The isomorphism problem for basic modules and the divisibility profile of the algebra of polynomials

Abstract

While mutual congeniality of bases is known to guarantee that basic modules from so-related bases are isomorphic, the question of what can be said about isomorphism of basic modules in general has remained open. We show that, for some algebras, basic modules may be non-isomorphic. We also show that it is possible, for some algebras, for all basic modules to be isomorphic, regardless of congeniality. In the process, and as a byproduct, we introduce the notion of domains of divisibility of modules over arbitrary rings. The mechanism employed here to differentiate non-isomorphic basic modules is by showing that they have different domains of divisibility. Domains of divisibility measure how divisible a module can be; for those cases when divisibility is equivalent to injectivity, domains of divisibility provide a way to gauge injectivity of modules as an alternative to the domains of injectivity and other mechanisms in the literature. We focus on the algebra of polynomials with one variable, and observe that F[x] has a full divisibility profile for any field F. When F is algebraically closed, we see that the direct product and the direct sum of all Pascal basic modules have the smallest divisibility domain. We also analyze the diversity of the family of basic modules as we explore how many of those divisibility domains correspond to basic modules. We show that, for an arbitrary infinite field F, F[x] has infinitely many pairwise non-isomorphic basic modules, the divisibility profile of F[x] is complete, and for an algebraically closed field F, the collection of basic modules has any subset S of F with a non-empty at most countable complement, the set $\{ x + \alpha | \alpha \notin S \}$ as a domain of divisibility for some basic F[x]-module.

1 Preliminaries and introduction

This paper serves at least two purposes: it is a paper about basic modules in which domains of divisibility are a significant tool, and a paper about divisibility profiles in which basic modules play an important role. An interesting subplot arises when, in the process of telling the story, we introduce, in Sect. 2.2, an operation on certain amenable bases over the algebra of polynomials on a single variable, which is commutative and associative modulo mutual congeniality.

In the first two subsections of this introduction we introduce the protagonists for both aspects of our story. The final subsection of this introduction outlines the remainder of the paper.

1.1 Basic modules, amenable bases, and the congeniality relation

Basic modules, a new kind of module over an infinite dimensional algebra A, were introduced in [5]. The name is an intentional pun inspired by the fact that the first step in the construction of these modules is to fix a suitable basis for A over which the entire construction revolves. Not all bases allow the construction to work and, for that reason, the notion of amenability of bases ${{\mathcal {B}}}$ of an infinite dimensional algebra A over a field F was also introduced in [5]. Amenability is a condition that permits the regular A-module structure on the direct sum $F^{({{\mathcal {B}}})}\cong A$ to be extended, in a natural way, to an A-module structure on the direct product $F^{{{\mathcal {B}}}}$. That module is hence denoted $F^{{{\mathcal {B}}}}=$ $_{{{\mathcal {B}}}}{{\mathcal {M}}}$ and referred to as the basic module induced by the basis ${{\mathcal {B}}}$. The study of amenability has been extended in several papers such as [7, 24, 25], and [26], etc. To be precise, we state next the pertinent definitions.

Definition 1.1

Given a field F and an infinite dimensional F-algebra A, a basis ${{\mathcal {B}}}$ of A is said to be left amenable if, for every $r \in A$, $[l_r]_{{{\mathcal {B}}}}$, the matrix representation of the F-linear map $l_r$ (left multiplication by r) with respect to the basis ${{\mathcal {B}}}$, is row-finite.

Remark 1.2

The burden of proof to assert the amenability of a basis ${{\mathcal {B}}}$ can be made lighter. For instance,

• (1) it suffices to check the requirement that $[l_r]$ be row finite only for elements r in some basis ${{\mathcal {C}}}$, not necessarily the same as ${{\mathcal {B}}}$, and

• (2) in the case of the algebra F[x], it is sufficient to check that $[l_x]$ is row finite.

Mutual congeniality, a condition also introduced in [5], guarantees that two basic modules are indeed isomorphic via an isomorphism which extends the identity map on the algebra A. We record that definition and other pertinent terminology next.

Definition 1.3

Given a field F, an infinite dimensional F-algebra A, and two bases ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$ of A, we say that ${{\mathcal {B}}}$ is congenial to ${{\mathcal {C}}}$ if, $[I]_{\mathcal B}^{{{\mathcal {C}}}}$, the matrix that translates representations with respect to ${{\mathcal {B}}}$ into representations with respect to ${{\mathcal {C}}}$, is row-finite. If ${{\mathcal {B}}}$ is congenial to ${{\mathcal {C}}}$ and ${{\mathcal {C}}}$ is congenial to ${{\mathcal {B}}}$, then we say that ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$ are mutually congenial. If C is not congenial to B while B is congenial to C, then B is said to be properly congenial to C. If neither ${{\mathcal {B}}}$ is congenial to ${{\mathcal {C}}}$ nor ${{\mathcal {C}}}$ is congenial to ${{\mathcal {B}}}$, then we say that ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$ are discordant.

Apart from when two amenable bases are mutually congenial, it has been open till now, not only whether two basic modules may be otherwise isomorphic (that is whether they can be isomorphic when they are induced by non-congenial bases), but also, on the opposite extreme, whether non-isomorphic basic modules can actually exist. In Sect. 2, we use the notion of the domain of divisibility of a module to show that various basic modules of the algebra of polynomials on one indeterminate are non-isomorphic because they do not share a common divisibility domain.

Remark 1.4

The following is a non-exhausting list of potentially feasible scenarios. There may exist

• (1) algebras over which all basic modules are isomorphic,

• (2) algebras having discordant amenable bases whose corresponding basic modules are isomorphic,

• (3) algebras such that basic modules are isomorphic only when their corresponding bases are mutually congenial, and

• (4) algebras over which basic modules are not isomorphic only when their corresponding bases are discordant.

1.2 Domains and profiles of divisibility

Recall that an element m of a right R-module M is said to be divisible if for every $r\in R$ which is not a right zero-divisor, there exists $m'\in M$ such that $m=rm'$. If every element of M is divisible, then M is called a divisible module. In other words, a right R-module M is divisible if and only if $M=rM$ whenever r is an element of the ring R which is not a right zero-divisor. Hence, a module M over an integral domain is divisible if $rM=M$ for all $r\in R-\{0\}$. In general, it is true that injective modules are divisible. The converse also holds for torsion-free modules over integral domains. It is also well-known that injective and divisible modules coincide over a Dedekind domain. For further details, we refer the reader [27]. The connection between divisibility and injectivity has remained an interesting topic until recently (see for example, [13, 17, 28]).

Following a pattern from several publications, such as [1,2,3,4, 8,9,10,11,12, 14,15,16, 18,19,21], etc., we approach divisibility in a gradual way; not only do we ask whether a module is divisible or not but, instead, we ask how divisible the module is. Our instrument for gauging this property is the domain of divisibility, defined next.

Definition 1.5

Given a module M over a ring R, the domain of divisibility of M is the subset $Div^{-1}(M)= \{r\in R \mid rM=M\}$.

The following properties of divisibility domains are straightforward:

Remark 1.6

• (1) Every domain of divisibility is closed under multiplication,

• (2) if R is a commutative ring and M is an R-module, then for any $l,k \in R$, $lk \in Div^{-1}(M)$ if and only if $l,k\in Div^{-1}(M)$. We will refer to any such a structure as a factorizable semigroup.

• (3) If R is a unique factorization domain, domains of divisibility are determined by the irreducible elements they contain. Domains of divisibility will actually be characterized by any chosen family of representatives of irreducible elements which is complete modulo associates.

• (4) For any family of modules $\{M_i|i \in I\}$, the domain of divisibility of both $\bigoplus _{i \in I} M_i$ and $\Pi _{i \in I}M_i$ is exactly $\bigcap _{i \in I} Div^{-1} (M_i)$.

Our narrative will be facilitated if we use suitable generating sets in lieu of domains of divisibility, that is the reason for introducing some additional notation and terminology next.

Definition 1.7

For an algebra of polynomials $A=F[x]$ and A-module M, the set of monic irreducible polynomials in $Div ^{-1}(M)$ will be denoted d(M).

As noticed above, in part (3) of remark 1.6, d(M) determines uniquely $Div^{-1}(M)$ since $Div^{-1}(M)$ is spanned, as a semigroup, by the union $F \cup d(M)$.

A modification of part (4) of Remark 1.6 holds:

Remark 1.8

For any family of modules $\{M_i|i \in I\}$,

$$\begin{aligned} d(\bigoplus _{i \in I} M_i) = d(\Pi _{i \in I}M_i) = \bigcap _{i \in I} d (M_i). \end{aligned}$$

Similarly as with the various types of domains in the literature, (such as injectivity, projectivity, flatness, weak-injectivity, etc), we can define the divisibility profile of a ring; that will be the subject our next definition.

Definition 1.9

Given a ring R, the collection of factorizable semigroups

$$\begin{aligned} div\ p(R) = \{Div^{-1}(M)| M \in R-mod\} \end{aligned}$$

is called the divisibility profile of R.

Remark 1.10

In light of our discussions so far, it is obvious that the divisibility profile of $R=F[x]$ may be identified with the family

$$\begin{aligned} dp(F[x])=\{d(M)| M \in F[x]-mod\}. \end{aligned}$$

For that reason, the expression divisibility profile of F[x] will be used interchangeably to refer to either $div\ p(F[x])$ or to dp(F[x]). The same is true of the expression domain of divisibility, which, in the context of F[x], will be used to refer to either $Div^{-1}(M)$ or to d(M). The benefit of the convenience that this flexibility will entail surpasses any small risk of confusion.

It is natural to ponder about the divisibility profile of an algebra. In particular, for F[x], one may ask which sets of irreducible polynomials induce domains of divisibility. We shall see, in Theorem 3.1, that when the field F is algebraically closed, every subset ${{\mathcal {A}}}$ of the family ${{\mathcal {L}}}$ of linear monic polynomials is a domain of divisibility.

1.3 Outline of the paper

Section 2 consists of three subsections. In Section 2.1, families of pairwise non-isomorphic basic modules which are induced by discordant bases are exhibited. A natural follow-up question is whether two non-isomorphic basic modules must be induced by discordant bases. This question is considered in Section  2.3 where we show that this is not true for the algebra of polynomials on one indeterminate. In order to achieve this goal, we show in Section 2.2, that the behaviour of the mutual congeniality classes of certain family of amenable bases enjoys a commutative semigroup structure. This commutative semigroup is isomorphic to the set of at-most-countable-complement subsets of F with the union operation.

We then consider, in Sect. 3, the divisibility profile of the algebra F[x] and see that, when F is algebraically closed, all sets of monic linear polynomials are domains of divisibility (Theorem 3.1). Furthermore, we show that all domains of divisibility with finite or at most countable complement correspond to basic modules (Theorem 3.5.)

In Remark 1.4, above, we listed four types of algebra whose existence would be interesting. One can easily think of many other types. In Sect. 4, we produce an example of an algebra of the first type in that remark, that is, an algebra over which all basic modules are isomorphic, regardless of the congeniality of their inducing bases. By showing that there are discordant amenable bases for that algebra, one sees that the algebra is also an example of type 2 in that same remark. It is not known whether types 3 and 4 are indeed feasible; they are mentioned here as proposed questions for future research.

2 Non-isomorphic basic modules over the algebra F[x]

2.1 Non-isomorphic basic modules from discordant bases

We use divisibility domains (Definition 1.5) as a tool to show that there are abundant pairwise non-isomorphic basic modules over F[x]. Considering that mutual congeniality of the inducing amenable bases assures isomorphy of the resulting basic modules, it makes sense to start our discussion by considering pairs of discordant amenable bases.

Let F be a field and $\alpha \in F$. In [5], a basis of F[x] of the type ${{\mathcal {B}}}_{\alpha }= \{ (x+\alpha )^n| n \ge 0 \}$ (with $\alpha \in F$) is called the Pascal basis generated by $x+\alpha$. Pascal bases were shown to be a family of mutually discordant simple bases. The definition of a simple basis is not relevant here, it just suffices to mention that they are an important class of amenable bases; the interested reader may look at [5] and [7] for further details on simple bases. Let us denote the basic module obtained from the Pascal basis ${{\mathcal {B}}}_{\alpha }$ by $M_{\alpha }$. We will refer to all such modules as Pascal basic modules.

Considering that congeniality is a key to isomorphy, a collection of discordant bases is a good place to check for non-isomorphic basic modules; we will show that this is indeed the case by showing that basic modules induced by distinct Pascal bases are pairwise non-isomorphic. It was shown in [5] that every Pascal basis is amenable and, when $\alpha \ne \beta$, the Pascal bases ${{\mathcal {B}}}_{\alpha }$ and ${{\mathcal {B}}}_{\beta }$ are discordant. To assert that the Pascal basic modules $M_{\alpha }$ and $M_{\beta }$ are indeed non-isomorphic, we calculate and compare their divisibility domains.

Proposition 2.1

Let $\alpha \in F$. Then the divisibility domain $Div^{-1}(M_{\alpha })$ of the basic module $M_{\alpha }$ is the factorizable semigroup generated by all irreducible polynomials over F[x] different from $x+\alpha$. In other words, $d(M_{\alpha })$ equals $\{x+\alpha \}^c$, the complement of the singleton $\{x+\alpha \}$ in the set of all monic irreducible polynomials over F[x].

Proof

Recall that the basic module $M_{\alpha }$ induced by the basis ${{\mathcal {B}}}_\alpha$ is, in fact, the module $F^{{{\mathcal {B}}}_\alpha }=$ $_{{{\mathcal {B}}}_\alpha }{{\mathcal {M}}}$. Therefore, $M_{\alpha }=\{\sum _{i=0}^{\infty }a_i(x+\alpha )^i|a_i\in F\}$. Then

$$\begin{aligned} (x+\alpha )M_{\alpha }=\left\{ \sum _{i=1}^{\infty }a_i(x+\alpha )^i|a_i\in F\right\} \ne M_{\alpha }. \end{aligned}$$

This means that $x+\alpha \notin Div^{-1}(M_{\alpha })$. Now let $\beta \ne \alpha$. Then the equation

$$\begin{aligned}{} & {} \sum _{i=0}^{\infty }a_i(x+\alpha )^i=(x+\beta )\sum _{j=0}^{\infty } b_j(x+\alpha )^j=(x+\beta -\alpha +\alpha )\sum _{j=0}^{\infty } b_j(x+\alpha )^j\\{} & {} \quad =[(\beta -\alpha )+(x+\alpha )]\sum _{j=0}^{\infty } b_j(x+\alpha )^j=\sum _{j=0}^{\infty } (\beta -\alpha )b_j(x+\alpha )^j+\sum _{j=0}^{\infty } b_j(x+\alpha )^{j+1} \end{aligned}$$

is solvable for all $b_j's$ since we obtain $a_0=(\beta -\alpha )b_0$ and $a_i=(\beta -\alpha )b_i+b_{i-1}$ for any $i\ge 1$. Hence, $x+\beta \in Div^{-1}(M_{\alpha })$ for any $\beta (\ne \alpha )$ in F.

Now let p(x) be any monic irreducible polynomial which is different from $x+\alpha$. Write $p(x)=P(x+\alpha )=\sum _{i=0}^N p_i(x+\alpha )^i$ for some $p_i\in F$ and $N\in {{\mathbb {N}}}$. Since $x+\alpha \not \mid p(x)$, $p(-\alpha )\ne 0$. Hence, $p_0\ne 0$. Write

$$\begin{aligned} \sum _{i=0}^{\infty }a_i(x+\alpha )^i=P(x+\alpha )\sum _{j=0}^{\infty } b_j(x+\alpha )^j. \end{aligned}$$

Then one will get solvable equations for all $b_j's$. Thus, $p(x)\in Div^{-1}(M_{\alpha })$.

Let $S_{\alpha }$ be the set of all irreducible polynomials in F[x] which are different from $x+\alpha$. Pick $p(x)\in Div^{-1}(M_{\alpha })$. Write $p(x)=q_1(x)\ldots q_n(x)$, where $q_i(x)$ is an irreducible polynomial in F[x] for each $i=1,\ldots , n.$ Since $Div^{-1}(M_{\alpha })$ is closed under factors, $q_i(x)\in Div^{-1}(M_{\alpha })$ for each $i=1,\ldots , n$. It follows that $q_i(x)\ne x+\alpha$ for each $i=1,\ldots , n$. Therefore, $q_i(x)\in S_{\alpha }$ for each $i=1,\ldots , n$. This implies that $p(x)\in <S_{\alpha }>$. $\square$

Corollary 2.2

Let $\alpha , \beta \in F$. If $\alpha \ne \beta$, then the basic modules $M_{\alpha }$ and $M_{\beta }$ are non-isomorphic.

Proof

Isomorphic modules have the same divisibility domain. $\square$

Notice that it was sufficient, for our purpose of proving the above corollary, to show that $Div^{-1}(M_{\alpha })$ contains every linear monic polynomial $x+\beta$, when $\beta \ne \alpha$, but does not contain $x+\alpha$. However, we could not resist the temptation to produce an entire characterization of the domain of divisibility in the proof of our Proposition because we found it exciting that we could do so without assuming that the field is algebraically closed. Things get more complicated when dealing with other more complex bases and therefore we have chosen to accept the simplifying assumption that all irreducible polynomials are linear (i.e., F is algebraically closed). We use this opportunity, however, to suggest that it would be interesting to pursue answers to these questions in the general case.

2.2 A semigroup structure on mutual-congeniality classes of amenable bases

Having shown that discordant bases may produce non-isomorphic basic modules over F[x], one may ask whether two non-isomorphic basic modules, must be induced by discordant bases. The results of this section show that this is not the case for F[x]. Aiming to define (Definition 2.5 and Example 2.6) a product $B_{\alpha }*B_{\beta }$ of two Pascal bases, $B_{\alpha }$ and $B_{\beta }$, we start by presenting an appropriate family that includes both the Pascal bases and their products.

Definition 2.3

• (1) A basis ${{\mathcal {B}}}$, for the algebra $A=F[x]$, is said to be monotone if ${{\mathcal {B}}}=\{b_i|i \ge 0\},$ with each $b_i$ monic and with degree $\deg b_i=i$, and such that, for $i \ge 0$, $b_i$ divides $b_{i+1}$.

• (2) The collection of linear monic polynomials

  $$\begin{aligned} Support({{\mathcal {B}}})=\left\{ x+\alpha _i=\frac{b_{i}}{b_{i-1}}|i \ge 1 \right\} \end{aligned}$$

  is called the support of ${{\mathcal {B}}}$.

• (3) The map $\varphi : {{\mathbb {Z}}}^+ \rightarrow F[x]$ such that $\varphi (i)=x + \alpha _i$ is said to be the inducing map of the monotone base; $\varphi$ is said to be recurring if, for every $i \in {{\mathbb {Z}}}^+$, there exists $j > i$ such that $\varphi (j)=\varphi (i)$.

• (4) We will call a monotone basis a blockchain of F[x] if its inducing map $\varphi$ is recurring.

Example 2.4

Every Pascal basis is a blockchain.

The following operation may be considered on monotone bases or on blockchains. In both settings, it will be instrumental to create amenable bases.

Definition 2.5

For two monotone bases (resp. blockchains) $B_{1}$ and $B_{2}$, with corresponding inducing maps $\varphi _1$ and $\varphi _2$, we define

$$\begin{aligned} B_{1}*B_{2} =\left\{ \prod _ {i=0} ^ {n}(\varphi _1(i)\varphi _2(i)) \mid n\ge 1\right\} \cup \left\{ \varphi _1(n+1)\prod _{i=0}^n(\varphi _1(i)\varphi _2(i)) \mid n\ge 0\right\} . \end{aligned}$$

Notice that this is also a monotone basis (resp. a blockchain) with inducing function $\varphi : {{\textbf{N}}} \rightarrow F[x]$ given by $\varphi (2n+1)=\varphi _1(n+1)$, and $\varphi (2m)=\varphi _2(m)$, for any $n \ge 0, m \ge 1$. We will refer to this basis as the product of the two original bases.

Example 2.6

Given two Pascal bases ${{\mathcal {B}}}_{\alpha }$ and ${{\mathcal {B}}}_{\beta }$ for the algebra F[x],

$$\begin{aligned} B_{\alpha }*B_{\beta } =\{(x+\alpha )^i(x+\beta )^i \mid i\ge 0\} \cup \{(x+\alpha )^{i+1}(x+\beta )^i \mid i\ge 0\} \end{aligned}$$

is a blockchain with inducing function $\varphi : {{\textbf{Z}}}^+ \rightarrow F[x]$ given by $\varphi (n) = x+\alpha$ if n is odd and $\varphi (n) = x+\beta$ if n is even.

It is easy to see that the product of two Pascal bases is a Pascal basis only when the two factors are identical. Also, for any Pascal basis ${{\mathcal {B}}}_{\alpha }$, ${{\mathcal {B}}}_{\alpha }*{{\mathcal {B}}}_{\alpha }={{\mathcal {B}}}_{\alpha }$. Notice that this product is neither commutative nor associative. However, as we will see, the operation $*$ is commutative and associative when considered up to mutual-congeniality. We must start with some preliminary results.

The following striking property of monotone bases is a key reason for our interest in this type of basis.

Lemma 2.7

If ${{\mathcal {B}}}=\{b_i | i \ge 0 \}$ is a monotone basis then, for every $n \in {{\mathbb {N}}}$, the subspaces of A generated by $\{ b_0, \dots , b_{n-1}\}$ and $\{b_n, b_{n+1}, \dots \}$ are. respectively, $F[x]_{< n}$ (the polynomials of degree less than n) and the ideal $I_n=<b_n>=b_nF[x]$, that consists of all multiples of $b_n$.

The following is a very important consequence of this lemma:

Proposition 2.8

• (1) Every monotone basis is amenable, and

• (2) for any monotone bases ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$ of F[x], ${{\mathcal {B}}}*{{\mathcal {C}}}$ is an amenable basis.

Proof

• (1) Let ${{\mathcal {B}}}=\{b_i | i \ge 0 \}$ be a monotone basis. As mentioned in Remark 1.2, it suffices to show that $[l_x]$ is row finite. In fact, the matrix is lower triangular because $xb_n$, being an element of $I_n=<b_n>=b_nF[x]$, is in the span of $\{ b_n, b_{n+1}, \dots \}$.

• (2) The second part follows from the first one because the product of two monotone bases is monotone.

$\square$

Theorem 2.1

Let ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$ be blockchains with the same support. Then ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$ are mutually congenial.

Proof

Say, ${{\mathcal {B}}}=\{b_i|i \ge 0 \}$ and ${{\mathcal {C}}}=\{c_j| j \ge 0\}$. By symmetry, it suffices to show that ${{\mathcal {B}}}$ is congenial to ${{\mathcal {C}}}$. To that effect, consider the representations of elements of ${{\mathcal {B}}}$ with respect to ${{\mathcal {C}}}=\{c_j|j \ge 0\}$. The goal is to show that, given any element $c_m \in {{\mathcal {C}}}$, a moment comes when no element of ${{\mathcal {B}}}$ uses $c_m$ in its representation. Repeated uses of recurrence allow us to find $n \in {{\mathbb {N}}}$ such that $b_n$ is a multiple of $c_{m+1}$. Let $c_{m+1}=(x+\alpha _1)^{n_1}(x+\alpha _2)^{n_2}\ldots (x+\alpha _k)^{n_k}$, where $n_1+\cdots +n_k=m+1$. Since ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$ have the same support, there exists a positive integer $N_1$ such that $(x+\alpha _1)^{n_1}|b_{N_1}$. Similarly, there exists a positive integer $N_2\ge N_1$ such that $(x+\alpha )^{n_1}(x+\alpha _2)^{n_2}|b_{N_2}$. Continuing in this way, $c_{m+1}|b_n$ for some $n=N_k$. Hence, $<b_n> \subseteq <c_{m+1}>$. So no element of $<b_n>$ uses $c_m$ in its representation with respect to ${{\mathcal {C}}}$. But $b_j$ is a multiple of $b_n$ for all $j \ge n$. This proves our claim. $\square$

Example 2.9

Notice that the result of Theorem 2.1 does not hold if one only assumes that the factors are just monotone instead of blockchains. Consider, for example, the discordant monotone bases $\{1\}\cup (x + \beta ){{\mathcal {B}}}_{\alpha }$ and $\{1\}\cup (x + \alpha ){{\mathcal {B}}}_{\beta }$, where $\alpha$ and $\beta$ are distinct scalars such that $\beta \ne \alpha \pm 1$, and ${{\mathcal {B}}}_{\alpha }$ and ${{\mathcal {B}}}_{\beta }$ are the corresponding Pascal bases.

Proof

We calculate $(x+\beta )(x+\alpha )^n$, a typical element of the basis $\{1\}\cup (x + \beta ){{\mathcal {B}}}_{\alpha }$, modulo $\{1\}\cup (x + \alpha ){{\mathcal {B}}}_{\beta }{\setminus } \{x + \alpha \}$ and show that it is not zero for all n. In other words, the coefficient for $x+\alpha$ for the representation with respect to $\{1\}\cup (x + \alpha ){{\mathcal {B}}}_{\beta }$ of $(x+\beta )(x+\alpha )^n$ is non-zero.

First,

$$\begin{aligned} (x+\alpha )^n = ((\alpha -\beta )+(x+\beta ))^n = \sum _{i=0}^n \left( {\begin{array}{c}n\\ i\end{array}}\right) (\alpha -\beta )^{n-i}(x+ \beta )^{i}. \end{aligned}$$

(1)

It follows that

$$\begin{aligned} (x+\beta )(x+\alpha )^n =((\beta - \alpha )+ (x+\alpha )\sum _{i=0}^n \left( {\begin{array}{c}n\\ i\end{array}}\right) (\alpha -\beta )^{n-i}(x+ \beta )^{i}, \end{aligned}$$

(2)

which is congruent, modulo $\{1\}\cup (x + \alpha ){{\mathcal {B}}}_{\beta }{\setminus } \{x + \alpha \}$, to

$$\begin{aligned}{} & {} (\beta -\alpha )\sum _{i=0}^n \left( {\begin{array}{c}n\\ i\end{array}}\right) (\alpha -\beta )^{n-i}(x+ \beta )^{i} + \left( {\begin{array}{c}n\\ 0\end{array}}\right) (\alpha - \beta )^n(x+\alpha ) \end{aligned}$$

(3)

$$\begin{aligned}{} & {} \equiv (\beta -\alpha )\sum _{i=1}^n \left( {\begin{array}{c}n\\ i\end{array}}\right) (\alpha -\beta )^{n-i}((\beta - \alpha )+(x+ \alpha ))(x+ \beta )^{i-1} \nonumber \\{} & {} \quad + \left( {\begin{array}{c}n\\ 0\end{array}}\right) (\alpha - \beta )^n(x+\alpha ). \end{aligned}$$

(4)

Similarly, the last expression is congruent to

$$\begin{aligned}{} & {} (\beta -\alpha )^2\sum _{i=2}^n \left( {\begin{array}{c}n\\ i\end{array}}\right) (\alpha -\beta )^{n-i}((\beta - \alpha )+(x+ \alpha ))(x+ \beta )^{i-1}\nonumber \\{} & {} \qquad \quad + \{\left( {\begin{array}{c}n\\ 0\end{array}}\right) (\alpha - \beta )^n(x+\alpha )+\left( {\begin{array}{c}n\\ 1\end{array}}\right) (\alpha - \beta )^{n-1}\}. \end{aligned}$$

(5)

Repeating this argument recursively, yields that the above expressions are congruent to

$$\begin{aligned} \sum _{i=0}^n\left( {\begin{array}{c}n\\ i\end{array}}\right) (\alpha -\beta )^{n-i}(x+\alpha ), \end{aligned}$$

(6)

whose coefficient for $x + \alpha$ equals the outcome of evaluating the polynomial $(x+1)^n$ at $\alpha - \beta$. The only way this could be zero is if $\beta = \alpha + 1$, which, by assumption, is not the case. $\square$

The following result is an immediate consequence of Theorem 2.1 and is the reason for the title of this subsection.

Theorem 2.2

The set of equivalence classes of blockchains modulo mutual congeniality is a commutative semigroup under an operation induced by the product $*$. This semigroup is easily seen to be isomorphic to the set of at most countable subsets of the field F, with the union of sets as its operation.

Proof

For any block chains ${{\mathcal {B}}}, {{\mathcal {C}}}$, and ${{\mathcal {D}}}$, the blockchains ${{\mathcal {B}}} *{{\mathcal {C}}}$ and ${{\mathcal {C}}} *{{\mathcal {B}}}$ have the same support, as do $({{\mathcal {B}}} *{{\mathcal {C}}}) *{{\mathcal {D}}}$ and ${{\mathcal {B}}} *({{\mathcal {C}}} *{{\mathcal {D}}})$. The result therefore follows the previous theorem. $\square$

2.3 Non-isomorphic basic modules from properly congenial bases

Having shown in Sect. 2.1 that there exist non-isomorphic basic modules stemming from discordant amenable bases, we show here that it is possible for properly congenial bases to induce non isomorphic modules. Our opening proposition sets the stage by providing the appropriate amenable bases.

Proposition 2.10

Let ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$ be two monotone bases for F[x]. Then ${{\mathcal {B}}}*{{\mathcal {C}}}$ is congenial to ${{\mathcal {B}}}$ and ${{\mathcal {C}}}$.

Proof

It is not hard to visualize the matrix that represents the identity linear transformation with respect to the bases ${{\mathcal {B}}}*{{\mathcal {C}}}$ and ${{\mathcal {C}}}$. Doing so will allow us to see that, when ordering the elements of ${{\mathcal {B}}}*{{\mathcal {C}}}$ in an appropriate way, the matrix will be lower echelon and therefore row finite. To this avail, list the elements of the basis ${{\mathcal {B}}}*{{\mathcal {C}}}$ with respect to the index i in such a way that, for $i\in {{\mathbb {Z}}}^+\cup \{0\}$, the elements $b_ic_i$ and $b_{i+1}c_i$ appear next to each other. Then, the $i-$th pair of elements are from the ideal $<c_i>$ and, therefore, by Lemma  2.7 their representations with respect to $\mathcal {C}$ use zero coefficients for $\{c_1, \dots ,c_{i-1}\}$, confirming our claim. Clearly, this argument also shows that ${{\mathcal {C}}}*{{\mathcal {B}}}$ is congenial to ${{\mathcal {B}}}$, since, by Theorem  2.1, ${{\mathcal {B}}}*{{\mathcal {C}}}$ and ${{\mathcal {C}}}*{{\mathcal {B}}}$ are mutually congenial. $\square$

Proposition 2.11

Let ${{\mathcal {B}}}$ be any monotone basis for F[x]. If $x+\delta \notin Support({{\mathcal {B}}})$, then $x+\delta \in Div^{-1}(M_{{{\mathcal {B}}}})$.

Proof

Let $b_i$ be the i-th element of the basis ${{\mathcal {B}}}$ for $i\ge 1$ and $b_0=1$. Take an element $x+\delta \notin Support({{\mathcal {B}}})$. Consider

$$\begin{aligned} (x+\delta )\sum _{i=0}^{\infty }\lambda _i b_i= x\sum _{i=0}^{\infty }\lambda _i b_i+ \delta \sum _{i=0}^{\infty }\lambda _i b_i, \end{aligned}$$

where $\lambda _i\in F$ for $i=0,1,\ldots$. If we let $x+\alpha _{i+1}=\frac{b_{i+1}}{b_i}$, then $xb_i=b_{i+1}-\alpha _{i+1}b_i$. Hence, we may write

$$\begin{aligned} (x+\delta )\sum _{i=0}^{\infty }\lambda _i b_i=\sum _{i=0}^{\infty } \lambda _i (\delta -\alpha _{i+1})b_i+\sum _{i=0}^{\infty } \lambda _i b_{i+1}. \end{aligned}$$

Now pick any element $\sum _{i=0}^{\infty }\beta _i b_i\in M_{{{\mathcal {B}}}}$, where $\beta _i\in F$ for all $i=0,1,\ldots$. Write $\sum _{i=0}^{\infty }\beta _i b_i=(x+\delta )\sum _{i=0}^{\infty }\lambda _i b_i$. It follows that $\beta _0=\lambda _0(\delta -\alpha _1)$, $a_1=\lambda _1(\delta -\alpha _2)+\lambda _0$ and so on. Since $x+\delta \notin Support({{\mathcal {B}}})$, all $\lambda _i$’s can be obtained from these equations. Thus, $(x+\delta )M_{{{\mathcal {B}}}}=M_{{{\mathcal {B}}}}$, which means that $x+\delta \in Div^{-1}(M_{{{\mathcal {B}}}})$. $\square$

Lemma 2.12

Let ${{\mathcal {B}}}=\{1, x+\alpha _1, (x+\alpha _1)(x+\alpha _2), (x+\alpha _1)(x+\alpha _2)(x+\alpha _3),\ldots \}$ be a monotone basis for F[x]. Then $x+\alpha _1\notin Div^{-1}(M_{{{\mathcal {B}}}})$.

Proof

Notice that $(x+\alpha _1)^2=(x+\alpha _1)(x+\alpha _2)+(\alpha _1-\alpha _2)(x+\alpha _1)$ is contained in the subspace generated by ${{\mathcal {B}}}{\setminus } \{1\}$ which is a proper subset of the basic module $M_{{{\mathcal {B}}}}$. It follows that $(x+\alpha _1)M_{{{\mathcal {B}}}}$ is properly contained in $M_{{{\mathcal {B}}}}$. Thus, $x+\alpha _1\notin Div^{-1}(M_{{{\mathcal {B}}}})$. $\square$

Theorem 2.3

Let ${{\mathcal {B}}}$ be a blockchain for F[x]. Then $x+\delta \in Div^{-1}(M_{{{\mathcal {B}}}})$ if and only if $x+\delta \notin Support({{\mathcal {B}}})$.

Proof

Let ${{\mathcal {B}}}_1=\{1, x+\alpha _1, (x+\alpha _1)(x+\alpha _2), (x+\alpha _1)(x+\alpha _2)(x+\alpha _3),\ldots \}$ be a blockchain of F[x], and let $x+\alpha _i\in Support({{\mathcal {B}}}_1)$. Define ${{\mathcal {B}}}_i=\{1, x+\alpha _i, (x+\alpha _i)(x+\alpha _1), (x+\alpha _i)(x+\alpha _1)(x+\alpha _2),\ldots \}$ for any $i\ne 1$. By Lemma 2.12, $x+\alpha _1\notin Div^{-1}(M_{{{\mathcal {B}}}_1})$. By a proof similar to that of Lemma 2.12, we have that $x+\alpha _i\notin Div^{-1}(M_{{{\mathcal {B}}}_i})$ for any non-negative integer i. Since $|{{\mathcal {B}}}_i\setminus {{\mathcal {B}}}_j|<\infty$ and $|{{\mathcal {B}}}_j\setminus {{\mathcal {B}}}_i|<\infty$ for any i, j, ${{\mathcal {B}}}_i$ and ${{\mathcal {B}}}_j$ are mutually congenial. This implies that all basic modules induced by the bases ${{\mathcal {B}}}_i$’s are isomorphic for each i. Hence, $x+\alpha _i\notin Div^{-1}(M_{{{\mathcal {B}}}_1})$. The necessity follows from Proposition 2.11. $\square$

Corollary 2.13

For distinct elements $\alpha , \beta \in F$, the basic modules $M_{\alpha }, M_{\beta }$, and $_{{{\mathcal {B}}} *{{\mathcal {C}}}}M$ are pairwise non isomorphic. As ${{\mathcal {B}}} *{{\mathcal {C}}}$ is properly congenial to ${{\mathcal {B}}}$ (and ${{\mathcal {C}}}$) one sees that non-isomorphic basic modules do not necessarily come from discordant bases.

Corollary 2.14

Let F be an algebraically closed field and $M_{{{\mathcal {B}}}}$ be the basic module induced by the blockchain ${{\mathcal {B}}}$. Then $Div^{-1}(M_{{{\mathcal {B}}}})$ is generated by the set of irreducible polynomials in F[x] not in $Support({{\mathcal {B}}})$.

Remark 2.15

Every factor appears infinitely many times in ${{\mathcal {B}}}$, and so it is a recurring sequence. Therefore, ${{\mathcal {B}}}$ is a blockchain. Any factors in the blockchain ${{\mathcal {B}}}$ can be reordered thanks to Theorem 2.1.

3 On the divisibility profile of F[x] and the diversity of basic modules

It is a natural pursuit, as seen in [1,2,3,4, 8,9,10,11,12, 14,15,16, 18,19,21, 26] etc. to try to understand the structure of a ring in terms of its different profiles.

A common pattern in these projects is the search for modules with smallest possible domain. Injectively poor modules exist for any ring but we not know whether subinjectively poor modules (also known as indigent modules) do as well. $F^*=F{\setminus } \{0\} \subset Div^{-1}(M)$ for all F[x]-module M, the smallest conceivable $Div^{-1}(M)$ would equal $F^*$.

Definition 3.1

Mimicking the nomenclature conventions prevalent in the literature so far, we will say that an F[x]-module M is divisibly poor if $Div^{-1}(M)=F^*$ (equivalently, $d(M) = \emptyset$).

Note that every nonzero ideal of F[x] is divisibly poor for any field F.

We will show that, over any field F, the divisibility profile (Definition 1.9) of F[x] consists of all possible sets of monic irreducibles (as defined in part (2) of Remark 1.6.)

Theorem 3.1

Let I be the set of all monic irreducible polynomials over a field F. Then dp(F[x]) equals ${{\mathcal {P}}}(I)$, the power set of I.

Proof

By virtue of the existence of injective modules, we know that there are modules with divisibility domain I. First pick any monic irreducible polynomial p(x) in F[x]. If q(x) is another monic irreducible in F[x], then $q(x)\in Div^{-1}(\frac{F[x]}{p(x)F[x]})$. Hence, $Div^{-1}(\frac{F[x]}{p(x)F[x]})$ is a factorizable monoid generated by the set $F^*\cup \{q(x)\in F[x]: q(x) \text { is monic irreducible and } q(x)\ne p(x)\}$. Now let S be a subset of I. Then $Div^{-1}(\oplus _{p(x)\notin S}\frac{F[x]}{p(x)F[x]})=<F^*\cup S>$. $\square$

We have the following special case for Pascal basic modules when the field is algebraically closed.

Theorem 3.2

The direct sum and the direct product of all Pascal basic modules are divisibly poor over F[x], where F is an algebraically closed field.

Proof

Proposition 2.1 shows that for every $\alpha \in F$, the basic module $M_{\alpha }$, induced by the Pascal basis ${{\mathcal {B}}}_{\alpha }$, has divisibility domain equal to the complement, $\{x + \alpha \}^c$, in the set of all monic irreducibles, of the singleton $\{x + \alpha \}$. This sets the stage to think of domains of divisibility in terms of their complements in the set of monic irreducibles. The injective modules are the case when the complement of the domain of divisibility is the empty set. Divisibly poor modules would be those for which the complement of its divisibility domain equals the set of all monic irreducibles. Consider a

set A of irreducibles with non-empty complement $S=A^c$ in the set of all monic irreducibles. Then, by

Remark 1.8, both modules $\bigoplus _{(x+\alpha ) \in S}M_{\alpha }$ and $\Pi _{(x+\alpha ) \in S}M_{\alpha }$ have divisibility domain

$$\begin{aligned} \bigcap _{(x+\alpha ) \in S}d(M_{\alpha })= \bigcap _{(x+\alpha ) \in S}\{x+ \alpha \}^c=\left( \bigcup _{(x+\alpha ) \in S}\{(x+\alpha )\}\right) ^c=S^c=A. \end{aligned}$$

$\square$

Remark 3.2

It seems appropriate to point out at this point that our decision to introduce the study of divisibility domains and profiles was not a capricious one. Keeping in mind that an original goal was to show that one could produce non-isomorphic basic modules, a wedge needed to be introduced to show that some basic modules are not isomorphic. Had we tried using the injectivity domains of [1], we would have found that the various Pascal basic modules would have been indistinguishable as they all have the same injectivity domain (the class of all singular F[x]-modules). In the terminology of [23], they are maximally injective because their injectivity domain is a co-atom in the injective profile of F[x]. The use of divisibility to draw the line was therefore legitimately justified since the tools already available did not suffice for the task.

Remark 3.3

Another interesting connection is to show that an F[x]-module can be divisibly poor and maximally injective. Indeed, F[x] itself is such a module.

In particular, a divisibly poor module need not be injectively poor.

Definition 3.4

A notion of diversity can be given for any family of modules ${{\mathcal {X}}} \subset R-Mod$ where one compares $div\ p(R)$ to $\{Div^{-1}(M)| M \in {{\mathcal {X}}}\}$. A family ${{\mathcal {X}}}$ would be fully diverse with respect to divisibility if $div\ p(R) = \{Div^{-1}(M)| M \in {{\mathcal {X}}}\}$.

We focus next on an analysis of the diversity of basic modules. We shall see that, when the field F is algebraically closed, the representation of basic modules in the divisibility profile of the algebra F[x] is quite extensive. The domains of divisibility of basic modules include all sets of monic linear polynomials with an at-most-countable complement.

For the finite case, $\{\alpha _1, \dots , \alpha _n\}$ the recursive product ${{\mathcal {B}}}_{\alpha _1} *\dots *{{\mathcal {B}}}_{\alpha _n}$ is an amenable basis which induces a basic module M having $d(M) = I {\setminus } \{x+\alpha _1, \dots , x+\alpha _n\}$.

Our following theorem handles the infinite case $\{\alpha _1, \dots , \alpha _n, \dots \}$

Theorem 3.5

Let F be an algebraically closed field. For any given (countable) sequence of elements $\{\alpha _1,\alpha _2,\ldots ,\alpha _n,\ldots \}$, there exists a basis ${{\mathcal {B}}}$ such that $Div^{-1}(M_{{{\mathcal {B}}}})$ is generated by the set of irreducible polynomials in F[x] not in the set $\{x+\alpha _1,x+\alpha _2,\ldots ,x+\alpha _n,\ldots \}$.

Proof

Let $\{\alpha _1,\alpha _2,\ldots ,\alpha _n,\ldots \}$ be a (necessarily countable) sequence of elements in the field F. The goal is to produce a blockchain with support $\{x+\alpha _1,x+\alpha _2,\ldots ,x+\alpha _n,\ldots \}$. It does not really matter how exactly one does it because any two such bases will be mutually congenial. So, envision, for example, a Cantor-counting inspired inducing sequence yielding

$$\begin{aligned}{} & {} \{x+\alpha _1\} \cup \{ x+\alpha _2, x+\alpha _1\} \cup \{ x+\alpha _3, x+\alpha _2, x+\alpha _1\} \cup \dots \\{} & {} \quad \cup \{x+\alpha _n, x+\alpha _{n-1}, \dots , x+\alpha _1\} \cup \{ x+\alpha _{n+1} \dots x+\alpha _1\} \cup \dots \cup \dots \end{aligned}$$

For (hopefully) obvious reasons, we find it appealing to name this basis

as it indeed acts as an infinite product of Pascal bases. The induced module M has $d(M) = I {\setminus } \{x+\alpha _1, x+\alpha _2, \dots \}$. Then the result follows from Theorem 2.1 and Corollary 2.14. $\square$

Remark 3.6

In summary, in this section we have shown that

• (1) for an arbitrary infinite field F, F[x] has infinitely many pairwise non-isomorphic basic modules,

• (2) for an algebraically closed field F, the divisibility profile of F[x] is complete, and

• (3) for an algebraically closed field F, the collection of basic modules is very diverse in the sense that for any subset S of F with a non-empty at most countable complement, the set $\{ x + \alpha | \alpha \notin S \}$ is a domain of divisibility for a basic F[x]-module. This still leaves open the question whether the family of basic modules is fully diverse (as defined in Remark 3.4). In particular, we do not know whether a basic F[x]-module can be injective or divisibly poor.

4 An Algebra for which all Basic Modules are isomorphic

Having shown in the previous section that there are abundant pairwise non-isomorphic basic modules over the algebra F[x], in this section, we show an example of an algebra for which the opposite extreme situation holds true. Namely, we present here an example of an algebra over which all basic modules are isomorphic, regardless of congeniality.

Let F be any field. Throughout this section, $I={{\mathbb {Z}}}^+$ and A will be the infinite dimensional F-algebra $\frac{F[x_i| i \in I]}{<x_j x_k | j, k \in I>}$. We will refer to ${{\mathcal {B}}}= \{1, x_1, x_2, x_3,... \}$ as the standard basis of A. Note that $x_i x_j=0$ for all $x_i,x_j \in {{\mathcal {B}}}$. If we consider any element $c \in A$, represented in terms of ${{\mathcal {B}}}$, as $c = \sum _{i=0}^{n} \alpha _i x_i$ (for $\alpha _i\in F, i \in \{0\} \cup I$, and where $x_0=1$), then we will denote $\alpha _0$ as $[c]_1$. With this terminology we can now characterize the amenable bases of A in terms of their representations with respect to the standard basis ${{\mathcal {B}}}$.

Lemma 4.1

Let ${{\mathcal {C}}}= \{c_1, c_2,c_3,...\}$ be a basis for A. Then ${{\mathcal {C}}}$ is amenable if and only if the set $S=\{c \in {{\mathcal {C}}}\mid [c]_1 \ne 0\}$ is finite.

Proof

$(\Rightarrow )$ Assume that, for some amenable basis ${{\mathcal {C}}}$, S is infinite. In this case, $x_ic=[c]_1x_i$ for every $c\in S$. This means that the rows of the matrix $[l_{x_i}]_{C}$ corresponding to the ${{\mathcal {C}}}$- support of $x_i$ are infinite, contradicting the amenability of ${{\mathcal {C}}}$.

$(\Leftarrow )$ Let ${{\mathcal {C}}}= \{c_1, c_2,c_3,...\}$ be a basis for A and assume that $S=\{c \in {{\mathcal {C}}}\mid [c]_1 \ne 0\}$ is finite. Pick an element $x_k\in {{\mathcal {B}}}$ satisfying that $x_k\ne 1$. Reordering for the sake of visual clarity, let us write ${{\mathcal {C}}}=S\sqcup S^{c}$, and observe that, clearly, the matrix below is row-finite:

$\square$

Before proving that all basic A-modules are isomorphic, we establish the value of this result by showing that not all amenable bases are mutually congenial (which would render the result trivial). To that effect, we give examples of amenable bases of the algebra A which are induced by properly congenial or discordant bases.

Example 4.2

We show that amenable bases for the algebra A can relate to each other in a variety of ways:

• (1) Consider ${{\mathcal {B}}}$, the standard basis, and ${{\mathcal {D}}}= \{1, x_1, x_1 + x_2, x_1 + x_2 + x_3, \ldots \}$. It is straightforward to see that they are both amenable and ${{\mathcal {D}}}$ is properly congenial to ${{\mathcal {B}}}$.

• (2) Consider ${{\mathcal {B}}}$, the standard basis, and ${{\mathcal {E}}}= \{1, x_1, x_2 + x_1, x_3 + x_1, \dots , x_i+ x_1 \ldots \}$. It is straightforward to see that they are both amenable and ${{\mathcal {B}}}$ and ${{\mathcal {E}}}$ are discordant. In particular, they show that this algebra is an example of one satisfying condition 2 from the introduction.

Now we will prove that every basic module over this particular algebra A is isomorphic to each other.

Theorem 4.3

Let ${{\mathcal {B}}}=\{1\}\cup \{x_i|i\in I \}$ be the standard basis and C be some other amenable basis for the algebra A. Then ${{\mathcal {M}}}_{{\mathcal {B}}}\cong {{\mathcal {M}}}_{{\mathcal {C}}}$.

Proof

Let ${{\mathcal {C}}}$ be an amenable basis for A. Extend ${{\mathcal {C}}}$ to a basis $\mathcal {C^*}={{\mathcal {C}}}\sqcup \mathcal {C'}$ for the vector space $F^{{{\mathcal {C}}}}$. Without loss of generality, we may assume that the subspace $<\mathcal {C'}>$ spanned by $\mathcal {C'}$ is contained in $F^{{{\mathcal {B}}}\setminus \{1\}}$, that is, the subspace $<\mathcal {C'}>$ does not contain constant terms. In a similar way, we may extend the standard basis ${{\mathcal {B}}}$ to a basis $\mathcal {B^*}={{\mathcal {B}}}\sqcup \mathcal {B'}$ for the vector space $F^{{{\mathcal {B}}}}$. Then we have that $<\mathcal {B'}>\subseteq F^{{{\mathcal {B}}}{\setminus } \{1\}}$.

Now we claim that $F^{{{\mathcal {B}}}}={{\mathcal {M}}}_{{{\mathcal {B}}}}\cong {{\mathcal {M}}}_{{{\mathcal {C}}}}=F^{{{\mathcal {C}}}}$. Note that the algebra $A=F^{\mathcal {(B)}}=F^{\mathcal {(C)}}$ is contained both in ${{\mathcal {M}}}_{{{\mathcal {B}}}}$ and ${{\mathcal {M}}}_{{{\mathcal {C}}}}$. Since $F^{\mathcal {(B)}}=F^{\mathcal {(C)}}$, $|B'|=|C'|$ and so there exists an F-isomorphism $\phi :<\mathcal {B'}>\longrightarrow <\mathcal {C'}>$.

Define $\psi :{{\mathcal {M}}}_{{{\mathcal {B}}}}\longrightarrow {{\mathcal {M}}}_{{\mathcal {C}}}$ via $\psi :A\oplus<\mathcal {B'}>\longrightarrow A\oplus <\mathcal {C'}>$, $\psi (a+b)=a+\phi (b)$, where $a\in A$ and $b\in <\mathcal {B'}>$. Obviously, $\psi$ is additive. Now we will show that it is also A-linear. Pick $\alpha \in A$. It is enough to show that $\phi (\alpha b)=\alpha \phi (b)$ for all $b\in <\mathcal {B'}>$. If $\alpha \in F$, then $\phi (\alpha b)=\alpha \phi (b)$ since $\phi$ is F-linear. Let $\alpha =x_i \in {{\mathcal {B}}}$. Then $\phi (\alpha b)=\phi (0)=0$. In addition, $\alpha \phi (b)=0$ because $x_ix_j=0$ for all $i,j\in {{\mathbb {Z}}}^+$. Hence, $\psi$ is an A-isomorphism. $\square$

Considering that all basic modules considered in the previous section were not injective (exhibiting various degrees of divisibility) it seemed natural to wonder whether the unique basic module of this section is injective. The following remark addresses that question and answers it in the negative.

Remark 4.4

The only basic module ${{\mathcal {M}}}_{{{\mathcal {B}}}}$ of the algebra A is not divisible since $x_i\notin Div^{-1}({{\mathcal {M}}}_{{{\mathcal {B}}}})$ for all i. Hence, in particular, ${{\mathcal {M}}}_{{{\mathcal {B}}}}$ is not an injective A-module.