@V{}V{\rho_{1}}V@V{\rho_{2}}V{}V@V{\ldots}V{}V@V{}V{\rho_{n-1}}V@V{\rho_{n}}V{}V\\\
{{dE}}^{\mathfrak{e}}_{1}@>{H_{1}}>{}>{{dE}}^{\mathfrak{e}}_{2}@>{H_{2}}>{}>\ldots
@>{}>{}>{{dE}}^{\mathfrak{e}}_{n-1}@>{H_{n-1}}>{}>{{dE}}^{\mathfrak{e}}_{n}@>{H_{n}}>{}>\ldots\end{CD}$

Taking direct limits of these directed systems, lead us to a natural map

$\rho:{\bf S}=\varinjlim{\bf
S}\to\varinjlim{{dE}}^{\mathfrak{e}}_{n}={{dE}}^{\mathfrak{e}},$

as claimed.

(5). This is similar to (4).

(6). The assignment

$(m+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}),{\bf
h}^{\mathfrak{e},n}_{{\mathbb{M}},z})\mapsto{\bf
h}^{\mathfrak{e},n}_{{\mathbb{M}},z}(m+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}))$

defines the scaler multiplication
$\frac{\psi^{\mathfrak{e}}_{n}({\mathbb{M}})}{\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})}\times{{D}{E}}^{\mathfrak{e}}_{n}\to\frac{\psi^{\mathfrak{e}}_{n}({\mathbb{M}})}{\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})}$.
Now, let
$m+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})\in\frac{\psi^{\mathfrak{e}}_{n}({\mathbb{M}})}{\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})},{\bf
h}^{\mathfrak{e},n}_{{\mathbb{M}},z}\in{{De}}^{\mathfrak{e}}_{n}$ and ${\bf
h}^{\mathfrak{e},n}_{{\mathbb{M}},w}\in{{dE}}^{\mathfrak{e}}_{n}$. The
assignment

$({\bf
h}^{\mathfrak{e},n}_{{\mathbb{M}},z},m+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}),{\bf
h}^{\mathfrak{e},n}_{{\mathbb{M}},w})\mapsto{\bf
h}^{\mathfrak{e},n}_{{\mathbb{M}},w}({\bf
h}^{\mathfrak{e},n}_{{\mathbb{M}},z}(m+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})))$

defines the scaler multiplication

${{De}}^{\mathfrak{e}}_{n}\times\frac{\psi^{\mathfrak{e}}_{n}({\mathbb{M}})}{\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})}\times{{dE}}^{\mathfrak{e}}_{n}\to\frac{\psi^{\mathfrak{e}}_{n}({\mathbb{M}})}{\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})}.$

Since
${{de}}^{\mathfrak{e}}_{n}={{dE}}^{\mathfrak{e}}_{n}\cap{{De}}^{\mathfrak{e}}_{n}$
is commutative, this induces the desired bimodule structure on
$\psi^{\mathfrak{e}}_{n}({\mathbb{M}})/\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})$.
∎

The following diagram summarizes the relation between the above rings, where
by $A\to B$ we mean $A$ is a subring of $B$:

$\textstyle\bigcup$${{de}}$$\textstyle\bigcup$${{De}}$$\textstyle\bigcup$${{dE}}$$\textstyle\bigcup$${{D}{E}}$$\textstyle\bigcup$${{de}}_{\ast}$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$${{de}}$$\textstyle\bigcup$${{De}}$$\textstyle\bigcup$${{dE}}$$\textstyle\bigcup$${{D}{E}}$$\textstyle\bigcup$${{de}}_{\ast}$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$$\textstyle\bigcup$

The following lemma says that for example for strongly semi-nice construction
$\bar{\mathbb{M}}$ we have some control over ${\rm End}_{\bf
R}({\mathbb{M}}_{\kappa})$; note that it only says it is not too large, but we
have the freedom to choose the ring ${\bf S}$ in order to make ${\rm
End}({\mathbb{M}}_{\lambda})$ have some elements with desirable properties.

######  Lemma 5.25.

Suppose
$\bar{\mathbb{M}}=\langle{\mathbb{M}}_{\alpha}:\alpha\leq\kappa\rangle$ is a
strongly semi-nice construction for $(\lambda,{\mathfrak{m}},S,\kappa)$, every
${\mathfrak{e}}\in{{\mathfrak{E}}}$ is simple and
${\mathbb{M}}:={\mathbb{M}}_{\kappa}$. The following assertions are hold:

  1. (i)

If $({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha,z}[{\mathbf{f}},{\mathfrak{e}}]$
holds, then ${\bf h}^{{\mathfrak{e}},n}_{z}$ is an endomorphism of
${\mathfrak{B}}^{\mathfrak{e}}_{n}({\mathbb{M}})$. Furthermore ${\bf
h}^{{\mathfrak{e}},n}_{z}\in{{dE}}^{\mathfrak{e}}_{n,\ast}$.

  2. (ii)

If $({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha,z}[{\mathbf{f}},{\mathfrak{e}}]$
holds and ${\mathbf{f}}$ is an automorphism of ${\mathbb{M}}$, then ${\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}},z}$ is an automorphism of
${\mathfrak{B}}^{\mathfrak{e}}_{n}({\mathbb{M}})$ and even of
${}^{+}{\mathfrak{B}}^{\mathfrak{e}}_{n}({\mathbb{M}})$.

  3. (iii)

${\rm End}^{{\mathfrak{e}},\omega}({\mathbb{M}})/{\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}})$ embedded into
${{dE}}^{\mathfrak{e}}$. Suppose in addition that $\lambda=\kappa$ and let
${\bf S}$ be a subring of ${\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}})/{\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}})$ of cardinality
$<\lambda$. There is a club $C$ of $\kappa$ such that for any $\alpha\in
C\setminus S$ large enough, the ring ${\bf S}$ is embedded into ${\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}}/{\mathbb{M}}_{\alpha})$.

  4. (iv)

Let ${\bf h}^{{\mathfrak{e}},n}[{\mathbb{M}}]$ denote the natural map ${\rm
End}^{{\mathfrak{e}},n}({\mathbb{M}})\to{\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}})$ and set171717See Definition 5.18
for the definition of $\hat{\bf f}_{n}.$

$\begin{array}[]{lr}{\bf E}_{n}:=\big{\\{}{\bf
h}^{{\mathfrak{e}},n}[{\mathbb{M}}](\hat{\bf f}_{n}):{\mathbf{f}}\in{\rm
End}({\mathbb{M}})\mbox{ and there are
}z_{n}({\mathbf{f}})\in{\mathbb{L}}^{\mathfrak{e}}_{n}\mbox{ and }\\\
\qquad\qquad\qquad\qquad\qquad\alpha_{n}({\mathbf{f}})<\kappa\text{ such that
}({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha_{n}({\mathbf{f}}),z_{n}({\mathbf{f}})}[{\mathbf{f}},{\mathfrak{e}}]\big{\\}}.\end{array}$

Then

     * •

${\rm End}^{{\mathfrak{e}},\omega}({\mathbb{M}})=\bigcup\limits_{n<\omega}{\bf
E}_{n}$,

     * •

${\bf E}_{n}\subseteq{\bf E}_{n+1}$,

     * •

$z_{n}({\mathbf{f}})$ is unique modulo
$\varphi^{\mathfrak{e}}_{\omega}({\mathbb{N}}^{\mathfrak{e}}_{n})$.

  5. (v)

${\bf E}_{n}$ is a subring of ${\rm End}^{\mathfrak{e},\omega}({\mathbb{M}})$
and the mapping $\hat{{\mathbf{f}}}_{n}\mapsto{\bf
h}^{\mathfrak{e},n}_{z_{n}({\mathbf{f}})}$ is a homomorphism from

$\begin{array}[]{lr}\\{\hat{{\mathbf{f}}}_{n}:&{\mathbf{f}}\in{\rm
End}({\mathbb{M}})\mbox{ and
}({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha_{n}({\mathbf{f}}),z_{n}({\mathbf{f}})}\mbox{
for some }\\\ &\alpha_{n}({\mathbf{f}})<\kappa,\
z_{n}({\mathbf{f}})\in{\mathbb{L}}^{\mathord{\mathrm{tr}}}_{n}\\}\end{array}$

into ${{dE}}^{\mathfrak{e}}_{n}$ with kernel ${\rm
End}^{{\mathfrak{e}},n}_{<\lambda}({\mathbb{M}})$, i.e.,

$\\{{\mathbf{f}}\in{\rm
End}^{{\mathfrak{e}},n}({\mathbb{M}}):z_{n}({\mathbf{f}})\in\varphi^{\mathfrak{e}}_{\omega}({\mathbb{N}}_{n})\\}.$

  6. (vi)

The ring ${\bf S}$ is naturally mapped into ${\rm End}({\mathbb{M}})$, for
each $\alpha\leq\omega$, there is a natural homomorphism from ${\rm
End}({\mathbb{M}})$ to ${\rm End}^{{\mathfrak{e}},\alpha}({\mathbb{M}})$,
where for $\alpha<\omega$ has a natural mapping to ${{dE}}$. In particular,
${\bf S}$ is naturally mapped into $dE^{\mathfrak{e}}$.

###### Proof.

(i). In view of Lemma 5.20(ii), ${\bf f}$ induces a homomorphism from
${\mathfrak{B}}^{\mathfrak{e}}_{n}({\mathbb{M}}_{1})\to{\mathfrak{B}}^{\mathfrak{e}}_{n}({\mathbb{M}}_{2})$,
and we conclude from Lemma 5.16 that
$z\in{\mathbb{L}}^{{\mathfrak{e}},\ast}_{n}$. By definition, ${\bf
h}^{{\mathfrak{e}},n}_{z}\in{{dE}}^{\mathfrak{e}}_{n,\ast}$.

(ii). For some formula $\psi(x,y)\in\mathcal{L}_{\mu,\kappa}^{pe}(\tau_{({\bf
R},{\bf S})})$, for all ${\mathbb{M}}$, and all $x,y\in\psi_{n}({\mathbb{M}})$
we have

${\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}},z}(x+\varphi_{\omega}({\mathbb{M}}))=y+\varphi_{\omega}({\mathbb{M}})\Leftrightarrow{\mathbb{M}}\models\psi(x,y).$

Now, we define $\psi^{\prime}(x,y)$ by the following role

${\mathbb{M}}\models\psi^{\prime}(x,y)\Leftrightarrow{\mathbb{M}}\models\psi(y,x).$

Since ${\bf f}$ is an automorphism, $\psi^{\prime}(x,y)$ satisfies the
assumptions (i)-(iii) of Lemma 5.22(3), and hence for some
$z^{\prime}\in{\mathbb{L}}^{{\mathfrak{e}}}_{n}$ and all
$x,y\in\psi_{n}({\mathbb{M}})$,

${\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}},z^{\prime}}(x+\varphi_{\omega}({\mathbb{M}}))=y+\varphi_{\omega}({\mathbb{M}})\Leftrightarrow{\mathbb{M}}\models\psi^{\prime}(x,y).$

It is now clear that ${\bf h}^{{\mathfrak{e}},n}_{{\mathbb{M}},z^{\prime}}$ is
the inverse of ${\bf h}^{{\mathfrak{e}},n}_{{\mathbb{M}},z}$. In particular,
${\bf h}^{{\mathfrak{e}},n}_{{\mathbb{M}},z}$ is an automorphism of
${\mathfrak{B}}^{\mathfrak{e}}_{n}({\mathbb{M}})$.

(iii)$+$(iv)$+$(v). For each $n$, set

$\begin{array}[]{lr}{\bf F}_{n}=\\{\hat{{\mathbf{f}}}_{n}:&{\mathbf{f}}\in{\rm
End}({\mathbb{M}})\mbox{ and
}({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha_{n}({\mathbf{f}}),z_{n}({\mathbf{f}})}\mbox{
for some }\\\ &\alpha_{n}({\mathbf{f}})<\kappa,\
z_{n}({\mathbf{f}})\in{\mathbb{L}}^{\mathord{\mathrm{tr}}}_{n}\\}.\end{array}$

We will show that:

  1. (1)

${\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}}_{\kappa})=\bigcup\limits_{n<\omega}{\bf
E}_{n}$.

  2. (2)

${\bf E}_{n}$ is a subring of ${\rm End}^{\mathfrak{e},\omega}({\mathbb{M}})$.

  3. (3)

the assignment $\hat{{\mathbf{f}}}_{n}\mapsto{\bf
h}^{\mathfrak{e},n}_{z_{n}({\mathbf{f}})}$ yields a homomorphism
$\varrho_{n}:{\bf F}_{n}\to{{dE}}^{\mathfrak{e}}_{n}$ with kernel

$\\{\hat{{\mathbf{f}}}_{n}\in{\rm
End}^{{\mathfrak{e}},n}({\mathbb{M}}):z_{n}({\mathbf{f}})\in\varphi^{\mathfrak{e}}_{\omega}({\mathbb{N}}_{n})\\}.$

  4. (4)

The assignment ${\bf
h}^{\mathfrak{e},n}[{\mathbb{M}}](\hat{{\mathbf{f}}}_{n})\mapsto{\bf
h}^{\mathfrak{e},n}_{z_{n}({\mathbf{f}})}$ induces a homomorphism ${\bf
E}_{n}\to{{dE}}^{\mathfrak{e}}_{n}$.

To see (1), clearly we have $\bigcup\limits_{n<\omega}{\bf E}_{n}\subset{\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}}_{\kappa})$. To see the reverse
inclusion take ${\bf g}\in{\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}}_{\kappa})$. Then for some $n<\omega$
we have ${\bf g}={\bf
h}^{\mathfrak{e},n}[{\mathbb{M}}](\hat{{\mathbf{f}}}_{n})$ where
${\mathbf{f}}\in{\rm End}^{{\mathfrak{e}},n}({\mathbb{M}})$. Here, we are
going to use the a strongly semi-nice construction. In view of this
assumption, we can find $\alpha_{n}({\mathbf{f}})<\lambda$ and
$z_{n}({\mathbf{f}})\in{\mathbb{L}}^{\mathord{\mathrm{tr}}}_{n}$ such that the
property
$({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha_{n}({\mathbf{f}}),z_{n}({\mathbf{f}})}$
holds. By definition ${\bf g}\in{\bf E}_{n}$. This completes the proof of
$(1)$.

$(2)$. This is clear and we leave it to the reader.

$(3)$. Note that
$z_{n}({\mathbf{f}}_{1}\pm{\mathbf{f}}_{2})=z_{n}({\mathbf{f}}_{1})\pm
z_{n}({\mathbf{f}}_{2})$ modulo $\varphi_{\omega}({\mathbb{N}}_{n})$. This
shows that $\varrho_{n}$ is additive:

$\varrho_{n}({\mathbf{f}}_{1}\pm{\mathbf{f}}_{2})={\bf
h}^{n}_{z_{n}({\mathbf{f}}_{1}\pm{\mathbf{f}}_{2})}={\bf
h}^{n}_{z_{n}({\mathbf{f}}_{1})}\pm{\bf
h}^{n}_{z_{n}({\mathbf{f}}_{2})}=\varrho_{n}({\mathbf{f}}_{1})+\varrho_{n}({\mathbf{f}}_{2}).$

Similarly, for any $r\in R$ we have
$\varrho_{n}(r{\mathbf{f}})=\varrho_{n}(r{\mathbf{f}})$. Here, we compute
$\ker(\varrho_{n})$:

$\varrho_{n}({\mathbf{f}})=0\Leftrightarrow{\bf
h}^{n}_{z_{n}({\mathbf{f}})}={\bf h}^{n}_{0}\Leftrightarrow
z_{n}({\mathbf{f}})\in\varphi_{\omega}({\mathbb{N}}_{n}),$

i.e., $\ker(\varrho_{n})=\\{\hat{\bf f}_{n}\in{\bf
F}_{n}:z_{n}({\mathbf{f}})\in\varphi_{\omega}({\mathbb{N}}_{n})\\}$, as
claimed.

$(4)$. This is trivial.

Items (iv) and (v) follow immediately. The assignment

${\bf h}^{\mathfrak{e},n}[{\mathbb{M}}](\hat{\bf f}_{n})\mapsto{\bf
h}^{\mathfrak{e},n}_{z_{n}({\mathbf{f}})}$

defines a homomorphism from ${\rm End}^{{\mathfrak{e}},\omega}({\mathbb{M}})$
into ${{dE}}^{\mathfrak{e}}$ with kernel

$\bigcup\limits_{n<\omega}\\{{\bf h}^{\mathfrak{e},n}[{\mathbb{M}}](\hat{\bf
f}_{n})\in{\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}}):z_{n}({\mathbf{f}})\in\varphi^{\mathfrak{e}}_{\omega}({\mathbb{N}}_{n})\\},$

which is included in ${\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}}).$ In sum, we have a
natural embedding from the ring ${\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}})/{\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}})$ into
${{dE}}^{\mathfrak{e}}$.

It remains to prove the moreover part of (iii). To this end, let ${\bf S}$ be
any subring of ${\rm End}^{{\mathfrak{e}},\omega}({\mathbb{M}})/{\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}})$ of cardinality
$<\lambda$. For each $s\in{\bf S}$ find $n_{s}<\omega$ and ${\bf f}_{s}\in{\rm
End}({\mathbb{M}})$ such that $s={\bf
h}^{\mathfrak{e},n}[{\mathbb{M}}](\hat{({\bf f}_{s})}_{n_{s}})+{\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}})$. We look at the
following club of $\kappa$:

$C_{s}:=\\{\alpha<\kappa:\mathop{\mathrm{Rang}}({\bf
f}_{s}\restriction{\mathbb{M}}_{\alpha})\subseteq{\mathbb{M}}_{\alpha}\\}.$

For each $\alpha\in C_{s}\setminus S$, we define an endomorphism
$\theta(s):{\mathbb{M}}/{\mathbb{M}}_{\alpha}\to{\mathbb{M}}/{\mathbb{M}}_{\alpha}$
by the following role:

$\theta(s)(x+{\mathbb{M}}_{\alpha})={\bf f}_{s}(x)+{\mathbb{M}}_{\alpha}.$

It is now natural to take $\bigcap\limits_{s\in{\bf S}}C_{s}$ as our required
club, but we need to do a little more. Indeed, it is not clear that if
$\theta$ is a homomorphism, as it may not preserve addition or multiplication.
To handle this, we shrink the above intersection further as follows. Given
$s_{1},\cdots,s_{m}\in{\bf S}$, we have

${\bf h}^{\mathfrak{e},n}[{\mathbb{M}}]((\hat{{\bf
f}}_{s_{1}+\cdots+s_{m}})_{n_{s_{1}+\cdots+s_{m}}})-{\bf
h}^{\mathfrak{e},n}[{\mathbb{M}}]((\hat{{\bf
f}}_{s_{1}})_{n_{s_{1}}})-\cdots-{\bf
h}^{\mathfrak{e},n}[{\mathbb{M}}]((\hat{{\bf f}}_{s_{m}})_{n_{s_{m}}})\in{\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}})$

and

${\bf h}^{\mathfrak{e},n}[{\mathbb{M}}]((\hat{{\bf f}}_{s_{1}\cdots
s_{m}})_{n_{s_{1}\cdots s_{m}}})-\left({\bf
h}^{\mathfrak{e},n}[{\mathbb{M}}]((\hat{{\bf
f}}_{s_{1}})_{n_{s_{1}}})\circ\cdots\circ{\bf
h}^{\mathfrak{e},n}[{\mathbb{M}}]((\hat{{\bf
f}}_{s_{m}})_{n_{s_{m}}})\right)\in{\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}}).$

Thus we can find $\alpha(s_{1},\cdots,s_{m})<\kappa$ such that

$\mathop{\mathrm{Rang}}(\left({\bf f}_{s_{1}+\cdots+s_{m}}-{\bf
f}_{s_{1}}-\cdots-{\bf
f}_{s_{m}}\right)\restriction\psi^{\mathfrak{e}}({\mathbb{M}}))\subseteq{\mathbb{M}}_{\alpha(s_{1},\cdots,s_{m})}$

and

$\mathop{\mathrm{Rang}}(\left({\bf f}_{s_{1}\cdots s_{m}}-\left({\bf
f}_{s_{1}}\circ\cdots\circ{\bf
f}_{s_{m}}\right)\right)\restriction\psi^{\mathfrak{e}}({\mathbb{M}}))\subseteq{\mathbb{M}}_{\alpha(s_{1},\cdots,s_{m})}.$

Let
$\alpha(\ast)=\mathop{\mathrm{sup}}\\{\alpha(s_{1},\cdots,s_{m}):m<\omega,~{}s_{1},\cdots,s_{m}\in{\bf
S}\\}<\kappa$. Then $C=\bigcap\limits_{s\in{\bf
S}}C_{s}\setminus(\alpha(\ast)+1)$ is as required. The key point is that for
$s_{1},\cdots,s_{m}\in S,$ and $\alpha\in C,$ modulo ${\mathbb{M}}_{\alpha}$,
we have

${\bf f}_{s_{1}+\cdots+s_{m}}={\bf f}_{s_{1}}+\cdots+{\bf f}_{s_{m}}$

and

${\bf f}_{s_{1}\cdots s_{m}}={\bf f}_{s_{1}}\circ\cdots\circ{\bf f}_{s_{m}}.$

(vi). Let $s\in S$. The notation ${\bf f}_{s}$ stands for the multiplication
map by $s$, e.g., ${\bf f}_{s}\in{\rm End}({\mathbb{M}})$ and it is defined by
${\bf f}_{s}(x)=xs$. The assignment $s\mapsto{\bf f}_{s}$ defines an embedding
$S\longrightarrow{\rm End}({\mathbb{M}})$. Let $\alpha<\omega$. The map ${\bf
f}\mapsto\hat{{\bf f}}_{n}$ yields a natural homomorphism ${\rm
End}({\mathbb{M}})\to{\rm End}^{{\mathfrak{e}},n}({\mathbb{M}})$. For
$\alpha:=\omega$, as ${\rm End}^{{\mathfrak{e}},\omega}({\mathbb{M}})$ is the
direct limit of $\langle{\rm
End}^{{\mathfrak{e}},n}({\mathbb{M}}):n<\omega\rangle$, and by the above
argument, we have a natural homomorphism ${\rm End}({\mathbb{M}})\to{\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}})$. Finally, the assignment ${\bf
h}^{\mathfrak{e},n}[{\mathbb{M}}](\hat{\bf f}_{n})\mapsto{\bf
h}^{n}_{z_{n}({\mathbf{f}})}$ defines a homomorphism ${\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}})\to{{dE}}$. ∎

######  Definition 5.26.

By ${\rm End}^{{\mathfrak{e}},n}_{cpt}({\mathbb{M}})$ we mean

$\left\\{h\in{\rm End}^{{\mathfrak{e}},n}:\mbox{the range of $h$ is compact
for }({\mathfrak{e}},n)\mbox{ in }{\mathbb{M}}_{\lambda}\right\\}.$

######  Remark 5.27.

In Lemma 5.25 we can replace ${\rm
End}^{{\mathfrak{e}},n}_{<\lambda}({\mathbb{M}})$ with ${\rm
End}^{{\mathfrak{e}},n}_{cpt}({\mathbb{M}})$ and drive the analogue statement.
Since this has no role in this paper, we leave the routine modification to the
reader.

######  Lemma 5.28.

Let ${\mathfrak{m}}:=(\mathcal{K},{\mathbb{M}}_{\ast},{{\mathfrak{E}}},{\bf
R},{\bf S},{\bf T})$ be a $\lambda$-context, where $\lambda$ is a regular
cardinal such that $\lambda=\lambda^{\aleph_{0}}>|{\bf R}|+|{\bf
S}|+\aleph_{0}+||{\mathbb{M}}_{*}||$ and for all
$\alpha<\lambda,~{}\alpha^{\aleph_{0}}<\lambda$. Then there is a bi-module
${\mathbb{M}}\geq_{\aleph_{0}}{\mathbb{M}}_{\ast}$ satisfying
$\|{\mathbb{M}}\|=\lambda=|\psi^{\mathfrak{e}}_{n}({\mathbb{M}})/\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})|$
such that ${\mathbb{M}}$ has few direct decompositions in the following sense:

  1. (i)

if ${\mathbb{M}}=\mathop{{\bigoplus}}\limits_{t\in J}{\mathbb{M}}_{t}$ and
${\mathfrak{e}}\in{{\mathfrak{E}}}$, then for all but finitely many $t\in J$
we have
$\bigvee\limits_{n}[\psi^{\mathfrak{e}}_{n}({\mathbb{M}}_{t})\subseteq\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}_{t})]$.
In particular,

$\bigvee\limits_{n}[\varphi^{\mathfrak{e}}_{n}({\mathbb{M}}_{t})=\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}_{t})],$

provided ${\mathfrak{e}}$ is simple.

  2. (ii)

Assume in addition that $|{\bf R}|+|{\bf S}|<2^{\aleph_{0}}$ and that
${\mathbb{M}}={\mathbb{K}}_{\alpha}\oplus{\mathbb{P}}_{\alpha}$ for
$\alpha<(|{\bf R}|+|{\bf S}|+\aleph_{0})^{+}$. Then for some
$\alpha_{0}<\alpha_{1}$ and some $n<\omega$ we have
$\psi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{\ell}})\subseteq\varphi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{1}})+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}),$
where $\ell<2$. In particular, if ${\mathfrak{e}}$ is simple, then

$\varphi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{1}})+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})=\varphi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{2}})+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}).$

  3. (iii)

${\rm End}^{\mathfrak{e},\omega}({\mathbb{M}})/{\rm
End}^{{\mathfrak{e}},\omega}_{<\lambda}({\mathbb{M}})$ has cardinality
$\leq|{\bf R}|+|{\bf S}|+\aleph_{0}$.

###### Proof.

Let $\langle{\mathbb{M}}_{\alpha}:\alpha\leq\lambda\rangle$ be a strongly
semi-nice construction for ${\mathfrak{m}}$ where $S\subset
S^{\lambda}_{\aleph_{0}}$ is stationary such that
$S^{\lambda}_{\aleph_{0}}\setminus S$ is stationary as well. Let
${\mathbb{M}}:={\mathbb{M}}_{\lambda}$. We show ${\mathbb{M}}$ is as required.

(i). Suppose not. Let ${\mathbb{M}}=\mathop{{\bigoplus}}\limits_{t\in
J}{\mathbb{M}}_{t}$ and ${\mathfrak{e}}\in{{\mathfrak{E}}}$ be a
counterexample to the claim. Without loss of the generality, $J=\omega$ and
that for each
$n<\omega,~{}\psi^{\mathfrak{e}}_{n}({\mathbb{M}}_{n})\nsubseteq\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}_{n})$.
Define ${\bf f}:{\mathbb{M}}\to{\mathbb{M}}$ in such a way that

${\bf f}(x)=\left\\{\begin{array}[]{ll}0&x\in{\mathbb{M}}_{2n},\\\
x&x\in{\mathbb{M}}_{2n+1}\end{array}\right.$

In other words, ${\bf f}$ is the natural projection from ${\mathbb{M}}$ onto
$\bigoplus\limits_{n<\omega}{\mathbb{M}}_{2n+1}$. Then for some
$n(\ast)<\omega$, $\alpha<\lambda$ and
$z\in{\mathbb{L}}^{\mathfrak{e}}_{n(\ast)}$ the property
$({\mathord{\mathrm{Pr}}}1)^{n(\ast)}_{\alpha,z}[{\bf{\bf f}},{\mathfrak{e}}]$
holds. Let $\bar{\mathbb{G}}^{\ast}=\langle{\mathbb{G}}^{\ast}_{n}:n\geq
n(\ast)\rangle$ be a decreasing sequence of additive subgroups of
$\varphi^{\mathfrak{e}}_{n(\ast)}({\mathbb{M}})$ as in Lemma 4.17. In
particular,

  1. (1)

If $n\geq n(\ast)$, ${\bf h}:{\mathbb{N}}^{\mathfrak{e}}_{n}\to{\mathbb{M}}$
and $z_{n}:=g_{n(\ast),n}(z)$, then ${\bf f}({\bf
h}(x^{\mathfrak{e}}_{n}))-{\bf h}(z_{n})\in{\mathbb{G}}_{n}^{\ast}$.

  2. (2)

If $z^{*}_{\ell}\in\mathbb{G}^{\ast}_{\ell},$ for $\ell\geq n(\ast),$ then
there exists $z^{*}\in\mathbb{G}^{\ast}_{n(\ast)}$ such that
$z^{*}-\sum\limits_{\ell=n(\ast)}^{n}z_{\ell}^{*}\in\varphi^{\mathfrak{e}}_{n+1}({\mathbb{M}})$.

Due to Lemma 4.19(1) we know that there are $k,m<\omega$ such that

  1. (3)

${\mathbb{G}}^{\ast}_{m}\subseteq\mathop{{\bigoplus}}\limits_{\ell<k}{\mathbb{M}}_{\ell}+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}).$

For each $n>n(\ast)+m+k$ pick some
$y_{n}\in\psi^{\mathfrak{e}}_{n}({\mathbb{M}}_{n})\setminus\varphi_{\omega}^{\mathfrak{e}}({\mathbb{M}}_{n})$.
Let also ${\bf h}_{n}:{\mathbb{N}}^{\mathfrak{e}}_{n}\to{\mathbb{M}}_{n}$ be a
bimodule homomorphism such that $y_{n}={\bf h}_{n}(x_{n}^{\mathfrak{e}})$. We
choose $n$ large enough. The assignment

$x\mapsto({\bf h}_{n}(x),{\bf
h}_{n+1}(g_{n}(x))\in{\mathbb{M}}_{n}\oplus{\mathbb{M}}_{n+1}\subset{\mathbb{M}}$

defines a map ${\bf h}:{\mathbb{N}}_{n}\to{\mathbb{M}}$. This unifies ${\bf
h}_{n+1}$ and ${\bf h}_{n}$. By symmetry, we may assume that $n$ is even. It
then follows from clause (3) that

${\bf h}(z_{n})=({\bf h}_{n}(z_{n}),{\bf
h}_{n+1}(z_{n+1}))\in\varphi_{\omega}^{\mathfrak{e}}({\mathbb{M}}_{n})\subset\varphi_{\omega}^{\mathfrak{e}}({\mathbb{M}})$

and

$y_{n}-{\bf h}(z_{n+1})=y_{n}-({\bf h}_{n+1}(z_{n+1}),{\bf
h}_{n+2}(z_{n+2}))\in\varphi_{\omega}^{\mathfrak{e}}({\mathbb{M}}_{n})\subset\varphi_{\omega}^{\mathfrak{e}}({\mathbb{M}}).$

It follows from the first one that ${\bf h}_{n}(z_{n})$ and ${\bf
h}_{n+1}(z_{n+1})$ are in $\varphi_{\omega}^{\mathfrak{e}}({\mathbb{M}})$. By
repetition, ${\bf h}_{n+2}(z_{n+2})$ is in
$\varphi_{\omega}^{\mathfrak{e}}({\mathbb{M}})$. We plug these in the second
containment to see

$y_{n}\in\varphi_{\omega}^{\mathfrak{e}}({\mathbb{M}}).$

This is a contradiction that we searched for it.

Here, we assume that $\mathfrak{e}$ is simple. According to Lemma 5.8,
$\psi^{\mathfrak{e}}_{n}({\mathbb{M}}_{t})=\varphi^{\mathfrak{e}}_{n}({\mathbb{M}}_{t})$
for all $n$ and $t\in J$. Hence for the $n$ as chosen above, we have

$\varphi^{\mathfrak{e}}_{n}({\mathbb{M}}_{t})=\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}_{t}),$

as required.

(ii). For each $\alpha<(|{\bf R}|+|{\bf S}|+\aleph_{0})^{+}$, let ${\bf
f}_{\alpha}$ be the projection onto ${\mathbb{K}}_{\alpha}$, i.e.,

${\bf
f}_{\alpha}(x)=\left\\{\begin{array}[]{ll}x&x\in{\mathbb{K}}_{\alpha},\\\
0&x\in{\mathbb{P}}_{\alpha}\end{array}\right.$

Pick $n_{\ast}(\alpha)<\omega$,
$z_{\ast}(\alpha)\in{\mathbb{L}}^{\mathfrak{e}}_{n_{\ast}(\alpha)}$ and
$\beta_{\ast}(\alpha)$ such that
$({\mathord{\mathrm{Pr}}}1)^{n_{\ast}(\alpha)}_{\beta_{\ast}(\alpha),z_{\ast}(\alpha)}[{\bf
f}_{\alpha},\mathfrak{e}]$ holds. We combine Lemma 4.17 along with Lemma
4.19(1) to find $m(\alpha)<\omega$ and a compact and decreasing sequence
$\langle\bar{\mathbb{G}}^{\ast}_{\ell}:\ell\geq m(\alpha)\rangle$ of additive
subgroups of $\varphi^{\mathfrak{e}}_{n(\ast)}({\mathbb{M}})$. Furthermore, in
the light of Lemma 4.21 we can assume that
$\mathbb{G}^{\ast}_{m(\alpha)}\subseteq\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}).$
There exists a stationary set $S$ of $\alpha<(|{\bf R}|+|{\bf
S}|+\aleph_{0})^{+}$ such that for some $n<\omega$ and some fixed
$z\in{\mathbb{L}}^{\mathfrak{e}}_{n}$, and for all $\alpha\in S,$ we have
$n_{\ast}(\alpha)=n$ and $z_{\ast}(\alpha)=z$. Let $\alpha_{0}<\alpha_{1}$ be
in $S$ and $\ell<2$. It follows that:

$\begin{array}[]{ll}y\in\psi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{\ell}})&\Rightarrow\exists{\bf
h}:{\mathbb{N}}^{\mathfrak{e}}_{n}\to{\mathbb{K}}_{\alpha_{\ell}},\quad{\bf
h}(x^{\mathfrak{e}}_{n})=y\\\ &\Rightarrow{\bf f}(y)-{\bf
h}(z_{n})\in{\mathbb{G}}^{\ast}_{n}\subset\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})\\\
&\Rightarrow y-{\bf
h}(z_{n})\in\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})\\\ &\Rightarrow
y\in\varphi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{\ell}})+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}).\end{array}$

We proved that
$\psi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{\ell}})\subseteq\varphi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{\ell}})+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})$.

Now suppose that ${\mathfrak{e}}$ is simple. In view of Lemma
5.8,$\psi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{\ell}})=\varphi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{\ell}})$.
It then follows that

$\varphi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{\ell}})+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})=\varphi^{\mathfrak{e}}_{n}({\mathbb{K}}_{\alpha_{1-\ell}})+\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}).$

(iii). We apply the notation introduced in Lemma 5.25. Recall that $E_{n}$ is
defined by the natural image of

$\begin{array}[]{lr}F_{n}:=\\{{\mathbf{f}}\restriction(\varphi_{n}^{\mathfrak{e}}({\mathbb{M}}_{\lambda})/\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}_{\lambda}):&{\mathbf{f}}\in{\rm
End}({\mathbb{M}})\mbox{ and
}({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha_{n}({\mathbf{f}}),z_{n}({\mathbf{f}})}\mbox{
for some }\\\ &\alpha_{n}(F)<\lambda,\
z_{n}({\mathbf{f}})\in{\mathbb{L}}^{\mathord{\mathrm{tr}}}_{n}\\}\end{array}$

in ${\rm End}^{\bar{\varphi},\omega}({\mathbb{M}})$. Also, the mapping
${\mathbf{f}}\mapsto\bf{h}^{n}_{z_{n}({\mathbf{f}})}$ induces a homomorphism
from $F_{n}$ into ${{dE}}^{\mathfrak{e}}_{n}$ with kernel ${\rm
End}^{{\mathfrak{e}},n}_{<\lambda}({\mathbb{M}})$. Since ${\rm
End}^{{\mathfrak{e}},\omega}({\mathbb{M}})=\bigcup\limits_{n<\omega}{\bf
E}_{n}$, it is enough to show that
$\\{\bf{h}^{n}_{z_{n}({\mathbf{f}})}:z_{n}({\mathbf{f}})\in{\mathbb{L}}^{\mathord{\mathrm{tr}}}_{n}\\}$
is of cardinality at most $\leq|{\bf R}|+|{\bf S}|+\aleph_{0}$. Thus it is
enough to show that the cardinality of ${\mathbb{N}}^{{\mathfrak{e}}}_{n}$ is
at most $\leq|{\bf R}|+|{\bf S}|+\aleph_{0}$. This holds, because
${\mathbb{N}}^{{\mathfrak{e}}}_{n}$ as an $({\bf R},{\bf S})$ bimodule is
finitely generated. ∎

######  Remark 5.29.

If we omit “$|{\bf R}|+|{\bf S}|<2^{\aleph_{0}}$”, we get by the same proof
weaker conclusions: with an “error term” which is included in a finitely
generated bimodule.

The following is supposed to be used together with any of the later lemmas
here as its conclusion is in their assumptions.

######  Lemma 5.30.

Let ${\bf R}$ be a ring which is not purely semisimple and let ${\bf S}:={\bf
T}:=\langle 1\rangle_{\bf R}$. Let ${\mathcal{K}}={\mathcal{K}}[{\bf R},\mu]$
be the family of $(\leq\mu_{1})$ generated and $(\leq\theta)$ presented
bimodules where $\mu_{1}^{\theta}\leq\mu$ and let ${\mathbb{M}}_{*}$ be a
bimodule of cardinality $<\mu$. Finally, let
${\mathfrak{E}}:={\mathfrak{E}}_{{\bf R},\mu}$ be the set of
$(<\kappa)$-simple non trivial ${\mathfrak{e}}\in{\mathfrak{E}}_{({\bf R},{\bf
S})}$ wherever $(||{\bf R}||+\aleph_{0})^{<\kappa}\leq\mu$. The following
assertions are true.

  1. (a)

There is ${\mathfrak{e}}\in{{\mathfrak{E}}}$ which is very nice.

  2. (b)

If $(|{\bf R}|+\aleph_{0})^{\aleph_{0}}\leq\mu$ then
${\mathfrak{e}}({\bar{\varphi}})\in{{\mathfrak{E}}}$ for every very nice
$\bar{\varphi}$.

  3. (c)

${\mathfrak{m}}=({\mathcal{K}},{\mathbb{M}}_{*},{\mathfrak{E}},{\bf R},{\bf
S},{\bf T})$ is a non trivial context which is simple.

###### Proof.

Since ${\bf R}$ is not purely semisimple, and in the light of Theorem 2.21, we
can find a sequence $\bar{\varphi}=\langle\varphi_{n}(x):n<\omega\rangle$ as
in Definition 5.2. Then
$\mathfrak{e}=\mathfrak{e}(\bar{\varphi})\in{\mathfrak{E}}$ is very nice. This
confirms (a). Items (b) and (c) are clear. ∎

######  Lemma 5.31.

Suppose ${\bf R}$ is a ring, $T$ is a complete first order theory of ${\bf
R}$-modules which is not superstable, ${\bf S}={\bf T}=\langle 1\rangle_{\bf
R}$ and $\mu\geq|{\bf R}|+\kappa$. Then there is a family
${\mathcal{K}}\cup\\{{\mathbb{M}}_{\ast}\\}$ of ${\bf R}$-modules with $\mu$
members such that:

  1. (a)

${\mathbb{M}}_{\ast}\oplus\mathop{{\bigoplus}}\limits_{t\in
I}{\mathbb{M}}_{t}$ is model of $T$ whenever
${\mathbb{M}}_{t}\in{\mathcal{K}}$, moreover

${\mathbb{M}}_{\ast}\prec_{\mathcal{L}({\tau_{{\bf
R}}})}{\mathbb{M}}_{*}\oplus\mathop{{\bigoplus}}\limits_{t\in
I}{\mathbb{M}}_{t}.$

  2. (b)

For any ${\mathbb{N}}\in{\mathcal{K}}$ we have $||{\mathbb{N}}||\leq\mu$.

  3. (c)

If $2^{\mu_{1}}\leq\mu$ and $\mu_{1}\geq||{\bf R}||+\kappa$, then every model
${\mathbb{N}}$ of $T$ satisfying

${\mathbb{M}}_{*}\prec_{\mathcal{L}({\tau_{{\bf
R}}})}{\mathbb{M}}_{*}\oplus{\mathbb{N}}$

belongs to ${\mathcal{K}}$.

  4. (d)

Let $2^{\mu_{1}}\leq\mu$ and $\mu_{1}\geq||{\bf R}||+\aleph_{0}$. Then every
appropriate sequence $\langle{\mathbb{N}}_{n},g_{n},x_{n}:n<\omega\rangle$
with $||{\mathbb{N}}_{n}||\leq\mu_{1}$ belongs to ${\mathcal{K}}$.

  5. (e)

Let ${\mathfrak{E}}$ be the set of $(<\aleph_{0})$-simple non trivial
${\mathfrak{e}}\in{\mathfrak{E}}_{({\bf R},{\bf S})}$ such that each
${\mathbb{N}}_{n}^{\mathfrak{e}}$ is in $\mathcal{K}$. Then
${\mathfrak{m}}=({\mathcal{K}},{\mathbb{M}}_{*},{\mathfrak{E}},{\bf R},{\bf
S},{\bf T})$ is a $\mu$-context; note that a bimodule for ${\mathfrak{m}}$ is
just an ${\bf R}$-module.

###### Proof.

Let ${\mathbb{M}}_{\ast}$ be any $\aleph_{1}$-saturated model of $T$ of size
$\leq\mu$ and set

${\mathcal{K}}:=\\{{\mathbb{N}}:{\mathbb{N}}\mbox{ is an ${\bf R}$--module
such that }{\mathbb{M}}_{\ast}\prec_{{\mathcal{L}}(\tau_{\bf
R})}{\mathbb{M}}_{\ast}\oplus{\mathbb{N}}\mbox{ and
}2^{\|{\mathbb{N}}\|}\leq\mu\\}.$

It is easily seen that $\mathcal{K}$ is as required. Since this has no role in
the paper, we leave the routine details to the reader. ∎

The following easy lemma plays an essential role in the sequel:

######  Lemma 5.32.

Let ${\mathfrak{m}}$ be a $\lambda$-context and ${\bf S}$ be a free ${\bf
T}$-module with a base $\\{c^{\ast}_{\beta}:\beta<\alpha\\}$. Let
$\bar{\varphi}$ be very nice, $\varphi_{n}=(\exists
y_{0},\ldots,y_{k_{n}-1})\varphi^{\prime}_{n}$ where

$\varphi^{\prime}_{n}=\bigwedge\limits^{m_{n}-1}_{\ell=0}[a^{n}_{\ell}x-\sum\limits_{i=0}^{k_{n}-1}b^{n}_{\ell,i}y_{i}].$

Let
${\mathfrak{e}}={\mathfrak{e}}({\bar{\varphi}})\in{\mathfrak{E}}^{\mathfrak{m}}$
and ${\mathbb{N}}_{n}={\mathbb{N}}^{\mathfrak{e}}_{n}$ (see Definition 5.6).
The following assertions hold:

  1. (1)

Let ${\mathbb{N}}_{n,0}$ be the ${\bf R}$-submodule of ${\mathbb{N}}_{n}$
generated by $\\{x,y_{i}:i<k_{m_{n}-1}\\}$. Then ${\mathbb{N}}_{n}$ is the
direct sum $\sum\limits_{\beta<\alpha}{\mathbb{N}}_{n,\beta}$ and
${\mathbb{N}}_{n,0}\stackrel{{\scriptstyle
h_{\beta}}}{{\cong}}{\mathbb{N}}_{n,\beta}$, as ${\bf R}$-modules, where
${\mathbb{N}}_{n,\beta}$ is the ${\bf R}$-module generated by
$\\{xc^{\ast}_{\beta}\\}\cup\\{y_{i}c^{\ast}_{\beta}:i<k_{m_{n}-1}\\}$ freely
except the equations $\varphi^{\prime}_{n}$ and $h_{0}$ is the identity.

  2. (2)

As a ${\bf T}$-module,
$\varphi_{n}({\mathbb{N}}_{n})/\varphi_{\omega}({\mathbb{N}}_{n})=\sum\limits_{\beta<\alpha}\varphi_{n}({\mathbb{N}}_{n,\beta})/\varphi_{\omega}({\mathbb{N}}_{n,\beta})$.

  3. (3)

For any $z\in{\mathbb{L}}^{{\mathfrak{e}}}_{n}$, there are
$z_{\beta}\in{\mathbb{N}}_{n,0}\cap{\mathbb{L}}^{{\mathfrak{e}}}_{n}\cap\varphi^{\mathfrak{e}}_{n}({\mathbb{N}}_{n,0})$
such that $z=\sum\limits_{\beta<\alpha}h_{\beta}(z_{\beta})$. Also, ${\bf
h}_{z}^{\mathfrak{e},n}=\sum\limits_{\beta<\alpha}{\bf
h}_{h_{\beta}(z_{\beta})}^{\mathfrak{e},n}$. In particular, $z$ is $n$-nice if
and only if each $z_{\beta}$ is $n$-nice.

  4. (4)

The rings ${{de}}^{\mathfrak{e}}_{n}$ and ${\bf S}$ generate
${{dE}}^{\mathfrak{e}}_{n}$. In fact each element of
${{dE}}^{\mathfrak{e}}_{n}$ has the form
$\sum\limits_{\beta<\alpha}x_{\beta}b_{\beta}$, where
$x_{\beta}\in{{de}}^{\mathfrak{e}}_{n}$. Also we have
${{dE}}^{\mathfrak{e}}_{n}={{de}}^{\mathfrak{e}}_{n}\mathop{{\otimes}}\limits_{\bf
T}{\bf S}$.

  5. (5)

Let $\mathcal{I}_{n}$ be a maximal ideal of ${{de}}^{\mathfrak{e}}_{n}$. Then
${\bf D}_{n}:={{de}}^{\mathfrak{e}}_{n}/\mathcal{I}_{n}$ is a field.

  6. (6)

Let ${\bf T}^{\prime}:={\bf T}/(\mathcal{I}_{n}\cap{\bf T})$, ${\bf
S}^{\prime}:={\bf S}/(\mathcal{I}_{n}\cap{\bf T})$ and let ${\mathbb{M}}$ be
from a strongly semi-nice construction. Then any set of equations on ${\bf S}$
which has a solution in ${\rm End}_{\bf R}({\mathbb{M}})$ has a solution in
${\bf D}_{n}\mathop{{\otimes}}\limits_{{\bf T}^{\prime}}{\bf S}^{\prime}$.

###### Proof.

$(1)$. Since ${\bf S}$ is free as a ${\bf T}$-module with the base
$\\{c^{\ast}_{\beta}:\beta<\alpha\\}$, we have ${\bf
S}=\bigoplus_{\beta<\alpha}{\bf T}c^{\ast}_{\beta}$. Now, we apply this
through the following natural identifications of ${\bf R}$-modules:

$\begin{array}[]{ll}{\bf R}x{\bf S}\oplus\bigoplus_{i<k_{m}-1}{\bf R}y_{i}{\bf
S}&=\bigoplus_{\beta<\alpha}{\bf
R}xc^{\ast}_{\beta}\oplus\bigoplus_{i<k_{m}-1}\bigoplus_{\beta<\alpha}{\bf
R}y_{i}c^{\ast}_{\beta}\\\ &=\bigoplus_{\beta<\alpha}\left({\bf
R}xc^{\ast}_{\beta}\oplus\bigoplus_{i<k_{m}-1}{\bf
R}y_{i}c^{\ast}_{\beta}\right).\end{array}$

It turns out from the previous displayed identification that

${\mathbb{N}}_{n}\cong\bigoplus_{\beta<\alpha}{\mathbb{N}}_{n,\beta},$

as an ${\bf R}$-module. Also for any $y\in{\mathbb{N}}_{n,0}$, we set
$h_{\beta}(y):=yc_{\beta}$. This yields an ${\bf R}$-module isomorphism
${\mathbb{N}}_{n,0}\stackrel{{\scriptstyle
h_{\beta}}}{{\cong}}{\mathbb{N}}_{n,\beta}$.

$(2)$. We apply Lemma 3.19 along with (1) to conclude that
$\varphi_{n}({\mathbb{N}}_{n})=\bigoplus_{\beta<\alpha}\varphi_{n}({\mathbb{N}}_{n,\beta})$.
Consequently,
$\varphi_{n}({\mathbb{N}}_{n})/\varphi_{\omega}({\mathbb{N}}_{n})=\bigoplus_{\beta<\alpha}\varphi_{n}({\mathbb{N}}_{n,\beta})/\varphi_{\omega}({\mathbb{N}}_{n,\beta})$.

$(3)$. Let $z\in{\mathbb{L}}^{{\mathfrak{e}}}_{n}$. Then,

$z\in\varphi_{n}({\mathbb{N}}_{n})=\bigoplus_{\beta<\alpha}\varphi^{\mathfrak{e}}_{n}({\mathbb{N}}_{n,\beta})\cong\bigoplus_{\beta<\alpha}h_{\beta}^{\prime\prime}(\varphi^{\mathfrak{e}}_{n}({\mathbb{N}}_{n,0})).$

Let $z_{\beta}\in{\mathbb{N}}_{n,0}$ be such that
$z=\sum\limits_{\beta<\alpha}h_{\beta}(z_{\beta})$. It is evident that
$z_{\beta}\in{\mathbb{N}}_{n,0}\cap{\mathbb{L}}^{\mathfrak{e}}_{n}\cap\varphi^{\mathfrak{e}}_{n}({\mathbb{N}}_{n,0})$.

Now suppose that ${\bf g}:{\mathbb{N}}_{n}\to{\mathbb{M}}$ is a bimodule
homomorphism. Then

$\begin{array}[]{ll}{\bf h}_{z}^{\mathfrak{e},n}({\bf
g}(x_{n})+\varphi_{\omega}({\mathbb{M}}))&={\bf
g}(z)+\varphi_{\omega}({\mathbb{M}})\\\ &=\left(\sum\limits_{\beta<\alpha}{\bf
g}(h_{\beta}(z_{\beta}))\right)+\varphi_{\omega}({\mathbb{M}})\\\
&=\sum\limits_{\beta<\alpha}\left({\bf
g}(h_{\beta}(z_{\beta}))+\varphi_{\omega}({\mathbb{M}})\right)\\\
&=\sum\limits_{\beta<\alpha}{\bf
h}_{h_{\beta}(z_{\beta})}^{\mathfrak{e},n}({\bf
g}(x_{n})+\varphi_{\omega}({\mathbb{M}})).\end{array}$

It follows that ${\bf h}_{z}^{\mathfrak{e},n}=\sum\limits_{\beta<\alpha}{\bf
h}_{h_{\beta}(z_{\beta})}^{\mathfrak{e},n}$.

$(4)$. Let $x\in{{de}}^{\mathfrak{e}}_{n}$ and $s\in{\bf S}$. Then
$xs\in{{dE}}^{\mathfrak{e}}_{n}$. This implies the existence of a ${\bf
T}$-linear map ${{de}}^{\mathfrak{e}}_{n}\times{\bf
S}\to{{dE}}^{\mathfrak{e}}_{n}$. By the universal property of tensor products,
there is a map $f:{{de}}^{\mathfrak{e}}_{n}\otimes_{{\bf T}}{\bf
S}\to{{dE}}^{\mathfrak{e}}_{n}$. Now, let $z\in{{dE}}^{\mathfrak{e}}_{n}$. By
clause (3), there are $z_{\beta}\in{{de}}^{\mathfrak{e}}_{n}$ and
$s_{\beta}\in{\bf S}$ such that $z=\sum\limits_{\beta}z_{\beta}s_{\beta}$.
Moreover, by its proof, we know that such a presentation is unique. This shows
that $f$ is an isomorphism. Up to this identification,
${{dE}}^{\mathfrak{e}}_{n}={{de}}^{\mathfrak{e}}_{n}\mathop{{\otimes}}\limits_{\bf
T}{\bf S}$. Thanks to Lemma 5.24 we know the rings ${{de}}^{\mathfrak{e}}_{n}$
and ${\bf S}$ commute with each other.

$(5)$. As $\mathcal{I}_{n}$ is a maximal ideal, it is clear that ${\bf D}_{n}$
is a division ring. Since the ring ${{de}}^{\mathfrak{e}}_{n}$ is commutative
(see Lemma 5.24(2)) we deduce that ${\bf D}_{n}$ is a field.

$(6)$. Let

  1. $(\ast)_{1}$

$\sum_{j}s_{ij}X_{i}^{j}=0,$

be a system of polynomial equations with parameters $s_{ij}\in S$ and
indeterminates $\\{X_{i}\\}$. Suppose these equations have a solution
${\mathbf{f}}\in{\rm End}({\mathbb{M}})$. This means that

  1. $(\ast)_{2}$

$\sum_{j}s_{ij}{\mathbf{f}}^{j}=0.$

There are $z_{n}({\mathbf{f}})\in{\mathbb{L}}^{\mathfrak{e}}_{n}$ and
$\alpha_{n}({\mathbf{f}})<\lambda$ such that
$({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha_{n}({\mathbf{f}}),z_{n}({\mathbf{f}})}[{\mathbf{f}},{\mathfrak{e}}]$.
Let $f_{n}:=\hat{{\bf f}}_{n}$. Then

  1. $(\ast)_{3}$

$\sum_{j}s_{ij}f^{j}_{n}=0.$

Recall from Lemma 5.25(v) that the natural mapping
$\varrho_{n}:\hat{{\mathbf{f}}}_{n}\mapsto{\bf h}^{n}_{z_{n}({\mathbf{f}})}$
is a homomorphism from

$\begin{array}[]{lr}\\{\hat{{\mathbf{f}}}_{n}:&{\mathbf{f}}\in{\rm
End}({\mathbb{M}})\mbox{ and
}({\mathord{\mathrm{Pr}}}1)^{n}_{\alpha_{n}({\mathbf{f}}),z_{n}({\mathbf{f}})}\mbox{
holds for some }\\\ &\alpha_{n}({\mathbf{f}})<\lambda,\
z_{n}({\mathbf{f}})\in{\mathbb{L}}^{\mathord{\mathrm{tr}}}_{n}\\}\end{array}$

into ${{dE}}^{\mathfrak{e}}_{n}$ with kernel included in ${\rm
End}^{\mathfrak{e},n}_{<\lambda}({\mathbb{M}})$. Let

$\pi:{\rm End}^{{\mathfrak{e}},n}({\mathbb{M}})\to\frac{{\rm
End}^{{\mathfrak{e}},n}({\mathbb{M}})}{{\rm
End}^{{\mathfrak{e}},n}_{<\lambda}({\mathbb{M}})}$

be the canonical map and let $g_{n}=\pi({\bf h}^{n}_{z_{n}({\mathbf{f}})})$.
Applying $\pi\circ\varrho_{n}$ to both sides of $(\ast)_{3}$, we get

  1. $(\ast)_{4}$

$\sum_{j}s_{ij}g^{j}_{n}=0.$

Since there is an embedding

$\rho_{n}:\frac{{\rm End}^{{\mathfrak{e}},n}({\mathbb{M}})}{{\rm
End}^{{\mathfrak{e}},n}_{<\lambda}({\mathbb{M}})}\hookrightarrow{{dE}}^{\mathfrak{e}}_{n},$

by setting $e_{n}:=\rho_{n}(g_{n}),$ we have

  1. $(\ast)_{5}$

$\sum_{j}s_{ij}e^{j}_{n}=0$.

In view of (4) we see that
${{dE}}^{\mathfrak{e}}_{n}={{de}}^{\mathfrak{e}}_{n}\mathop{{\otimes}}\limits_{\bf
T}{\bf S}$. Let

$\sigma:{{de}}^{\mathfrak{e}}_{n}\mathop{{\otimes}}\limits_{\bf T}{\bf
S}\to{\bf D}_{n}\mathop{{\otimes}}\limits_{{\bf T}^{\prime}}{\bf S}^{\prime}$

be the natural map induced by ${{de}}^{\mathfrak{e}}_{n}\twoheadrightarrow{\bf
D}_{n}$ and ${\bf S}\twoheadrightarrow{\bf S}^{\prime}$. Set
$t_{n}:=\sigma(e_{n})$. By applying $\sigma$ to $(\ast)_{5}$ we have

  1. $(\ast)_{6}$

$\sum_{j}s_{ij}t^{j}_{n}=0$.

This essentially says that the polynomial equations from $(\ast)_{1}$ with
parameters in $S$ have a solution in ${\bf
D}_{n}\mathop{{\otimes}}\limits_{{\bf T}^{\prime}}{\bf S}^{\prime}$ as well.
This is what we want to prove. ∎

In what follows we will use the following two consequences of Lemma 5.32:

######  Corollary 5.33.

Suppose that the following three items holds:

  1. (a)

${\bf R}$ is a ring which is not pure semisimple and let ${\bf T}$ be the
subring of ${\bf R}$ generated by $1$, i.e., ${\bf
T}\cong{\mathbb{Z}}/n{\mathbb{Z}}$, where $n:=\operatorname{char}({\bf R})$
which is not necessarily prime.

  2. (b)

${\bf S}$ is a ring containing ${\bf T}$ such that $({\bf S},+)$ is a free
${\bf T}$-module and suppose that for every $s\in{\bf S}\setminus\\{0_{\bf
S}\\}$ for some ${\mathbb{N}}\in{\mathcal{K}}\cup\\{{\mathbb{M}}_{*}\\}$ we
have ${\mathbb{N}}s\not=\\{0_{\mathbb{N}}\\}$, and

  3. (c)

$\lambda=\mathop{\mathrm{cf}}(\lambda)>||{\bf R}||+||{\bf S}||+\aleph_{0}$ and
$\alpha<\lambda\Rightarrow|\alpha|^{\aleph_{0}}<\lambda$.

Then we can find an ${\bf R}$-module ${\mathbb{M}}$ of cardinality $\lambda$,
and a homomorphism $\bf{h}$ from ${\bf S}$ into ${\rm End}({\mathbb{M}})$ such
that:

  1. (d)

$\mathop{\mathrm{Ker}}({\bf h})=\\{0\\}$.

  2. (e)

If $\Sigma$ is a set of equations with parameters in ${\bf S}$ such that ${\bf
h}(\Sigma)$ is solvable in ${\rm End}_{\bf R}({\mathbb{M}})$, then $\Sigma$ is
solvable in ${\bf D}\otimes{\bf S}$ for some field ${\bf D}$.

  3. (f)

If $s\in{\bf S}\setminus\\{0_{{\bf S}}\\}$ and ${\mathbb{N}}\in{\mathcal{K}}$
is such that ${\mathbb{N}}s\not=\\{0_{\mathbb{N}}\\}$, then the image of
${\mathbb{M}}$ under ${\bf h}(s)$ has cardinality $\lambda$.

######  Corollary 5.34.

Suppose ${\bf S}$ is a ring extending $\mathbb{Z}$ such that $({\bf S},+)$ is
free, and let ${\bf R}$ be a ring which is not pure semisimple. Let ${\bf D}$
be a field such that $p:=\operatorname{char}({\bf D})|\operatorname{char}({\bf
R})$ and set ${\mathbb{Z}}_{p}:={\mathbb{Z}}/p{\mathbb{Z}}$ if $p>0$, and
${\mathbb{Z}}_{p}:={\mathbb{Z}}$ otherwise. Let $\Sigma$ be a set of equations
over ${\bf S}$ which is not solvable in ${\bf
D}\mathop{{\otimes}}_{{\mathbb{Z}}_{p}}({\bf S}/p{\bf S})$. Then for
${\mathbb{M}}$ which is strongly nicely constructed, $\Sigma$ is not solvable
in ${\rm End}({\mathbb{M}})$.

Recall that a module is $\aleph_{0}$-free if each of its finitely generated
submodules are free. This yields the following statement:

######  Remark 5.35.

In Corollary 5.34, if $({\bf S},+)$ is an $\aleph_{0}$-free ${\bf T}$-module,
the similar conclusions are hold.

##  6\. All things together: Kaplansky test problems

We are now ready to answer Kaplansky test problems.

######  Notation 6.1.

For the rest of this section we assume that ${\bf R}$ is a ring which is not
pure semisimple.

######  Theorem 6.2.

Let ${\mathfrak{m}}$ be a non trivial $\lambda$-context with ${\bf S}={\bf
T}=\langle 1_{\bf R}\rangle_{\bf R}$ where
$\lambda=\mathop{\mathrm{cf}}(\lambda)>|\alpha|^{\aleph_{0}}+||{\mathfrak{m}}||$
for $\alpha<\lambda$. Then there are ${\bf R}$-modules ${\mathbb{M}}$,
${\mathbb{M}}_{1}$ and ${\mathbb{M}}_{2}$ of cardinality $\lambda$ such that:

  1. (1)

${\mathbb{M}}\oplus{\mathbb{M}}_{1}\cong{\mathbb{M}}\oplus{\mathbb{M}}_{2}$,
and ${\mathbb{M}}_{1}\not\cong{\mathbb{M}}_{2}$.

  2. (2)

${\mathbb{M}}_{1}$, ${\mathbb{M}}_{2}$, ${\mathbb{M}}_{1}\oplus{\mathbb{M}}$
and ${\mathbb{M}}_{2}\oplus{\mathbb{M}}$ are $\leq_{\aleph_{0}}$–extensions of
${\mathbb{M}}^{\ast}$.

  3. (3)

${\mathbb{M}}_{1}\equiv_{\mathcal{L}_{\infty,\lambda}}{\mathbb{M}}_{2}$.

######  Remark 6.3.

Adopt the notation of Theorem 6.2.

  1. (a)

Note that (1) becomes trivial if we remove the “of cardinality $\lambda$”. To
see this, take ${\mathbb{M}}$, ${\mathbb{M}}_{1}$ and ${\mathbb{M}}_{2}$ to be
free ${\bf R}$-modules with

$\|{\mathbb{M}}\|>\|{\mathbb{M}}_{1}\|>\|{\mathbb{M}}_{2}\|\geq|{\bf
R}|+\aleph_{0}.$

  2. (b)

Recall that
${\mathbb{M}}_{1}\equiv_{\mathcal{L}_{\infty,\lambda}}{\mathbb{M}}_{2}$ means
for every sentence $\sigma\in\mathcal{L}_{\infty,\lambda},$

${\mathbb{M}}_{1}\models\sigma\iff{\mathbb{M}}_{2}\models\sigma.$

###### Proof.

$(1)$. Let ${\bf T}$ be the subring of ${\bf R}$ which 1 (the unit) generates.

Step A): Here, we introduce the auxiliary ring ${\bf S}$:

Let ${\bf S}:=\frac{{\bf
T}\langle{\mathcal{X}},{\mathcal{W}}_{1},{\mathcal{Y}},{\mathcal{W}}_{2}\rangle}{I}$
where ${\bf
T}\langle{\mathcal{X}},{\mathcal{W}}_{1},{\mathcal{Y}},{\mathcal{W}}_{2}\rangle$
is the skew polynomial ring in noncommuting variables
$\\{{\mathcal{X}},{\mathcal{W}}_{1},{\mathcal{Y}},{\mathcal{W}}_{2}\\}$ with
coefficients in the commutative ring ${\bf T}$, and $I$ is its two-sided ideal
generated by:

$(\ast)$:

${\mathcal{X}}{\mathcal{X}}={\mathcal{X}}$,
${\mathcal{Y}}{\mathcal{Y}}={\mathcal{Y}}$,
${\mathcal{X}}{\mathcal{W}}_{1}{\mathcal{W}}_{2}={\mathcal{X}}$,
${\mathcal{Y}}{\mathcal{W}}_{2}{\mathcal{W}}_{1}={\mathcal{Y}}$,
${\mathcal{X}}{\mathcal{W}}_{1}{\mathcal{Y}}={\mathcal{X}}{\mathcal{W}}_{1}$,
$(1-{\mathcal{X}})(1-{\mathcal{Y}})=1-{\mathcal{X}}$,
${\mathcal{Y}}{\mathcal{X}}={\mathcal{Y}}$,
${\mathcal{Y}}{\mathcal{W}}_{2}{\mathcal{X}}={\mathcal{Y}}{\mathcal{W}}_{2}$.

In other words, ${\bf S}$ is the ring generated by ${\bf
T}\cup\\{{\mathcal{X}},{\mathcal{W}}_{1},{\mathcal{Y}},{\mathcal{W}}_{2}\\}$
extending ${\bf T}$ freely except equations $(\ast)$ (to understand these
equations see the definition of ${\mathbb{M}}^{\otimes}$ as a bimodule below).

Step B): Let $\alpha<\beta<\gamma$ be additively indecomposable
ordinals181818Recall that an ordinal $\gamma$ is additively indecomposable if
for all ordinals $\alpha,\beta<\gamma$ we have $\alpha+\beta<\gamma.$ and let
${\mathbb{M}}$ be an ${\bf R}$-module. We define a new bimodule
${\mathbb{M}}^{\otimes}$ related to ${\mathbb{M}}$ and ordinals
$\alpha,\beta,\gamma$.

For $i<\gamma$, let ${\mathbb{M}}\stackrel{{\scriptstyle
h_{i}}}{{\cong}}{\mathbb{M}}^{\otimes}_{i}$ (where
${\mathbb{M}}_{i}^{\otimes}$ is an ${\bf R}$-module) and set
${\mathbb{M}}^{\otimes}:=\mathop{{\bigoplus}}\limits_{i<\gamma}{\mathbb{M}}_{i}^{\otimes}$.
We expand ${\mathbb{M}}^{\otimes}$ to an $({\bf R},{\bf S})$-bimodule. To this
end, we take $x\in{\mathbb{M}}^{\otimes}_{i}$. Due to the axioms of bimodules,
it is enough to define
$\\{h_{i}(x){\mathcal{X}},h_{i}(x){\mathcal{Y}},h_{i}(x){\mathcal{W}}_{1},h_{i}(x){\mathcal{W}}_{2}\\}$.
We define these via the following rules:

$h_{i}(x){\mathcal{X}}:=\left\\{\begin{array}[]{ll}h_{i}(x)&\qquad
i\geq\alpha,\\\ 0&\qquad i<\alpha,\end{array}\right.$
$h_{i}(x){\mathcal{Y}}:=\left\\{\begin{array}[]{ll}h_{i}(x)&\qquad
i\geq\beta,\\\ 0&\qquad i<\beta,\end{array}\right.$
$h_{i}(x){\mathcal{W}}_{1}:=\left\\{\begin{array}[]{ll}h_{j}(x)&\mbox{ if for
some }\epsilon,\ i=\alpha+\epsilon<\gamma,\ j=\beta+\epsilon<\gamma,\\\
0&\mbox{ otherwise,}\end{array}\right.$

and

$h_{i}(x){\mathcal{W}}_{2}:=\left\\{\begin{array}[]{ll}h_{j}(x)&\mbox{ if for
some }\epsilon,\ i=\beta+\epsilon<\gamma,\ j=\alpha+\epsilon<\gamma,\\\
0&\mbox{ otherwise.}\end{array}\right.$

Let $\bar{\mathbb{M}}=\langle{\mathbb{M}}_{\alpha}:\alpha\leq\kappa\rangle$ be
a strongly semi-nice construction. Recall that semi-nice construction is a
consequence of Section 3, and its strong form was constructed in Section 4.

Let ${\mathbb{P}}:={\mathbb{M}}_{\kappa}$ and let ${}_{\bf R}{\mathbb{P}}$ be
${\mathbb{P}}$ as an ${\bf R}$-module.

Step C): Here, we define the ${\bf R}$-modules ${\mathbb{M}}$,
${\mathbb{M}}_{1}$ and ${\mathbb{M}}_{2}$ of cardinality $\lambda$ such that
${\mathbb{M}}\oplus{\mathbb{M}}_{1}\cong{\mathbb{M}}\oplus{\mathbb{M}}_{2}$.

To this end, recall from the second step that every element of ${\bf S}$ may
be considered as an endomorphism of ${}_{\bf R}{\mathbb{P}}$. Set ${}_{\bf
R}{\mathbb{M}}^{1}:=({}_{\bf R}{\mathbb{P}}){\mathcal{X}}$ and ${}_{\bf
R}{\mathbb{M}}_{1}:=({}_{\bf R}{\mathbb{P}})(1-{\mathcal{X}})$. We conclude
from the formula ${\mathcal{X}}{\mathcal{X}}-{\mathcal{X}}=0$ that ${}_{\bf
R}{\mathbb{M}}^{1}\cap{}_{\bf R}{\mathbb{M}}_{1}=0$. Let us use from the
formula ${\mathcal{X}}(1-{\mathcal{X}})=1$ that

${}_{\bf R}{\mathbb{P}}={}_{\bf
R}{\mathbb{P}}({\mathcal{X}}+(1-{\mathcal{X}}))={}_{\bf
R}{\mathbb{P}}{\mathcal{X}}+(1-{\mathcal{X}}){}_{\bf R}{\mathbb{P}}={}_{\bf
R}{\mathbb{M}}^{1}+{}_{\bf R}{\mathbb{M}}_{1}={}_{\bf
R}{\mathbb{M}}^{1}\oplus{}_{\bf R}{\mathbb{M}}_{1}.$

Let ${}_{\bf R}{\mathbb{M}}^{2}:=({}_{\bf R}{\mathbb{P}}){\mathcal{Y}}$ and
${}_{\bf R}{\mathbb{M}}_{2}:=({}_{\bf R}{\mathbb{P}})(1-{\mathcal{Y}})$. In
the same vein, the above formulas lead us to the following decomposition

${}_{\bf R}{\mathbb{P}}={}_{\bf R}{\mathbb{M}}^{2}\oplus{}_{\bf
R}{\mathbb{M}}_{2}.$

In view of the equation
${\mathcal{X}}{\mathcal{W}}_{1}{\mathcal{Y}}={\mathcal{X}}{\mathcal{W}}_{1}$,
we have

${\mathbb{M}}^{1}{\mathcal{W}}_{1}={}_{\bf
R}{\mathbb{P}}{\mathcal{X}}{\mathcal{W}}_{1}={}_{\bf
R}{\mathbb{P}}{\mathcal{X}}{\mathcal{W}}_{1}{\mathcal{Y}}\subset{}_{\bf
R}{\mathbb{P}}{\mathcal{Y}}={\mathbb{M}}^{2}.$

This yields a homomorphism from ${\mathbb{M}}^{1}$ to ${\mathbb{M}}^{2}$,
defined by the help of the following assignment

$a\mapsto a{\mathcal{W}}_{1}.$

Similarly, ${\mathcal{W}}_{2}$ provides a homomorphism from ${\mathbb{M}}^{2}$
onto ${\mathbb{M}}^{1}$, defined via

$b\mapsto b{\mathcal{W}}_{2}.$

Thanks to the equations
${\mathcal{Y}}{\mathcal{W}}_{2}{\mathcal{W}}_{1}={\mathcal{Y}}$ and
${\mathcal{Y}}{\mathcal{X}}={\mathcal{Y}}$, it is easily seen that the
multiplication maps by ${\mathcal{W}}_{1}$ and ${\mathcal{W}}_{2}$ are inverse
to each other. This provides an isomorphism from ${\mathbb{M}}^{1}$ onto
${\mathbb{M}}^{2}$, so let ${}_{\bf R}{\mathbb{M}}:={}_{\bf
R}{\mathbb{M}}^{1}\cong{}_{\bf R}{\mathbb{M}}^{2}$.

Step D): One has ${}_{\bf R}{\mathbb{M}}_{1}\not\cong{}_{\bf
R}{\mathbb{M}}_{2}$.

Assume towards contradiction that ${}_{\bf R}{\mathbb{M}}_{1}\cong{}_{\bf
R}{\mathbb{M}}_{2}$. Thus there are $f_{1}:{}_{\bf
R}{\mathbb{M}}_{1}\to{}_{\bf R}{\mathbb{M}}_{2}$ and $f_{2}:{}_{\bf
R}{\mathbb{M}}_{2}\to{}_{\bf R}{\mathbb{M}}_{1}$ such that $f_{1}f_{2}=1$ and
$f_{2}f_{1}=1$. Recall that ${}_{\bf R}{\mathbb{P}}={}_{\bf
R}{\mathbb{M}}^{1}\oplus{}_{\bf R}{\mathbb{M}}_{1}={}_{\bf
R}{\mathbb{M}}^{2}\oplus{}_{\bf R}{\mathbb{M}}_{2}$. Define
${\mathcal{Z}}_{1}\in{\rm End}_{\bf R}({}_{\bf R}{\mathbb{P}})$ by applying
the following assignment

$(a,b)\in{}_{\bf R}{\mathbb{M}}^{1}\oplus{}_{\bf
R}{\mathbb{M}}_{1}\mapsto(a{\mathcal{W}}_{1},f_{1}(b))\in{}_{\bf
R}{\mathbb{M}}^{2}\oplus{}_{\bf R}{\mathbb{M}}_{2}.$

In the same vein, define ${\mathcal{Z}}_{2}\in{\rm End}_{\bf R}({}_{\bf
R}{\mathbb{P}})$ via

$(a,b)\in{}_{\bf R}{\mathbb{M}}^{2}\oplus{}_{\bf
R}{\mathbb{M}}_{2}\mapsto(a{\mathcal{W}}_{2},f_{2}(b))\in{}_{\bf
R}{\mathbb{M}}^{1}\oplus{}_{\bf R}{\mathbb{M}}_{1}.$

Clearly,
${\mathcal{Z}}_{1}{\mathcal{Z}}_{2}={\mathcal{Z}}_{2}{\mathcal{Z}}_{1}=1=\mathop{\mathrm{id}}_{\mathbb{P}}$.
It is also easy to check that:

$\begin{array}[]{ll}{\mathcal{X}}{\mathcal{Z}}_{1}={\mathcal{X}}{\mathcal{Z}}_{1}{\mathcal{Y}},\\\
(1-{\mathcal{X}}){\mathcal{Z}}_{1}=(1-{\mathcal{X}}){\mathcal{Z}}_{1}(1-{\mathcal{Y}}),\\\
{\mathcal{Y}}{\mathcal{Z}}_{2}={\mathcal{Y}}{\mathcal{Z}}_{2}{\mathcal{X}},\\\
(1-{\mathcal{Y}}){\mathcal{Z}}_{2}=(1-{\mathcal{Y}}){\mathcal{Z}}_{2}(1-{\mathcal{X}}).\end{array}$

We use just one very simple non trivial
${\mathfrak{e}}\in{{\mathfrak{E}}}^{{\mathfrak{m}}}$. According to Lemma 4.17,
there are $n(\ast)<\omega$, $z_{1}$ and
$z_{2}\in{\mathbb{L}}^{{\mathfrak{e}},\ast}_{n(\ast)}$ such that the equations
above hold in the endomorphism ring of Abelian group
$\varphi^{\mathfrak{e}}_{n(\ast)}({\mathbb{M}})/\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}})$
for any bimodule ${\mathbb{M}}$,191919Recall from Lemma 5.8 that
$\varphi^{\mathfrak{e}}_{n}({\mathbb{M}})\equiv\psi^{\mathfrak{e}}_{n}({\mathbb{M}})$
holds for all $n<\omega.$ when we replace ${\mathcal{Z}}_{1}$ (resp.
${\mathcal{Z}}_{2}$) by ${\bf
h}^{{\mathfrak{e}},n(\ast)}_{{\mathbb{M}},z_{1}}$ (resp. ${\bf
h}^{{\mathfrak{e}},n(\ast)}_{{\mathbb{M}},z_{2}}$) and interpret
${\mathcal{X}},{\mathcal{Y}},{\mathcal{W}}_{1},{\mathcal{W}}_{2}\in{\bf S}$
naturally. This holds in particular for the bimodule ${\mathbb{M}}^{\otimes}$
we defined in Step B). So, the following equations hold:

$\begin{array}[]{ll}\mathcal{X}{\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}}^{\otimes},z_{1}}=\mathcal{X}{\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}}^{\otimes},z_{1}}\mathcal{Y},\\\
(1-\mathcal{X}){\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}}^{\otimes},z_{1}}=(1-\mathcal{X}){\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}}^{\otimes},z_{1}}(1-\mathcal{Y}),\\\
\mathcal{Y}{\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}}^{\otimes},z_{2}}=\mathcal{Y}{\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}}^{\otimes},z_{2}}\mathcal{X},\\\
(1-\mathcal{Y}){\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}}^{\otimes},z_{2}}=(1-\mathcal{Y}){\bf
h}^{{\mathfrak{e}},n}_{{\mathbb{M}}^{\otimes},z_{2}}(1-\mathcal{X}).\end{array}$

These equations in turn define certain decompositions of
$\varphi^{\mathfrak{e}}_{n}({\mathbb{M}}^{\otimes})/\varphi^{\mathfrak{e}}_{\omega}({\mathbb{M}}^{\otimes})$
which yield to the following isomorphism

$\frac{\varphi^{\mathfrak{e}}_{n}\bigl{(}\sum_{i<\beta}{\mathbb{M}}^{\otimes}_{i}\bigl{)}}{\varphi^{\mathfrak{e}}_{\omega}\bigl{(}\sum_{i<\beta}{\mathbb{M}}^{\otimes}_{i}\bigr{)}}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\frac{\varphi^{\mathfrak{e}}_{n}\bigl{(}\sum_{i<\alpha}{\mathbb{M}}^{\otimes}_{i}\bigl{)}}{\varphi^{\mathfrak{e}}_{\omega}\bigl{(}\sum_{i<\alpha}{\mathbb{M}}^{\otimes}_{i}\bigr{)}}.$

The cardinality of left (resp. right) hand side is $|\beta|$ (resp.
$|\alpha|$). Thus if we choose $|\beta|>|\alpha|$, we get a contradiction that
we searched for it.

$(2)$. This is in the proof of $(1)$.

$(3)$. This follows from (2) and Lemma 3.18. ∎

We now prove the existence of ${\bf R}$-modules with the Corner pathology. We
present such a thing by applying Corollary 5.33.

######  Theorem 6.4.

Let $m(\ast)>2$ and let $\lambda>|{\bf R}|$ be a cardinal of the form
$\lambda=\left(\mu^{\aleph_{0}}\right)^{+}$. Then there is an ${\bf R}$-module
${\mathbb{M}}$ of cardinality $\lambda$ such that:

${\mathbb{M}}^{n}\cong{\mathbb{M}}\quad\emph{ if and only if }\quad
m(\ast)-1\mbox{ divides }n-1.$

###### Proof.

We divide the proof into nine steps.

Step A) We first introduce rings ${\bf S}_{0}$ and ${\bf S}$. The ring ${\bf
S}_{0}$ is incredibly easy compared to ${\bf S}$ and ${\bf S}$ is essentially
the ring of endomorphisms we would like.

To this end, let ${\bf T}$ be the subring of ${\bf R}$ which 1 generates. Let
${\bf S}_{0}$ be the ring extending ${\bf T}$ generated by
$\\{{\mathcal{X}}_{0},\ldots,{\mathcal{X}}_{m(\ast)-1},{\mathcal{W}},{\mathcal{Z}}\\}$
freely except the equations:

$(\ast)_{1}$:

${\mathcal{X}}^{2}_{\ell}={\mathcal{X}}_{\ell}$,
${\mathcal{X}}_{\ell}{\mathcal{X}}_{m}=0$ ($\ell\neq m$),
$1={\mathcal{X}}_{0}+\ldots+{\mathcal{X}}_{m(\ast)-1}$,
${\mathcal{X}}_{\ell}{\mathcal{W}}{\mathcal{X}}_{m}=0$ for $\ell+1\neq m\
\mod\ m(\ast)$,
${\mathcal{W}}^{m(\ast)}=1$,
${\mathcal{Z}}^{2}=1$,
${\mathcal{X}}_{0}{\mathcal{Z}}(1-X_{0})={\mathcal{X}}_{0}{\mathcal{Z}}$,
$(1-{\mathcal{X}}_{0}){\mathcal{Z}}{\mathcal{X}}_{0}=(1-{\mathcal{X}}_{0}){\mathcal{Z}}$.

The meaning of these equations will become clear when we use them.

Similarly, we define ${\bf S}$, but in addition we require $\sigma=0$, where
$\sigma$ is a term in the language of rings, when ${}_{\bf
D}{\mathbb{M}}^{\ast}\sigma=0$ for every field $\bf D$ and where ${}_{\bf
D}{\mathbb{M}}^{*}$ is the $({\bf D},{\bf S}_{0})$\- bimodule as defined
below. Now “${\bf S}$ is a free ${\bf T}$-module” will be proved later.

For an integer $m$, the notation $[-\infty,m)$ stands for $\\{n:n\mbox{ is an
integer }<m\\}$ and if $\eta\in{}^{[-\infty,m)}\omega$, then we set
$\eta\restriction
k:=\eta\restriction[-\infty,\mathop{\mathrm{min}}\\{m,k\\})$.

We look at the following sets:

$\begin{array}[]{lll}W_{0}:=&\big{\\{}\eta:&\eta\mbox{ is a function with
domain of the form }[-\infty,n)\\\ &&\mbox{and range
}\subseteq\\{1,\ldots,m(\ast)-1\\},\mbox{ and such that}\\\ &&\mbox{for every
small enough }m\in{\mathbb{Z}},\ \eta(m)=1\big{\\}},\end{array}$

and $W_{1}:=W_{0}\times\\{0,\ldots,m(\ast)-1\\}$.

Let $\bf D$ be a field such that if ${\bf T}$ is finite and of cardinality
$n$, then $\operatorname{char}({\bf D})$ divides $n$. So ${\bf D}\otimes{\bf
S}$ is the ring extending ${\bf D}$ by adding ${\bf
D},{\mathcal{X}}_{0},\ldots,{\mathcal{X}}_{m(\ast)-1},{\mathcal{W}},{\mathcal{Z}}$
as non commuting variables freely except satisfying the equation in
$(\ast)_{1}$ and if $||{\bf T}||$ is finite we divide ${\bf S}$ by $p{\bf S}$
where $p:=\operatorname{char}({\bf D})$. So there is a homomorphism $g_{\bf
D}$ from ${\bf S}$ to ${\bf D}\otimes{\bf S}$ such that $\\{0_{{\bf
S}}\\}=\cap\\{\mathop{\mathrm{Ker}}(g_{\bf D}):{\bf D}\emph{ as above}\\}$.

Let ${\mathbb{M}}^{\ast}={}_{\bf D}{\mathbb{M}}^{\ast}$ be the left $\bf
D$-module freely generated by

$\\{x_{\eta,\ell}:\eta\in W_{0},\ \ell<m(\ast)\\}.$

We make ${}_{\bf D}{\mathbb{M}}^{\ast}$ to a right $({\bf
D}\mathop{{\otimes}}{\bf S}_{0})$-module by defining $xz$ for $x\in{}_{{\bf
D}}{\mathbb{M}}^{*}$ and $z\in{\bf S}_{0}$. It is enough to deal with
$z\in\\{{\mathcal{X}}_{m}:m<m(*)\\}\cup\\{{\mathcal{Z}},{\mathcal{W}}\\}$. Let
$x=\sum\limits_{\eta,\ell}a_{\eta,\ell}x_{\eta,\ell}$ where $(\eta,\ell)$ vary
on $W_{0}$, $a_{\eta,\ell}\in{\bf D}$ and $\\{(\eta,\ell):a_{\eta,\ell}\neq
0\\}$ is finite. We define

$(\sum\limits_{\eta,\ell}a_{\eta,\ell}x_{\eta,\ell})z:=\sum\limits_{\eta,\ell}a_{\eta,\ell}(x_{\eta,\ell}z).$

Here, we define the action of
$\\{{\mathcal{X}}_{m},{\mathcal{Z}},{\mathcal{W}}\\}$ on $x_{\eta,\ell}$ as
follows:

$\begin{array}[]{rcl}x_{\eta,\ell}{\mathcal{X}}_{m}&:=&\left\\{\begin{array}[]{ll}x_{\eta,\ell}&\mbox{if
}\ell=m,\\\ 0&\mbox{if }\ell\neq m,\end{array}\right.\\\ \\\
x_{\eta,\ell}{\mathcal{Z}}&:=&\left\\{\begin{array}[]{ll}x_{\eta{}^{\frown}\\!\langle\ell\rangle,0}&\mbox{if
}\ell>0,\\\ x_{\eta\restriction[-\infty,n-1),\eta(n-1)}&\mbox{if
}\ell=0,\mbox{ and
}(-\infty,n)=\mathop{\mathrm{Dom}}(\eta),\end{array}\right.\\\ \\\ and\\\
x_{\eta,\ell}{\mathcal{W}}&:=&x_{\eta,m}\ \mbox{ when }m=\ell+1\ \mod\
m(\ast).\end{array}$

We get a $({\bf D},{\bf S})$-bimodule as the identities in the definition of
${\bf S}_{0}$ and ${\bf S}$ holds. If ${\bf D}={\bf T}$, some by inspection
(those of $(\ast)_{1}$), the rest by the choice of ${\bf S}$. If ${\bf
D}\neq{\bf T}$, by the restriction on ${\bf D}$. Let

${}_{\bf
D}{\mathbb{M}}^{\ast}_{\ell}:=\\{\sum\limits_{\eta}d_{\eta,\ell}x_{\eta,\ell}:\
\eta\in W_{0}\mbox{ and }d_{\eta,\ell}\in{\bf D}\\}.$

So clearly ${}_{\bf
D}{\mathbb{M}}^{\ast}=\mathop{{\bigoplus}}\limits_{\ell=0}^{m(\ast)-1}{}_{\bf
D}{\mathbb{M}}^{\ast}_{\ell}$.

Step B) In this step we introduce an ${\bf R}$-module ${\mathbb{M}}$ such that
${\mathbb{M}}^{m(\ast)-1}\cong{\mathbb{M}}$.

Let ${\mathbb{P}}$ be the $({\bf R},{\bf S})$-bimodule constructed in Lemma
4.7 for $\lambda$. Here, we used the existence of seminice construction and
its strong version.

We look at ${\mathbb{P}}_{\ell}:={\mathbb{P}}{\mathcal{X}}_{\ell}$. Let $i\neq
j$. We use the formula ${\mathcal{X}}_{i}{\mathcal{X}}_{j}=0$ to observe that
${\mathbb{P}}_{i}\cap{\mathbb{P}}_{j}=0$. Since
$\sum\limits^{m(\ast)-1}_{\ell=0}{\mathcal{X}}_{\ell}=1$ we have

${}_{{\bf
R}}{\mathbb{P}}=\sum\limits^{m(\ast)-1}_{\ell=0}{\mathbb{P}}{\mathcal{X}}_{\ell}=\mathop{{\bigoplus}}\limits^{m(\ast)-1}_{\ell=0}{}_{{\bf
R}}{\mathbb{P}}_{\ell}.$

Suppose $\ell+1\neq m\ \mod\ m(\ast)$. We combine the formula
${\mathcal{X}}_{\ell}{\mathcal{W}}{\mathcal{X}}_{m}=0$ with the formula
$\sum\limits^{m(\ast)-1}_{\ell=0}{\mathcal{X}}_{\ell}=1$ to observe that

${\mathbb{P}}_{\ell}{\mathcal{W}}={\mathbb{P}}{\mathcal{X}}_{\ell}{\mathcal{W}}({\mathcal{X}}_{0}+\ldots+{\mathcal{X}}_{m(\ast)})={\mathbb{P}}_{\ell}{\mathcal{W}}{\mathcal{X}}_{\ell+1}\subset{\mathbb{P}}_{\ell+1},$

i.e., ${\mathcal{W}}:{}_{\bf R}{\mathbb{P}}_{\ell}\twoheadrightarrow{}_{\bf
R}{\mathbb{P}}_{\ell+1}$ is surjective. Here, we use the relation
${\mathcal{W}}^{m(\ast)}=1$ to equip ${\mathcal{W}}$ as an embedding from
${\mathbb{P}}$ onto ${\mathbb{P}}$. Hence, ${\mathcal{W}}\restriction{}_{\bf
R}{\mathbb{P}}_{\ell}$ is an isomorphism from ${}_{\bf R}{\mathbb{P}}_{\ell}$
onto ${}_{\bf R}{\mathbb{P}}_{\ell+1}$. So

${}_{\bf R}{\mathbb{P}}_{m(\ast)-1}\cong\ldots\cong{}_{\bf
R}{\mathbb{P}}_{1}\cong{}_{\bf R}{\mathbb{P}}_{0}.$

Thanks to the formulas
${\mathcal{X}}_{0}{\mathcal{Z}}(1-X_{0})={\mathcal{X}}_{0}{\mathcal{Z}}$ and
$\sum\limits^{m(\ast)-1}_{\ell=0}{\mathcal{X}}_{\ell}=1$ we conclude that
${\mathcal{Z}}$ maps ${}_{\bf R}{\mathbb{P}}_{0}$ into

${\mathbb{P}}{\mathcal{X}}_{0}{\mathcal{Z}}={\mathbb{P}}{\mathcal{X}}_{0}{\mathcal{Z}}(1-X_{0})={\mathbb{P}}{\mathcal{X}}_{0}{\mathcal{Z}}({\mathcal{X}}_{1}+\ldots+{\mathcal{X}}_{m(\ast)-1})\subset\mathop{{\bigoplus}}\limits^{m(\ast)-1}_{\ell=1}{\mathbb{P}}{\mathcal{X}}_{\ell}=\mathop{{\bigoplus}}\limits^{m(\ast)-1}_{\ell=1}{\mathbb{P}}_{\ell}\cong{\mathbb{P}}_{0}.$

By the same vein, ${\mathcal{Z}}$ maps
$\mathop{{\bigoplus}}\limits^{m(\ast)-1}_{\ell=1}{\mathbb{P}}_{\ell}$ into
${}_{\bf R}{\mathbb{P}}_{0}$. We are going to use the formula
${\mathcal{Z}}^{2}=1$ to exemplifies ${\mathcal{Z}}$ with the following
isomorphism

$\mathop{{\bigoplus}}\limits^{m(\ast)-1}_{\ell=1}{}_{\bf
R}{\mathbb{P}}_{\ell}\cong{}_{\bf R}{\mathbb{P}}_{0}.$

But, we have just shown

$\mathop{{\bigoplus}}\limits^{m(\ast)-1}_{\ell=1}{}_{\bf
R}{\mathbb{P}}_{\ell}\cong({}_{\bf R}{\mathbb{P}}_{0})^{m(\ast)-1}.$

This completes the proof of Step B).

So, it is enough to show

$(*)_{2}$:

$1<k<m(\ast)-1\quad\Rightarrow\quad{}_{\bf
R}{\mathbb{P}}^{k}_{0}\not\cong{}_{{\bf R}}{\mathbb{P}}_{0}$.

Assume $k$ is a counterexample. The desired contradiction will be presented in
Step I), see below. To this end we need some preliminaries.

Step C) There exist a field ${\bf D}$ and ${\mathcal{Y}}\in{\bf
D}\mathop{{\otimes}}\limits_{\bf T}{\bf S}$ satisfying the following
equations:

$(*)_{3}$ :

${\mathcal{Y}}\restriction{}_{\bf D}{\mathbb{M}}^{\ast}_{0}$ is an isomorphism
from ${}_{\bf D}{\mathbb{M}}^{\ast}_{0}$ onto
$\mathop{{\bigoplus}}\limits_{\ell=1}^{k}{}_{\bf
D}{\mathbb{M}}^{\ast}_{\ell}$,

${\mathcal{Y}}\restriction\mathop{{\bigoplus}}\limits_{\ell=1}^{k}{}_{\bf
D}{\mathbb{M}}^{\ast}_{\ell}$ is an isomorphism from
$\mathop{{\bigoplus}}\limits_{\ell=1}^{k}{}_{\bf D}{\mathbb{M}}^{\ast}_{\ell}$
onto ${}_{\bf D}{\mathbb{M}}^{\ast}_{0}$,

${\mathcal{Y}}\restriction\mathop{{\bigoplus}}\limits^{m(\ast)}_{\ell=k+1}{}_{\bf
D}{\mathbb{M}}^{\ast}_{\ell}$ is the identity, and

${\mathcal{Y}}^{2}=1$.

Indeed, recall from ${}_{\bf R}{\mathbb{P}}^{k}_{0}\cong{}_{{\bf
R}}{\mathbb{P}}_{0}$ that ${}_{\bf R}{\mathbb{P}}$ is equipped with an
endomorphism ${\mathbf{f}}$ such that:

$(*)_{4}$ :

${\mathbf{f}}\restriction{}_{\bf R}{\mathbb{P}}_{0}$ is an isomorphism from
${}_{\bf R}{\mathbb{P}}_{0}$ onto ${}_{\bf
R}{\mathbb{P}}_{1}\oplus\ldots\oplus{}_{\bf R}{\mathbb{P}}_{k}$,

${\mathbf{f}}\restriction(_{\bf R}{\mathbb{P}}_{1}\oplus\ldots\oplus{}_{\bf
R}{\mathbb{P}}_{k})$ is an isomorphism from ${}_{\bf
R}{\mathbb{P}}_{1}\oplus\ldots\oplus{}_{\bf R}{\mathbb{P}}_{k}$ onto ${}_{\bf
R}{\mathbb{P}}_{0}$,

${\mathbf{f}}\restriction{}_{\bf R}{\mathbb{P}}_{j}$, for $k<j<m(\ast)$, is
the identity, and

${\mathbf{f}}^{2}=1$.

Assuming $({\bf S},+)$ is a free ${\bf T}$-module, according to Corollary
5.33, there is a field $\bf D$ with the property that
$p:=\operatorname{char}({\bf D})$ divides $n:=\operatorname{char}({\bf R})$,
when $n>0$, and there exists ${\mathcal{Y}}\in{\bf
D}\mathop{{\otimes}}\limits_{\bf T}{\bf S}$ satisfying the the desired
equations. This completes the proof of Step C).

In sum, we are left to show that $({\bf S},+)$ is a free ${\bf T}$-module.
Note that each of
$\\{1_{S},{\mathcal{X}}_{0},\ldots,{\mathcal{X}}_{m(\ast)},{\mathcal{W}},{\mathcal{Z}\\}}$
map each generator $x_{\eta,\ell}$ to another generator or zero, so this
applies to any composition of them, in fact, in quite specific way for $\rho$,
$\nu\in\\{1,\ldots,m(\ast)-1\\}^{<\omega}$, $k$, $m<\omega$. Let
${\mathcal{Y}}^{k,m}_{\rho,\nu}$ be a monomial operator in the generators of
${\bf S}$ such that:

$x_{\eta,\ell}{\mathcal{Y}}^{k,m}_{\rho,\nu}=\left\\{\begin{array}[]{ll}x_{\eta_{1}{}^{\frown}\\!\nu,m}&\mbox{
if }\ell=k,\mbox{ and }\eta=\eta_{1}{}^{\frown}\\!\rho\mbox{ for some
}\eta_{1}\in W_{0}\\\ 0&\mbox{ otherwise.}\end{array}\right.$

In order to define these operators, first we define
$Y^{k,k}_{\rho,\langle\rangle}$ and ${\mathcal{Y}}^{k,k}_{\langle\rangle,\nu}$
by induction on $\mathop{\rm\ell g}(\rho)$ and $\mathop{\rm\ell g}(\nu)$
respectively. Also, negative powers of ${\mathcal{W}}$ can be defined, because
${\mathcal{W}}$ is invertible. Now, we let

${\mathcal{Y}}^{k,m}_{\rho,\nu}:=Y^{k,k}_{\rho,\langle\rangle}{\mathcal{Y}}_{\langle\rangle,\nu}^{k,k}{\mathcal{W}}^{m-k}.$

Step D): Set $\Omega=\\{{\mathcal{Y}}^{k,m}_{\rho,\nu}:\
\rho,\nu\in[1,m(\ast))^{<\omega},\ k,m\in[0,m(\ast))$ and if $\mathop{\rm\ell
g}(\rho)>0$ and $\mathop{\rm\ell g}(\nu)>0$, then $\rho(0)=\nu(0)]\\}.$ Then

  1. $\boxtimes_{1}$

$\Omega$ generates ${\bf S}$ as a ${\bf T}$-module

First note that if ${\mathcal{Y}}^{k,m}_{\rho_{1},\nu_{1}}$ is not in the
family, then for some $(\varrho,\rho,\nu)$ we have $Y^{k,m}_{\rho,\nu}$
belongs to the family and
$(\rho,\nu)=(\varrho^{\frown}\rho_{1},\varrho{}^{\frown}\\!\nu_{1})$. Hence
${\mathcal{Y}}^{k,m}_{\rho,\nu}={\mathcal{Y}}^{k,m}_{\rho_{1},\nu_{1}}$. So,
$\Omega$ generates all of ${\mathcal{Y}}^{k,m}_{\rho,\nu}$’s and thus we can
use them.

Let ${\bf S}^{\prime}$ be the sub-${\bf T}$-module of ${\bf S}$ generated by
${\bf T}$ and the family above. Now ${\bf T}\subseteq{\bf S}^{\prime}$ because

  1. $(\dagger)_{1}$

$1=\sum\\{Y_{\langle\rangle,\langle\rangle}^{m,m}:m=0,\ldots,m(*)-1\\}.$

To see this, we evaluate both sides at $x_{\eta,\ell}$. The right hand side of
$(\dagger)_{1}$ is equal to

$\begin{array}[]{ll}x_{\eta,\ell}\sum\\{Y_{\langle\rangle,\langle\rangle}^{m,m}:m=0,\ldots,m(*)-1\\}&=x_{\eta^{\frown}\langle\rangle,\ell}\sum\\{Y_{\langle\rangle,\langle\rangle}^{m,m}:m=0,\ldots,m(*)-1\\}\\\
&=x_{\eta^{\frown}\langle\rangle,\ell}Y_{\langle\rangle,\langle\rangle}^{\ell,\ell}\\\
&=x_{\eta^{\frown}\langle\rangle,\ell}\\\ &=x_{\eta,\ell}.\end{array}$

Since the left hand side of $(\dagger)_{1}$ is the identity, we get the
desired equality.

Next we show ${\mathcal{X}}_{m}\in{\bf S}^{\prime}$ for each $m<m(*)$. It
suffices to show that

  1. $(\dagger)_{2}$

${\mathcal{X}}_{m}={\mathcal{Y}}^{m,m}_{\langle\rangle,\langle\rangle}.$

In order to see $(\dagger)_{2}$, we evaluate both sides at $x_{\eta,\ell}$.
Let $n=\mathop{\rm\ell g}(\eta)+1$. First, assume that $\ell=m$. Then, the
right hand side of $(\dagger)_{2}$ is equal to

$x_{\eta,\ell}{\mathcal{Y}}^{m,m}_{\langle\rangle,\langle\rangle}=x_{\eta,m}{\mathcal{Y}}^{m,m}_{\langle\rangle,\langle\rangle}=x_{\eta,m},$

which is equal to $x_{\eta,\ell}{\mathcal{X}}_{m}$. Now, we show the claim
when $\ell\neq m$. In this case both sides of $(\dagger)_{2}$ are equal to
zero, e.g., $(\dagger)_{2}$ is valid.

In order to show ${\mathcal{W}}\in{\bf S}^{\prime}$ we bring the following
claim:

  1. $(\dagger)_{3}$

${\mathcal{W}}=\sum\\{{\mathcal{Y}}^{\ell,m}_{\langle\rangle,\langle\rangle}:\ell=m+1\mod\
m(*)\\}.$

As before, it is enough to evaluate both sides of $(\dagger)_{3}$ at
$x_{\eta,\ell}$. Let $m$ be such that $\ell\equiv_{m(\ast)}m+1$. The right
hand side of $(\dagger)_{3}$ is equal to

$\begin{array}[]{ll}x_{\eta,\ell}\sum\\{{\mathcal{Y}}^{\ell,m}_{\langle\rangle,\langle\rangle}:\ell=m+1\mod\
m(*)\\}&=x_{\eta,\ell}{\mathcal{Y}}^{\ell,m}\\\ &=x_{\eta,m}.\end{array}$

By definition, this is equal to the left hand side of $(\dagger)_{3}$.

Finally, we claim that ${\mathcal{Z}}\in{\bf S}^{\prime}$. Indeed, it suffices
to prove that

  1. $(\dagger)_{4}$

${\mathcal{Z}}=\\{{\mathcal{Y}}^{0,m}_{\langle
m\rangle,\langle\rangle}:m=1,\ldots,m(*)-1\\}+\sum\\{{\mathcal{Y}}^{m,0}_{\langle\rangle,\langle
m\rangle}:m=1,\ldots,m(*)-1\\}.$

Again we evaluate both sides of $(\dagger)_{4}$ at $x_{\eta,\ell}$. Let
$n=\mathop{\rm\ell g}(\eta)+1$ and first suppose that $\ell=0.$ Then, the
right hand side of $(\dagger)_{4}$ is equal to

$\begin{array}[]{ll}x_{\eta,0}\\{{\mathcal{Y}}^{0,m}_{\langle
m\rangle,\langle\rangle}:m=1,\ldots,m(*)-1\\}&+x_{\eta,0}\sum\\{{\mathcal{Y}}^{m,0}_{\langle\rangle,\langle
m\rangle}:m=1,\ldots,m(*)-1\\}\\\ &=x_{\eta,0}\\{{\mathcal{Y}}^{0,m}_{\langle
m\rangle,\langle\rangle}:m=1,\ldots,m(*)-1\\}\\\
&=x_{\eta{\restriction}{n-1}{}^{\frown}\\!\langle\eta(n)\rangle,0}{\mathcal{Y}}^{0,\eta(n)}\\\
&=x_{\eta{\restriction}{n-1}{}^{\frown}\\!\langle\eta(n)\rangle,\eta(n)}.\end{array}$

This is equal to the left hand side of $(\dagger)_{4}$. Suppose now that
$\ell>0$. Then the right hand side is equal to
$x_{\eta{}^{\frown}\\!\langle\ell\rangle,0}$. By definition this is equal to
the left hand side of $(\dagger)_{4}$.

So far, we have proved that the subring ${\bf S}^{\prime}$ includes ${\bf
T},{\mathcal{X}}_{m}(m<m(*)),{\mathcal{Z}}$ and ${\mathcal{W}}$ and is a sub
${\bf T}$-module. To finish this step, it suffices to prove that ${\bf
S}^{\prime}$ is closed under product. For this, it is enough to prove that
${\bf S}^{\prime}$ includes the product
${\mathcal{Y}}^{k^{1},m^{1}}_{\rho_{1},\nu_{1}}{\mathcal{Y}}^{k^{2},m^{2}}_{\rho_{2},\nu_{2}}$
of any two members of $\Omega$.

Now if $m^{1}\neq k^{2}$ or no one of $\nu_{1},\rho_{2}$ is an end-segment of
the other then this is not the case. So easily the product is
$Y^{k^{1},m^{2}}_{\rho_{3},\nu_{3}}$ where:

  * •

if $\nu_{1}=\rho_{2}$, then $(\rho_{3},\nu_{3})=(\rho_{1},\nu_{2})$,

  * •

if $\nu_{1}=(\varrho^{1})^{\frown}\rho_{2}$, then
$(\rho_{3},\nu_{3})=(\rho_{1},(\varrho^{1})^{\frown}\nu_{2})$,

  * •

if $\rho_{2}=(\varrho^{1})^{\frown}\nu_{1}$, then
$(\rho_{3},\nu_{3})=((\varrho^{1})^{\frown}\nu_{1},\nu_{2})$.

So $\boxtimes_{1}$ holds, which completes the proof of Step D).

Step E) The set $\Omega$ from Step D) is a free basis of ${\bf S}$ as a ${\bf
T}$-module.

Assume that
${\mathcal{X}}=\sum\\{a^{k,m}_{\nu,\rho}{\mathcal{Y}}^{k,m}_{\nu,\rho}:\rho,k,\nu,m\\}=0$
for some finite set $\\{(\rho,\nu,k,m):a^{k,m}_{\nu,\rho}\neq 0\\}$. We have
to show that $a^{k,m}_{\nu,\rho}=0$, for all such $\rho,k,\nu,m$.

Let $k^{*}<\omega$ and $\rho^{*}\in\\{0,\ldots,m(*)-1\\}^{<\omega}$ be such
that $\mathop{\rm\ell g}(\rho^{*})>\mathop{\mathrm{sup}}\\{\mathop{\rm\ell
g}(\rho):a^{k,n}_{\rho,\nu}\neq 0\\}$. We look at
$x_{\rho^{*},k^{*}}{\mathcal{X}}$. Let $A(k^{*},\rho^{*})$ be the set of all
$\nu$ in the finite sequence above where
$\rho^{*}=\rho^{*}_{\nu}$${}^{\frown}\nu$, for some $\rho^{*}_{\nu}.$ Then

$\begin{array}[]{ll}x_{\rho^{*},k^{*}}{\mathcal{X}}&=\sum\limits_{\nu,\rho,k,m}a^{k,m}_{\nu,\rho}(x_{\rho^{*},k^{*}}{\mathcal{Y}}^{k,m}_{\nu,\rho})\\\
&=\sum\limits_{\nu,\rho,m}\\{a^{k^{*},m}_{\nu,\rho}(x_{\rho^{*},k^{*}}{\mathcal{Y}}^{k^{*},m}_{\nu,\rho}):\nu\in
A(k^{*},\rho^{*})\\}\\\
&=\sum\limits_{\nu,\rho,m}\\{a^{k^{*},m}_{\nu,\rho}x_{(\rho^{*}_{\nu})^{\frown}\rho,m}:\nu\in
A(k^{*},\rho^{*})\\}\\\ &=0.\end{array}$

Recall that ${\mathbb{M}}^{\ast}$ is the left $\bf D$-module freely generated
by $x_{\eta,\ell}$’s, so as we are free in choosing $k^{*},\rho^{*},$ we can
easily show that for any tuple $(\rho,\nu,k,n)$ in the finite set above, for
some suitable choice of $\rho^{*}$, $a^{k,n}_{\rho,\nu}$ is the only component
of $x_{(\rho^{*}_{\nu})^{\frown}\rho,m}$ in the sum above, and hence
$a^{k,n}_{\rho,\nu}=0$. This completes the proof of Step E).

Before we continue, let us introduce some notations and definitions. Clearly,
${\mathcal{Y}}$ can be represented as a finite sum of the form:

${\mathcal{Y}}=\sum\\{d^{k,m}_{\rho,\nu}{\mathcal{Y}}^{k,m}_{\rho,\nu}:k,m,\rho,\nu\\}.$

Choose $n(*)$ large enough such that
$n(*)>\mathop{\mathrm{max}}\\{\mathop{\rm\ell g}(\rho),\mathop{\rm\ell
g}(\nu):d^{k,m}_{\rho,\nu}\neq 0\\}$. For any
$u\subseteq\\{(\eta,\ell):\eta\in W_{0},\ \ell<m(\ast)\\}$, let
${\mathbb{N}}_{u}$ be the sub ${\bf D}$-module of ${}_{\bf
D}{\mathbb{M}}^{\ast}$ generated by $\\{x_{\eta,\ell}:(\eta,\ell)\in u\\}$.

For $\eta\in w_{0}$ and $\ell=1,\ldots,m(\ast)$ set ${}_{\bf
D}{\mathbb{M}}^{\ast}_{[1,k]}:=\mathop{{\bigoplus}}\limits^{k}_{j=1}{\mathbb{M}}^{\ast}_{\ell}$,
and let

$\begin{array}[]{ll}w_{\eta,\ell}&:=\\{(\nu,m)|(\nu,m)=(\eta,\ell)\mbox{ or
}\eta^{\frown}\langle\ell\rangle\trianglelefteq\nu\mbox{ and }m<m(*)\\},\\\
u_{\ell}&:=\\{(\eta,m)|m=\ell,\ \eta\in w_{0}\\},\\\ w^{m}_{\eta,\ell},&:=\
w_{\eta,\ell}\cap u_{m},\\\
w^{[1,n]}_{\eta,\ell}&:=w_{\eta,\ell}\cap\bigcup\limits_{m\in[1,n]}u_{m}.\end{array}$

Step F) For any $\eta\in w_{0}$ and $\ell<m(\ast)$, there is a finite subset
$u\subseteq w^{0}_{\eta,\ell}$ such that

  1. $(\alpha)$

${\mathcal{Y}}$ maps ${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus u}$ into
${\mathbb{N}}_{w_{\eta,\ell}}$ in fact into
${\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}$

  2. $(\beta)$

if $v$ is finite and $u\subseteq v\subseteq w^{0}_{\eta,\ell}$ then
${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus v}{\mathcal{Y}}$ is a ${\bf
D}$-vector subspace of ${\mathbb{N}}_{w^{0}_{\eta,\ell}}{\mathcal{Y}}$ of
cofinite dimension.

  3. $(\gamma)$

for any finite subsets $v_{1}$, $v_{2}$ of $w^{0}_{\eta,\ell}$ extending $u$
the following equalities are true:

$\begin{array}[]{l}\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}}/{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}})-\dim({\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}/({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}{\mathcal{Y}}))\\\
=\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}}/{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}})-\dim({\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}/({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}}{\mathcal{Y}})),\end{array}$

where the dimensions are computed as ${\bf D}$-vector spaces.

Indeed, the case $\ell=0$ is easy. Recalling the representation of
${\mathcal{Y}}$ and the choice of $n(\ast)<\omega$, if we set

$u=\\{(\nu,m)\in w_{\eta,\ell}:\
|\mathop{\mathrm{Dom}}(\nu)\setminus\mathop{\mathrm{Dom}}(\eta)|<n(\ast)\\},$

then we have

$(\nu,m)\in w_{\eta,\ell}\setminus u\quad\Rightarrow\quad
x_{\nu,m}{\mathcal{Y}}\in{\mathbb{N}}_{w_{\eta,\ell}}.$

Note that $u$ is finite and clause $(\alpha)$ holds. As
${\mathcal{Y}}{\mathcal{Y}}=1$ etc., we can show that $(\beta)$ holds. For
checking the equalities in clause $(\gamma)$, let $v_{3}:=v_{1}\cup v_{2}$.
Due to the transitivity of equality, it is enough to prove the required
equality for pairs $(v_{1},v_{3})$ and $(v_{2},v_{3})$. This enable us to
assume without loss of generality that $v_{1}\subseteq v_{2}$. Also, we
stipulate $v_{0}=\emptyset$.

Now

${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}}\subseteq{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}\subseteq{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
w_{0}}={\mathbb{N}}_{w^{0}_{\eta,\ell}},$

and

$\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}}/{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{i}})=|v_{i}|.$

Manipulating the equalities in clause $(\alpha)$, they are equivalent to

  1. (*)

$\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}}/{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}})-\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}}/{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}})=$

$\dim({\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}/({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}{\mathcal{Y}}))-\dim({\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}/({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}}{\mathcal{Y}})).$

Since ${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}}\subseteq{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}\subseteq{\mathbb{N}}_{w^{0}_{\eta,\ell}}$, the left hand side of $(*)$
is equal to

$\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}/{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus v_{2}}).$

Also, ${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}}{\mathcal{Y}}\subseteq{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}{\mathcal{Y}}\subseteq{\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}$ (the first
inclusion holds as ${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}}\subseteq{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus v_{1}}$ and the second
one holds by clause $(\beta)$ and $u\subseteq v_{1}\cap v_{2}$). Hence the
right hand side of $(\ast)$ is equal to

$\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}{\mathcal{Y}}/{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}}{\mathcal{Y}}).$

Hence $(\ast)$ is reduced in proving

  1. $(\ast)^{\prime}$

$\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}/{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{2}})=\dim({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}/({\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus v_{2}}{\mathcal{Y}}))$.

Since ${\mathcal{Y}}$ is an isomorphism from
${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus v_{2}}$ onto
${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus v_{2}}{\mathcal{Y}}$, it implies the
validity of $(\ast)^{\prime}$. This completes the proof.

Let ${\bf n}_{\eta,\ell}$ be the natural number so that for every large enough
finite subset $v\subseteq w_{\eta,\ell}$

${\bf
n}_{\eta,\ell}:=\dim\left(\frac{{\mathbb{N}}_{w^{0}_{\eta,\ell}}}{{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v}}\right)-\dim\left(\frac{{\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}}{{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v}{\mathcal{Y}}}\right).$

In view of Step F), this is a well-defined notion, which does not depend on
the choice of $v$.

Step G) If $\eta$, $\nu\in w_{0}$ and $\ell\in\\{0,1,\ldots,m(\ast)-1\\}$,
then ${\bf n}_{\eta,\ell}={\bf n}_{\nu,\ell}$.

To see this, we define a function $f:w_{\eta,\ell}\longrightarrow
w_{\nu,\ell}$ by $f(\eta{}^{\frown}\\!\rho,m)=(\nu{}^{\frown}\\!\rho,m)$ for
any $\rho\in\\{0,\ldots,m(*)-1\\}^{<\omega}$. This function is one-to-one and
onto, and it induces an isomorphism from
${\mathbb{N}}_{w^{[0,m(\ast))}_{\eta,\ell}}$ onto
${\mathbb{N}}_{w_{\nu,\ell}^{[0,m(\ast))}}$. It almost commutes with all of
our operations (the problems are “near” $\eta$ and $\nu$). So choose
$v_{1}\subseteq w^{0}_{\eta,\ell}$ large enough, as required in the definition
of ${\bf n}_{\eta,\ell}$, ${\bf n}_{\nu,\ell}$ and to make
${\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v_{1}}{\mathcal{Y}}\subseteq{\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}$, and let
$v_{2}=f(v_{1})$. Now check the desired claim.

So, we shall write ${\bf n}_{\ell}$ instead of ${\bf n}_{\nu,\ell}$. By Step
G), this is well-defined.

Step H) The following equations are valid:

${\bf n}_{\ell}=\left\\{\begin{array}[]{ll}0&\mbox{if }\ell\in[1,k]\\\ {\bf
n}_{0}+{\bf n}_{1}+\ldots+{\bf n}_{m(\ast)-1}&\mbox{if
}\ell\in[k+1,m(\ast))\emph{ or }\ell=0\end{array}\right.$

To prove this, first note that
$w_{\eta,\ell}=\\{(\eta,\ell)\\}\cup\bigcup\limits_{m<m(*)}w_{\eta{}^{\frown}\\!\langle\ell\rangle,m}$.
In order to apply Step F), we fix a pair $(\eta,\ell)$ and we choose a finite
subset $v\subseteq w^{0}_{\eta,\ell}$ large enough as there. Let
$(\eta{}^{\frown}\\!\langle\ell\rangle,m)$ be such that

$\\{(\eta,\ell)\\}\cup\\{(\eta{}^{\frown}\\!\langle\ell\rangle,m):\
m<m(\ast)\\}\subseteq v.$

Now we compute ${\bf n}_{\eta,\ell}$ which is equal to ${\bf n}_{\ell}$, and
${\bf n}_{\eta{}^{\frown}\\!\langle\ell\rangle,m}$ which is equal to $={\bf
n}_{m}$ for $m<m(*)$ and we shall get the equality.

Now set

${\bf
n}^{1}_{v,\eta,\ell}:=\dim\left(\frac{{\mathbb{N}}_{w^{0}_{\eta,\ell}}}{{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v}}\right)-\sum\limits_{m<m(*)}\dim\left(\frac{{\mathbb{N}}_{w^{0}_{\eta{}^{\frown}\\!\langle\ell\rangle,m}}}{{\mathbb{N}}_{w^{0}_{\eta{}^{\frown}\\!\langle\ell\rangle,m}\setminus
v}}\right).$

By definition, we have
${\mathbb{N}}_{w^{0}_{\eta,\ell}}=\bigoplus\limits_{m<m(*)}{\mathbb{N}}_{w^{0}_{\eta{}^{\frown}\\!\langle\ell\rangle,m}}\oplus{\bf
D}x_{\eta,\ell}$. Using this equality, we can easily see that

  1. $(\otimes_{1})$

${\bf n}^{1}_{v,\eta,\ell}$ is equal to $1$ when $\eta\in w_{0},\ell<m(\ast)$
and $v\subseteq w_{0}$ is finite large enough (we can replace $v$ by $v\cap
w^{0}_{\eta,\ell}$ of course).

In particular, the following definition makes sense:

  1. $(\otimes_{2})$

${\bf
n}^{2}_{v,\eta,\ell}:=\dim\left(\frac{{\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}}{{\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}\setminus
v}}\right)-\sum\limits_{m<m(*)}\dim\left(\frac{{\mathbb{N}}_{w^{[1,k]}_{\eta{}^{\frown}\\!\langle\ell\rangle,m}}}{{\mathbb{N}}_{w^{[1,k]}_{\eta{}^{\frown}\\!\langle\ell\rangle,m}\setminus
v}}\right).$

Suppose $\eta\in w_{0},\ell<m(\ast)$ and $v\subseteq w_{0}$ is finite and
large enough. In view of definition, and as the argument of $(\otimes_{1})$ we
present the following three implications:

$(\otimes_{2.1})\quad\ell=0\ \Rightarrow\ {\bf n}^{2}_{v,\eta,\ell}=0$,

$(\otimes_{2.2})\quad\ell\in[1,k]\ \Rightarrow\ {\bf n}^{2}_{v,\eta,\ell}=1$,

$(\otimes_{2.3})\quad\ell\in[k+1,m(*))\Rightarrow\ {\bf
n}^{2}_{v,\eta,\ell}=0$.

Next, we show that

  1. $(\otimes_{3})$

${\bf n}^{1}_{v,\eta,\ell}-{\bf n}^{2}_{v,\eta,\ell}={\bf
n}_{\ell}-\sum\limits_{m<m(*)}{\bf n}_{m}$.

Indeed, we have

$\begin{array}[]{ll}{\bf n}^{1}_{v,\eta,\ell}-{\bf
n}^{2}_{\nu,\eta,\ell}&={\rm
dim}(\frac{{\mathbb{N}}_{w^{0}_{\eta,\ell}}}{{\mathbb{N}}_{w^{0}_{\eta,\ell}\setminus
v}})-{\rm
dim}(\frac{{\mathbb{N}}_{w^{[1,k]}_{\eta,\ell}}}{w^{[1,k]}_{\eta,\ell}\setminus
v})\\\ &-\sum\limits_{m<n(*)}{\rm
dim}(\frac{{\mathbb{N}}_{w^{0}_{\eta{}^{\frown}\\!\langle\ell\rangle,m}}}{{\mathbb{N}}_{w^{0}_{\eta^{{}^{\frown}\\!}{\langle\ell\rangle,m}}\smallsetminus
v}})\\\ &+\sum\limits_{m<n(*)}{\rm
dim}(\frac{{\mathbb{N}}_{w^{0}_{\eta^{{}^{\frown}\\!}{\langle\ell\rangle,m}}}}{{\mathbb{N}}_{w^{[1,k]}_{\eta^{{}^{\frown}\\!}{\langle\ell\rangle,m}}\smallsetminus
v}})\\\ &={\bf n}_{\eta,\ell}-\sum\limits_{\ell<m(*)}{\bf
n}_{\eta^{{}^{\frown}\\!}{\langle\ell\rangle,m}}\\\ &={\bf
n}_{\ell}-\sum\limits_{m<m(*)}{\bf n}_{m}.\end{array}$

Now we combine $(\otimes_{1})$, $(\otimes_{2.1})$ and $(\otimes_{2.2})$ along
with $(\otimes_{2.3})$ to deduce the following formula.

  1. $(\otimes_{4})$

${\bf n}^{1}_{v,\eta,\ell}-{\bf
n}^{2}_{v,\eta,\ell}=\left\\{\begin{array}[]{ll}0&\mbox{if }\ell\in[1,k]\\\
1&\mbox{otherwise }\end{array}\right.$

Step H) follows from $(\otimes_{3})+(\otimes_{4})$.

Step I) In this step, we drive our desired contradiction. To see this, recall
from Step H) that

$\begin{array}[]{ll}\sum\limits^{m(\ast)-1}_{\ell=0}{\bf n}_{\ell}&=({\bf
n}_{0}+\ldots+{\bf n}_{m(*)-1})\\\ &+\sum\limits^{k}_{\ell=1}({\bf
n}_{0}+\ldots+{\bf n}_{m(\ast)-1}-1)+\sum\limits^{m(\ast)-1}_{\ell=k+1}({\bf
n}_{0}+\ldots+{\bf n}_{m(\ast)-1})\\\
&=m(\ast)\sum\limits^{m(\ast)-1}_{\ell=0}{\bf n}_{\ell}-k.\end{array}$

So,

$-(m(\ast)-1)\left(\sum^{m(\ast)-1}_{\ell=1}{\bf n}_{\ell}\right)=-k,$

i.e.,

$\sum^{m(\ast)-1}_{\ell=1}{\bf n}_{\ell}=\frac{k}{m(\ast)-1}.$

Now, the left hand side is an integer, while as $1<k<m(\ast)-1$, the right
hand side is not an integer, a contradiction.

This completes the proof. ∎

An additional outcome is a slight improvement of the above pathological
property.

######  Corollary 6.5.

Assume $\lambda=(\mu^{\aleph_{0}})^{+}>|{\bf R}|$ and let $0<m_{1}<m_{2}-1$.
Then there is an ${\bf R}$-module ${\mathbb{M}}$ of cardinality $\lambda$ such
that:

${\mathbb{M}}^{n^{1}}\cong{\mathbb{M}}^{n^{2}}\mbox{ if and only if
}m_{1}<n^{1}\ \&\ m_{1}\leq n^{2}\ \&\ [(m_{2}-m_{1})|(n^{1}-n^{2})]$

( $|$ means divides).

###### Proof.

Let $m(\ast):=m_{2}-m_{1}$ and let ${\bf S}_{1}$ be the ring constructed in
the proof of Theorem 6.4 with respect to $m(\ast)$. Let also
${\mathbb{M}}^{\ast}$ be the $({\bf R},{\bf S}_{1})$-bimodule constructed
there. In particular, it is equipped with a structure of bimodule over $({\bf
R},{\bf S}_{1})$. Choose cardinals

$\lambda_{m_{1}}>\lambda_{m_{1}-1}>\ldots>\lambda_{0}>\|{\mathbb{M}}^{*}\|+\|{\bf
R}\|+\|{\bf S}_{1}\|+\aleph_{0}.$

Let $I:=\bigcup\limits_{m\leq m_{1}}\prod\limits_{\ell=0}^{m-1}\lambda_{\ell}$
and define a function $f:I\longrightarrow I$ by the following rules:

$f(\eta):=\left\\{\begin{array}[]{ll}\eta\restriction k&\mbox{ if }\eta\in
I,\mbox{ and }\mathop{\rm\ell g}(\eta)=k+1\\\ \eta&\mbox{ if
}\eta=\langle\rangle.\end{array}\right.$

Let ${\mathbb{M}}^{\otimes}$ be a bimodule which is
$\mathop{\oplus}\limits_{\eta\in I}{\mathbb{M}}^{\otimes}_{\eta}$ with
${\mathbb{M}}^{\otimes}_{\eta}\cong{\mathbb{M}}^{\ast}$ for each $\eta\in I$
and let
$h_{\eta}:{\mathbb{M}}^{\ast}\longrightarrow{\mathbb{M}}^{\otimes}_{\eta}$ be
such an isomorphism.

For every endomorphism ${\mathcal{Y}}$ of ${\mathbb{M}}^{\ast}$ we define an
endomorphism ${\mathcal{Y}}^{\otimes}\in{\rm End}_{{\bf
R}}({\mathbb{M}}^{\otimes})$ of ${\mathbb{M}}^{\otimes}$ as an ${\bf
R}$-module as follows. We are going to define the action of
${\mathcal{Y}}^{\otimes}$ on each ${\mathbb{M}}^{\otimes}_{\eta}$ ($\eta\in
I$). To this end, we take $x\in{\mathbb{M}}^{\otimes}_{\eta}$ and define the
action via:

$x{\mathcal{Y}}^{\otimes}:=h_{f(\eta)}\left((h^{-1}_{\eta}(x){\mathcal{Y}})\right).$

Let ${\bf S}^{\otimes}$ be the subring of the ring of endomorphisms of
${\mathbb{M}}^{\otimes}$ as an ${\bf R}$-module generated by

$\\{1^{\otimes},{\mathcal{X}}^{\otimes}_{0},{\mathcal{X}}^{\otimes}_{1},\ldots,{\mathcal{X}}^{\otimes}_{m(\ast)},{\mathcal{W}}^{\otimes},{\mathcal{Z}}^{\otimes}\\}.$

Note also that $1^{\otimes}$ is not a unit in ${\bf S}^{\otimes}$. Now we
continue as in the proof of Theorem 6.2, with ${\mathbb{M}}^{\otimes}$ here
instead of ${\mathbb{M}}$ there, and ${\bf S}^{\otimes}$ here instead of ${\bf
S}$ there. We leave the details to the reader. ∎

######  Theorem 6.6.

Let $\lambda=(\mu^{\aleph_{0}})^{+}>|{\bf R}|$ be a regular cardinal. Then
there are ${\bf R}$-modules ${\mathbb{M}}_{1}$ and ${\mathbb{M}}_{2}$ of
cardinality $\lambda$ such that:

  1. (1)

${\mathbb{M}}_{1}$, ${\mathbb{M}}_{2}$ are not isomorphic,

  2. (2)

${\mathbb{M}}_{1}$ is isomorphic to a direct summand of ${\mathbb{M}}_{2}$,

  3. (3)

${\mathbb{M}}_{2}$ is isomorphic to a direct summand of ${\mathbb{M}}_{1}$.

###### Proof.

Let ${\bf T}$ be the subring of ${\bf R}$ which $1$ generates. As before, we
need to choose a ring ${\bf S}$ (essentially the ring of endomorphisms we
would like).

Step A): To make things easier, we first introduce a ring ${\bf S}_{0}$ which
is easy compared to ${\bf S}$.

Let $A_{1}$ (resp. $A_{-1})$ be the set of even (resp. odd) integers and let
$f$ be the following function:

$f(i):=\left\\{\begin{array}[]{ll}i+1&\mbox{if }i\geq 0,\\\ i-1&\mbox{if
}i<0.\end{array}\right.$

Thus, $f$ maps $A_{1}$ (resp. $A_{-1}$) into $A_{-1}$ (resp. $A_{1}$). Also,
$A_{1}\setminus\mathop{\mathrm{Rang}}(f\restriction A_{-1})=\\{0\\}$ and
$A_{-1}\setminus\mathop{\mathrm{Rang}}(f\restriction A_{1})=\\{-1\\}$. Let $i$
vary on the integers. Let ${\bf S}_{0}$ be the ring extending ${\bf T}$
generated freely by
$\\{{\mathcal{X}}_{1},{\mathcal{X}}_{-1},{\mathcal{W}}_{1},{\mathcal{W}}_{-1},{\mathcal{Z}}_{1},{\mathcal{Z}}_{-1}\\}$.
Let ${\bf D}$ be a field such that if $\|{\bf T}\|$ is finite, then the
characteristic of ${\bf D}$ is finite and divides $\|{\bf T}\|$. Then, we set
${\bf S}_{\bf D}^{*}={\bf D}\mathop{{\otimes}}\limits_{\bf T}{\bf S}_{0}$ (see
Theorem 6.4).

We are going to define a right $({\bf D}\mathop{{\otimes}}\limits_{\bf T}{\bf
S}_{0})$– module ${\mathbb{M}}^{\ast}_{{\bf D}^{\ast}}$. We first define a
left ${\bf D}$-module structure over ${\mathbb{M}}:=\sum\\{{\bf
D}x_{i}:i\in{\mathbb{Z}}\\}$ with
$b(\sum\limits_{i}a_{i}x_{i})=\sum\limits_{i}(ba_{i})x_{i}$ where $a_{i}$ and
$b$ are in ${\bf D}$. As mentioned, we like to make ${\mathbb{M}}$ a right
$({\bf D}\mathop{{\otimes}}\limits_{\bf T}{\bf S}_{0})$–module which will be
called ${\mathbb{M}}^{*}_{\bf D}$. Let $x\in{\mathbb{M}}$ and $c\in{\bf
D}\mathop{{\otimes}}\limits_{\bf T}{\bf S}_{0}$. To define $xc$, as ${\bf D}$
and ${\bf S}_{0}$ commute, it is enough to define it for $x=x_{i}$ and
$c\in\\{{\mathcal{X}}_{1},{\mathcal{X}}_{-1},{\mathcal{W}}_{1},{\mathcal{W}}_{-1},{\mathcal{Z}}_{1},{\mathcal{Z}}_{-1}\\}$.
So let

$x_{i}{\mathcal{X}}_{1}:=\left\\{\begin{array}[]{ll}x_{i}&\mbox{if }i\in
A_{1},\\\ 0&\mbox{if }i\in A_{-1},\end{array}\right.$
$x_{i}{\mathcal{X}}_{-1}:=\left\\{\begin{array}[]{ll}0&\mbox{if }i\in
A_{1},\\\ x_{i}&\mbox{if }i\in A_{-1},\end{array}\right.$
$x_{i}{\mathcal{W}}_{1}:=x_{f(i)},$
$x_{i}{\mathcal{W}}_{-1}:=\left\\{\begin{array}[]{ll}x_{f^{-1}(i)}&\mbox{if
}i\in{\mathop{\mathrm{Rang}}(f)},\\\ 0&\mbox{otherwise,}\end{array}\right.$
$x_{i}{\mathcal{Z}}_{1}:=\left\\{\begin{array}[]{ll}x_{i}&\mbox{if }i\in
A_{1}\cap({\mathbb{Z}}\setminus\\{0\\}),\\\
0&\mbox{otherwise,}\end{array}\right.$

and

$x_{i}{\mathcal{Z}}_{-1}:=\left\\{\begin{array}[]{ll}x_{i}&\mbox{if }i\in
A_{-1}\cap({\mathbb{Z}}\setminus\\{0\\}),\\\
0&\mbox{otherwise.}\end{array}\right.$

These make ${\mathbb{M}}^{\ast}_{\bf D}$ a right ${\bf S}^{*}_{\bf D}$–module.
Let $g^{\ast}_{{\bf D}}$ be the natural ring homomorphism from ${\bf S}_{0}$
into ${\bf S}^{*}_{\bf D}$. By using $g^{\ast}_{{\bf D}}$, the ${\bf
S}^{*}_{\bf D}$–module ${\mathbb{M}}^{\ast}_{\bf D}$ becomes a right ${\bf
S}_{0}$-module.

Step B): In this step we define the auxiliary ring ${\bf S}$.

Let ${\bf S}$ be the ring (with 1, associative but not necessarily
commutative) extending ${\bf T}$ generated by ${\mathcal{X}}_{1}$,
${\mathcal{X}}_{-1}$, ${\mathcal{W}}_{1}$, ${\mathcal{W}}_{-1}$,
${\mathcal{Z}}_{1}$, ${\mathcal{Z}}_{-1}$ freely except the following
equations (to understand them see below):

  1. $(\ast)$

$\sigma=0$ if $\sigma$ is a term212121i.e. in the language of rings, in the
variables
${\mathcal{X}}_{1},{\mathcal{X}}_{-1},{\mathcal{W}}_{1},{\mathcal{W}}_{-1},{\mathcal{Z}}_{1},{\mathcal{Z}}_{-1}$,
${\mathbb{M}}^{\ast}_{\bf D}\sigma=0$ for ${\mathbb{M}}^{\ast}_{\bf D}$ as
defined in Step A), for every field ${\bf D}$ such that if ${\bf T}$ is of
finite cardinality $n$, then $\operatorname{char}({\bf D})|n$.

Let us state some of the relations, which follow from $(\ast)$, and we will
use them more in the rest of our argument.

$(\star)_{1}$:

${\mathcal{X}}^{2}_{1}={\mathcal{X}}_{1}$,
${\mathcal{X}}^{2}_{-1}={\mathcal{X}}_{-1}$,
${\mathcal{X}}_{1}+{\mathcal{X}}_{-1}=1$,
${\mathcal{X}}_{1}{\mathcal{X}}_{-1}={\mathcal{X}}_{-1}{\mathcal{X}}_{1}=0$
${\mathcal{Z}}^{2}_{1}={\mathcal{Z}}_{1}$,
${\mathcal{Z}}_{1}{\mathcal{X}}_{1}={\mathcal{Z}}_{1}={\mathcal{X}}_{1}{\mathcal{Z}}_{1}$
${\mathcal{X}}_{-1}{\mathcal{W}}_{1}=({\mathcal{X}}_{-1}{\mathcal{W}}_{1}){\mathcal{X}}_{1}{\mathcal{Z}}_{1}$
${\mathcal{X}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}=({\mathcal{X}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}){\mathcal{X}}_{-1}$
${\mathcal{X}}_{-1}{\mathcal{W}}_{1}{\mathcal{W}}_{-1}={\mathcal{X}}_{-1}$ and
${\mathcal{X}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}{\mathcal{W}}_{1}={\mathcal{Z}}_{1}={\mathcal{X}}_{1}{\mathcal{Z}}_{1}$.

Let us show, for example, that ${\mathcal{X}}_{1}+{\mathcal{X}}_{-1}=1$. We
need to show the left hand side is the identity map when we consider it as an
endomorphism of ${\mathbb{M}}^{\ast}_{\bf D}$. To this end, we evaluate
${\mathcal{X}}_{1}+{\mathcal{X}}_{-1}$ at any generator of
${\mathbb{M}}^{\ast}_{\bf D}$, say $x_{i}$. Recall that

$x_{i}{\mathcal{X}}_{1}=\left\\{\begin{array}[]{ll}x_{i}&\mbox{if }i\in
A_{1},\\\ 0&\mbox{if }i\in A_{-1},\end{array}\right.$
$x_{i}{\mathcal{X}}_{-1}=\left\\{\begin{array}[]{ll}0&\mbox{if }i\in A_{1},\\\
x_{i}&\mbox{if }i\in A_{-1},\end{array}\right.$

Thus, $x_{i}({\mathcal{X}}_{1}+{\mathcal{X}}_{-1})=x_{i}$, as claimed. The
other relations will follow in the same way.

Step C): Here, we introduce the ${\bf R}$-modules ${\mathbb{M}}_{1}$ and
${\mathbb{M}}_{2}$ of cardinality $\lambda$ such that:

  1. (i)

${\mathbb{M}}_{1}$ is isomorphic to a direct summand of ${\mathbb{M}}_{2}$ and

  2. (ii)

${\mathbb{M}}_{2}$ is isomorphic to a direct summand of ${\mathbb{M}}_{1}$.

Let $\langle{\mathbb{M}}_{\alpha}:\alpha\leq\lambda\rangle$ be a strongly
semi-nice construction for $(\lambda,{{\mathfrak{m}}},S,\lambda)$ and let
${\mathbb{M}}:={\mathbb{M}}_{\lambda}$. Recall that such a thing exists. Let
${\mathbb{M}}_{1}:={\mathbb{M}}{\mathcal{X}}_{1}$ and
${\mathbb{M}}_{-1}:={\mathbb{M}}{\mathcal{X}}_{-1}$. Since
${\mathcal{X}}_{1}{\mathcal{X}}_{-1}={\mathcal{X}}_{-1}{\mathcal{X}}_{1}=0$,
we have ${\mathbb{M}}_{1}\cap{\mathbb{M}}_{-1}=0$. Thanks to the formula
${\mathcal{X}}_{1}+{\mathcal{X}}_{-1}=1$,

${\mathbb{M}}={\mathbb{M}}({\mathcal{X}}_{1}+{\mathcal{X}}_{-1})={\mathbb{M}}{\mathcal{X}}_{1}+{\mathbb{M}}{\mathcal{X}}_{-1}={\mathbb{M}}_{1}+{\mathbb{M}}_{-1}={\mathbb{M}}_{1}\oplus{\mathbb{M}}_{-1}.$

Then ${\mathbb{M}}_{1}$ and ${\mathbb{M}}_{-1}$ are equipped with ${\bf
R}$-module structure, and there is the identification
${\mathbb{M}}={\mathbb{M}}_{1}\oplus{\mathbb{M}}_{-1}$. We shall show that
${\mathbb{M}}_{1}$, ${\mathbb{M}}_{-1}$ are as required in Theorem 6.6 (with
respect to ${\mathbb{M}}_{1}$ and ${\mathbb{M}}_{2}$).

The relations ${\mathcal{Z}}^{2}_{1}={\mathcal{Z}}_{1}$ and
${\mathcal{Z}}_{1}{\mathcal{X}}_{1}={\mathcal{Z}}_{1}={\mathcal{X}}_{1}{\mathcal{Z}}_{1}$
imply that

${\mathbb{M}}_{1}={\mathbb{M}}_{1}(1-{\mathcal{Z}}_{1})\oplus{\mathbb{M}}_{1}{\mathcal{Z}}_{1},$

i.e., ${\mathbb{M}}_{1}{\mathcal{Z}}_{1}$ is a direct summand of
${\mathbb{M}}_{1}$. Since
${\mathcal{X}}_{-1}{\mathcal{W}}_{1}=({\mathcal{X}}_{-1}{\mathcal{W}}_{1}){\mathcal{X}}_{1}{\mathcal{Z}}_{1}$,
we have

${\mathbb{M}}_{-1}{\mathcal{W}}_{1}={\mathbb{M}}{\mathcal{X}}_{-1}{\mathcal{W}}_{1}={\mathbb{M}}({\mathcal{X}}_{-1}{\mathcal{W}}_{1}){\mathcal{X}}_{1}{\mathcal{Z}}_{1}\subset{\mathbb{M}}{\mathcal{X}}_{1}{\mathcal{Z}}_{1}={\mathbb{M}}_{1}{\mathcal{Z}}_{1}.$

Thus, ${\mathcal{W}}_{1}$ maps ${\mathbb{M}}_{-1}$ into
${\mathbb{M}}_{1}{\mathcal{Z}}_{1}$. By using the formula
${\mathcal{X}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}=({\mathcal{X}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}){\mathcal{X}}_{-1}$
we observe that

${\mathbb{M}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}={\mathbb{M}}{\mathcal{X}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}={\mathbb{M}}{\mathcal{X}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}{\mathcal{X}}_{-1}\subset{\mathbb{M}}{\mathcal{X}}_{-1}={\mathbb{M}}_{-1}.$

In other words, ${\mathcal{W}}_{-1}$ maps ${\mathbb{M}}_{1}{\mathcal{Z}}_{1}$
into ${\mathbb{M}}_{-1}$. We are going to combine the formula
${\mathcal{X}}_{-1}{\mathcal{W}}_{1}{\mathcal{W}}_{-1}={\mathcal{X}}_{-1}$
along with
${\mathcal{X}}_{1}{\mathcal{Z}}_{1}{\mathcal{W}}_{-1}{\mathcal{W}}_{1}={\mathcal{Z}}_{1}={\mathcal{X}}_{1}{\mathcal{Z}}_{1}$
to deduce that the multiplication maps by ${\mathcal{W}}_{1}$ and
${\mathcal{W}}_{-1}$ are the inverse of each other. This implies that
${\mathbb{M}}_{-1}$ is isomorphic to a direct summand of ${\mathbb{M}}_{1}$
(as left ${\bf R}$-modules). Similarly, we obtain:

${\mathbb{M}}_{-1}={\mathbb{M}}_{-1}(1-{\mathcal{Z}}_{-1})\oplus{\mathbb{M}}_{-1}{\mathcal{Z}}_{-1}.$

So, ${\mathbb{M}}_{-1}{\mathcal{Z}}_{-1}$ is a direct summand of
${\mathbb{M}}_{-1}$ and ${\mathbb{M}}_{-1}{\mathcal{Z}}_{-1}$ is isomorphic to
${\mathbb{M}}_{1}$. By the same argument, ${\mathbb{M}}_{1}$ is isomorphic to
a direct summand of ${\mathbb{M}}_{2}$ as left ${\bf R}$-modules. In summary,
we showed that

  1. $\bullet$

${\mathbb{M}}_{1}\cong{\mathbb{M}}_{1}(1-{\mathcal{Z}}_{1})\oplus{\mathbb{M}}_{-1}$
and

  2. $\bullet$

${\mathbb{M}}_{-1}\cong{\mathbb{M}}_{-1}(1-{\mathcal{Z}}_{-1})\oplus{\mathbb{M}}_{1}.$

This completes the proof of Step C).

It remains to show that ${\mathbb{M}}_{1}\not\cong{\mathbb{M}}_{-1}$. Suppose
on the contrary that they are isomorphic, and we get a contradiction, which is
presented at Step H) below. Let
$\mathfrak{e}\in{\mathfrak{E}}^{{\mathfrak{m}}}$ be simple.

Step D): There is a solution ${\mathcal{Y}}\in{{dE}}^{\mathfrak{e}}_{n}$ to
the following equations:

$(\ast)_{2}$:

${\mathcal{X}}_{1}{\mathcal{Y}}{\mathcal{X}}_{-1}={\mathcal{X}}_{1}{\mathcal{Y}}$,
${\mathcal{X}}_{-1}{\mathcal{Y}}{\mathcal{X}}_{1}={\mathcal{X}}_{-1}{\mathcal{Y}}$,
${\mathcal{Y}}{\mathcal{Y}}=1$.

To see this, recall that ${\mathbb{M}}_{1}\cong{\mathbb{M}}_{-1}$ and
${\mathbb{M}}={\mathbb{M}}_{1}\oplus{\mathbb{M}}_{-1}$. Let $h$ be an
isomorphism from ${\mathbb{M}}_{1}$ onto ${\mathbb{M}}_{-1}$. Define
${\mathbf{f}}:{\mathbb{M}}\to{\mathbb{M}}$ by

$(a,b)\in{\mathbb{M}}={\mathbb{M}}_{1}\oplus{\mathbb{M}}_{-1}\implies{\mathbf{f}}(a,b)=(h^{-1}(b),h(a)).$

So, ${\mathbf{f}}\in{\rm End}_{\bf R}({\mathbb{M}})$ and it satisfies
${\mathbf{f}}\restriction{\mathbb{M}}_{1}=h$ and
${\mathbf{f}}\restriction{\mathbb{M}}_{-1}=h^{-1}$. The member in
${{dE}}^{\mathfrak{e}}_{n}$ which ${\mathbf{f}}$ induces solves the equations
in $(\ast)_{2}$. This completes the proof of Step D).

In the light of Corollary 5.33, it is enough to prove the following two items:

  1. (a)

in ${\bf D}\mathop{{\otimes}}\limits_{\bf T}{\bf S}_{0}$ there is no solution
to $(\ast)_{2}$, in particular, there is no such ${\mathcal{Y}}$. Note that
${\bf S}^{*}_{{\bf D}}$ have the same characteristic as ${\bf D}$.

  2. (b)

${\bf S}_{0}$ is a free ${\bf T}$–module.

Clearly ${\bf S}$ is a ${\bf T}$-module, generated by the set of monomials in

$\\{{\mathcal{X}}_{1},{\mathcal{X}}_{-1},{\mathcal{W}}_{1},{\mathcal{W}}_{-1},{\mathcal{Z}}_{1},{\mathcal{Z}}_{-1}\\}.$

Our aim is to show that ${\bf S}$ is a free ${\bf T}$-module; in fact we shall
exhibit explicitly a free basis. For $\ell\in\\{1,-1\\}$, $k\in{\mathbb{Z}}$,
$n\geq 0$, $n\geq-k$, we define an endomorphism ${\mathbf{f}}^{\ell}_{k,n}$ of
${\mathbb{M}}^{\ast}_{\bf D}$ by

${\mathbf{f}}^{\ell}_{k,n}(x_{i}):=\left\\{\begin{array}[]{ll}x_{f^{k}(i)}&\mbox{
if }f^{-n}(i)\mbox{ is well defined and }x_{i}\in A_{\ell},\\\
0&\mbox{{otherwise}}\end{array}\right.$

It is easy to see that it is an endomorphism of ${\mathbb{M}}^{\ast}_{\bf D}$
as a left ${\bf D}$-module. We will define a monomial
${\mathcal{Y}}^{\ell}_{k,n}$, in the following way. For every monomial
$\sigma$ let $\sigma^{0}$ be $1=\mathop{\mathrm{id}}_{{\mathbb{M}}^{\ast}_{\bf
D}}$ and remember $n\geq-k$ so $n+k\geq 0$. Now set

${\mathcal{Y}}^{\ell}_{k,n}={\mathcal{X}}_{\ell}({\mathcal{W}}_{-1})^{n}{\mathcal{W}}^{n+k}_{1}.$

It is easy to see that the operation of ${\mathcal{Y}}^{\ell}_{k,n}$ on
${\mathbb{M}}^{\ast}_{\bf D}$ by right multiplication, is equal to
${\mathbf{f}}^{\ell}_{k,n}$.

Let

$\mathcal{G}:=\\{{\mathcal{Y}}^{\ell}_{k,n}:(\ell,k,n)\in w\\}$

where $w=\\{(\ell,k,n):\ \ell\in\\{1,-1\\},\ k\in{\mathbb{Z}},\ n\geq 0,\
k+n\geq 0\\}.$

In the next step, we show that $\mathcal{G}$ generates ${\bf S}$ as a ${\bf
T}$-module.

Step E): The set $\mathcal{G}$ generates ${\bf S}$ as a ${\bf T}$-module.

It is enough to show that for every monomial $\sigma$, some equation
$\sigma=\sum a^{\ell}_{n,k}{\mathcal{Y}}^{\ell}_{k,n}$ holds in ${\bf S}$,
where $\\{(\ell,n,k):a^{\ell}_{n,k}\neq 0\\}$ is finite and
$a^{\ell}_{k,n}\in{\bf T}$, i.e., it holds in the ring of endomorphism of
${\mathbb{M}}^{\ast}_{\bf D}$. We prove this by induction on the length of the
monomial $\sigma$.

If the length is zero, $\sigma$ is 1. Recall from $(\star)_{1}$ that
$1={\mathcal{X}}_{1}+{\mathcal{X}}_{-1}$. By definition,
${\mathcal{X}}_{\ell}={\mathcal{Y}}^{\ell}_{0,0}$; so
$1={\mathcal{Y}}^{1}_{0,0}+{\mathcal{Y}}^{-1}_{0,0}$ as required.

If the length is $>0$, by the induction hypothesis it is enough to prove the
following:

  1. $(\ast)$

Let
$\tau\in\\{{\mathcal{X}}_{1},{\mathcal{X}}_{-1},{\mathcal{W}}_{1},{\mathcal{W}}_{-1},{\mathcal{Z}}_{1},{\mathcal{Z}}_{-1}\\}$.
Then ${\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}\tau$ is equal to some
$\sum\limits_{\ell,k,n}a^{\ell}_{k,n}{\mathcal{Y}}^{\ell}_{k,n}$.

Indeed, it is enough to check equality on the generators of
${\mathbb{M}}^{\ast}_{\bf D}$, that is the $x_{i}$’s. The proof of $(\ast)$ is
divided into three cases:

Case 1: ${\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{W}}_{\ell}$ is:

$\begin{array}[]{rcl}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast)+1,n(\ast)}&\mbox{ if
}&\ell=1\\\ {\mathcal{Y}}^{\ell(\ast)}_{k(\ast)-1,n(\ast)}&\mbox{ if
}&\ell=-1\mbox{ and }\ k(\ast)+n(\ast)>0,\\\
{\mathcal{Y}}^{\ell(\ast)}_{k(\ast)-1,n(\ast)+1}&\mbox{ if }&\ell=-1\mbox{ and
}\ k(\ast)+n(\ast)=0.\end{array}$

First assume that $\ell=1$. Then

$\begin{array}[]{ll}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{W}}_{\ell}&={\mathcal{X}}_{1}({\mathcal{W}}_{-1})^{n}{\mathcal{W}}^{n+k}_{1}{\mathcal{W}}_{1}\\\
&={\mathcal{X}}_{1}({\mathcal{W}}_{-1})^{n}{\mathcal{W}}^{n+k+1}\\\
&={\mathcal{Y}}^{\ell(\ast)}_{k(\ast)+1,n(\ast)}.\end{array}$

Now assume that $\ell=-1$ and $k(\ast)+n(\ast)>0$. If
$i\in\mathop{\mathrm{Rang}}(f)$, then
$x_{i}{\mathcal{W}}_{1}{\mathcal{W}}_{-1}=x_{i}$. From this,

$x_{i}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{W}}_{\ell}=x_{i}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast)-1,n(\ast)}.$

If $i\notin\mathop{\mathrm{Rang}}(f)$, then $i\in\\{0,1\\}$. It is easy to see
that

$x_{0}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{W}}_{-1}=x_{0}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast)-1,n(\ast)}$

and

$x_{-1}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{W}}_{-1}=x{-1}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast)-1,n(\ast)}.$

Thus, the functions
${\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{W}}_{-1}$ and
${\mathcal{Y}}^{\ell(\ast)}_{k(\ast)-1,n(\ast)}$ are the same. Finally, assume
that $\ell=-1$ and $k(\ast)+n(\ast)=0$. Then

$\begin{array}[]{ll}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{W}}_{-1}&={\mathcal{X}}_{-1}({\mathcal{W}}_{-1})^{n}{\mathcal{W}}^{n+k}_{1}{\mathcal{W}}_{-1}\\\
&={\mathcal{X}}_{-1}({\mathcal{W}}_{-1})^{n}{\mathcal{W}}_{-1}\\\
&={\mathcal{X}}_{-1}({\mathcal{W}}_{-1})^{n+1}\\\
&={\mathcal{Y}}^{\ell(\ast)}_{k(\ast)-1,n(\ast)+1}.\end{array}$

This completes the argument.

Case 2: ${\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{X}}_{\ell}$ is:

$\mbox{zero }\quad\mbox{ if }\quad[\ell(\ast)=\ell\iff k(\ast)\mbox{ odd}],$
${\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}\mbox{if
}\quad[\ell(\ast)=\ell\iff k(\ast)\mbox{ even}].$

The proof of this is similar to Case 1.

Case 3: ${\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}{\mathcal{Z}}_{\ell}$ is:

$\begin{array}[]{rcl}{\mathcal{Y}}^{\ell(\ast)}_{k(\ast),n(\ast)}&\mbox{ if
}&n(\ast)+k(\ast)>0\mbox{ and }[\ell(\ast)=\ell\iff k(\ast)\mbox{ even}],\\\
Y^{\ell(\ast)}_{k(\ast),n(\ast)+1}&\mbox{ if }&n(\ast)+k(\ast)=0\mbox{ and
}[\ell(\ast)=\ell\iff k(\ast)\mbox{ even}],\\\ \mbox{zero}&\mbox{ if
}&[\ell(\ast)=\ell\iff k(\ast)\mbox{ odd}].\end{array}$

It is enough to check equality on the generators of ${\mathbb{M}}^{\ast}_{{\bf
D}}$, that is the $x_{i}$’s. The proof is again similar to the proof of Case
1.

Step F): The set $\mathcal{G}=\\{{\mathcal{Y}}^{\ell}_{k,n}:(\ell,k,n)\in
w\\}$ generates ${\bf S}$ freely as a ${\bf T}$-module.

In the light of Step E) we see that $\mathcal{G}$ generates ${\bf S}$ as a
${\bf T}$-module. Towards contradiction suppose that
$0=\sum\\{a^{\ell}_{k,n}{\mathcal{Y}}^{\ell}_{k,n}:(\ell,k,n)\in w\\}$ in
${\bf S}$ where $w\subseteq w^{\ast}$ is finite, $a^{\ell}_{k,n}\in{\bf T}$
and not all of them are zero. If $n=|{\bf T}|$ is finite, we take the field
$\bf D$ such that $\operatorname{char}({\bf D})|n$, and some $a^{\ell}_{k,n}$
is not zero in ${\bf D}$. Hence, $0=\sum\\{a^{\ell}_{k,n}:(\ell,k,n)\in w\\}$
as an endomorphism of ${\mathbb{M}}^{\ast}_{\bf D}$, a left ${\bf D}$-module
where $a^{\ell}_{k,n}\in{\bf D}$ and $w\subseteq w^{\ast}$ is finite. We shall
prove that $a^{\ell}_{k,n}=0$ for every $(\ell,k,n)\in w$.

If $i\in A_{1}$ and $i\geq 0$, then

$\begin{array}[]{rlc}0&=x_{i}(\sum\limits_{(\ell,k,n)\in
w}a^{\ell}_{k,n}{\mathcal{Y}}^{\ell}_{k,n})\\\ &=\sum\limits_{(\ell,k,n)\in
w}a^{\ell}_{k,n}(x_{i}{\mathcal{Y}}^{\ell}_{k,n})\\\
&=\sum\limits_{(\ell,k,n)\in w}\\{a^{\ell}_{k,n}x_{i+k}:\ell=1,\ \mbox{ and
}n\leq i\\}\\\ &=\sum\limits_{j\geq 0}(\sum\limits_{(1,k,n)\in
w}\\{a^{1}_{k,n}:\ \ i\geq n,\ i+k=j\\})x_{j}\\\ &=\sum\limits_{j\geq
0}(\sum\limits_{(1,j-i,n)\in w}\\{a^{1}_{j-i,n}:\ i\geq n,\
\\})x_{j}.\end{array}$

Hence for every $i\in A_{1}$, $i\geq 0$ and $j\geq 0$ we have

$(\ast)^{a}_{i,j}$:

$0=\sum\\{a^{1}_{j-i,n}:\ n\geq 0,\ n\leq i\mbox{ and }n+(j-i)\geq 0\\}$.

Similarly, for $i\in A_{-1}$, $i\geq 0$ (equivalently, $i>0$ as $i\in A_{-1}\
\Rightarrow\ i\neq 0$) and $j\geq 0$ we can prove

$(\ast)^{b}_{i,j}$:

$0=\sum\\{a^{-1}_{j-i,n}:\ n\geq 0,\ n\leq i\mbox{ and }n+(j-i)\geq 0\\}$.

Similarly, for $i\in A_{1}$, $i<0$,

$\begin{array}[]{rlc}0&=x_{i}(\sum\limits_{(\ell,k,n)\in
w}a^{\ell}_{k,n}{\mathcal{Y}}^{\ell}_{k,n})\\\ &=\sum\limits_{(\ell,k,n)\in
w}a^{\ell}_{k,n}(x_{i}{\mathcal{Y}}^{\ell}_{k,n})\\\ &=\sum\limits_{(1,k,n)\in
w}\\{a^{1}_{k,n}x_{i-k}:\mbox{ }-i>n\\}\\\
&=\sum\limits_{j<0}(\sum_{(1,i-j,n)\in w}\\{a^{1}_{i-j,n}:\mbox{
}n<-i\\})x_{j}.\end{array}$

Hence for every $i\in A_{1}$, $i<0$ and $j<0$ we have

$(\ast)^{c}_{i,j}$:

$0=\sum\\{a^{1}_{i-j,n}:n\geq 0$ and $n+(i-j)\geq 0$ and $n<-i\\}$.

Similarly, for every $i\in A_{-1}$, $i<0$ and $j<0$

$(\ast)^{d}_{i,j}$:

$0=\sum\\{a^{-1}_{i-j,n}:\ n\geq 0\mbox{ and }n+(i-j)\geq 0\mbox{ and
}n<-i\\}$.

Choose, if possible, $(k,m)$ such that:

  1. (1)

$(1,k,m)$ belongs to $w$,

  2. (2)

$a^{1}_{k,m}\neq 0$,

  3. (3)

$m$ is minimal under (1)+(2).

Note that $m\geq 0$ by the definition of $w$. First assume that $m$ is even.
Let $i=m$ and $j=i+k$. So $i\in A_{1}$ (being even), $i\geq 0$ and $j=m+k$ is
$\geq 0$ as $(1,k,m)\in w$. In the equation $(\ast)^{a}_{i,j}$ the term
$a^{1}_{k,m}$ appears in the sum, and for every other term
$a^{1}_{k_{1},m_{1}}$ which appears in the sum, we have $m_{1}<m$ (and
$k_{1}=k$), and hence by (3) above is zero. So it follows that $a^{1}_{k,m}$
is zero, a contradiction.

If $m$ is odd, we get a similar contradiction using $(\ast)^{c}_{i,j}$. Let
$i=-m-1$ and $j=i-k$. Note that $m\geq 0$, hence $i<0$ and $i$ is even as $m$
is odd, so $i\in A_{1}$. Also $j=i-k=-m-1-k\leq-1<0$ (recalling $k+m\geq 0$ as
$(1,k,m)\in w^{*}$). In the equation $(\ast)^{c}_{i,j}$, the term
$a^{1}_{i-j,n}=a^{1}_{k,n}$ appears in the sum if and only if $0\leq
n<-i=m+1$, and $n+(i-j)=n+k\geq 0$ (but if the later fails, $a^{1}_{k,m}$ is
not defined) so $a^{1}_{k,m}$ appears, and if another term
$a^{1}_{k_{1},m_{1}}$ occurs then $m_{1}\leq m$ (and $k_{1}=k$). Hence
$m_{1}<m$ and so $a^{1}_{k_{1},m_{1}}=0$. Necessarily $a^{1}_{k,m}$ is zero, a
contradiction.

So $a^{1}_{k,n}=0$ whenever it is defined.

Similarly $a^{-1}_{k,n}=0$ whenever it is defined (use
$(\ast)^{b}_{i,j}+(\ast)^{d}_{i,j}$). So, ${\bf S}_{0}$ is a free module over
${\bf T}$, as required.

Step G): In this step we get a contradiction that we searched for it. This
will show that ${\mathbb{M}}_{1}\not\cong{\mathbb{M}}_{-1}$.

To this end, recall that there are finitely many nonzero
$a^{\ell}_{k,n}\in{\bf D}$ such that

$\boxtimes$:

${\mathcal{Y}}=\sum\left\\{a^{\ell}_{k,n}{\mathcal{Y}}^{\ell}_{k,n}:\ n\geq
0\mbox{ and }k+n\geq 0\mbox{ and }\ell\in\\{1,-1\\}\right\\}.$

Let $n(\ast)<\omega$ be such that

$a^{\ell}_{k,n}\neq 0\quad\Rightarrow\quad|k|,n<n(\ast).$

For $\ell=1,-1$ let

$\begin{array}[]{lr}{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{\ell}:=\big{\\{}\sum\limits_{i\geq
0}d_{i}x_{i}:&d_{i}\in{\bf D},\mbox{ all but finitely many of
}d_{i}^{\prime}s\mbox{ are zero }\\\ &\mbox{ and }d_{i}\neq 0\ \Rightarrow\
i\in A_{\ell}\big{\\}},\\\ {\mathbb{M}}^{\rm
neg}_{\ell}:=\big{\\{}\sum\limits_{i<0}d_{i}x_{i}:&d_{i}\in{\bf D},\mbox{ all
but finitely many of }d_{i}^{\prime}s\mbox{ are zero }\\\ &\mbox{ and
}d_{i}\neq 0\ \Rightarrow\ i\in A_{\ell}\big{\\}}.\end{array}$

Clearly as a ${\bf D}$-module

${\mathbb{M}}^{\ast}_{\bf
D}={\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}\oplus{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}\oplus{\mathbb{M}}^{\rm
neg}_{1}\oplus{\mathbb{M}}^{\rm neg}_{-1}.$

Let
${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{\ell}:={\mathcal{Y}}\restriction{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{\ell}$
and ${\mathcal{Y}}^{\rm
neg}_{\ell}:={\mathcal{Y}}\restriction{\mathbb{M}}^{\rm neg}_{\ell}$ for
$\ell\in\\{1,-1\\}$.

Now each ${\mathcal{Y}}^{\ell}_{k,n}$ maps
${\mathbb{M}}^{\mathord{\mathrm{pos}}}={\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}\oplus{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$
to itself, and ${\mathbb{M}}^{\rm neg}={\mathbb{M}}^{\rm
neg}_{1}\oplus{\mathbb{M}}^{\rm neg}_{-1}$ to itself, and hence by $\boxtimes$
above also ${\mathcal{Y}}$ does it. According to $(\ast)_{2}$ from Step B) we
have
${\mathcal{X}}_{1}{\mathcal{Y}}{\mathcal{X}}_{-1}={\mathcal{X}}_{1}{\mathcal{Y}}$.
Hence

${\mathbb{M}}_{1}{\mathcal{Y}}={\mathbb{M}}{\mathcal{X}}_{1}{\mathcal{Y}}={\mathbb{M}}{\mathcal{X}}_{1}{\mathcal{Y}}{\mathcal{X}}_{-1}\subset{\mathbb{M}}{\mathcal{X}}_{-1}={\mathbb{M}}_{-1},$

i.e., ${\mathcal{Y}}$ maps ${\mathbb{M}}_{1}$ to ${\mathbb{M}}_{-1}$. Thus, by
the previous sentence, ${\mathcal{Y}}$ maps
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}$ into
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$, and ${\mathbb{M}}^{\rm neg}_{1}$
into ${\mathbb{M}}^{\rm neg}_{-1}$, i.e.,
${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}$ (resp. ${\mathcal{Y}}^{\rm
neg}_{1}$) is into ${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$ (resp.
${\mathbb{M}}^{\rm neg}_{-1}$).

Similarly by $(\ast)_{2}$ we have
${\mathcal{X}}_{-1}{\mathcal{Y}}{\mathcal{X}}_{1}={\mathcal{X}}_{-1}{\mathcal{Y}}$,
hence ${\mathcal{Y}}$ maps ${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$ into
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}$ and ${\mathbb{M}}^{\rm neg}_{-1}$
into ${\mathbb{M}}^{\rm neg}_{1}$. Also, the mappings
${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}$,
${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{-1}$, ${\mathcal{Y}}^{\rm neg}_{1}$,
${\mathcal{Y}}^{\rm neg}_{-1}$ are endomorphisms of ${\bf D}$-modules. As
${\mathcal{Y}}^{2}=1$ (again by $(\ast)_{2}$) we conclude that
${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}$ and
${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{-1}$ are the inverse of each other,
so both of them are isomorphisms. Similarly for ${\mathcal{Y}}^{\rm neg}_{1}$
and ${\mathcal{Y}}^{\rm neg}_{-1}$.

Let

$\begin{array}[]{lr}{\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}:=\\{\sum\limits_{i>0}d_{i}x_{i}:&d_{i}\in{\bf
D},\mbox{ all but finitely many $d_{i}$'s are zero }\\\ &\mbox{and }d_{i}\neq
0\ \Rightarrow\ i\in A_{1}\\}.\end{array}$

Clearly ${\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}$ is a sub ${\bf D}$-module
of ${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}$. Note that
$x_{0}\in{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}\setminus{\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}$.
This yields the difference between ${\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}$
and ${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}$.

Let

$\begin{array}[]{lr}{\mathbb{N}}:=\\{\sum\limits_{i>n(\ast)}d_{i}x_{i}:&d_{i}\in{\bf
D},\mbox{ all but finitely many of }d_{i}^{\prime}s\mbox{ are zero }\\\
&\mbox{and }d_{i}\neq 0\ \Rightarrow\ i\in A_{1}\\}.\end{array}$

Let
$H^{\mathord{\mathrm{pos}}}:{\mathbb{M}}_{1}^{\mathord{\mathrm{stp}}}\longrightarrow{\mathbb{M}}_{1}^{\rm
neg}$ be defined by $x_{i}H^{\mathord{\mathrm{pos}}}=x_{-i}$ and $H^{\rm
neg}:{\mathbb{M}}^{\rm
neg}_{-1}\longrightarrow{\mathbb{M}}^{\mathord{\mathrm{stp}}}_{-1}$ be defined
by $x_{i}H^{\rm neg}=x_{-i}$. Both of them are isomorphisms of ${\bf
D}$-modules. Note that ${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}$ is an
isomorphism from ${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}$ onto
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$ and
$H^{\mathord{\mathrm{pos}}}{\mathcal{Y}}^{\rm neg}_{1}H^{\rm neg}$ is an
isomorphism from ${\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}$ onto
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$. Note that

${\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}\stackrel{{\scriptstyle
H^{\mathord{\mathrm{pos}}}}}{{\longrightarrow}}{\mathbb{M}}^{\rm
neg}_{1}\stackrel{{\scriptstyle{\mathcal{Y}}^{\rm
neg}_{1}}}{{\longrightarrow}}{\mathbb{M}}^{\rm
neg}_{-1}\stackrel{{\scriptstyle H^{\rm
neg}}}{{\longrightarrow}}{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}.$

We show that

${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}\restriction{\mathbb{N}}=(H^{\mathord{\mathrm{pos}}}{\mathcal{Y}}^{\rm
neg}_{1}H^{\rm neg})\restriction{\mathbb{N}}.$

To see this, it is enough to check equality on the generators of
${\mathbb{N}}$, that is over the $x_{i}$’s where $i$ is even and it is larger
than $n(\ast)$. In particular, by choosing $n(\ast)$ large enough, we may
assume that $i\gg 0$. Recall that

${\mathcal{Y}}=\sum\left\\{a^{\ell}_{k,n}{\mathcal{Y}}^{\ell}_{k,n}:\ n\geq
0\mbox{ and }k+n\geq 0\mbox{ and }\ell\in\\{1,-1\\}\right\\}.$

By definition $x_{i}{\mathcal{X}}_{-1}=0$. Then

$\begin{array}[]{ll}x_{i}(H^{\mathord{\mathrm{pos}}}{\mathcal{Y}}^{\rm
neg}_{1}H^{\rm
neg})&=x_{-i}(\sum\\{a^{\ell}_{k,n}{\mathcal{Y}}^{\ell}_{k,n}\\})H^{\rm
neg}\\\ &=\sum
a^{\ell}_{k,n}x_{-i}({\mathcal{X}}_{1}{\mathcal{W}}_{-1}^{n}{\mathcal{W}}^{n+k}_{1})H^{\rm
neg}\\\ &=\sum
a^{1}_{k,n}(x_{-i}{\mathcal{W}}_{-1}^{n}{\mathcal{W}}^{n+k}_{1})H^{\rm neg}\\\
&=\sum a^{1}_{k,n}(x_{f^{-n}(-i)}{\mathcal{W}}^{n+k}_{1})H^{\rm neg}\\\ &=\sum
a^{1}_{k,n}(x_{f^{n+k}(f^{-n}(-i))})H^{\rm neg}\\\ &=\sum
a^{1}_{k,n}(x_{f^{k}(-i)})H^{\rm neg}\\\ &=\sum a^{1}_{k,n}(x_{-i-k)})H^{\rm
neg}\\\ &=\sum a^{1}_{k,n}x_{i+k}.\end{array}$

On the other hand,
$(x_{i}){\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}\restriction{\mathbb{N}}=\sum
a^{1}_{k,n}x_{i+k}$. Thus we have
${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}\restriction{\mathbb{N}}=(H^{\mathord{\mathrm{pos}}}{\mathcal{Y}}^{\rm
neg}_{1}H^{\rm neg})\restriction{\mathbb{N}}$, as claimed.

Let
${\mathbb{N}}^{\ast}:=\mathop{\mathrm{Rang}}({\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}\restriction{\mathbb{N}})$.
Then
${\mathbb{N}}^{\ast}=\mathop{\mathrm{Rang}}((H^{\mathord{\mathrm{pos}}}{\mathcal{Y}}^{\rm
neg}_{1}H^{\rm neg})\restriction{\mathbb{N}})$. So, as
${\mathcal{Y}}^{\mathord{\mathrm{pos}}}_{1}$ is an isomorphism from
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}$ onto
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$, and
${\mathbb{N}}\subseteq{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}$ we have that
${\mathbb{N}}^{\ast}$ is a sub $\bf D$-module of
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$. This means that
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}/{\mathbb{N}}^{\ast}$ is isomorphic
to ${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}/{\mathbb{N}}$ (as $\bf
D$-modules).

But $H^{\mathord{\mathrm{pos}}}{\mathcal{Y}}^{\rm neg}_{1}H^{\rm neg}$ is an
isomorphism from ${\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}$ onto
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}$ and
${\mathbb{N}}\subseteq{\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}$, and it maps
${\mathbb{N}}$ onto ${\mathbb{N}}^{\ast}$ (see above), so
${\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}/{\mathbb{N}}$ is isomorphic to
${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}/{\mathbb{N}}^{\ast}$. By the
previous paragraph we get

${\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}/{\mathbb{N}}\cong{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{-1}/{\mathbb{N}}^{\ast}\cong{\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}/{\mathbb{N}}.$

On the one hand, ${\mathbb{M}}^{\mathord{\mathrm{pos}}}_{1}/{\mathbb{N}}$ is a
free $\bf D$-module, as $\\{x_{2i}+{\mathbb{N}}:0\leq 2i\leq n(\ast)\\}$ is a
free basis for it and also
${\mathbb{M}}^{\mathord{\mathrm{stp}}}_{1}/{\mathbb{N}}$ is a free $\bf
D$-module with the base $\\{x_{2i}+{\mathbb{N}}:0<2i\leq n(\ast)\\}$. On the
other hand, the number of their generators differ by $1$. This is a
contradiction that we searched for it.

The theorem follows. ∎

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