are equal to x_1, and 0 ≤β≤ℵ_0 is the number of elements that are equal to x_2. We call such a representation α X_1 + β X_2 a form
  for y (based on generating set {x_1, x_2}). Clearly the equality y = α x_1 + β x_2 also holds. 
  We say that α X_1 + β X_2 is an infinite form if at least one of α, β is infinite, and a finite form otherwise. Depending on H, an element may have both infinite and finite forms. In <cit.>, a characterization is provided for when a two-generated ℵ_0-monoid has realization to hereditary rings. Namely,




 <cit.>
  A non-cyclic ℵ_0-monoid H generated by two elements x_1, x_2 is isomorphic to V^ℵ_0(R) for a
   hereditary ring R, if and only if the following conditions hold for 1 ≤ i ≠ j ≤ 2.
   
  
  *   n x_i + ℵ_0 x_j = ℵ_0 x_i + ℵ_0 x_j with n finite implies ℵ_0 x_j = ℵ_0 x_i +ℵ_0 x_j and x_i ∈(x_j).
  
  *  If x_i ∉(x_j) and m x_i + ℵ_0 x_j = n x_i + ℵ_0 x_j with m, n finite, then there exist finite k, k' such that m x_i + k x_j = n x_i + k' x_j.
  
  *   An element of H cannot have both finite and infinite forms.
  




We note that a two-generated ℵ_0-monoid H cannot simultaneously be realized as both a PDFG and a not-PDFG ring. We examine the properties of those with realizations as not-PDFG rings; refer to Propositions <ref> and <ref> for detailed discussions.
To deepen our comprehension of a two-generated ℵ_0-monoid H = ⟨x_1, x_2⟩_ℵ_0, we introduce the concept of type for these monoids, which hinges on whether x_i belongs to  (x_j) or not. Consequently, three distinct possibilities emerge, each of which we scrutinize. We establish that if H possesses a realization, it can only conform to one of these types. However, the converse remains uncertain; see Theorem <ref> for further insights. 

A ring R is called von Neumann regular (VNR) if, for every element a∈ R, there exists an element x ∈ R such that axa=a. 
When R is VNR, (R) becomes particularly interesting as it forms a refinement monoid. Recall that a monoid H is considered a refinement if a + b = c + d implies 
that a = x + y, b = z + t, c = x + z, and d = y + t, for some 
x, y, z, t ∈ H. This situation can be represented in the form of a square:



    c     d 
a     x     y 
b     z     t 




For years, as discussed in <cit.> for a comprehensive survey, the question of whether a refinement-reduced monoid has a realization to a VNR remained open. Finally, in <cit.>, P. Ara, J. Bosa, and P. Enrique provided an affirmative answer to this question in the case where the monoid is finitely generated.
In Section 4, we provide a complete understanding of when a two-generated ℵ_0-monoid H = ⟨x_1, x_2⟩_ℵ_0 has a realization to VNR hereditary rings. 
It is not difficult to see that this is equivalent to ⟨x_1, x_2⟩ being a refinement monoid and H being braiding over add(x_1 + x_2). We provide 
another equivalent characterization by introducing additional conditions to Theorem <ref>; for further details,  see Theorem <ref>.
















  
Acknowledgments.
The author was supported by the Austrian Science Fund (FWF): P 36742-N.  She  would like to thank  Alfred Geroldinger for the mentoring during running this project and also Roozbeh Hazrat and Gene Abrams 
for introducing some papers. 







§  REALIZATION OF ℵ_0-MONOIDS: SOME BACKGROUND
 
Before delving into the realization problem, we gain a deeper understanding of 
 ^ℵ_0(R), where R is a ring, viewed as a monoid. Since ^ℵ_0(R) is a reduced monoid with 
order-unit F:= R^(ℵ_0), it is isomorphic to (S) for a hereditary ring S, as shown by Bergman and Dicks' realization. However, as we will see in the next lemma, there is a monoid isomorphism between ^ℵ_0(R) and ((F_R)).

Thinking of elements of F as columns, we can identify (F_R) with the column finite matrix  ring _ℵ_0(R), consisting of ℵ_0 ×ℵ_0 matrices, where each column has only finitely many nonzero entries.


For a ring R, let S := _κ(R). Then, ^ℵ_0 (R) ≅(S) as an isomorphism of monoids.
 In particular, when R is a VNR, S ≅(R), the ring of row and column finite matrix rings over R.
 


For a given right R-module M, let (M_R) be the class of modules, such as N, where N is isomorphic to a direct summand of a finite direct sum of M. There exists an equivalence between the full subcategory (M) and finitely generated projective End(M)-module, given by Hom(M, -) or, conversely, by - ⊗_End(M) M (see <cit.>). Now, for F := R^(ℵ_0), it is evident that (F) is precisely the class of countably generated projective modules.
 Therefore,
 the map ϕ : ^ℵ_0(R) →(S) defined by ϕ([P]) = [Hom_R(R^(ℵ_0), P)] is an isomorphism of monoids. The rest of the lemma follows from <cit.>.





  If we rephrase the braiding concept from <cit.> for ℵ_0-monoids, we get:

 <cit.> 
  Let X be a submonoid of an ℵ_0-monoid H.
  
  
  * Two families (x_i)_i ≥ 0 and (y_i)_i ≥ 0 in X are braided if there exist indexed partitions (I_n)_n ≥ 0 and (J_n )_n ≥ 0 of
   ℵ_0 with I_n, J_n finite, and families (u_n )_n ≥ 0 and (v_n)_n ≥ 0 in X such that v_0 = 0, and 
    
    ∑_i ∈ I_n x_i = v_n + u_n,
              and    ∑_j ∈ J_n y_j = v_n+1 + u_n    for all  n ≥ 0.

  
  * H is braided over X if H = ⟨ X ⟩_ℵ_0, and all families (x_i)_n ≥ 0, (y_j)_n ≥ 0 in X 
  with ∑_i ∈ℵ_0 x_i  =  ∑_j ∈ℵ_0  y_j are  braided.
  




In this article, while working with ℵ_0-monoids, our objective is to offer a clearer understanding of <cit.> and introduce a slightly different concept of braiding,
 from what is defined in Definition <ref>. The proof follows a similar structure to that of <cit.>, but we specify in Definition <ref> that 
 I_n and J_n are intervals in ℵ_0, particularly in the form { k | m_1 ≤ k ≤ m_2 }, where m_1, m_2 ∈ℵ_0.



 
Suppose that the ℵ_0-monoid H has a realization. Then, there exists an element x such that whenever {x_i | i ≥ 0} and {y_i | i ≥ 0} are families in (x), then

    ∑_j ∈ℵ_0 x_i = ∑_j ∈ℵ_0 y_i

if and only if there exist integers 0 = m_0 < m_1 < ⋯ and 0 ≤ n_0 < n_1 < ⋯ and a sequences u_0, u_1, u_2, ⋯ and 
v_0 = x_0, v_1, v_2, ⋯ in (x) such that

    ∑_i = 0^n_0 y_i = v_0 + u_0  
       ∑_i = m_j + 1^m_j + 1 x_i = u_ j  + v_j + 1,  for every  j ≥ 0, and
       ∑_i = n_j + 1^n_j + 1 y_i = u_j + 1  + v_ j + 1,  for every  j ≥ 0.




⟸ is clear. 


For the other side, 
let R be the realization of H, and 
ϕ : H →^ℵ_0(R)
 be an ℵ_0-isomorphism and 


    ∑_j ∈ℵ_0 x_i = ∑_j ∈ℵ_0 y_i Eq 2.1



Take x to be the element such that ϕ(x) = [R]. 
Then clearly whenever t ∈(x) and  ϕ(t) = [X], then X  is finitely generated. 
 For each i ≥ 0, let A_i and B_i be modules such that   [A_i] = ϕ(x_i) and [B_i ]= ϕ(y_i). Following <ref> and using the fact that 
ϕ is an ℵ_0-isomorphism, we  conclude that  



    ⊕_i ≥ 0 A_i ≅⊕_i ≥ 0 B_i Eq 2.2



Let  f:  ⊕_i ≥ 0  A_i →⊕_i ≥ 0 B_i be an isomorphism and show its inverse with 
 g:  ⊕_i ≥ 0  B_i →⊕_i ≥ 0 A_i. 
We have m_0 = 0. Since f(A_0) is
 finitely generated, there exists n_0 such that f(A_0) ≤⊕_i = 0^n_0 B_i. Since A_0 is a direct summand of ⊕_i ≥ 0 A_i, so is its
  image in ⊕_i = 0^n_0 B_i, see <cit.>. Thus, ⊕_i = 0^n_0 B_i = f(A_0) ⊕ T^y_0, for some R-module T^y_0, and so 



    ⊕_i = 0^n_0 g(B_i) = A_0 ⊕ g(T^y_0).  Eq 2.3
 

Then put u_0 := ϕ^-1([T^y_0]). 
Note that  g is  isomorphim and so g(B_i) ≅ B_i and also 
ϕ is a monoid isomorphism and so 
we get to the first equation asked in the theorem, that is ∑_i = 0^n_0 y_i = v_0 + u_0, where v_0 := x_0. 



By a similar argument, since ⊕_i = 0^n_0 g(B_i) is finitely generated submodule of ⊕_i ≥ 0 A_i, 
 there exist
 m_1 > m_0 = 0 and R-module T^x_1 such that 
⊕_i = 0^m_1 A_i = ⊕_i = 0^n_0 g(B_i) ⊕ T_1 ^x. 
 Therefore,
 ⊕_i = 0^m_1 A_i = A_0 ⊕ g(T^y_0) ⊕ T^x_1,
 and so ⊕_i = m_0 + 1^m_1 A_i ≅ g(T^y_0) ⊕ T^x_1,
  noting that m_0 = 0. Since g is injective,  g(T^y_0) ≅ T^y_0 and consequently 
 u_0 = ϕ^-1([g(T^y_0)]). Now put v_1 := ϕ^-1([T^x_1]).  Thus:


    ∑_i = m_0 + 1^m_1 x_i = u_ 0  + v_1



Now, by induction, suppose that for k ≥ 1, integers n_0, ⋯, n_k-1 and m_0, ⋯, m_k, and elements u_0, ⋯, u_k-1, v_0, ⋯, v_k ∈(x), and a finitely generated module T^x_k are found, such that for them all equations given in the theorem hold, and moreover, v_k = ϕ^-1([T^x_k]),

    ⊕_i=0^m_k A_i = g(⊕_i=0^n_k-1 B_i) ⊕ T^x_k Eq 2.4

Now, we find integers n_k, m_k+1, and u_k, v_k+1∈add(x) satisfying the conditions of the theorem.
 Since f(⊕_i = 0^m_k  A_i) is finitely generated, there exists n_k > n_k - 1 and finitely generated module T^y_k such that 

    ⊕_i = 0^n_k  B_i = f(⊕_i = 0^m_k  A_i) ⊕ T^y_k Eq 2.5



Then u_k:= ϕ^- 1 ( [T^y_k] ) and from <ref>, we conclude that 

    ⊕_i = 0^n_k  B_i = ⊕_i = 0^n_k - 1 B_i ⊕ T^x_k  ⊕ T^y_k


Consequently, ∑_i = n_k-1 + 1^n_k y_i = u_k + v_k. 

Now, similarly, since g (⊕_i = 0^n_k  B_i ) is finitely generated, 
there exists m_k + 1 > m_k and finitely generated module T^x_k + 1  such that

 
    ⊕_i = 0^m_k + 1  A_i = g(⊕_i = 0^n_k B_i) ⊕ T^x_k + 1
 


That is we set v_k + 1:= ϕ^- 1 ( [T^x_k+ 1] ) 
and using <ref>
we conclude  that
∑_i = m_k + 1^m_k + 1 x_i = u_k + v_k + 1. 








The element x given in Theorem <ref> is called a braiding element. The proof of Theorem <ref>
is showing that 
[R] is a braiding element in ^ℵ_0 (R). 



As a corollary, we have:

 
  If x is a braiding element and ℵ_0 z_1 = ℵ_0 z_2 for some z_1, z_2 ∈(x), then (z_1) = (z_2). 



  Let ℵ_0 z_1 = ℵ_0 z_2. Then, by Theorem <ref>, considering all x_i in
   Theorem <ref> as z_1 and all y_i  as z_2, we find n_0 ≥  0 and u_0 such that
  
    (n_0 + 1) z_2 = ∑_i = 0^n_0 z_2 = z_1 + u_0.

  That is, z_1 ∈(z_2).  By a similar argument, we have  z_2 ∈(z_1). Consequently, (z_1) = (z_2). 







We need to define and recall the following concepts: 


  * If R is a ring whose projectives are direct sums of finitely generated ones, we say R is  PDFG. 

  * We call   an ℵ_0-monoid H  hereditary if there exists a hereditary ring R such that H ≅^ℵ_0 (R). 

  * Recalling 
 from  <cit.>, if  
H is an  ℵ_0-monoid and x ∈ H, then H is called braided over (x), whenever:


 
  * ( x) is a generating set.
 
  * When ever  ∑_i ∈ℵ_0 x_i  =  ∑_i ∈ℵ_0 y_i, there exist 
 finite partitions (I_μ)_μ∈ℵ_0 and (J_μ)_μ∈ℵ_0 of ℵ_0, and families (u_μ)_μ∈ℵ_0 and (v_μ)_μ∈ℵ_0 in (x) such that v_0=0, and 
    
    ∑_i ∈ I_μ x_i = v_μ + u_μ,
              and    ∑_j ∈ J_μ y_j = v_μ+1 + u_μ    for all μ∈ℵ_0.






Clearly, when x is a braiding element and H = ⟨(x)  ⟩_ℵ_0, then H is braided 
over (x) according to Part (4) of the above definition. The converse is also true:


 
For an ℵ_0-monoid H, the followings are equivalent:

    
  * H has a realization to a PDFG ring.
    
  * There is a braiding element x such that (x) is a generating set for   H.
    
  * There is an element x ∈ H such that H is braided  over (x). 
    
  * H is hereditary. 



 

    
  (1)  (2). 
    Let H have a realization to a PDFG ring. Then we have an ℵ_0-monoid isomorphism, denoted by f, from H to (R) for a PDFG ring R. 
    Let x := f^-1 ([R]). Then, since 
    [R] is a braiding element in ^ℵ_0 (R), so is      
     x  in H. Moreover, since f is an ℵ_0-morphism, so is f^-1, and therefore:
    
    H     = f^-1( ^ℵ_0 (R)) 
        = f^-1(  ⟨ ([R])  ⟩ _ℵ_0) 
        =   ⟨ (f^-1([R]))  ⟩_ℵ_0
        =  ⟨(x)  ⟩_ℵ_0.

    Where, the second equality holds because R is a PDFG. This proves (2).
    
    
  (2)  (3) is clear. 
    
     
  (3)  (4)  is   <cit.>. 
     
    
  (4)  (1). It is known that over a hereditary ring, every projective is a direct sum of finitely generated ones,  see for example <cit.>.  
  





Note that x is an order-unit in add(x), and according to  Bergman's realization, the monoid isomorphism
from the reduced commutative monoid M with order unit x to 
(R)
 can take x  to 
[R]. Consequently, when H is braided over add(x), then by 
<cit.>, we arrive at an isomorphism of ℵ_0-monoids H and 𝒱^ℵ_0(R) such that x maps to [R]. This means, as we see in the proof of (1)  (2) in the above theorem, x is a braiding element.

















§ TYPE OF ℵ_0-MONOIDS


We begin this section with the realization of cyclic ℵ_0-monoids.
When M is a module and  0 ≤α≤ℵ_0, then α M denotes 
direct sum of α copies of M. 



For a cyclic ℵ_0-monoid H := ⟨ x ⟩_ℵ_0, the following are equivalent:
 

  * H has a realization. 

  * H is a hereditary. 

  * ℵ_0 x ≠ nx for every n < ℵ_0.




 
  (1)  (3). 
Let H := ⟨ x ⟩ _ℵ_0 be a cyclic ℵ_0-monoid, R be a realization of H, and f: H → V^ℵ_0(R) be an isomorphism of ℵ_0-monoids.
There exists a right R-module N such that f(x) = [N]. Consequently, every element of V^ℵ_0(R) is equal to [α N], for some α≤ℵ_0, and therefore N is a finitely generated module. If ℵ_0 x = nx for some finite integer n, then ℵ_0 N ≅ nN, which is not possible as N is finitely generated. 


  (3)  (2).  Suppose that (3) holds. The submonoid of H generated by x, denoted by ⟨ x ⟩, is a reduced, commutative monoid with x as an order-unit. 
Thus, by Bergman's realization, there is a hereditary ring R such that ⟨ x ⟩ has a realization to
 V(R). Let g: ⟨ x ⟩→ V(R) be an isomorphism of monoids. Define g̅: H → V^ℵ_0(R) such that g̅(ℵ_0 x): = ℵ_0 g(x) and g̅(n x): =  g(nx), for n < ℵ_0. Using the hypothesis (3), g̅ is well-defined. Moreover, since R is hereditary, all countably generated projective modules are direct sums of finitely generated ones. Thus, g̅: H → V^ℵ_0(R) is an isomorphism.


  (2)  (1)  is trivial.  
  


Due to the above lemma, from now on in this section, all ℵ_0-monoids (including V^ℵ_0(R)) are assumed to be non-cyclic and two-generated.


Let us start with the following fact: 

 
 H cannot be realized simultaneously to  both a PDFG and a non-PDFG ring.




We need to show that if R is PDFG and S is not PDFG, then ^ℵ_0(R) and ^ℵ_0(S) are not isomorphic. Let f: ^ℵ_0(R) →^ℵ_0(S) be an ℵ_0-isomorphism, and let P_1 and P_2 be finitely generated projective modules such that ^ℵ_0(R) = ⟨{ [P_1], [P_2]}⟩_ℵ_0. Let P'_1 and P'_2 be S-modules such that f(P_1) = [P'_1] and f(P_2) = [P'_2], respectively. Then ^ℵ_0(S) = ⟨{ [P'_1], [P'_2] }⟩_ℵ_0 and P'_1, P'_2 are both finitely generated, as shown by <cit.>.
 This implies that S has to be a PDFG ring.





If P is a projective module, the trace of P is

    (P) ∑_f ∈(P,R)(f).

Recall that (P) is an idempotent ideal of R, and that (P) is the least among all ideals I such that P=PI, see for example <cit.> and <cit.>.



 
Suppose that H is generated by x_1 and x_2 and has realization to a non-PDFG ring. 
Then  there exist 1 ≤  i ≠  j ≤ 2  such that:

    
  * x_j ∉(x_i).
    
  * ℵ_0 x_i + x_j =ℵ_0 x_i.


  * x:= x_1 + x_2 is not a braiding element. 

    
  * x_i ∈(x_j) if  and only if ℵ_0 x_j = ℵ_0  x_i.

    
  * If x_i ∉(x_j) and  α X_i + β X_j and α' X_i + β' X_j are two forms of an element, then  α, α' are both finite or infinite.




Suppose that
H has realization to V^ℵ_0 (R), where R is a  non-PDFG ring and f: H → V^ℵ_0 (R) is an ℵ_0-monoid isomorphism. 
Then { f(x_1), f(x_2) } is a generating set 
for  V^ℵ_0 (R). 
Then  [R] =n  f(x_1)  +m f(x_2), for some  finite integers n, m  ≥ 0. 
Since R is not a PDFG, there exists a projective module P such that P is not direct sum of finitely generated ones.  On the other hand, [P]  must be
sum of  some (finite or infinite) copies of f(x_1) and  f(x_2). This implies that f(x_1) or f(x_2) can not be finitely generated. Suppose that f(x_2) is not finitely generated. We set j: = 2 and i:= 1. Then: 



 
  * We have   [R] =n  f(x_i)
and consequently f(x_i) is finitely generated and f(x_j) is not in add(f(x_i)). Consequently x_j ∉add(x_i). 

    
    
  * f(x_j) is class isomorphism of a pojective module  and  so f(x_j) + y = ℵ_0 [R], for some y ∈^ℵ_0 (R). Now from <cit.>, 
we conclude that f(x_j) + ℵ_0 [R] = ℵ_0 [R].  Replacing [R]  with 
n f(x_i), we conclude that  f(x_j) + ℵ_0 f(x_i) = ℵ_0 f(x_i)
and so x_j +   ℵ_0  x_i= ℵ_0  x_i. 
    
    

  * Since H = ⟨{x_1, x_2}⟩_ℵ_0, and x_1, x_2 ∈add(x),  where x:= x_1 + x_2, it is clear that H = ⟨add(x) ⟩_ℵ_0. Then, using
  Lemma <ref>, we conclude that x is not a braiding element.


  * Let x_i ∈add(x_j). Then m x_j = x_i + y, for some y ∈ H and finite integer m ≥ 1. 
Consequently,  m f(x_j) = f(x_i) + f(y). Now,  following equivalence condition of (c) and (d) in <cit.>, we conclude that 
ℵ_0 ( m f(x_j)), which is equal to  ℵ_0  f(x_j),   represents the  free module ℵ_0 R. Therefore,  bearing in mind that [R] = n f( x_i), we have  ℵ_0  f(x_j) =  ℵ_0   [R] = ℵ_0 f(x_i ).
Consequently ℵ_0  x_j =   ℵ_0  x_i.  


Conversely, let ℵ_0  x_j =   ℵ_0  x_i. Then we have ℵ_0  f( x_j) =   ℵ_0  f(x_i). 
Suppose that  P_i and P_j are projective R-modules that 
 [P_i] =  f(x_i) and [P_j] = f(x_j). Therefore, we have an isomorphism  
g: ⊕ _i ≥ 1 P_j →⊕ _i ≥ 1 P_i. Since P_i is finitely generated R-module, there are y_1, ⋯ y_k ∈ P_i such that P_i = y_1 R + ⋯ + y_k R. For each  y_i, there exists m_i ∈⊕ _i ≥ 1 P_j 
such that g(m_i) = y_i. Now each m_i has finite support and we conclude that there exists an epimorphism from a  direct sum of finitely many copies of P_j, say m copies, 
 to P_i.  Then 
 P_i is projective and so this epimorphism 
is split and we have that m P_j ≅  P_i ⊕ Y, for some 
projective module Y.  That means  m f(x_j)  =  f(x_i)  +  [Y]  and so  x_i ∈(x_j). 

    
    
  * We have 
 [P_i] =  f(x_i) and [P_j] = f(x_j), for some 
projectives P_i and P_j.
Suppose that α x_i+ β x_j = α' x_i + β' x_j, thus
 α P_i ⊕β P_j ≅α' P_i   ⊕β' P_j. 
 Using Part (4), ℵ_0 P_i ℵ_0 P_j, and since ℵ_0 P_i is free, we conclude that ℵ_0 P_j is not free. Therefore, the trace of P_j is not equal to R, see <cit.>.
 Consequently, by tensoring α P_i ⊕β P_j ≅α' P_i   ⊕β' P_j
 with R/ I where I = (P_j), we have that α (P_i / P_i I ) ≅α' (P_i / P_i I ). Since P_i
and consequently P_i / P_i I  is finitely generated module, we conclude that α and α' are both finite or both infinite.




    Let H be a noncyclic ℵ_0-monoid generated by two elements. Then H is classified into the following types:

    
        
  * Type (1): There exists a generating set {x_1, x_2} such that x_1 ∉(x_2) and  x_2 ∉(x_1).

        
  * Type (2): There exists a generating set {x_1, x_2} such that x_i ∈(x_j) for 
some i ≠ j, but x_j  ∉(x_i).

        
  * Type (3): There exists a generating set {x_1, x_2} such that  x_i ∈(x_j) for every 1 ≤ i, j ≤ 2. Equivalently 
        (x_1) = (x_2). 
    







 
Suppose that  H   is of Type (i), for 1 ≤ i ≤ 3. 
If H has realization, then it is not of Type  j ≠ i. 







It is enough to prove the following: 



  (i) If H is of Type (3), then it cannot be of Type (1) or (2). 
 
 
  (ii) If  H is of Type  (2), then it can not be of Type (1). 
 

 [itemsep=5pt]

  Proof of (i): 
If H is of Type (3), then there exists a generating set {x_1, x_2} such that (x_1) = (x_2). 
Then there is only one infinite form regarding {x_1, x_2}, see Part (2) of <cit.>. 
Consequently, if h ∈ H and ℵ_0 x_i ≤ h, then h = ℵ_0 x_1 = ℵ_0 x_2. 
Now let {y_1, y_2} be another generating set. 
For each 1 ≤ i ≤ 2, there exist 0 ≤α_i, β_i ≤ℵ_0 such that 
 y_i = α_i x_1 + β_i x_2. 
Since y_i is nonzero, for each i, either α_i or β_i is nonzero. 
Therefore, ℵ_0 y_i = ℵ_0 x_1; in particular, this implies ℵ_0 y_1 = ℵ_0 y_2.

Now let R be a realization of H. Since (x_1) = (x_2), by Part (1) of Proposition <ref>, R is  a PDFG ring. So, 
by Corollary  <ref>, there is a braiding element, say x ∈ H. 
There exist 0 ≤ m_1, m_2 ≤ℵ_0 such that
 x = m_1 x_1 + m_2 x_2. 
 Then since x is nonzero, one of m_1 or m_2 is nonzero. 
So either x_1 or x_2 is in (x). Since (x_1) = (x_2), we conclude that (x_1) = (x_2) ⊆(x). 

For each 1 ≤ i ≤ 2, there exist 0 ≤γ_i, δ_i ≤ℵ_0 such that 
 x_i = γ_i y_1 + δ_i y_2. 
If both γ_1, γ_2 are zero, then H = < x_1, x_2>_ℵ_0⊆ < y_2 >_ℵ_0 and consequently, H would be cyclic, which is a contradiction. 
Therefore, one of γ_1 or γ_2 is nonzero. That means y_1 ∈(x_1) or y_1 ∈(x_2). Thus, y_1 ∈(x). 
Similarly, at least one of β_1 or β_2 is nonzero, and so y_2 ∈(x).  So both y_1, y_2 ∈(x). 

Finally, we are able to use Corollary <ref> and from ℵ_0 y_1 = ℵ_0 y_2 and y_1, y_2 ∈(x), we conclude that (y_1) = (y_2). 

  Proof of (ii): 
Suppose that H is of Type (2) with the generating set {x_1, x_2} such that x_1 ∈(x_2) and x_2 ∉(x_1).
Let H also be of Type (1). Then, there exists a generating set {y_1, y_2} with y_1 ∉(y_2) and y_2 ∉(y_1).
Let R be realization of H and  f: H →^ℵ_0 (R) be an ℵ_0-monoid isomorphism. There exist projective R-modules M_1, M_2, N_1, N_2
such that  [M_i] = f(x_i) and [N_i] = f(y_i), for every 1 ≤ i ≤ 2.
   Then  {[M_1], [M_2]} and  {[N_1], [N_2]} are  generating sets for  ^ℵ_0 (R). 
Moreover, since x_1 ∈(x_2), there exist finite n_1 ≠ 0 and also  1 ≤ n_2 ≤ℵ_0  and 
0 ≤ n_3 ≤ℵ_0  such that n_1 x_2 = n_2 x_1 + n_3 x_2.
Now, let's consider two cases: whether H has a realization to a PDFG or not.

Firstly, suppose that H has a realization to a non-PDFG.
As R is not a PDFG, either M_1 or M_2 is not finitely generated. Thus, [R] = n [M_i] for some 1 ≤ i ≤ 2 and finite n ≥ 1, implying that M_i must be finitely generated.
From 
 n_1 x_2 = n_2 x_1 + n_3 x_2, we have  n_1 f(x_2) = n_2 f(x_1) + n_3 f(x_2). As n_2 is nonzero, if M_1 is not finitely generated, then M_2 is not finitely generated either. This is a contradiction because for 1 ≤ i ≤ 2, [R] = n [M_i] implies that M_i must be finitely generated. This shows that M_1 is finitely generated, and M_2 is not. Therefore, [R] = n [M_1]. Consequently, for x := f^- 1 ([R]),  we have (x_1) = (x).

Since R is not a PDFG ring,  one of N_i is not finitely generated. Thus there exists 1 ≤ i ≤ 2 such that 
 x = n' y_i for some finite n'. Without loss of generality, we suppose that x = n' y_1. Thus, (y_1) = (x) = (x_1).

Let α X_2 + β X_1 be a form of y_2 with respect to {x_1, x_2}. Consequently 
y_2 = α x_2 + β x_1.  
 We claim that β = 0. Otherwise, x_1 ∈(y_2), and so (x_1) ⊆(y_2), which is a contradiction since y_1 ∈(x_1). 
 Consequently,
 y_2 = α x_2.
  That is, (x_2) ⊆(y_2). Now, y_1 ∈(x_1) ⊆(x_2) implies that y_1 ∈(y_2), which is a contradiction.  


Secondly, suppose that H has a realization to a PDFG ring. In this case, all M_i and N_i are finitely generated.
Therefore, for each 1 ≤ i ≤ 2, there exist finite numbers α_i and β_i such that x_i = α_i y_1 + β_i y_2.
If both α_1 and α_2 are zero, then x_i = β_i y_2. 
Given that { x_1, x_2 } is a generating set, this implies that {y_2} is a generating set for 
 H. However, this contradicts the fact that H is non-cyclic.
 Therefore, at least one of α_1 and α_2, and similarly one of β_1 and β_2, is nonzero.
From the equation n_1 x_2 = n_2 x_1 + n_3 x_2, we conclude that n_3 is nonzero; otherwise, (x_1) = (x_2), leading to a contradiction.
Recall that n_2  is non-zero and we have the equation:


    n_1 (α_2 y_1 + β_2 y_2) = n_2 (α_1 y_1 + β_1 y_2) + n_3 (α_2 y_1 + β_2 y_2)


If β_2 = 0, then β_1 is nonzero, and from the above equation, we conclude that y_2 ∈(y_1), which is a contradiction. 
Consequently, β_2 ≠ 0. Similarly, α_2 ≠ 0.

Thus, both y_1 and y_2 are in (x_2). Since H is not cyclic and cannot be generated solely by x_1, 
we conclude that x_2 appears in a form of either y_1 or y_2 regarding to 
generating  set {x_1, x_2}. Therefore, (x_2) ⊆(y_i) for either i = 1 or i = 2. In any case, this is a contradiction, 
as both y_1 and y_2 belong to (x_2) and therefore lie in (y_i).
 


 





§ REALIZATION TO HEREDITARY VNR RINGS 




The aim of this section is understanding when H has realization to a VNR hereditary ring. 
Let M be a reduced commutative monoid.
We write x ≤ y whenever y = x + z for some z ∈ H.
 An element p ∈ M is called prime if
 whenever p ≤ x+ y, then p ≤ x or p ≤ y. If every nonzero element of M can be 
 written as a sum of prime elements, then M is called 
 primely generated.
  By <cit.>, every finitely generated refinement monoid is primely generated.
  Recall that M  is separative if the following cancellation of elements holds for every a, b, c ∈ M:
 
    a + c = b + c   and  c ∈(a) ∩(b)    a = b


Moreover, c ∈ M is called cancelable if for every a, b ∈ M:

    a + c = b + c  a = b


When H = ⟨x, y⟩_ℵ_0 is an ℵ_0-monoid, by H_f:= ⟨ x, y ⟩, we refer to the submonoid generated by x and y. In the next lemma, we examine properties of H_f when it is a refinement monoid. 


 
Let H = ⟨x, y⟩_ℵ_0 such that   H_f is refinement. Then:


  * Suppose that   n,   m, k, k' are finite integers such that 
n > m and n x + k y  = m x + k' y, then (n - m + 1) x + k y  =  x + k' y. 

  * If H is of Type (1) and n_1 x + m_1 y   = n_2 x  +  m_2 y, where all n_i , m_i  are nonzero finite integers. Then 
n_1 x = n_2 x and m_1 y = m_2 y. 

  * If
 x ∈(y), then x is cancelable.  Moreover, if H is of Type (2), then  x ≤ y. 





  * By [22, Theorem 4.5], every primely generated refinement monoid is separative. In
particular every finitely generated refinement monoid is separative. This shows easily that (1) holds. 


  * Suppose that H is of Type (1) and 
n_1 x + m_1 y   = n_2 x  +  m_2 y, where all n_i , m_i  are nonzero finite integers. 
Since H_f is refinement, this 
implies that n_1 x = x' + y', m_1 y = z' + t', n_2 x = x' + z', and m_2 y = y' + t', for x', y', z', t' ∈ H_f. 
This situation can be represented in the form of a square:



    n_2 x      m_2 y 
n_1 x     x'     y' 
m_1 y     z'     t' 





Each x', y', z', t' has a form regarding to {x, y}, and since H is of Type (1), there exist finite integers k, k' such that x' = kx and t' = k'y. We present the following table:





    n_2 x      m_2 y 
n_1 x     kx     0 
m_1 y     0     k' y 




This implies that n_1 x = kx = n_2 x and m_1 y = k' y = m_2 y. 


  * As we mentioned in Part (1), H_f is separative, so when x ∈add(y), it has to be cancelable. Now suppose that H is of Type (2) and ny = mx + n'y, where n, m are nonzero finite integers.
  Since  (x) ≠(y), n' must be nonzero. 
Therefore, we can suppose that n is the smallest possible nonzero integer such that ny = mx + n'y, for nonzero finite integers m, n, n'. We claim n = 1. If not,
 we write (n - 1)y + y = mx + n'y. Then, since H_f is a refinement monoid, there exist x', y', z', t' such that
 
 


    m x      n' y 
(n - 1) y     x'     y' 
y     z'     t' 


 

That is mx = x' + z', n'y = y' + t', (n - 1) y = x' + y', and y = z' + t'.
Since y ∉add(x) and mx = x' + z', if x' is nonzero, then in  a  form of x' regarding to {x, y} only x can appear. 
But then, due to the minimality of n, and since (n - 1) y = x' + y' and add(x) ≠add(y), in a form of y',  y should appear. 
This means x' = 0, due to the minimality of n. But then z' has to be nonzero and is equal to kx, for some finite k ≥ 1. 
Now, again, the fact that add(x) ≠add(y) implies that t' is nonzero. However, this is in contradiction with the minimality of n. 
That means n = 1.  



A strong source of examples of hereditary  VNR rings with interesting properties can be found in rings bases on  weighted graphs and separated graphs.
 In the next few results, we will understand when a two-generated ℵ_0-monoid can be realized to 
  algebras based on these graphs. Let us recall the definitions.



 <cit.>  
A weighted graph E=(E^0,E^, E^1,r,s,w) consists of three countable sets, E^0 called vertices, E^ structured edges and E^1 edges and
 maps s,r:E^→ E^0,  and  a weight map w:E^→ℕ such that E^1= ∐_α∈ E^{α_i | 1≤ i ≤ w(α)}, i.e., 
 for any α∈ E^, with w(α)=k, there are k distinct elements {a_1,…,α_k}, and E^1 is the disjoint union of  all such sets for all α∈ E^. 


We refer to <cit.> for weighted Leavitt path algebra constructed by the above graph. This algebra is denoted by _R(E,w).
Before recalling the main theorem from <cit.>, let us recall that it is always supposed that the graph is row finite, that means for every vertex v,
s^-1 (v) is a finite set. 




  <cit.>    Let K be a field and E be a  weighted graph. Let M_E be the abelian monoid generated by {v | v ∈ E^0 } subject to the relations 

    n_vv=∑_{α∈ E^| s(α)=v } r(α),

for every v∈ E^0 that emits edges, where n_v= max{w(α)|α∈ E^,   s( α)=v}. Then there is a natural monoid isomorphism (_K(E,w))≅ M_E. Furthermore, if E is finite, then _K(E,w) is hereditary. 





When E is a graph, we represent the number of edges from vertex u to v as d(u, v).
Let E be a weighted graph with two vertices, u and v: 


    @=150pt   
      u @/^3.5pc/@[r]^a_d(u, v) |cm@>@/^2.4pc /@.>[r] |cm@>@/^1.0 pc/[r]^a_1 |cm@>@.@(l,d) @(ur,dr)^γ_1@(r,d)^γ_2@(dr,dl)^γ_3@(l,u)^γ_d(u, u)   
    v   @/^1.0 pc/[l]^β_1|cm@>@/^2.5pc /@.>[l] |cm@>@/^3.5pc/[l]^β_d(v, u) |cm@>@.@(l,d) @(ur,dr)^τ_1@(r,d)^τ_2@(dr,dl)^τ_3@(l,u) ^(0.6) τ_d(v, v)Figure 1



 
M_E is the monoid generated by u, v and with the following relations:


  
  * n_u u = d(u, u) u + d(u, v) v

  
  * n_v u = d(v, v) u + d(v, u) v




Where,  n_u = max{ w(a) | a ∈{ a _i  |  1 ≤ i ≤ d(u, v)  }∪{γ_i |  1 ≤ i ≤ d(u, u)  } and 
 n_v = max{ w(a) | a ∈{β _i  |  1 ≤ i ≤ d(v, u)  }∪{τ_i |  1 ≤ i ≤ d(v, v)  } 



Recall that (E,w)  is called acyclic if there is no cycle in (E,w). 
An edge in (E,w)  is unweighted if its weight is 1 and  weighted otherwise.
E^1_w denotes the set of all weighted edges in (E,w).









The following statements for H are equivalent:

    
  * H has a realization to a regular weighted Leavitt path algebra with two vertices.
    
  * H is hereditary, and there exists a generating set {x_1, x_2} such that 1 ≤ i ≠ j ≤ 2 and x_i = nx_j (resp., nx_i = x_j), for some finite n ≥ 1. Moreover, all other relations are derived from x_i = nx_j (resp., nx_i = x_j).
 

In this case, for every generating set {y_1, y_2}, there is only one infinite form and H is of Type (3).  



(1)  (2).  Let H ≅_ℵ_0 ^ℵ_0(_K(E,w)), where _K(E,w) is regular weighted Leavitt path algebra. 
Then, using <cit.>, there is an isomorphism of monoids, f: (_K(E,w)) →(x), where x is a
 braiding element in H. 
Now we claim that E has two vertices u, v such that 
 d(u, u) = 0 = d(v, v) and one of the following holds:

    
  (a) d(u, v) = 0 and if E^1_w is nonempty, then d(v, u) = 1.
    
  (b) d(v, u) = 0 and if E^1_w is nonempty, then d(u, v)  = 1.


Before proving the claim, recall that if u, v ∈ E^0 and there is a path p such that s(p) = u and r(p) = v, then we write u ≥ v. 
If u∈ E^0, then T(u):={v∈ E^0 | u≥ v} is called the tree of u. If X⊆ E^0, we define T(X):=⋃_v∈ XT(v). 
Recall that (E,w) is called well-behaved if Conditions (LPA1), (LPA2), (LPA3), (W1), and (W2) from <cit.>,  as explained in the following, hold.
Following <cit.>, _K(E,w) is VNR if and only if (E,w) is acyclic and well-behaved. 




Proof of claim: 

Since  _K(E,w) is VNR,  it is acyclic, so d(u, u) = 0 = d(v, v), and either d(v, u) = 0 or d(u, v) = 0. 

As we mentioned, (E,w) satisfies (LPA1) and  (W2),  that is:


    
  (LPA1): Any vertex x ∈ E_0 emits at most one weighted edge.
   
    
  (W2): There is no n≥ 1 and paths p_1,…,p_n,q_1,…,q_n in (E,w) such 
    that r(p_i)=r(q_i) (1≤ i≤ n), s(p_1)=s(q_n), s(p_i)=s(q_i-1) (2≤ i ≤ n) and for any 1≤ i ≤ n, the
     first letter of p_i is a weighted edge, the first letter of q_i is an unweighted edge, and the last letters of p_i and q_i are distinct.
    

  If d(u, v) = 0, then there is no a_i, and following (LPA1), there is at most one 1 ≤ i ≤ d(v, u) such that β_i is weighted. That means   E^1_w has at most one element. 
  Then since there is no loop, the only path that starts with a weighted edge could be β_i.  
  If E^1_w   is nonempty with weighted edge β_i, and   β_l is unweighted edge, where l ≠ i. 
  Then we have two 
  paths β_i and β_l contradicting (W2).
 That means (a) should hold.
 Similarly, if d(v, u) = 0, then (b) holds.
So we have proved the claim. 
Now  suppose that (a)  holds. 
Then, by Remark <ref>, M_E is the monoid generated only by the relation, either 

    n_vu = v

if E^1_w is nonempty or 

    u = d(u, v)v

if E^1_w is empty.

Then {x_1 := f(v), x_2 := f(u)} is a generating set for H and n := d(u, v) or n := n_v satisfying condition (2). The argument is similar if (b) holds. 


(2)  (1).
Assuming that assumption (2) holds and a generating set {x_1, x_2} exists such that
 for n ≥ 1 and 1 ≤ i ≠ j ≤ 2, we have x_i = nx_j or nx_i = x_j. 
 Utilizing Remark <ref>, we construct the graph (E, w) based on any of these relations such that H_f and M_E are isomorphic as a monoid. 
 Specifically, we suppose that d(u, u) = d(v, v) = d(v, u) = 0. If x_i = nx_j, then in Figure 1, we suppose that the graph is unweighted. Moreover, if nx_i = x_j, then in Figure 1, the graph has only one arrow a_1, which is weighted with weight n. Note that by  <cit.>, add(x_1 + x_2) is equal to H_f, and so by <cit.>, (ℒ_K(E,w)) is ℵ_0-isomorphic to H. Now we need to show that ℒ_K(E,w) is regular. Since d(u, u) = d(v, v) = d(v, u) = 0, then (E,w) is acyclic, and clearly, in any case, whether E is weighted or not, (LPA1) and (W2) hold. It remains to check conditions (LPA2), (LPA3), and (W1) as explained below:



    
  (LPA2): Any vertex x ∈ T(r(E^1_w)) emits at most one edge.
    Since E^1_w is either empty or contains only one weighted vertex, then 
     r(E^1_w) is either empty or contains only v. The same holds for T(r(E^1_w)), so it is empty or contains only v. Since d(v, v) = 0 = d(v, u), (LPA2) simply holds. 
     
    


    
  (LPA3): If two weighted edges e,f∈ E^1_w are not in line, then T(r(e))∩ T(r(f))=∅.
    Recall that  two edges e and f are in line if they are equal or there is a path from r(e) to s(f) or 
   from r(f) to s(e). Thus, clearly, (LPA3) holds; indeed, there is at most one weighted edge.
    
    
  (W1): No cycle in (E,w) is based at a vertex v∈ T(r(E^1_w)).
    This automatically holds since (E,w) is acyclic and there is no cycle in (E,w).


For the last part of the theorem, observe that whenever x_1 = nx_2 or nx_1 = x_2, then (x_1) =  (x_2). 
 Thus, H is of Type (3). Therefore, using Theorem <ref> and <cit.>, there exists only one infinite form with respect to any generating set.





In the special case of a graph with weights of 1 (or unweighted), this construction yields the usual Leavitt path algebras (LPAs).
Recall that a vertex v in a graph is called regular if  1 ≤| s^-1 (v) | < ∞ and 
M_E of a graph is the commutaive monoid generated by { a_v | v ∈ E_0} and with relations 

    a_v = ∑ _e ∈ s^-1 (v) r (e)

for each regular vertex v ∈ E. 
One may wonder if a two-generated ℵ_0-monoid 
 H can have a realization to a (weighted) LPA  of a graph with only one vertex and countably many loops:


 Let E be the infinite rose graph:


    ∙@<.@(l,d)  @(ur,dr)^e_1@(r,d)^e_2@(dr,dl)^e_3   E




According to <cit.>,
 LPA of E, denoted by L_K(E), is a purely infinite simple ring, and some properties of (L_K(E)) are understood in
 <cit.>. We use <cit.> to see that (R) is exactly M_F, where F is the row finite equivalence of E:





    v_0  @(l,u) ^g_1[r]       v_1 [r]  @/^1.0 pc/[l]^f_1|cm@>   
     v_2 [r] [r]  @/^1.5 pc/[ll]^f_2|cm@>   v_3 [r]  
    @/^2.0pc /@.>[lll] |cm@>   ⋯   F

 
 
Consequently, M_F is generated by {a_v_i| i ≥ 0} such that a_v_i = a_v_0 + a_v_i + 1 for every i ≥ 0. In particular, it is not finitely generated.
For a better understanding of M_F, note that v_1 = v_0 + v_2 = (v_0 + v_1) + v_2 = v_0 + v_2 + v_1 = v_1 + v_1.

If Z is the ring of integers considered to be a monoid with addition and z is a distinct element from the elements of Z, then Z⊔z denotes the monoid with an operation compatible with the operation of Z, where z has the property that z + x = x + z = x for every x ∈Z or x = z.

We can see that M_F is the monoid isomorphic to Z⊔z. To see this, define f: M_F →Z⊔z such that f(v_i) = i - 1. Then f(n v_0) = -n, and by this, it is easy to see that f is a monoid isomorphism. 



To conclude this section and provide a characterization of realizable two-generated monoids to VNR rings, we recall separated graphs:








 


 
  <cit.>
A separated graph is a pair (E,C) where E is a graph,  C=
_v∈ E^ 0 C_v, and
C_v is a partition of s^-1(v) (into pairwise disjoint nonempty
subsets) for every vertex v. (In case v is a sink, we take C_v
to be the empty family of subsets of s^-1(v).)

If all the sets in C are finite, we shall say that (E,C) is a finitely separated graph.

The constructions we introduce revert to existing ones in case C_v= {s^-1(v)}
for each non-sink v∈ E^0. We refer to a non-separated graph or a trivially separated graph in that situation.

 


If all the sets in C
 are finite, we shall say that (E,C)
 is a finitely separated graph.
Given a finitely separated graph (E,C),
 we define the monoid of the separated graph (E,C)
 to be the commutative monoid given by generators and relations as follows:


    M(E, C) := ⟨ a_v := ∑_e ∈ X a_r(e) for every  X ∈ C_v, v ∈ E^0 ⟩



 For the definition of Leavitt path algebra of separated graphs denoted by L_K (E, C), K is a field,  we refer to <cit.>. 
 We recall from <cit.> that  (L_K (E, C)) ≅ M(E,C). 
Recall that  Given a directed graph E =(E_0,E_1,s,r), we have:


  (i) Similar to the case of weighted graphs, we define a pre-order on E_0  (the path-way pre-order) by v ≤  w 
 if and only if there is a directed path γ in E  with s(γ) = w and r(γ) = v.


  (ii)  Let ∼ be the equivalence relation on the set E_0 defined, for 
every v, w ∈  E_0, by  v ∼  w  if  v ≤  w  and w ≤ v.
 Set I = E_0/ ∼ , so that the preorder ≤ on E_0
 induces a partial order on I.
 We will also denote by ≤
 this partial order on I. Thus, denoting by [v] 
the class of v∈ E_0,  we have 
[v] ≤ [w] if and only if v ≤  w.
 We will often refer to [v]  as the strongly connected component of v.


  (iii) We say that E is strongly connected if every two vertices of E_0
 are connected through a directed path, i.e., if I is a singleton.


  <cit.> 
Let (E, C) be
a finitely separated graph and let (I,≤ ) be the partially ordered set associated to the pre-ordered set 
( E_0 , ≤ ). We say that 
( E , C ) is adaptable if I is finite, and there exist a 
partition I = I_free∪  I_reg, 
and a family of subgraphs {E_p}_ p∈ I 
 of E
 such that the following conditions are satisfied: 
 
 
  *  E_0 =  _p ∈ I E^0 _p , where E_p is a strongly connected row-finite graph 
 if p ∈  I_reg 
and E^0 _p = {v^p }  is a single vertex if p ∈  I_free.


  *  For p ∈ I_reg and w ∈  E^0 _p , we have
 that  | C_w |   = 1  and | s_ E_p ^- 1 (w) |≥ 2. Moreover,
all edges departing from w either belong to the graph E_p or connect w to a vertex
u ∈  E^0_q, with q < p  in I .


  *  For p ∈  I_free, we have that
s^-1 (v^p) = ∅ if and only if p is minimal in I.
 If p is not
minimal, then there is a positive integer k( p)
 such that  C_v^p = {X_1 ^(p) , . . . , X_k( p) ^(p)}.
Moreover, each X_i ^(p) 
is of the form

    X_i ^(p) = {α ( p, i),  β ( p, i, 1), β( p, i, 2), ⋯ , β( p, i, g( p, i))},

for some  g( p, i) ≥  1, where α ( p, i)  is a loop, i.e., 
s(α ( p, i)) = r (α( p, i)) = v^p , 
 and r (β ( p, i, t)) ∈  E ^0_q  for q < p in I.
 Finally, we have  E^ 1_p = {α( p, 1), . . . , α( p, k( p))}.
 



We understand those with two vertices in the following example:


Consider  the following graph: 


    @=150pt   
      u @/^3.5pc/@[r]^α_d(u, v) |cm@>@/^2.4pc /@.>[r] |cm@>@/^1.0 pc/[r]^α_1 |cm@>@.@(l,d) @(ur,dr)^γ_1@(r,d)^γ_2@(dr,dl)^γ_3@(l,u)^γ_d(u, u)   
    v   @/^1.0 pc/[l]^β_1|cm@>@/^2.5pc /@.>[l] |cm@>@/^3.5pc/[l]^β_d(v, u) |cm@>@.@(l,d) @(ur,dr)^τ_1@(r,d)^τ_2@(dr,dl)^τ_3@(l,u) ^(0.6) τ_d(v, v)
    
     G







To make G adaptable, we have the following possibilities:

    
  * | I | = 1. In this case, both
     d(u, v) and d(v, u) are nonzero. Also, by Definition <ref><ref>, I_free = ∅ and I_reg = { [u] = [v] }. Then E_[u]^0 = { u, v } and, by
      Definition <ref><ref>, | C_u | = | C_v | = 1, d(u, u) + d(u, v)≥ 2 and d(v, u) + d(v, v)≥ 2. Thus M_E is the monoid graph of a non-separated one generated by a_u, a_v and relations
    
    a_u   = d(u, v) a_v  + d(u, u) a_u, 
    
            a_v   = d(v, u) a_u  + d(v, v) a_v.


    
  * | I | = 2. In this case, we may assume that d(v, u) = 0. Then the following cases happen:
    

  * d(u, v) = 0. Thus, by  Definition <ref><ref>, both u and v are regular and consequently, by  Definition <ref><ref>,
         we have u = d(u, u) u and v = d(v, v) v, where
          d(u, u), d(v, v)≥ 2.

  * d(u, v) ≠ 0. Then u > v and so v is minimum. Both E^0_u and E^0_v have only one element. Then the four cases happen:
          
         
                 

  * u and v are both regular. Then by  Definition <ref><ref>,
             our graph is non-separated and d(u, u), d(v, v)≥ 2. Thus, the relations are:
            
    a_u   = d(u, v) a_v  + d(u, u) a_u, 
    
                    a_v   = d(v, v) a_v.


  * u is regular and v is free. Since s^-1(u) is nonempty, and v is minimal, by  Definition <ref><ref>, u ∈ E^0_u. 
            Then by  Definition <ref><ref>, d(u, u)≥ 2 and 
            d(v, v) = 0. Consequently, the relation is:
            
    a_u   = d(u, v) a_v  + d(u, u) a_u,  with  d(u, u)≥ 2.


  * Let u be free and v be regular. If v ∈ E^0_u, then  Definition <ref><ref>
             implies that 
            there should be an arrow from v to u which is a contradiction. Then v ∈ E^0_v, and so by 
             Definition <ref><ref>, 
             d(v, v)≥ 2 and a_v = d(v, v) a_v. 
            Since u is not minimal, by  Definition <ref><ref>, there exists a positive integer k(u) such that d(u, u) = k(u). 
             Also, we have relations a_u = a_u + i a_v, for every 1 ≤ i ≤ k(u).

  * Let u, v be free. Since s^-1 (u) ≠∅ and v is minimum in I, we conclude by  Definition <ref><ref>
             that u ∈ E^0_u. Then 
            from E^0_v = { v} and  Definition <ref><ref>, 
            we conclude that  
            d(v, v) = 0 and there exists k(u) ≥ 1 and relations
            
    a_u =  a_u + i a_v,  for every  1 ≤ i  ≤ k(u).

        
    








The example helps to understand two-generated ℵ_0-monoid  with characterization to VNR hereditary rings:  


 
Let H be a non-cyclic  ℵ_0-monoid generated by { x_1, x_2}. Then the following statements are equivalent:


  * H has a realization to a hereditary VNR  ring.

  * H_f:= ⟨ x_1, x_2 ⟩  =  (x_1 + x_2)  is a refinement monoid and
H is braiding over H_f. 


  * 

  *   n x_i + ℵ_0 x_j = ℵ_0 x_i + ℵ_0 x_j with n finite implies 
ℵ_0 x_j = ℵ_0 x_i +ℵ_0 x_j and x_i ∈ (x_j)
  and if H is of Type (2), then x_i ≤ x_j.

  *  Whenever    n ≥  m and n x_i + k x_j = m x_i + k' x_j, then (n - m + 1) x_i + k x_j =  x_i + k' x_j. 

  * If x_i ∉(x_j),  n ≥  m and m x_i + ℵ_0 x_j = n x_i + ℵ_0 x_j, then there exist k , k' such that n x_i + k x_j = m x_i + k' x_j.
   In this case: 


  (a) If  H is of Type (2), then 
x_j is cancelable  and consequently when 
 k > k'  then (n - m + 1) x_i + (k - k' ) x_j =  x_i and if  k' >  k, then (n - m + 1) x_i  =  x_i + (k' - k ) x_j. 

  (b) If H is of Type (1), then n x_i  =  m x_i and k x_j =  k' x_j. 
    

  *   An element of H cannot have both finite and infinite forms.

 




 
  (1)  (2). 
 Suppose that H has a realization to a hereditary VNR ring R,
then  H_f:= ⟨ x_1, x_2 ⟩≅(R)
and H is braiding over H_f, by <cit.>.  
Thus H_f is a refinement monoid.
 
  (2)  (3).   By Corollary <ref>, H is hereditary. 
Now if n x_i + ℵ_0 x_j = ℵ_0 x_i + ℵ_0 x_j with n finite, then by 
Theorem <ref>, x_i ∈(x_j) and if H is of Type (2), using   Part (3) of 
Lemma <ref>, we see that x_i ≤ x_j.  This shows (i) holds.
Part  (ii) is exactly Part (1) of  Lemma <ref>.  
Now let 
 x_i ∉(x_j),  n ≥  m and m x_i + ℵ_0 x_j = n x_i + ℵ_0 x_j, 
then  by Theorem <ref>, there exist k , k' such that n x_i + k x_j = m x_i + k' x_j.
Then if H is of Type (2), then (a) follows from the fact that x_j is  cancelable based on 
 Part (3) of 
Lemma <ref>. Part (b) is Part (2) of Lemma <ref>. Part (iv) follows 
 from 
Theorem <ref>. 

  

  (3)  (1). Suppose that (3) holds. Then  by 
Theorem <ref>, H is hereditary and using <cit.> to conclude (1),
it is enough to  show that add(x_1 + x_2) is  refinement. 

We consider two cases:



  T1: If H is of Type (1) and n_1 x_1 + m_1 x_2 = n_2 x_1 + m_2 x_2, then from n_1 x_1 + ℵ_0 x_2 = n_2 x_1 + ℵ_0 x_2 and Part (2)(iib), 
we have that x_1 = (| n_1 - n_2 | + 1) x_1 and 
 x_2 = (| m_1 - m_2 | + 1) x_2. Let k_1, k_2 be the smallest positive integers such that x_1 = k_1 x_1 and x_2 = k_2 x_2. 
We claim that add(x_1 + x_2) is a monoid graph of a separated graph with two vertices u, v such that 1 = | C_v | = | C_u |.

To see this, note that x_i = k_i^m x_i, for every 1 ≤ i ≤ 2 and m ≥ 1. If x_i = n x_i,  where n ≠ 1, 
then we write n = m_j k_i^j + ⋯ + m_1 k_i + m_0, for some positive integers m_j, ⋯, m_0 and j ≥ 1. Then n x_i = (m_j + ⋯ + m_0) x_i. We do the same for n' := m_j + ⋯ + m_0, if n' ≠ 1
 is not equal to k_i. By repeating this process if it is necessary and bearing in mind that k_i is the minimum one such that x_i = k_i x_i, we conclude that add(x_1 + x_2) is generated by { x_1, x_2} and relations x_i = k_i x_i.


  T2: If H is of Type (2), or (3), then for every a, b ∈⟨ x_1, x_2⟩, we have either a ≤ b or  b ≤ a. Also by condition of (3), we see that H_f 
is separative and consequently by <cit.>,  it is refinement. 






















hyperalphaabbr


