Gorenstein-projective and semi-Gorenstein-projective modules. II

TL;DR

This paper investigates the structure of 3-dimensional local modules over a specific algebra, revealing a family of semi-Gorenstein-projective modules that are not torsionless, expanding understanding of module properties in algebra.

Contribution

It provides a detailed classification of 3-dimensional local modules over algebra A(q), highlighting new semi-Gorenstein-projective modules not torsionless, especially when q has infinite order.

Findings

Existence of a family of semi-Gorenstein-projective modules not torsionless

Characterization of modules over algebra A(q) with infinite order q

Extension of previous results from Part I to a broader class of modules

Abstract

Let k be a field and q a non-zero element of k. In Part I, we have exhibited a 6-dimensional k-algebra A = A(q) and we have shown that if q has infinite multiplicative order, then A has a 3-dimensional local module which is semi-Gorenstein-projective, but not torsionless, thus not Gorenstein-projective. This Part II is devoted to a detailed study of all the 3-dimensional local A-modules for this particular algebra A. If q has infinite multiplicative order, we will encounter a whole family of 3-dimensional local modules which are semi-Gorenstein-projective, but not torsionless.

Equations

$$x^2,y^2,z^2,yz,xy+qyx,xz-zx,zy-zx.$$

$$U(a,b,c)=U(a:b:c)=\Lambda (a,b,c)+\mathrm{soc}\Lambda$$

$$M(a,b,c)=M(a:b:c)={}_{\Lambda }\Lambda /U(a,b,c).$$

$$\bullet$$

$$H$$

$$\displaystyle\bigl{(}a\,x+qb\,y-\tfrac{a}{a+b}c\,z\bigr{)}\bigl{(}a\,x+b\,y+c\,z\bigr{)}$$

$$z(ax-ay+cz)$$

$$\displaystyle\bigl{(}a\,x+b\,y+c\,z\bigr{)}\bigl{(}a\,x+q^{-1}b\,y-\tfrac{a+q^{-1}b}{a}c\,z\bigr{)}$$

$$(by+cz)z$$

$$\displaystyle\bigl{(}a\,x+$$

$$=abxy+acxz+qabyx-\frac{a}{a+b}aczx-\frac{a}{a+b}bczy$$

$$\displaystyle=\ ab\,(xy+q\,yx)+\bigl{(}1-\tfrac{a}{a+b}-\tfrac{b}{a+b}\bigr{)}ac\,zx\ =\ 0.$$

$$\Omega M(a,b,c)\simeq \{\begin{array}{cccccc}M(\omega (a,b,c))& & \text{if}& a\ne 0,a+b\ne 0,& & (1)\\ M(0,0,1)& & \text{if}& a\ne 0,a+b=0,c\ne 0,& & (2)\\ \Lambda (x-y)\oplus \Lambda zx& & \text{if}& a\ne 0,a+b=0,c=0,& & (3)\\ M(0,1,0)& & \text{if}& a=0,b\ne 0,& & (4)\\ \Lambda z\oplus \Lambda yx& & \text{if}& a=0,b=0.& & (5)\end{array}$$

$$\bullet$$

$$■$$

$$0\to M(1,b^{\prime },c^{\prime })\to {}_{\Lambda }\Lambda \to M(1,b,c)\to 0$$

$$\bullet$$

$$0\to {\mathrm{Hom}}_{\overline{\Lambda }}(I,J)\to {\mathrm{Hom}}_{\overline{\Lambda }}(\overline{\Lambda },J)\to {\mathrm{Hom}}_{\overline{\Lambda }}(k^2,J)\to {\mathrm{Ext}}_{\overline{\Lambda }}^1(I,J)\to 0.$$

$$\Omega M^{\prime }(a,b,c)\simeq \{\begin{array}{cccccc}{M}^{\prime }({\omega }^{\prime }(a,b,c))& & \text{if}& a\ne 0,& & (1)\\ M^{\prime }(0,0,1)& & \text{if}& a=0,bc\ne 0,& & (2)\\ y\Lambda \oplus zx\Lambda & & \text{if}& a=0,c=0,& & (3)\\ z\Lambda \oplus yx\Lambda & & \text{if}& a=0,b=0.& & (4)\end{array}$$

$$■$$

$$0\to M^{\prime }(1,b,c)\to {\Lambda }_{\Lambda }\to M^{\prime }(1,b^{\prime },c^{\prime })\to 0$$

$$\bullet$$

$$M^{\prime }(1,b,0)=(x+qby)\Lambda =m_{-qb}\Lambda$$

$$\bullet$$

$$■$$

$$M(1,b,c{)}^{\ast }=M^{\prime }(({\omega }^{\prime }{)}^2(1,b,c)).$$

$$M^{\prime }(1,b,c{)}^{\ast }=M({\omega }^2(1,b,c)).$$

$$\begin{array}{c}0\to M(1,b,c)\to {}_{\Lambda }\Lambda \to M({\omega }^{\prime }(1,b,c))\to 0,\\ 0\to M({\omega }^{\prime }(1,b,c))\to {}_{\Lambda }\Lambda \to M(({\omega }^{\prime }{)}^2(1,b,c))\to 0\end{array}$$

Full text

Gorenstein-projective and

semi-Gorenstein-projective modules. II

Claus Michael Ringel, Pu Zhang

Abstract: Let $\scriptstyle k$ be a field and $\scriptstyle q$ a non-zero element of $\scriptstyle k$. In Part I, we have exhibited a 6-dimensional $\scriptstyle k$-algebra $\scriptstyle\Lambda=\Lambda(q)$ and we have shown that if $\scriptstyle q$ has infinite multiplicative order, then $\scriptstyle\Lambda$ has a 3-dimensional local module which is semi-Gorenstein-projective, but not torsionless, thus not Gorenstein-projective. This Part II is devoted to a detailed study of all the 3-dimensional local $\scriptstyle\Lambda$-modules for this particular algebra $\scriptstyle\Lambda$. If $q$ has infinite multiplicative order, we will encounter a whole family of 3-dimensional local modules which are semi-Gorenstein-projective, but not torsionless.

Key words and phrases. Gorenstein-projective module, semi-Gorenstein-projective module, torsionless module, extensionless module, reflexive module, $\scriptstyle t$-torsionfree module, $\scriptstyle\mho$-quiver. 2010 Mathematics Subject classification. Primary 16G10, 16G50. Secondary 16E05, 16E65, 20G42. Supported by NSFC 11431010

1. Introduction. (1.1) We refer to our previous paper [RZ1] as Part I. As in Part I, let $k$ be a field, and $q$ a non-zero element of $k$. We consider again the $k$-algebra $\Lambda=\Lambda(q)$ generated by $x,y,z$ with relations

[TABLE]

The algebra $\Lambda$ is a 6-dimensional local algebra with basis $1,x,y,z,yx,zx$. Its socle is $\operatorname{soc}\Lambda=\operatorname{rad}^{2}\Lambda=\Lambda yx\oplus\Lambda zx.$ If not otherwise stated, all the modules considered will be left $\Lambda$-modules.

We follow the terminology used in Part I. In particular, we denote by $\mho M$ the cokernel of a minimal left $\operatorname{add}(\Lambda)$-approximation of $M$. In addition, we introduce the following definitions. We say that a module $M$ is extensionless if $\operatorname{Ext}^{1}(M,\Lambda)=0$. An indecomposable semi-Gorenstein-projective module will be said to be pivotal provided it is not torsionless. An indecomposable $\infty$-torsionfree module will be said to be pivotal provided it is not extensionless. Thus, a module $M$ is semi-Gorenstein-projective if and only if $\Omega^{t}M$ is extensionless for all $t\geq 0$; a torsionless module $M$ is reflexive if and only if $\mho M$ is torsionless (see Part I (2.4)); a module $M$ is $\infty$-torsionfree if and only if $\mho^{t}M$ is reflexive for all $t\geq 0$; and $M$ is Gorenstein-projective if and only if $M$ is both semi-Gorenstein-projective and $\infty$-torsionfree.

(1.2) We are interested in the semi-Gorenstein-projective and the $\infty$-torsionfree modules and will exhibit those which are 3-dimensional. We recall that a finite length module is said to be local provided its top is simple. Thus, a local module is indecomposable; and if $R$ is a left artinian ring, then a left $R$-module $M$ is local if and only if $M$ is a quotient of an indecomposable projective module. A consequence of our study is the following assertion Proposition. Let $M$ be a non-zero module of dimension at most $3$. If $M$ is semi-Gorenstein-projective, then all the modules $\Omega^{t}M$ with $t\geq 0$ are $3$-dimensional and local. If $M$ is $\infty$-torsionfree, then all the modules $\mho^{t}M$ with $t\geq 0$ are $3$-dimensional and local. In particular, if $M$ is Gorenstein-projective, then all the modules $\Omega^{t}M$ and $\mho^{t}M$ with $t\geq 0$ are $3$-dimensional and local. (1.3) The text restricts the attention to the 3-dimensional local modules. The starting point of our investigation are two observations. The first one: Proposition 1. A module of dimension at most $3$ is annihilated by $\operatorname{rad}^{2}\Lambda$, thus it is a module of Loewy length at most $2$. The second observation is: Proposition 2. An indecomposable $3$-dimensional torsionless module is local. The proof of Proposition 1 will be given in (2.6), the proof of Proposition 2 in (2.7). (1.4) The 3-dimensional local modules. We identify $(a,b,c)\in k^{3}\setminus\{0\}$ with $ax+by+cz$ and denote by $(a\,\mathpunct{:}b\,\mathpunct{:}c)$ the 1-dimensional subspace of $k^{3}$ generated by $(a,b,c)$. The left ideal

[TABLE]

has dimension 3, and we obtain the left $\Lambda$-module

[TABLE]

Clearly, $M(a,b,c)$* is a $3$-dimensional local module and the modules $M(a,b,c)$, $M(a^{\prime},b^{\prime},c^{\prime})$ are isomorphic if and only if $(a\,\mathpunct{:}b\,\mathpunct{:}c)=(a^{\prime}\,\mathpunct{:}b^{\prime}\,\mathpunct{:}c^{\prime}).$* Let us add that the definition of $M(a,b,c)$ implies that $\Omega M(a,b,c)\simeq U(a,b,c)$, this will be used throughout the text.

Conversely, any $3$-dimensional local module is isomorphic to a module of the form $M(a,b,c)$. In order to see this, one should look at the factor algebra $\overline{\Lambda}$ of $\Lambda$ modulo $\operatorname{soc}\Lambda=\operatorname{rad}^{2}\Lambda$, thus $\overline{\Lambda}$ is the $k$-algebra generated by $x,y,z$ with relations all monomials of length $2$. The $\Lambda$-modules of Loewy length at most 2 are just the modules annihilated by all monomials of length $2$, thus the $\overline{\Lambda}$-modules. It is clear that the modules $M(a,b,c)=\overline{\Lambda}/(a\,\mathpunct{:}b\,\mathpunct{:}c)$ are representatives of the 3-dimensional local $\overline{\Lambda}$-modules. According to Proposition 1, all the 3-dimensional $\Lambda$-modules are $\overline{\Lambda}$-modules, thus the modules $M(a,b,c)$ are representatives of the 3-dimensional local $\Lambda$-modules. (1.5) The following theorem characterizes the modules of dimension at most 3 which have some relevant properties. We write $o(q)$ for the multiplicative order of $q$. Theorem. An indecomposable module $M$ of dimension at most $3$ is

*$\bullet$ *torsionless if and only if $M$ is simple or isomorphic to $\Lambda(x-y)$, to $\Lambda z$, to a module $M(1,b,c)$ with $b\neq-q$, to $M(0,1,0)$ or to $M(0,0,1)$;

*$\bullet$ *extensionless if and only if $M$ is isomorphic to a module $M(1,b,c)$ with $b\neq-1$;

*$\bullet$ *reflexive if and only if $M$ is isomorphic to a module $M(1,b,c)$ with $b\neq-q^{i}$ for $i=1,2$;

*$\bullet$ *Gorenstein-projective if and only if $M$ is isomorphic to a module $M(1,b,c)$ with $b\neq-q^{i}$ for $i\in\mathbb{Z}$;

*$\bullet$ *semi-Gorenstein-projective if and only if $M$ is isomorphic to a module $M(1,b,c)$ with $b\neq-q^{i}$ for $i\leq 0$;

$\bullet$ $\infty$-torsionfree if and only if $M$ is isomorphic to a module $M(1,b,c)$ with $b\neq-q^{i}$ for $i\geq 1$;

*$\bullet$ *pivotal semi-Gorenstein-projective if and only if $o(q)=\infty$ and $M$ is isomorphic to a module $M(1,-q,c)$;

*$\bullet$ *pivotal $\infty$-torsionfree if and only if $o(q)=\infty$ and $M$ is isomorphic to a module $M(1,-1,c)$.

For the proof of the Theorem, see (7.9). Looking at the Theorem, the reader will be aware that in the context considered here, the relevant modules of dimension at least 3 are the modules $M(1,b,c)$ with $b,c\in k.$ Nearly all the modules mentioned in Theorem are of this kind, the only exceptions are four isomorphism classes of torsionless modules, namely $\Lambda(x-y)$, $\Lambda z$, $M(0,1,0)$ and $M(0,0,1)$.

(1.6) As we have seen in (1.4), the set of isomorphism classes of the 3-dimensional local modules can be identified in a natural way with the projective plane $\mathbb{P}^{2}=\mathbb{P}(\operatorname{rad}\Lambda/\operatorname{rad}^{2}\Lambda)$, with the element $(a\,\mathpunct{:}b\,\mathpunct{:}c)\in\mathbb{P}^{2}$ corresponding to the module $M(a,b,c)$.

We use homogeneous coordinates in order to highlight elements and subsets of $\mathbb{P}^{2}$ (or the corresponding modules):

[TABLE]

As Theorem (1.5) shows, of special interest is the affine subspace $H$ of $\mathbb{P}^{2}$ given by the points $(1\,\mathpunct{:}b\,\mathpunct{:}c)$ with $b,c\in k$. As we will see in section 7, $H$ is a union of $\Omega\mho$-components, and the set of 3-dimensional Gorenstein-projective modules is always a (proper) subset of $H$. A module $M$ in $H$ is torsionless if and only if it does not belong to the line $T=\{(1\,\mathpunct{:}(-q)\,\mathpunct{:}c)\mid c\in k\}$, and is extensionless if and only if it does not belong to the line $E=\{(1\,\mathpunct{:}(-1)\,\mathpunct{:}c)\mid c\in k\}$ (see (6.1) and (5.1), respectively):

[TABLE]

In case the multiplicative order $o(q)$ of $q$ is infinite, $H$ is the set of the $3$-dimensional modules which are semi-Gorenstein-projective or $\infty$-torsionfree; the line $E$ consists of the pivotal semi-Gorenstein-projective modules in $H$; the line $T$ of the pivotal $\infty$-torsionfree modules in $H$.

Let us emphasize: There are $3$-dimensional pivotal semi-Gorenstein-projective modules if and only if there are $3$-dimensional pivotal $\infty$-torsionfree modules if and only if the multiplicative order of $q$ is infinite.

(1.7) The algebra $\Lambda=\Lambda(q)$ was exhibited in Part I in order to present in case $o(q)=\infty$ a module $M$ which is not torsionless, such that $M$ and its $\Lambda$-dual $M^{*}$ both are semi-Gorenstein-projective: namely the module $M=M(1,-q,0)$ with $M^{*}=M^{\prime}(1,-q,0)$. Now we see: Let $o(q)=\infty$ and assume that $M$ is a module of dimension at most $3$. Then both $M$ and $M^{*}$ are semi-Gorenstein-projective, whereas $M$ is not reflexive, if and only if $M$ is isomorphic to a module of the form $M(1,-q,c)$ with $c\in k$. In this case $M$ is not even torsionless and all the modules $M(1,-q,c)^{*}$ with $c\in k$ are isomorphic. Thus, we encounter a 1-parameter family of pairwise non-isomorphic semi-Gorenstein-projective left modules $M$ such that their $\Lambda$-dual modules $M^{*}$ are isomorphic and semi-Gorenstein-projective, see (9.5).

(1.8) The modules $M(1,b,0)$ with $\alpha\in k$ have been studied already in Part I (there, they have been denoted by $M(-b)$). Theorem (1.5) shows that these modules are quite typical for the behavior of the modules $M(1,b,c)$. Namely: The module $M(1,b,c)$ is Gorenstein-projective (or semi-Gorenstein-projective, or $\infty$-torsionfree, or torsionless, or extensionless) if and only if $M(1,b,0)$ has this property.

(1.8) Outline of the paper. Section 2 provides some preliminary results. Here, the main target is to show that any module of length at most 3 has Loewy length at most 2. In section 3 we collect some formulae which show that certain products of elements in $\Lambda$ are zero. Sections 4 to 7 deal with the 3-dimensional local left $\Lambda$-modules, section 8 with the 3-dimensional local right $\Lambda$-modules. Section 9 discusses the $\Lambda$-duality. The final section 10 provides an outline of the general frame for this investigation: the study of semi-Gorenstein-projective and $\infty$-torsionfree modules over local algebras with radical cube zero. There is an appendix which provides a diagrammatic description of the 3-dimensional indecomposable left $\Lambda$-modules.

2. Some left ideals and some right ideals of $\Lambda$. (2.1) Lemma. The left ideal $\Lambda(a,b,c)$ is $2$-dimensional if and only if $a+b=0$ and $ac=0.$ We have $\operatorname{soc}\Lambda(1,-1,0)=\Lambda yx$ and $\operatorname{soc}\Lambda(0,0,1)=\Lambda zx$. Proof. An easy calculation shows that $\operatorname{soc}\Lambda(1,-1,0)=\Lambda yx$ and $\operatorname{soc}\Lambda(0,0,1)=\Lambda zx$. Thus, the left ideals $\Lambda(0,0,1)$ and $\Lambda(1,-1,0)$ are $2$-dimensional.

Now, let $L=\Lambda(a,b,c)$ be any left ideal. If $a\neq 0$, then $yx\in L$ since $y(a,b,c)=ayx$.

First, assume that $a+b\neq 0.$ Then $z(a,b,c)=(a+b)zx$ shows that $zx\in L$. We know already that for $a\neq 0,$ also $yx\in L$. If $a=0,$ then $b\neq 0$. Thus $x(a,b,c)=-qbyx+czx$ shows that also in this case $yx\in L.$ Thus $L$ cannot be 2-dimensional.

Next, assume that $ac\neq 0.$ Since $a\neq 0,$ we know that $yx\in L$. Since $c\neq 0,$ we use $x(a,b,c)=-qbyx+czx$ in order to see that $zx\in L.$ Again, $L$ cannot be 2-dimensional. $\hfill\square$ (2.2) Let $L$ be a $2$-dimensional left ideal, different from $\operatorname{soc}\Lambda$. Then either $L\subseteq U(1,-1,0)$ and then $\operatorname{soc}L=\Lambda yx$ and $L$ is isomorphic to $\Lambda(x-y)$ or else $L\subseteq U(0,0,1)$ and then $\operatorname{soc}L=\Lambda zx$ and $L$ is isomorphic to $\Lambda z$. Proof: There is an element $(a,b,c)+w\in L$, with $(a,b,c)\neq 0$ and $w\in\operatorname{soc}\Lambda$. Since $\operatorname{rad}\Lambda((a,b,c)+w)=\operatorname{rad}\Lambda(a,b,c)$, also $L^{\prime}=\Lambda(a,b,c)$ is $2$-dimensional and $L\subseteq L^{\prime}+\operatorname{soc}\Lambda=U(a,b,c)$. According to (2,1), $(a\,\mathpunct{:}b\,\mathpunct{:}c)$ is equal to $(1\,\mathpunct{:}(-1)\,\mathpunct{:}0)$ or to $(0\,\mathpunct{:}0\,\mathpunct{:}1).$ Of course, $L$ and $L^{\prime}$ are isomorphic as (left) modules. $\hfill\square$ (2.3) Lemma. There is no $3$-dimensional torsionless module with simple socle. Proof. Assume that $U$ is a $3$-dimensional torsionless module with simple socle. Then $U$ is a submodule of $\Lambda$. It is a proper submodule, thus of Loewy length at most 2. Therefore, $U$ is the sum of two 2-dimensional left ideals $L\neq L^{\prime}$ with $\operatorname{soc}L=\operatorname{soc}L^{\prime}.$ Now we use (2.2). If $L,L^{\prime}$ have socle equal to $\Lambda yx$, then $U=L+L^{\prime}=U(1,-1,0)$. If $L,L^{\prime}$ have socle equal to $\Lambda zx$, then also $U=L+L^{\prime}=U(0,0,1)$. In both cases $\operatorname{soc}\Lambda\subseteq U,$ a contradiction. $\hfill\square$ (2.4) Any $3$-dimensional left ideal contains $\operatorname{soc}\Lambda$. (2.5) The $3$-dimensional left ideals are the subspaces $U(a,b,c)$. They have the following structure: $U(1,-1,0)=\Lambda(1,-1,0)\oplus\Lambda zx$; $U(0,0,1)=\Lambda(0,0,1)\oplus\Lambda yx$; and if $a+b\neq 0$ or $ac\neq 0$, then $U(a,b,c)=\Lambda(a,b,c)$ is a local module (in particular, indecomposable). Proof. The left ideals $U(a,b,c)$ are 3-dimensional. Conversely, let $U$ be a 3-dimensional left ideal of $\Lambda$. Since $\operatorname{soc}\Lambda$ is contained in $U$, there is an element $(a,b,c)\neq 0$ with $(a,b,c)\in U$, thus $U=U(a,b,c)$.

If $a+b=0$ and $ac=0$, then $(a\,\mathpunct{:}b\,\mathpunct{:}c)$ is equal to $(1\,\mathpunct{:}(-1)\,\mathpunct{:}0)$ or to $(0\,\mathpunct{:}0\,\mathpunct{:}1)$. By (2.1), we have $U(1,-1,0)=\Lambda(1,-1,0)\oplus\Lambda zx$ and $U(0,0,1)=\Lambda(0,0,1)\oplus\Lambda yx.$ If $a+b\neq 0$ or $ac\neq 0$, then $U(a,b,c)=\Lambda(a,b,c)$ is a local module, thus indecomposable. $\hfill\square$ (2.6) Proposition. Any module of dimension at most $3$ has Loewy length at most $2$. Proof. Let $M$ be a module of dimension at most 3. If $M$ is not local, then clearly $M$ has Loewy length at most 2. If $\dim M\leq 2$, then $M$ is of course local. Thus, we can assume that $M$ is 3-dimensional and local and therefore a factor module of $\Lambda$, say $M=\Lambda/U$. According to (2.4), $\operatorname{soc}\Lambda\subseteq U$, thus $M$ is annihilated by $\operatorname{soc}\Lambda$, and therefore $M$ has Loewy length at most 2. $\hfill\square$ (2.7) Lemma. Any indecomposable torsionless module $M$ of dimension at most $3$ is local and isomorphic to a left ideal of $\Lambda$. If $\dim M=3$, then $M$ is of the form $U(a,b,c)$. Proof. Let $M$ be indecomposable and torsionless. If $\dim M\leq 2$, then $M$ is of course local and isomorphic to a left ideal. Thus we can assume that $\dim M=3.$

Since $M$ is torsionless, there is a set of non-zero maps $u_{i}\>M\to{}_{\Lambda}\Lambda$ (say with index set $I$) such that $\bigcap_{i\in I}K_{i}=0$, where $K_{i}$ is the kernel of $u_{i}.$

If $K_{i}=0$ for some $i$, then already $u_{i}$ is an embedding (thus $M$ is isomorphic to a left ideal). In particular, if the socle of $M$ is simple, then we must have $K_{i}=0$ for some $i$. Thus, we can assume that the socle of $M$ is not simple. Therefore $M$ has to be a local module and we have a surjective map $\pi\>{}_{\Lambda}\Lambda\to M$.

It remains to look at the case where $\dim K_{i}=1$ or $2$ for all $i$. Since the only 2-dimensional submodule of $M$ is its radical, we have $\bigcap_{i\in I^{\prime}}K_{i}=0,$ where $I^{\prime}$ is the set of indices $i$ with $\dim K_{i}=1$. But then $K_{i}\cap K_{j}=0$ for some $i\neq j$ in $I^{\prime}$. This shows that we can assume that $I=\{1,2\}$ and that $K_{1},K_{2}$ are different 1-dimensional submodules of $M$.

Now $u_{i}$ provides an isomorphism from $M/K_{i}$ onto a (2-dimensional) left ideal of $\Lambda$. Since $M/K_{i}$ is indecomposable, (2.2) shows that $M/K_{i}$ is isomorphic to $\Lambda(1,-1,0)$ or to $\Lambda(0,0,1)$. Let $K^{\prime}_{i}=\operatorname{Ker}(u_{i}\pi)$ for $i=1,2$.

If $M/K_{i}\simeq\Lambda(1,-1,0)$, then $K^{\prime}_{i}$ is equal to $\Lambda(x+qy)+\Lambda z$, since $\Lambda(0,0,1)$ is annihilated by $x+qy$ and by $z$. Similarly, if $M/K_{i}\simeq\Lambda(0,0,1)$, then $K^{\prime}_{i}$ is equal to $\Lambda(x+qy)+\Lambda z$. Thus one of $M/K_{i}$ has to be isomorphic to $\Lambda(1,-1,0)$, the other one to $\Lambda(0,0,1)$ and $\operatorname{Ker}(\pi)=K_{1}^{\prime}\cap K^{\prime}_{2}=U(0,0,1).$ It follows that $M\simeq{}_{\Lambda}\Lambda/\operatorname{Ker}(\pi)={}_{\Lambda}\Lambda/U(0,0,1).$ But ${}_{\Lambda}\Lambda/U(0,0,1)$ is isomorphic to the left ideal $\Lambda(1,-1,1)=U(1,-1,1)$.

We have shown that $M$ is isomorphic to a left ideal, thus of the form $U(a,b,c)$, see (2.5). Since we assume that $M$ is indecomposable, (2.5) asserts that $M$ is local. $\hfill\square$

We need to know also the right ideals $(a,b,c)\Lambda$. Note that $U(a,b,c)$ is always a twosided ideal and it will be pertinent to denote $U(a,b,c)$ by $U^{\prime}(a,b,c)$, if we consider it as a right ideal (thus as a right module). (2.8) The right ideals $(a,b,c)\Lambda$. If $a\neq 0$ or $bc\neq 0$, then $(a,b,c)\Lambda=U(a,b,c)$ is $3$-dimensional. The right ideals $(0,1,0)\Lambda$ and $(0,0,1)\Lambda$ are $2$-dimensional with $\operatorname{soc}\,(0,1,0)\Lambda=yx\Lambda$ and $\operatorname{soc}\,(0,0,1)\Lambda=zx\Lambda.$ Proof: Let $V=(a,b,c)\Lambda$. First, let $a\neq 0$. Then $zx$ belongs to $V$, since $(a,b,c)z=azx.$ Also $yx\in V$, since $(a,b,c)y=-qayx+czx$. Second, assume that $a=0$ and $bc\neq 0.$ Then $(0,b,c)y=czx$ shows that $zx\in V$, and $(0,b,c)x=byx+czx$ shows that also $yx\in V.$ $\hfill\square$ (2.9) If a $3$-dimensional indecomposable right module $N$ is torsionless, then it is isomorphic to a right ideal, thus to $U^{\prime}(a,b,c)$ for some $(a,b,c)\neq 0.$ Proof. Let $N$ be a 3-dimensional indecomposable torsionless right module. As in (2.7) one shows that $N$ is isomorphic to a right ideal, using (2.8) instead of (2.2). It remains to see that all 3-dimensional right ideals are of the form $U^{\prime}(a,b,c)$. Here, one has to copy the proof of (2.5).

3. The transformations $\omega$ and $\omega^{\prime}$. If $(a\,\mathpunct{:}b\,\mathpunct{:}c)$ is different from $(1\,\mathpunct{:}(-1)\,\mathpunct{:}0)$ and $(0\,\mathpunct{:}0\,\mathpunct{:}1)$, then (2.5) shows that $U(a,b,c)$ is a $3$-dimensional local module, thus of the form $M(a^{\prime}\,\mathpunct{:}b^{\prime}\,\mathpunct{:}c^{\prime})$. In order to describe in which way $(a^{\prime}\,\mathpunct{:}b^{\prime}\,\mathpunct{:}c^{\prime})$ depends on $(a\,\mathpunct{:}b\,\mathpunct{:}c)$, we will need the transformations $\omega$ and $\omega^{\prime}$. We start with some equalities in $\Lambda$. (3.1) Formulae. Let $a,b,c\in k.$ Then

[TABLE]

Proof of the equality (1):

[TABLE]

The proof of the remaining equalities is similar. $\hfill\square$

(3.2) In case $a+b\neq 0$, let $\omega(a,b,c)=(a,qb,-\tfrac{a}{a+b}c).$ In case $a^{\prime}\neq 0,$ let $\omega^{\prime}(a^{\prime},b^{\prime},c^{\prime})=(a^{\prime},q^{-1}b^{\prime},-\tfrac{a^{\prime}+q^{-1}b^{\prime}}{a^{\prime}}c^{\prime}).$

Proposition. The transformation $\omega$ provides a bijection from the set $\{(a,b,c)\in k^{3}\mid a(a+b)\neq 0\}$ onto the set $\{(a^{\prime},b^{\prime},c^{\prime})\in k^{3}\mid a^{\prime}(a^{\prime}+q^{-1}b^{\prime})\neq 0\},$ with inverse $\omega^{\prime}$. Proof. Let $a(a+b)\neq 0$. Then $(a^{\prime},b^{\prime},c^{\prime})=\omega(a,b,c)$ is defined and $a^{\prime}=a\neq 0$, and $a^{\prime}+q^{-1}b^{\prime}=a+q^{-1}qb=a+b\neq 0.$ Thus $\omega$ maps $\{(a,b,c)\in k^{3}\mid a(a+b)\neq 0\}$ into $\{(a,b,c)\in k^{3}\mid a(a+q^{-1}b)\neq 0.$ Similarly, $\omega^{\prime}$ maps $\{(a^{\prime},b^{\prime},c^{\prime})\in k^{3}\mid a^{\prime}(a^{\prime}+q^{-1}b^{\prime})\neq 0$ into $\{(a,b,c)\in k^{3}\mid a(a+b)\neq 0.$ It is easy to check that $\omega^{\prime}\omega(a,b,c)=(a,b,c)$ for $a(a+b)\neq 0$ and that $\omega\omega^{\prime}(a^{\prime},b^{\prime},c^{\prime})=(a^{\prime},b^{\prime},c^{\prime})$ for $a^{\prime}(a^{\prime}+q^{-1}b^{\prime})\neq 0.$ $\hfill\square$

4. The isomorphism class of $U(a,b,c)\simeq\Omega M(a,b,c)$. (4.1) Proposition. Let $(a,b,c)\neq 0.$ Then

[TABLE]

Proof: If $a=0$ and $b=0,$ then $U(a,b,c)=U(0,0,1)$. If $a+b=0$ and $c=0$, then $U(a,b,c)=U(1,-1,0)$. According to (2.3), $U(0,0,1)=\Lambda z\oplus\Lambda yx$ and $U(1,-1,0)=\Lambda(x-y)\oplus\Lambda zx$, This shows (5) and (3). In this way, we have considered all triples $(a,b,c)$ with $a+b=0$ and $ac=0$.

Thus, let $a+b\neq 0$ or $ac\neq 0$. By (2.5), $U(a,b,c)=\Lambda(a,b,c)$ is local and we look at the surjective map $\phi\>{}_{\Lambda}\Lambda\to U(a,b,c)$ which sends $1$ to $(a,b,c)$.

Let $a+b\neq 0.$ According to formula (1) of (3.1), $\Lambda(a,b,c)$ is annihilated by $\omega(a,b,c)$, thus $M(\omega(a,b,c)))={}_{\Lambda}\Lambda/\Lambda(\omega(a,b,c))$ maps onto $\Lambda(a,b,c)$. Since the modules $M(\omega(a,b,c))$ and $\Lambda(a,b,c)$ both have dimension 3, we see that $U(a,b,c)=\Lambda(a,b,c)$ is isomorphic to $M(\omega(a,b,c))$. This yields (1) and (4) (namely, if $a=0,$ and $b\neq 0$, we have $\omega(0,b,c)=(0,qb,0)$).

Finally, we show (2). For $c\neq 0$, the module $U(1,-1,c)$ is isomorphic to $M(0,0,1)$. Now we use in the same way formula (2) of (3.1). $\hfill\square$ The following picture outlines the position of the partition of $\mathbb{P}^{2}$ which is used in the Proposition.

[TABLE]

(4.2) Corollary. The syzygy functor $\Omega$ provides a bijection from the set of isomorphism classes of modules $M(a,b,c)$ with $a(a+b)\neq 0$ onto the set of isomorphism classes of modules $M(a^{\prime},b^{\prime},c^{\prime})$ with $a^{\prime}(a^{\prime}+q^{-1}b^{\prime})\neq 0$ and we have $\Omega M(a,b,c)=M(\omega(a,b,c))$ for $a(a+b)\neq 0$. Proof. This follows directly from Propositions (3.2) and (4.1). $\hfill\square$

5. The extensionless modules $M(a,b,c)$. (5.1) Proposition. The module $M(a,b,c)$ is extensionless if and only if $a(a+b)\neq 0.$ For the proof, we need some preparations. (5.2) Lemma. The following conditions are equivalent:

(i) The module $M(a,b,c)$ is extensionless.

(ii) The inclusion map $\iota\>U(a,b,c)\to{}_{\Lambda}\Lambda$ is a left $\operatorname{add}(\Lambda)$-approximation.

(iii) $U(a,b,c)=\Lambda(a,b,c)$* and the inclusion map $\iota\>\Lambda(a,b,c)\to{}_{\Lambda}\Lambda$ is a left $\operatorname{add}(\Lambda)$-approximation.*

(iv) The subspace $U(a,b,c)$ is indecomposable both as a left module and as a right module, and the image of every homomorphism ${}_{\Lambda}U(a,b,c)\to{}_{\Lambda}\Lambda$ is contained in $U(a,b,c)$.

Proof. The equivalence of (i) and (ii) follows from Part I, Lemma 2.1.

(ii) $\implies$ (iii): We assume (ii). If $U(a,b,c)=U_{1}\oplus U_{2}$ with $U_{1},U_{2}$ both non-zero, then a minimal left $\operatorname{add}(\Lambda)$-approximation $U(a,b,c)\to\Lambda^{t}$ is the direct sum of minimal left $\operatorname{add}(\Lambda)$-approximations $U_{1}\to\Lambda^{t_{1}}$ and $U_{2}\to\Lambda^{t_{2}}$, thus $t=t_{1}+t_{2}\geq 2.$ This shows that $U(a,b,c)$ is indecomposable. According to (2.5), this means that $U(a,b,c)=\Lambda(a,b,c)$.

(iii) $\implies$ (iv). Since $\Lambda(a,b,c)$ is a local module, it is indecomposable. Thus $U(a,b,c)=\Lambda(a,b,c)$ implies that $U(a,b,c)$ considered as a left module is indecomposable. Given any homomorphism $\phi\>U(a,b,c)\to{}_{\Lambda}\Lambda$, (iii) provides $\lambda\in\Lambda$ with $\phi(a,b,c)=(a,b,c)\lambda\in(a,b,c)\Lambda\subseteq U(a,b,c).$ Now assume that $(a,b,c)\Lambda$ is a proper subset of $U(a,b,c).$ Let $w\in\operatorname{soc}\Lambda$. Since $\Lambda w$ is simple, there is a homomorphism $\phi\>\Lambda(a,b,c)\to\Lambda$ with $\phi(a,b,c)=w$ and (iii) asserts that $w=\phi(a,b,c)=(a,b,c)\lambda$ for some $\lambda\in\Lambda$. This shows that $\operatorname{soc}\Lambda\subseteq(a,b,c)\Lambda$ and therefore $U(a,b,c)=(a,b,c)\Lambda$. In particular, $U(a,b,c)$ is indecomposable also as a right $\Lambda$-module.

(iv) $\implies$ (ii). Let $\phi\>U(a,b,c)\to{}_{\Lambda}\Lambda$ be a homomorphism. Since $U(a,b,c)$ is indecomposable as a left module, we have $U(a,b,c)=\Lambda(a,b,c)$. Since $U(a,b,c)$ is indecomposable as a right module, we have $U(a,b,c)=(a,b,c)\Lambda$. According to (iv), $\phi(a,b,c)\in U(a,b,c)=(a,b,c)\Lambda,$ thus $\phi(a,b,c)=(a,b,c)\lambda=r_{\lambda}\iota(a,b,c)$ for some $\lambda\in\Lambda$, where $r_{\lambda}\>{}_{\Lambda}\Lambda\to{}_{\Lambda}\Lambda$ is the right multiplication by $\lambda$. Since the left module $U(a,b,c)=\Lambda(a,b,c)$ is generated by $(a,b,c)$, the equality $\phi(a,b,c)=r_{\lambda}\iota(a,b,c)$ implies that $\phi=r_{\lambda}\iota.$ $\hfill\square$ (5.3) Lemma. Let $R$ be a ring and $X$ a left $R$-module. If $\phi\>{}_{R}R\to X$ is an $R$-module homomorphism and $w\in R$ annihilates $X$, then $Rw\subseteq\operatorname{Ker}\phi.$ Corollary. Let $L$ be a left ideal of $R$ and $X$ an $R$-module annihilated by $w_{1},\dots,w_{t}\in R$. The image of any map $R/L\to X$ is a factor module of $R/(L+Rw_{1}+\cdots Rw_{t})$. Proof. Let $\phi\>R/L\to X$ be a homomorphism. Let $\pi\>R\to R/L$ be the canonical projection. By construction, $L$ is contained in $\operatorname{Ker}(\phi\pi)$. By the lemma, also the left ideals $Rw_{i}$ are contained in $\operatorname{Ker}(\phi\pi).$ Thus $L+Rw_{1}+\cdots+Rw_{t}\subseteq\operatorname{Ker}(\phi\pi).$ $\hfill\square$

(5.4) Proof of Proposition (5.1). According to (5.2), $M(a,b,c)$ is extensionless if and only if condition (iv) is satisfied. We look at all the elements $(a\,\mathpunct{:}b\,\mathpunct{:}c)\in\mathbb{P}^{2},$ using the partition of $\mathbb{P}^{2}$ into the subsets (1) to (5) as in (4.1).

The cases (3) and (5): Both $U(1,-1,0)$ and $U(0,0,1)$ are decomposable as left modules, see (2.5). Case (4): According to (4.1), $U(0,1,c)\simeq M(0,1,0)$. Obviously, $M(0,1,0)$ has $\Lambda z$ as a factor module, thus there is a homomorphism $U(0,1,c)\to{}_{\Lambda}\Lambda$ with image $\Lambda z$ and $\Lambda z\not\subseteq U(0,1,c)$. The case (2) is similar: (4.1) shows that $U(1,-1,c)\simeq M(0,0,1)$, and $M(0,0,1)$ maps onto $\Lambda z$; thus there is a homomorphism $U(1,-1,c)\to{}_{\Lambda}\Lambda$ with image $\Lambda z$ and $\Lambda z\not\subseteq U(1,-1,c)$. This shows that none of the modules $M(a,b,c)$ with $a(a+b)=0$ is extensionless. It remains to consider the case (1). Thus, assume that $a(a+b)\neq 0.$ Let $(1,b^{\prime},c^{\prime})=\omega(1,b,c)$, thus $b^{\prime}=qb.$ We want to show that the conditions (iv) of (5.2) are satisfied. According to (2.5) and (2.8), $U(a,b,c)$ is indecomposable both as a left module and as a right module, It remains to show that the image of every homomorphism ${}_{\Lambda}U(a,b,c)\to{}_{\Lambda}\Lambda$ is contained in $U(a,b,c)$. (a) The only left ideal isomorphic to $U(1,b,c)$ is $U(1,b,c)$ itself. Proof. The 3-dimensional left ideals are of the form $U(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})$, for some $(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})\neq 0$, see (2.5). Assume that $U(1,b,c)\simeq U(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})$. We have $U(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})\simeq\Omega M(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})$ and by (4.1) we must be in case (1), namely $a^{\prime\prime}\neq 0$ and $a^{\prime\prime}+b^{\prime\prime}\neq 0$. In particular, we may assume that $a^{\prime\prime}=1$ and (4.1)(1) asserts that $\Omega M(1,b^{\prime\prime},c^{\prime\prime})=M(\omega(1,b^{\prime\prime},c^{\prime\prime})).$ The isomorphy $M(\omega(1,b,c))\simeq M(\omega(1,b^{\prime\prime},c^{\prime\prime}))$ implies that the triples $\omega(1,b,c)$ and $\omega(1,b^{\prime\prime},c^{\prime\prime})$ yield the same element in $\mathbb{P}^{2},$ and since the first coordinate of both triples is equal to 1, we have $\omega(1,b,c)=\omega(1,b^{\prime\prime},c^{\prime\prime}).$ Since $1+b\neq 0$ and $1+b^{\prime\prime}\neq 0,$ we use (3.2) in oder to conclude that $(1,b,c)=(1,b^{\prime\prime},c^{\prime\prime}).$ (b) The left ideal $\Lambda z$ is not a factor module of $U(1,b,c)$. The proof uses Corollary (5.3) for the left ideal $L=U(1,b^{\prime},c^{\prime})$ and the module $X=\Lambda z$ which is annihilated by $y$ and $z$. Namely, on the one hand, we have $U(1,b,c)\simeq\Omega M(1,b,c)\simeq M(\omega(1,b,c))=M(1,b^{\prime},c^{\prime})=\Lambda/U(1,b^{\prime},c^{\prime})=\Lambda/L.$ On the other hand, $\operatorname{rad}\Lambda=\Lambda(x+b^{\prime}y+c^{\prime}z)+\Lambda y+\Lambda z\subseteq U(1,b^{\prime},c^{\prime})+\Lambda y+\Lambda z\subseteq\operatorname{rad}\Lambda$ shows that $L+\Lambda y+\Lambda z=\operatorname{rad}\Lambda$. Therefore, (5.3) asserts that the image of any homomorphism $U(1,b,c)\to\Lambda z$ is a factor module of $\Lambda/\operatorname{rad}\Lambda$, thus simple or zero. (c) The left ideal $\Lambda(x-y)$ is not a factor module of $U(1,b,c)$. Again, we use Corollary (5.3) for $L=U(1,b^{\prime},c^{\prime})$ and now for $X=\Lambda(x-y)$. Note that $\Lambda(x-y)$ is annihilated by $x-qy$ and $z$. We recall from (b) that $U(1,b,c)\simeq\Lambda/L.$ And we have $\operatorname{rad}\Lambda=\Lambda(x+b^{\prime}y+c^{\prime}z)+\Lambda(x-qy)+\Lambda z$, since $b^{\prime}=qb\neq-q$. Therefore, we also have $U(1,b^{\prime},c^{\prime})+\Lambda(x-qy)+\Lambda z=\operatorname{rad}\Lambda$, and (5.3) asserts that the image of any homomorphism $U(1,b,c)\to\Lambda z$ is simple or zero. Any homomorphism $\phi\>U(1,b,c)\to{}_{\Lambda}\Lambda$ maps into $U(1,b,c)$. Proof. According to (b) and (c), the image $I$ of $\phi$ is not of dimension $2$. If the image $I$ is of dimension $3$, then (a) shows that $I$ is equal to $U(1,b,c)$. Of course, if $I$ is of dimension at most $1$, then $I\subseteq\operatorname{soc}\Lambda\subseteq U(1,b,c).$ $\hfill\square$ (5.5) Corollary. If $M(a,b,c)$ is extensionless, then $\Omega M(a,b,c)\simeq M(\omega(a,b,c))$. Proof. This follows directly from (5.1) and the case (1) of (4.1). $\hfill\square$.

6. The torsionless modules $M(a,b,c)$. (6.1) Proposition. The module $M(a,b,c)$ is torsionless if and only if either $a(a+q^{-1}b)\neq 0$ or else $a=0$ and $bc=0$ (so that $(a\,\mathpunct{:}b\,\mathpunct{:}c)$ is equal to $(0\,\mathpunct{:}1\,\mathpunct{:}0)$ or to $(0\,\mathpunct{:}0\,\mathpunct{:}1)$). In order to prove (6.1), we consider the possible cases separately. First, we consider the modules $M(a,b,c)$ with $a\neq 0.$ In section 5 we have seen that $M(1,b,c)$ is extensionless if and only if $b\neq-1$, and then $\Omega M(1,b,c)\simeq M(\omega(1,b,c))$. There is the following corresponding assertion concerning the torsionless modules (see also (7.1)). (6.2) The module $M(1,b,c)$ is torsionless if and only if $b\neq-q$, and in this case $\mho M(1,b,c)\simeq M(\omega^{\prime}(1,b,c))$. Proof. Let $b\neq-q$. Then $\omega^{\prime}(1,b,c)=(1,q^{-1}b,c^{\prime})$ for some $c^{\prime}$. According to (5.1) and (5.5), $M(1,q^{-1}b,c^{\prime})$ is extensionless and $\Omega M(1,q^{-1}b,c^{\prime})\simeq M(1,b,c)$, since $\omega(1,q^{-1}b,c^{\prime})=\omega\omega^{\prime}(1,b,c)=(1,b,c).$ This shows that $M(1,b,c)$ is torsionless and that $\mho M(1,b,c)\simeq M(\omega^{\prime}(1,b,c))$.

Conversely, we consider $M(1,-q,c)$ and assume, for the contrary, that $M(1,-q,c)$ is torsionless. According to (2.7), this means that $M(1,-q,c)$ is isomorphic to a left ideal $U(a^{\prime},b^{\prime},c^{\prime})=\Omega M(a^{\prime},b^{\prime},c^{\prime}).$ According to (4.1), we must be in the case $a^{\prime}+b^{\prime}\neq 0$ and $a^{\prime}\neq 0$. We can assume that $a^{\prime}=1$, thus $1+b^{\prime}\neq 0$. We have $\Omega M(1,b^{\prime},c^{\prime})\simeq M(\omega(1,b^{\prime},c^{\prime}))=M(1,qb^{\prime},c^{\prime\prime})$ for some $c^{\prime\prime}$. Since $M(1,-q,c)\simeq\Omega M(1,b^{\prime},c^{\prime})\simeq M(1,qb^{\prime},c^{\prime\prime}),$ we see that $(1,-q,c)=(1,qb^{\prime},c^{\prime\prime})$, thus $b^{\prime}=-1.$ But this is a contradiction to $1+b^{\prime}\neq 0.$ $\hfill\square$ (6.3) For $M=M(0,1,0)$ and $M(0,0,1)$, there is no monomorphism $M\to{}_{\Lambda}\Lambda$ which is an $\operatorname{add}(\Lambda)$-approximation. Proof. Let $M$ be equal to $M(0,1,0)$ or to $M(0,0,1)$. Assume that there is a monomorphism $u\>M\to{}_{\Lambda}\Lambda$ which is an $\operatorname{add}(\Lambda)$-approximation. The image $u(M)$ is a 3-dimensional left ideal, thus of the form $U(a,b,c)$ for some $(a,b,c)\neq 0$, see (2.7). The implication (ii) $\implies$ (iv) in (5.2) asserts that any homomorphism $U(a,b,c)\to{}_{\Lambda}\Lambda$ maps into $U(a,b,c)$.

Obviously, both modules $M(0,1,0)$ and $M(0,0,1)$ have a factor module isomorphic to $\Lambda z,$ thus there is a surjective homomorphism $U(a,b,c)\to\Lambda z,$ and therefore $\Lambda z\subseteq U(a,b,c)$. But $\Lambda z$ is an indecomposable module of length 2, and $U(a,b,c)\simeq M$ is a local module of length 3 with socle of length 2. A local module of length 3 with socle of length 2 has no indecomposable submodule of length 2, thus we obtain a contradiction. $\hfill\square$

(6.4) Proposition. The modules $M(0,b,c)$ with $bc\neq 0$ are not torsionless. Proof. Let $M=M(0,b,c)$ with $bc\neq 0$ and assume that $M$ is torsionless. According to (2.7), this means that $M\simeq U(a^{\prime},b^{\prime},c^{\prime})\simeq\Omega M(a^{\prime},b^{\prime},c^{\prime})$ for some triple $(a^{\prime},b^{\prime},c^{\prime}),$ and (2.5) asserts that $a^{\prime}+b^{\prime}\neq 0$ or $a^{\prime}c^{\prime}\neq 0$. Now we use (4.1) and have to distinguish the three cases (1), (2) and (4). Case (1) means that $a^{\prime}+b^{\prime}\neq 0$ and $a^{\prime}\neq 0$, then $\Omega M(a^{\prime},b^{\prime},c^{\prime})\simeq M(\omega(a^{\prime},b^{\prime},c^{\prime}))$ and the first component of $\omega(a^{\prime},b^{\prime},c^{\prime})$ is $a^{\prime}$, thus non-zero. But then $M(\omega(a^{\prime},b^{\prime},c^{\prime}))$ cannot be isomorphic to $M(0,b,c)$. Case (4) means that $a^{\prime}=0$ and $b^{\prime}\neq 0$. Then $\Omega M(a^{\prime},b^{\prime},c^{\prime})\simeq M(0,1,0)$, thus not isomorphic to $M(0,b,c)$ with $bc\neq 0.$ Finally, there is the case (2) with $a^{\prime}+b^{\prime}=0$ and $a^{\prime}c^{\prime}\neq 0$. Then $\Omega M(a^{\prime},b^{\prime},c^{\prime})\simeq M(0,0,1)$, again not isomorphic to $M(0,b,c)$ with $bc\neq 0.$ In all cases, we get a contradiction. $\hfill\square$ (6.5) Proposition. If $M$ is equal to $M(0,1,0)$ or $M(0,0,1)$, then $M$ is torsionless and the module $\mho M$ has Loewy length $3$. Since $\mho M$ is indecomposable and non-projective, it is not torsionless. Proof. The modules $M$ of the form $M(0,1,0)$ and $M(0,0,1)$ are torsionless, since (4.1), (4) and (2) assert that $M(0,1,0)\simeq\Omega M(0,1,0)$ and that $M(0,0,1)\simeq\Omega M(1,-1,1)$. According to (5.2), in both cases there is no inclusion map $M\to\Lambda$ which is an $\operatorname{add}(\Lambda)$-approximation. Thus, a minimal left $\operatorname{add}(\Lambda)$-approximation of $M$ is an injective map $M\to\Lambda^{t}$ with $t\geq 2.$ This shows that $\mho M$ has dimension $6t-3$ and its top has dimension $t$. According to Part I (3.2), $\mho M$ is indecomposable and not projective. The Loewy length of $\mho M$ has to be 3. [Namely, an indecomposable module with Loewy length at most $2$ and top of dimension $t\geq 2$ has dimension at most $4t-1$, since it is a proper factor module of $\overline{\Lambda}^{t}$. But $6t-3\leq 4t-1$ implies $t\leq 1,$ a contradiction.] An indecomposable non-projective module of Loewy length 3 cannot be torsionless. $\hfill\square$

(6.6) We finish this section by reformulating the results concerning the modules of the form $M(0,b,c)$ in terms of $\Omega\mho$-components. Here, we will exhibit the structure of all the $\Omega\mho$-components containing modules of the form $M(0,b,c)$. We have to distinguish between the modules $M(0,1,0)$ and $M(0,0,1)$ and the modules $M(0,b,c)$ with $bc\neq 0$, thus lying on the dashed line $A^{\prime}=\{(0\,\mathpunct{:}b\,\mathpunct{:}c)\mid bc\neq 0\}$:

[TABLE]

The modules in $A^{\prime}$ are singletons (that is, components of type $\mathbb{A}_{1}$) in the $\Omega\mho$-quiver. And, there are the following two $\Omega\mho$-components of the form $\mathbb{A}_{2}$:

[TABLE]

(If $M$ is an indecomposable module, then we represent $[M]$ in the $\Omega\mho$-quiver usually just by a circle $\circ$. We use a bullet $\bullet$ in case we know that $M$ is torsionless and extensionless, a black square $\scriptscriptstyle\blacksquare$ in case we know that $M$ is extensionless, but not torsionless; and a black lozenge $\scriptstyle\blacklozenge$ in case we know that $M$ is torsionless, but not extensionless.)

7. The modules $M(1,b,c)$ and proof of Theorem (1.5). We consider now the affine subspace $H$ of $\mathbb{P}^{2}$ given by the points $(1\,\mathpunct{:}b\,\mathpunct{:}c)$ with $b,c\in k$ and the corresponding modules $M(1,b,c)$. We recall that $o(q)$ denotes the multiplicative order of $q$. (7.1) We have seen in (4.2) that $\Omega$ provides a bijection from the set of modules $M(1,b,c)$ with $b\neq-1$ onto the set of modules $M(1,b^{\prime},c^{\prime})$ with $b^{\prime}\neq-q$. The sections 5 and 6 strengthen this bijection as follows: If $b\neq-1$, then the exact sequence

[TABLE]

with $(1,b^{\prime},c^{\prime})=\omega(1,b,c)$ is an $\Omega\mho$-sequences (here, $(1,b^{\prime},c^{\prime})$ is an arbitrary triple with $b^{\prime}\neq-q,$ and $(1,b,c)=\omega^{\prime}(1,b^{\prime},c^{\prime})$). We obtain in this way all the $\Omega\mho$-sequences involving modules of the form $M(1,b,c)$.

(7.2) Reformulation. The neighborhood of $M(1,b,c)$ in the $\Omega\mho$-quiver looks as follows:

[TABLE]

and $M(1,b,c)$ is a singleton in the $\Omega\mho$-quiver if $q=1$ and $b=-1$. (7.3) The module $M(1,b,c)$ is semi-Gorenstein-projective if and only if $b\neq-q^{t}$ for all $t\leq 0$. The module $M(1,b,c)$ is $\infty$-torsionfree if and only if $b\neq-q^{t}$ for all $t\geq 1$. Proof: $M(1,b,c)$ is semi-Gorenstein-projective if and only if $\omega^{s}(1,b,c)\notin E$ for all $s\geq 0.$ Since $\omega^{s}(1,b,c)=(1,q^{s}b,c_{s})$ for some $c_{s}\in k$, we see that $M(1,b,c)$ is semi-Gorenstein-projective if and only if $1+q^{s}\neq 0$ for all $s\geq 0$, thus if and only if $q^{-s}\neq-b$ for all $s\geq 0$. Write $t=-s.$

Similarly, $M(1,b,c)$ is $\infty$-torsionfree if and only if $\omega^{-s}(1,b,c)\notin T$ for all $s\geq 0,$ thus if and only if $1+q^{-1}q^{-s}b\neq 0$ for all $s\geq 0$, if and only if $-b\neq q^{s+1}$ for all $s\geq 0.$ Write $t=s+1$. $\hfill\square$ Corollary. The module $M(1,b,c)$ is Gorenstein-projective if and only if $b\neq-q^{t}$ for all $t\in\mathbb{Z}.$

(7.4) Any module $M(1,0,c)$ with $c\in k$ is Gorenstein-projective with $\Omega$-period $1$ or $2$. Proof. According to (6.2), the modules $M(1,0,c)$ are extensionless and torsionless. Since $\omega(1,0,c)=(1,0,-c)$, we see that $M(1,0,0)$ has $\Omega$-period 1, and $M(1,0,c)$ with $c\neq 0$ has $\Omega$-period $2$ in case the characteristic of $k$ is different from $2$, otherwise its $\Omega$-period is also $1$. $\hfill\square$

(7.5) Proposition. If $o(q)=\infty$, then any module of the form $M(1,b,c)$ is semi-Gorenstein-projective or $\infty$-torsionfree (whereas the modules of the form $M(0,b,c)$ are never semi-Gorenstein-projective nor $\infty$-torsionfree). Proof. The first assertion follows immediately from (7.3), the additional assertion in the bracket is a consequence of (5.1), (6.4) and (6.5). $\hfill\square$

(7.6) Proposition. If $M(1,b,c)$ belongs to an $\Omega\mho$-component of the form $\mathbb{A}_{n}$, then $o(q)=n.$ Proof. We consider an $\Omega\mho$-component of type $\mathbb{A}_{n}$, say containing a module $M$ which is not torsionless. Since $M$ belongs to $T$, we have $M=M(1,-q,c)$ and the component consists of the modules $M,\ \Omega M,\ \dots,\ \Omega^{n-1}M$. In particular, $\omega^{n-1}(1,-q,c)$ belongs to $E$. Now $\Omega^{n-1}M=M(\omega^{n-1}(1,-q,c))=M(1,-q^{n},c^{\prime})$ for some $c^{\prime}$. Since $\Omega^{n-1}M$ is not extensionless, $(1,-q^{n},c^{\prime})$ belongs to $E$, thus $-q^{n}=-1$. This shows that $q^{n}=1.$ Finally, for $1\leq t<n$, we have $q^{t}\neq 1,$ since otherwise $\omega^{t-1}(1,-q,c)$ would belong to $E$. $\hfill\square$ Corollary. If $o(q)=\infty$, then all the $\Omega\mho$-components in $H$ are cycles or of type $\mathbb{Z},$ or $-\mathbb{N}$, or $\mathbb{N}$. Thus, any module in $H$ is semi-Gorenstein-projective or $\infty$-torsionfree. For $o(q)=\infty$, there are the following $\Omega\mho$-components of the form $-\mathbb{N}$ and $\mathbb{N}$:

[TABLE]

with arbitrary elements $c_{0},d_{1}\in k$ and $c_{t+1}=-\frac{1}{1-q^{t}}c_{t}$ for $t\geq 1$, whereas $d_{t+1}=-(1-q^{-t})d_{t}$ for $t\geq 0$. Of course, $(1,-q,c_{1})\in T$ and $(1,-1,d_{0})\in E$, thus the module $M(1,-q,c_{1})$ is pivotal semi-Gorenstein-projective, whereas $M(1,-1,d_{0})$ is pivotal $\infty$-torsionfree. (7.7) The case that $q$ has finite multiplicative order. Now let $o(q)=n<\infty$. Then the modules $M(1,-q^{t},c)$ with $0\leq t<n$ and $c\in k$ belong to $\Omega\mho$-components of the form $\mathbb{A}_{n}$. These $\Omega\mho$-components look as follows:

[TABLE]

with an arbitrary element $c_{1}\in k$ and $c_{t+1}=-\frac{1}{1-q^{t}}c_{t}$ for $1\leq t<n$ (of course, $(1,-1,c_{n})\in E$ and $(1,-q,c_{1})\in T$).

Corollary (7.3) asserts that the remaining modules $M(1,b,c)$ (those with $-b\notin q^{\mathbb{Z}}$) are Gorenstein-projective.

(7.9) Proof of Theorem (1.5). Torsionless modules: According to (2.7), an indecomposable torsionless module is isomorphic to a left ideal. Of course, $k$ is torsionless. According to (2.2), a 2-dimensional indecomposable left ideal is isomorphic to $\Lambda(x-y)$ or $\Lambda z$. According to (2.3), a 3-dimensional indecomposable torsionless module has to be local, thus it is of the form $M(a,b,c)$, and (6.1) says that $a(a+q^{-1}b)\neq 0$ or else $M(a,b,c)$ is equal to $M(0,1,0)$ or to $M(0,0,1)$. Extensionless modules: We show: An indecomposable module $M$ of dimension at most $3$ with simple socle is not extensionless.

Of course, $\operatorname{Ext}^{1}(k,\Lambda)\neq 0$, since otherwise we would have $\operatorname{Ext}^{1}(X,\Lambda)=0$ for all modules $X$.

Let $I$ be an indecomposable module of length 2. A projective cover of $I$ as an $\overline{\Lambda}$-module provides an exact sequence $0\to k^{2}\to\overline{\Lambda}\to I\to 0$. We apply $\operatorname{Hom}_{\overline{\Lambda}}(-,J)$, where $J=\operatorname{rad}\Lambda$. We obtain the exact sequence

[TABLE]

Now, $\dim\operatorname{Hom}_{\overline{\Lambda}}(I,J)\geq\dim\operatorname{Hom}_{\overline{\Lambda}}(k,J)=2,$ $\dim\operatorname{Hom}_{\overline{\Lambda}}(\overline{\Lambda},J)=\dim J=5,$ and finally $\dim\operatorname{Hom}_{\overline{\Lambda}}(k^{2},J)=4,$ thus $\dim\operatorname{Ext}_{\overline{\Lambda}}^{1}(I,J)\geq 1.$ This shows that there exists a non-split exact sequence $\epsilon\>0\to J@>u>>E\to I\to 0$ with some $\overline{\Lambda}$-module $E$. The inclusion map $\iota\>J\to\Lambda$ yields an induced exact sequence $\epsilon^{\prime}\>0\to\Lambda\to E^{\prime}\to I\to 0$. Assume that $\epsilon^{\prime}$ splits. Then we obtain a map $v\>E\to\Lambda$ such that $vu=\iota$. Now $E$ is an $\overline{\Lambda}$-module, thus of Loewy length at most $2$. Therefore $v\>E\to\Lambda$ maps into $\operatorname{rad}\Lambda=J$, thus $v=\iota v^{\prime}$ for some $v^{\prime}\>E\to J$. But $\iota v^{\prime}u=vu=\iota$ implies that $v^{\prime}u$ is the identity map of $E$, thus $\epsilon$ splits, a contradiction. The exact sequence $\epsilon^{\prime}$ shows that $\operatorname{Ext}_{\Lambda}^{1}(I,\Lambda)\neq 0.$ Thus $I$ is not extensionless.

A similar proof shows that $\operatorname{Ext}^{1}(V,\Lambda)\neq 0$ for any 3-dimensional module $V$ with simple socle. Again, we use that $V$ is an $\overline{\Lambda}$-module (see (1.3) Proposition 1), thus we start with an exact sequence $0\to k^{5}\to\overline{\Lambda}^{2}\to V\to 0.$

This completes the proof that an indecomposable module $M$ of dimension at most $3$ with simple socle is not extensionless. The remaining indecomposable modules of dimension at most 3 are the modules of the form $M(1,b,c)$. According to (5.1) $M(1,b,c)$ is extensionless if and only if $b\neq-1.$

Reflexive modules: We recall from Part I that a module $M$ is reflexive if and only if both $M$ and $\mho M$ are torsionless. We show: A module $M$ with simple socle is not reflexive. Assume that $M$ has simple socle and is torsionless. Since $M$ has simple socle, there is an embedding $M\to{}_{\Lambda}\Lambda,$ say with cokernel $Q$. The elements $yx$ and $zx$ cannot both belong to $u(M)$, since the socle of $u(M)$ is simple. If $yx\notin u(M)$, then $yxQ\neq 0$, otherwise $zxQ\neq 0$. Let $f\>M\to{}_{\Lambda}\Lambda^{t}$ be a minimal left $\operatorname{add}(\Lambda)$-approximation; its cokernel is $\mho M$. There is $u^{\prime}\>{}_{\Lambda}\Lambda^{t}\to\Lambda$ with $u^{\prime}f=u.$ The map $u^{\prime}$ has to be surjective, since otherwise $u^{\prime}$ would vanish on the socle of ${}_{\Lambda}\Lambda^{t}$. This implies that the map $\mho M\to Q$ induced by $u^{\prime}$ is also surjective. Since $\mho M$ is indecomposable, non-projective and not annihilated by $\operatorname{rad}^{2}\Lambda$, $\mho M$ cannot be torsionless.

Let us assume that $M$ is reflexive and $\dim M\leq 3$. It follows that $M$ has to be a torsionless module with $\dim M=3.$ Since also $\mho M$ has to be torsionless, (6.5) shows that the cases $M(0,1,0)$ and $M(0,0,1)$ are not possible, thus $M$ is of the form $M(1,b,c)$ with $b\neq-q.$ Using (6.2) and (6.1), we see that we also must have $b\neq-q^{2}.$ Conversely, the same references show that all the modules $M(1,b,c)$ with $b\neq-q^{i}$ for $i=1,2$ are reflexive. Semi-Gorenstein-projective and $\infty$-torsionfree modules. The semi-Gorenstein-projective modules are extensionless, the $\infty$-torsionfree modules are reflexive. The previous considerations therefore show that we only have to consider the modules of the form $M(1,b,c)$. (7.3) provides the conditions on $b$ so that $M(1,b,c)$ is semi-Gorenstein-projective, $\infty$-torsionfree, or Gorenstein-projective.

If $M(1,b,c)$ is pivotal semi-Gorenstein-projective, then $M(1,b,c)$ is not torsionless, thus $b=-q.$ If $M(1,-q,c)$ is semi-Gorenstein-projective, then $-q\neq-q^{-s}$ for all $s\geq 0$, thus $q^{s+1}\neq 1$ for all $s\geq 0$. This means that $o(q)=\infty.$ Of course, there is also the converse: if $o(q)=\infty$, then $M(1,-q,c)$ is pivotal semi-Gorenstein-projective.

A similar argument shows that $M(1,b,c)$ is pivotal $\infty$-torsionfree if and only if $o(q)=\infty$ and $b=-1.$ $\hfill\square$ Remark. It seems worthwhile to note that the set of modules $M(1,b,c)$ with $b,c\in k$ is a union of $\Omega\mho$-components.

8. Right modules. Recall that we write $U^{\prime}(a,b,c)$ instead of $U(a,b,c)$, if we consider $U(a,b,c)$ as a right ideal and that $M^{\prime}(a,b,c)=\Lambda_{\Lambda}/U^{\prime}(a,b,c)$. (8.1) Proposition. Let $(a,b,c)\neq 0.$ Then

[TABLE]

Proof. We have $\Omega M^{\prime}(a,b,c)=U^{\prime}(a,b,c)_{\Lambda}$. According to (2.8), $U^{\prime}(a,b,c)_{\Lambda}=(a,b,c)\Lambda$ if $a\neq 0$ or $bc\neq 0$, and $U^{\prime}(0,1,0)=y\Lambda\oplus zx\Lambda$, $U^{\prime}(0,0,1)=z\Lambda\oplus yx\Lambda.$

Consider the map $\pi\>\Lambda_{\Lambda}\to U^{\prime}(a,b,c)$ defined by $\pi(1)=(a,b,c)$. We assume that $a\neq 0$ or $bc\neq 0$, thus $\pi$ is surjective. If $a\neq 0$, the formula (3.1) (3) asserts that $\omega^{\prime}(a,b,c)$ is in the kernel of $\pi$, thus $\pi$ yields an epimorphism $M^{\prime}(\omega^{\prime}(a,b,c))=\Lambda_{\Lambda}/\omega^{\prime}(a,b,c)\Lambda\to U^{\prime}(a,b,c)$. Since this is a map between 3-dimensional modules, it has to be an isomorphism.

If $a=0$ and $bc\neq 0$, we use formula (3.1) (4) in order to get similarly an isomorphism $M^{\prime}(0,0,1)=\Lambda_{\Lambda}/(0,0,1)\Lambda\to U^{\prime}(0,b,c)$. $\hfill\square$

(8.2) If a $3$-dimensional indecomposable right module $N$ is torsionless and no embedding $N\to\Lambda_{\Lambda}$ is a left $\operatorname{add}(\Lambda_{\Lambda})$-approximation, then $\mho N$ has Loewy length $3$ and is not torsionless. Proof. Let $\phi\>N\to\Lambda_{\Lambda}^{t}$ be a minimal left $\operatorname{add}(\Lambda_{\Lambda})$-approximation of $N$. Since $N$ is torsionless, $\phi$ is a monomorphism. By assumption, we must have $t\geq 2.$ It follows that the cokernel $\mho N$ of $\phi$ is an indecomposable right $\Lambda$-module of length $6t-3$ with top of length $t$. But an indecomposable right $\Lambda$-module of Loewy length at most $2$ with top of length $t\geq 2$ is a right $\overline{\Lambda}$-module of length at most $4t-1$. Thus $6t-3\leq 4t-1$, therefore $2t\leq 2$, thus $t\leq 1$, a contradiction. This shows that $\mho N$ has Loewy length equal to $3$. Of course, $\mho N$ is not projective. Since an indecomposable non-projective torsionless right $\Lambda$-module has Loewy length at most $2$, we see that $\mho N$ cannot be torsionless. $\hfill\square$

(8.3) The right modules $M^{\prime}(0,b,c)$. The only right module of the form $M^{\prime}(0,b,c)$ which is torsionless is $M^{\prime}(0,0,1).$ The right module $\mho M^{\prime}(0,0,1)$ has Loewy length $3$ and thus it is not torsionless. No right module of the form $M^{\prime}(0,b,c)$ is extensionless. Proof. Let $N=M^{\prime}(0,b,c)$.

(a) If $N$ is torsionless, then $b=0$ (thus $(0\,\mathpunct{:}b\,\mathpunct{:}c)=(0\,\mathpunct{:}0\,\mathpunct{:}1)$). Namely, According to (2.9), $M^{\prime}(0,b,c)$ arises as a right ideal and (8.1) shows that this happens only for $b=0$.

(b) No embedding $M^{\prime}(0,0,1)\to\Lambda_{\Lambda}$ is a left $\operatorname{add}(\Lambda_{\Lambda})$-approximation. Proof. Let $\phi\>M^{\prime}(0,0,1)\to\Lambda_{\Lambda}$ be an embedding. According to (2.9), the image of $\phi$ is of the form $U^{\prime}(0,b,c)$ with $bc\neq 0.$ Now $M^{\prime}(0,0,1)$ has a factor module isomorphic to $(0,0,1)\Lambda$, thus there is $f\>M^{\prime}(0,0,1)\to\Lambda_{\Lambda}$ with image $(0,0,1)\Lambda$. If $\phi$ is a left $\operatorname{add}(\Lambda_{\Lambda})$-approximation, then there exists $f^{\prime}:\Lambda_{\Lambda}\to\Lambda_{\Lambda}$ with $f=f^{\prime}\phi$. The homomorphism $f^{\prime}$ is the left multiplication by some element $\lambda$ in $\Lambda$. If $\lambda$ belongs to $\operatorname{rad}\Lambda$, then the image of $f^{\prime}\phi$ is contained in $\operatorname{rad}^{2}\Lambda=\operatorname{soc}\Lambda$. If $\lambda$ is invertible, then the image of $f^{\prime}\phi$ is 3-dimensional. In both cases, we get a contradiction, since the image of $f$ is $(0,0,1)\Lambda$, thus $2$-dimensional and not contained in $\operatorname{soc}\Lambda$.

(c) It follows from (8.2) that $\mho M^{\prime}(0,0,1)$ has Loewy length 3 and is not torsionless.

(d) A right module of the form $M^{\prime}(0,b,c)$ is never extensionless: either $\Omega M^{\prime}(0,b,c)$ is decomposable, or else $\Omega M^{\prime}(0,b,c)=M^{\prime}(0,0,1)$ and according to (b), no embedding $M^{\prime}(0,0,1)\to\Lambda_{\Lambda}$ is a left $\operatorname{add}(\Lambda_{\Lambda})$-approximation. $\hfill\square$ Reformulation. The right modules $M^{\prime}(0,1,c)$ are singletons in the $\Omega\mho$-quiver. The right module $M^{\prime}(0,0,1)$ belongs to an $\Omega\mho$-component of the form $\mathbb{A}_{2}$:

[TABLE]

(8.4) The right modules $M^{\prime}(1,b,c)$ with $c\neq 0$. Proposition. Let $c\neq 0.$ The right module $M^{\prime}(1,b,c)$ is torsionless if and only if $b\neq-1$, and then $\mho M^{\prime}(1,b,c)=M^{\prime}(\omega(1,b,c))$. Let $c^{\prime}\neq 0.$ The right module $M^{\prime}(1,b^{\prime},c^{\prime})$ is extensionless if and only if $b^{\prime}\neq-q$, and then $\Omega M^{\prime}(1,b^{\prime},c^{\prime})=M^{\prime}(\omega^{\prime}(1,b^{\prime},c^{\prime}))$. Remark. If $b\neq-1$ and $c\neq 0$, then $\omega(1,b,c)=(1,b^{\prime},c^{\prime})$ with $b^{\prime}\neq-q$ and some $c^{\prime}\neq 0$. If $b^{\prime}\neq-q$, then $\omega^{\prime}(1,b^{\prime},c^{\prime})=(1,b,c)$ with $b\neq-1$ and some $c\neq 0.$ Thus, the proposition provides $\Omega\mho$-sequences

[TABLE]

with $b\neq-1$ and $b^{\prime}\neq-q$ (and both $c,c^{\prime}$ being non-zero). Any triple $(1,b,c)$ with $b\neq-1$ and $c\neq 0$ occurs on the left and given $(1,b,c)$, then we have $(1,b^{\prime},c^{\prime})=\omega(1,b,c)$ on the right. Any triple $(1,b^{\prime},c^{\prime})$ with $b^{\prime}\neq-q$ and $c^{\prime}\neq 0$ occurs on the right and given $(1,b^{\prime},c^{\prime})$, then we have $(1,b,c)=\omega^{\prime}(1,b^{\prime},c^{\prime})$ on the left.

Proof of Proposition. We follow closely the proof of (5.1) and (6.1). We always assume that $c\neq 0$. As in (5.2) one sees that $M^{\prime}(1,b,c)$ is extensionless if and only if the image of every homomorphism $U^{\prime}(1,b,c)\to\Lambda_{\Lambda}$ is contained in $U^{\prime}(1,b,c)$. (a) The module $M^{\prime}(1,-q,c)$ is not extensionless. Proof. According to (8.1), we have $U^{\prime}(1,-q,c^{\prime})\simeq\Omega M^{\prime}(1,-q,c^{\prime})\simeq M^{\prime}(\omega^{\prime}(1,-q,c^{\prime}))=M^{\prime}(1,-1,0)$ for all $c^{\prime}\in k.$ Thus, there is a homomorphism $U^{\prime}(1,-q,0)\to\Lambda_{\Lambda}$ with image $U^{\prime}(1,-q,0)$ and this image $U^{\prime}(1,-q,0)$ is not contained in $U^{\prime}(1,-q,c).$ (b) If $b\neq-q,$ then the module $M^{\prime}(1,b,c)$ is extensionless. For the proof, we need three assertions (b1), (b2) (b3). Note that (8.1) asserts that $U^{\prime}(1,b,c)\simeq\Omega M^{\prime}(1,b,c)\simeq M^{\prime}(\omega^{\prime}(1,b,c))=M^{\prime}(1,q^{-1}b,c^{\prime})$, where $\omega^{\prime}(1,b,c)=(1,q^{-1}b,c^{\prime})$. (b1) The only right ideal isomorphic to $U^{\prime}(1,b,c)$ is $U^{\prime}(1,b,c)$ itself. Proof. Let $V$ be a right ideal of $\Lambda_{\Lambda}$ which is isomorphic to $U^{\prime}(1,b,c)$, say $V=U^{\prime}(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})$ for some triple $(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})$. By (8.1), we have $U(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})\simeq\Omega M^{\prime}(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})=M^{\prime}(a^{\prime\prime},q^{-1}b^{\prime\prime},d)$, where $\omega^{\prime}(a^{\prime\prime},b^{\prime\prime},c^{\prime\prime})=(a^{\prime\prime},q^{-1}b^{\prime\prime},d)$ for some $d$. We must have $a^{\prime\prime}\neq 0$, since $M(a^{\prime\prime},q^{-1}b^{\prime\prime},d)\simeq U^{\prime}(1,b,c)\simeq M^{\prime}(1,q^{-1}b,c^{\prime})$. Thus, we may assume that $a^{\prime\prime}=1$ and then $M^{\prime}(1,q^{-1}b^{\prime\prime},d)\simeq M^{\prime}(1,q^{-1}b,c^{\prime})$ implies that $(1,q^{-1}b^{\prime\prime},d)=(1,q^{-1}b,c^{\prime})$. In particular, we have $b^{\prime\prime}=b\neq-q$. The equality $\omega^{\prime}(1,b^{\prime\prime},c^{\prime\prime})=\omega^{\prime}(1,b,c)$ yields $(1,b^{\prime\prime},c^{\prime\prime})=(1,b,c)$, see Proposition (3.2). Therefore $V=U(1,b^{\prime\prime},c^{\prime\prime})=U(1,b,c)$. (b2) The right ideal $z\Lambda$ is not a factor module of $U^{\prime}(1,b,c)$. Proof. The right ideal $z\Lambda$ is annihilated by $x-y$ and $z$, thus Corollary (5.3) asserts that the image $I$ of any homomorphism $M^{\prime}(1,b^{\prime},c^{\prime})\to z\Lambda$ is a factor module of $\Lambda/((1,b,c)\Lambda+(x-y)\Lambda+z\Lambda).$ Now $(x+by+cz)\Lambda+(x-y)\Lambda+z\Lambda=\operatorname{rad}\Lambda$, since $b\neq-1$, thus $I$ is simple or zero. (b3) The right ideal $y\Lambda$ is not a factor module of $U^{\prime}(1,b,c)$. Proof. The right ideal $y\Lambda$ is annihilated by $y$ and $z$, thus Corollary (5.3) asserts that the image $I$ of any homomorphism $M^{\prime}(1,b^{\prime},c^{\prime})\to y\Lambda$ is a factor module of $\Lambda/((1,b,c)\Lambda+y\Lambda+z\Lambda).$ Now $(x+by+cz)\Lambda+y\Lambda+z\Lambda=\operatorname{rad}\Lambda$, since $b\neq-1$, thus $I$ is simple or zero.

The assertions (b1), (b2) and (b3) show: if $\phi$ is any homomorphism $U^{\prime}(1,b,c)\to\Lambda_{\Lambda}$ and its image $I$ is of dimension at least 2, then $I$ is contained in $U^{\prime}(1,b,c)$. Of course, if $I$ is 1-dimensional, then $I$ is contained in $\operatorname{soc}\Lambda_{\Lambda}$ and $\operatorname{soc}\Lambda_{\Lambda}\subseteq U^{\prime}(1,b,c)$. Thus, we have obtained a proof of (b). In addition, (8.1) asserts that $\Omega M^{\prime}(1,b,c)\simeq M^{\prime}(\omega^{\prime}(1,b,c))$. (c) If $b\neq-1$, then $M^{\prime}(1,b,c)$ is torsionless and $\mho M^{\prime}(1,b,c)=M^{\prime}(\omega(1,b,c))$. Proof. Let $\omega(1,b,c)=(1,b^{\prime},c^{\prime}).$ Then $b^{\prime}=qb\neq-q$, and $\omega^{\prime}(1,b^{\prime},c^{\prime})=\omega^{\prime}\omega(1,b,c)=(1,b,c)$ by Proposition (3.2). According to (8.1), we have $\Omega M^{\prime}(1,b^{\prime},c^{\prime})\simeq M^{\prime}(\omega^{\prime}(1,b^{\prime},c^{\prime}))=M^{\prime}(1,b,c)$. This shows that $M^{\prime}(1,b,c)$ is torsionless.According to (b), the module $M^{\prime}(\omega(1,b,c))$ is extensionless, thus $\mho M^{\prime}(1,b,c)=M^{\prime}(1,b^{\prime},c^{\prime})=M^{\prime}(\omega(1,b,c))$. (d) The modules $M^{\prime}(1,-1,c)$ are not torsionless. Proof. Assume, for the contrary, that $M^{\prime}(1,-1,c)$ is torsionless, thus isomorphic to $U^{\prime}(a^{\prime},b^{\prime},c^{\prime})$ for some $(a^{\prime},b^{\prime},c^{\prime})$. According to (8.1), we must have $a^{\prime}\neq 0$, thus we can assume that $a^{\prime}=1,$ and $(1,-1,c)=\omega^{\prime}(1,b^{\prime},c^{\prime})=(1,q^{-1}b^{\prime},-(1+q^{-1}b^{\prime})c^{\prime})$. It follows that $b^{\prime}=-q$ and therefore $c=-(1+q^{-1}b^{\prime})c^{\prime}=0,$ a contradiction.

This completes the proof of (8.4). $\hfill\square$

Reformulation. The neighborhood of $M^{\prime}(1,b,c)$ with $c\neq 0$ in the $\Omega\mho$-quiver looks as follows:

[TABLE]

and $M^{\prime}(1,b,c)$ is a singleton in the $\Omega\mho$-quiver if $q=1$ and $b=-1$. Note that we want to use a fixed index set $\mathbb{P}^{2}$ both for the (left) modules $M(a\,\mathpunct{:}b\,\mathpunct{:}c)$ and the right modules $M^{\prime}(a\,\mathpunct{:}b\,\mathpunct{:}c)$, Since we have drawn the dashed arrows in the $\Omega\mho$-quiver of the left $\Lambda$-modules from right to left, we now have drawn the dashed arrows in the $\Omega\mho$-quiver of the right $\Lambda$-modules from left to right. As in section 7, we see that the $\Omega\mho$-components of the modules $M^{\prime}(1,b,c)$ with $c\neq 0$ are cycles, or of type $\mathbb{Z},\mathbb{N}$ or $-\mathbb{N}$ in case $o(q)=\infty$, and cycles or of type $\mathbb{Z}$ or $\mathbb{A}_{n}$ in case $o(q)=n<\infty.$ For $o(q)=\infty$, the right modules $M^{\prime}(1,-1,c)$ with $c\neq 0$ are pivotal semi-Gorenstein-projective, and the right modules $M^{\prime}(1,-q,c)$ with $c\neq 0$ are pivotal $\infty$-torsionfree.

(8.5) The right modules $M^{\prime}(1,b,0)$. The right modules $M^{\prime}(1,b,0)$ have been considered already in Part I: these are just the right ideals $m_{\alpha}\Lambda$, where $m_{\alpha}=x-\alpha y$. Namely, we have

[TABLE]

for all $b\in k.$ (Proof: We have $M^{\prime}(1,b,0)=\Lambda_{\Lambda}/U^{\prime}(1,b,0)=\Lambda_{\Lambda}/(x+by)\Lambda\simeq(x+qby)\Lambda$, where we use that $(x+qby)(x+by)=0$ and that both right ideals $(x+by)\Lambda$ and $(x+qby)\Lambda$ are 3-dimensional, see (2.8).) Let us recall the results presented in Part I using the present notation: If $b\notin-q^{\mathbb{Z}}$, then $M^{\prime}(1,b,0)$ is Gorenstein-projective and its $\Omega\mho$-component looks as follows:

[TABLE]

In particular, if $o(q)=n$, then these $\Omega\mho$-components are cycles with $n$ vertices, whereas for $o(q)=\infty$, one obtains $\Omega\mho$-components of type $\mathbb{Z}$.

For $o(q)=\infty$, there are three remaining $\Omega\mho$-components:

[TABLE]

These $\Omega\mho$-components are of type $\mathbb{N},\mathbb{A}_{2}$ and $-\mathbb{N}$, respectively.

For $2\leq n=o(q)<\infty$, there are two remaining $\Omega\mho$-components, one is of type $\mathbb{A}_{2}$, the other of type $\mathbb{A}_{n}$:

[TABLE]

In case $q=1$, there is only one additional $\Omega\mho$-component (of type $\mathbb{A}_{2}$), namely

[TABLE]

(8.6) Similar to Theorem (1.5), here is the summary which characterizes the right modules of dimension at most 3 with relevant properties. Theorem. An indecomposable right module $N$ of dimension at most $3$ is

*$\bullet$ *torsionless if and only if $N$ is simple or isomorphic to $y\Lambda$, to $z\Lambda$, to a module $M^{\prime}(1,b,c)$ with $b\neq-1$, to $M^{\prime}(1,-1,0)$ or to $M^{\prime}(0,0,1).$

*$\bullet$ *extensionless if and only if $N$ is isomorphic to a module $M^{\prime}(1,b,c)$ with $b\neq-q$;

*$\bullet$ *reflexive if and only if $M$ is isomorphic to a module $M^{\prime}(1,b,c)$ with $b\neq-q^{i}$ for $i=-1,0$;

*$\bullet$ *Gorenstein-projective if and only if $N$ is isomorphic to a module $M^{\prime}(1,b,c)$ with $b\neq-q^{i}$ for $i\in\mathbb{Z}$;

*$\bullet$ *semi-Gorenstein-projective if and only if $N$ is isomorphic to a module $M^{\prime}(1,b,c)$ with $b\neq-q^{i}$ for $i\geq 0$ or to a module $M^{\prime}(1,-1,c)$ with $c\neq 0$;

$\bullet$ $\infty$-torsionfree if and only if $N$ is isomorphic to a module $M^{\prime}(1,b,c)$ with $b\neq-q^{i}$ for $i\leq 0$;

*$\bullet$ *pivotal semi-Gorenstein-projective if and only if $o(q)=\infty$ and $N$ is isomorphic to a module $M^{\prime}(1,-1,c)$ with $c\neq 0$;

*$\bullet$ *pivotal $\infty$-torsionfree if and only if $o(q)=\infty$ and $N$ is isomorphic to a module $M^{\prime}(1,-q,c)$.

$\hfill\square$ Whereas the set of modules $M(1,b,c)$ with $b,c\in k$ is a union of $\Omega\mho$-components, the right modules behave differently: as we have seen already in Part I, 7.2, the $\Omega\mho$-component containing the right module $M(1,-1,0)$ consists of $M(1,-1,0)$ and the 9-dimensional right module $\mho M(1,-1,0)$.

9. The $\Lambda$-dual of $M(1,b,c)$ and $M^{\prime}(1,b,c)$. We need the following (of course well-known) Lemma. (9.1) Lemma. Let $R$ be a ring and $w\in R$. If any left-module homomorphism $Rw\to{}_{R}R$ maps $w$ into $wR$, then $\operatorname{Hom}(Rw,{}_{R}R)\simeq wR$ as right $R$-modules. Proof. Let $u\>Rw\to{}_{R}R$ be the inclusion map. We have $\operatorname{Hom}(Rw,{}_{R}R)=uR,$ since for any homomorphism $f\>Rw\to{}_{R}R$, there is $\lambda\in R$ with $f(w)=w\lambda$, thus $f=u\lambda.$ Now $I=\{r\in R\mid wr=0\}$ is a right ideal and $R_{R}/I\simeq wR$ as right modules (an isomorphism is given by the map $R_{R}\to wR$ defined by $1\mapsto w$). Since $I=\{r\in R\mid ur=0\}$, we have in the same way $R_{R}/I\simeq uR,$ and therefore $wR\simeq R_{R}/I\simeq uR=\operatorname{Hom}(Rw,{}_{R}R).$ $\hfill\square$

(9.2) Lemma. If $(1,b,c)$ is different from $(1,-1,0)$, then $M^{\prime}(1,b,c)\simeq\operatorname{Tr}M(1,b,c)$ and $M(1,b,c)\simeq\operatorname{Tr}M^{\prime}(1,b,c)$. Proof. We have $U^{\prime}(1,b,c)=(1,b,c)\Lambda$, and since $(1,b,c)\neq(1,-1,0)$, we also have $U(1,b,c)=\Lambda(1,b,c)$. By definition, $M(1,b,c)={}_{\Lambda}\Lambda/U(1,b,c)$, thus $M(1,b,c)$ is the cokernel of the right multiplication $r_{(1,b,c)}\>{}_{\Lambda}\Lambda\to{}_{\Lambda}\Lambda$ and $\operatorname{Tr}M(1,b,c)$ is the cokernel of the left multiplication $l_{(1,b,c)}\>\Lambda_{\Lambda}\to\Lambda_{\Lambda}$, thus isomorphic to $\Lambda_{\Lambda}/(1,b,c)\Lambda=\Lambda_{\Lambda}/U^{\prime}(1,b,c)$. $\hfill\square$ (9.3) Proposition. If $b\notin\{-q,-q^{2}\}$, then $M(1,b,c)$ is reflexive and

[TABLE]

*If $b\notin\{-1,-q^{-1}\}$, then $M^{\prime}(1,b,c)$ is reflexive and *

[TABLE]

Proof. According to (7.1), we have the following two $\Omega\mho$-sequences:

[TABLE]

(the first one, since $\omega^{\prime}(1,b,c)=(1,b^{\prime},c^{\prime})$ with $b^{\prime}=q^{-1}b\neq-1;$ the second one, since $(\omega^{\prime})^{2}(1,b,c)=(1,b^{\prime\prime},c^{\prime\prime})$ with $b^{\prime\prime}=q^{-2}b\neq-1$) This implies that $M(1,b,c)$ is reflexive and that $X=\mho^{2}M(1,b,c)=M((\omega^{\prime})^{2}(1,b,c))$ is a module with $\operatorname{Ext}^{i}(X,\Lambda)=0$ for $i=1,2.$ According to Part I, Lemma 2.5, we have $\operatorname{Tr}X=(\Omega^{2}X)^{*}.$ On the one hand, $\Omega^{2}X=\Omega\mho M(1,b,c)=M(1,b,c)$. On the other hand, (9.2) shows that $\operatorname{Tr}X=\operatorname{Tr}M((\omega^{\prime})^{2}(1,b,c))=M^{\prime}((\omega^{\prime})^{2}(1,b,c))$, since $(\omega^{\prime})^{2}(1,b,c)=(1,q^{-2}b,c^{\prime\prime})$ for some $c^{\prime\prime}$ and $q^{-2}b\neq-1$. This yields the first assertion. The second can be shown in the same way, or just by applying the $\Lambda$-duality to $M(1,b,c)^{*}=M^{\prime}((\omega^{\prime})^{2}(1,b,c))$. $\hfill\square$ (9.4) Proposition. For all $b,c\in k$,

[TABLE]

In particular, for all $b,c\in k$, the right module $M(1,b,c)^{*}$ is again 3-dimensional and local. Whereas $(\omega^{\prime})^{2}$ is a bijection from $\{(1,b,c)\mid b\notin\{-q,-q^{2}\}\}$ onto $\{(1,b,c)\mid b\notin\{-1,-q^{-1}\}\}$, we should stress that $(\omega^{\prime})^{2}(1,-q,c)=(1,-q^{-1},0)$ and that $(\omega^{\prime})^{2}(1,-q^{2},c)=(1,-1,0)$ for all $c\in k.$ Thus, (9.3) combines the first assertion of (9.2) with the corresponding assertion for the remaining cases, namely:

[TABLE]

for all $c\in k$.

Proof of Proposition. According to (9.2), we only have to consider the cases where $b=-q$ or $b=-q^{2}.$ Case 1. Let $b=-q.$ As we have seen in (6.2), the module $M(1,-q,c)$ is not torsionless. Now obviously, there is a surjective homomorphism $M(1,-q,c)\to\Lambda(1,-1,0)$ with kernel $zM(1,-q,c)$. It follows that $zM(1,-q,c)$ is contained in the kernel of every homomorphism $M(1,-q,c)\to{}_{\Lambda}\Lambda$ and therefore $M(1,-q,c)^{*}=(\Lambda(1,-1,0))^{*}$. Now, $(\Lambda(1,-1,0))^{*}\simeq(1,-1,0)\Lambda=U^{\prime}(1,-1,0)$, as shown in Part I, 6.5. On the other hand, according to (8.1), we have $U^{\prime}(1,-1,0)=\Omega M^{\prime}(1,-1,0)=M^{\prime}(\omega^{\prime}(1,-1,0))$ and $\omega^{\prime}(1,-1,0)=(1,-q^{-1},0).$ Case 2: $b=-q^{2}$ and $o(q)=2.$ The assumption $o(q)=2$ means that $q=-1\neq 1$, in particular, the characteristic of $k$ is different from $2$, and we have $b=-1.$ Since $q=-1$ and the characteristic of $k$ is different from $2$, (4.1) asserts that

[TABLE]

On the other hand, we have

[TABLE]

We claim that any homomorphism $\Lambda(1,1,-2c)\to{}_{\Lambda}\Lambda$ maps $(1,1,-2c)$ into $(1,1,-2c)\Lambda$. Namely, let $\phi\>\Lambda(1,1,-2c)\to{}_{\Lambda}\Lambda$ be a homomorphism. Now $\Lambda(1,1,-2c)$ is 3-dimensional, thus equal to $U(1,1,-2c)$, and ${}_{\Lambda}\Lambda/U(1,1,-2c)\simeq M(1,1,-2c).$ According to (5.1), the module $M(1,1,-2c)$ is extensionless, since $1+1\neq 0$. The implication (i) to (iv) in (5.2) shows that $\phi(1,1,-2c)\in(1,1,-2c)\Lambda$.

Since any homomorphism $\Lambda(1,1,-2c)\to{}_{\Lambda}\Lambda$ maps $(1,1,-2c)$ into $(1,1,-2c)\Lambda$, Lemma (9.0) implies that the right modules $(\Lambda(1,1,-2c))^{*}$ and $(1,1,-2c)\Lambda$ are isomorphic, thus $M(1,-1,c)^{*}\simeq M^{\prime}(1,-1,0).$ Case 3. $b=-q^{2}$ and $o(q)\geq 3$. There is the $\Omega\mho$-sequence

[TABLE]

for some $c^{\prime}$ (here we use that $q^{2}\neq 1$). The $\Lambda$-dual of $\epsilon$ is the exact sequence

[TABLE]

Since $q^{2}\neq 1$, proposition (9.3) asserts that $M(1,-q^{3},c^{\prime})^{*}=M^{\prime}(1,-q,c^{\prime\prime})$ for some $c^{\prime\prime}$. Altogether we see that

[TABLE]

where the final isomorphism is due to (8.1). $\hfill\square$ (9.5) The algebra $\Lambda=\Lambda(q)$ with $o(q)=\infty$ was exhibited in Part I in order to present a module $M$ which is not torsionless, such that $M$ and $M^{*}$ both are semi-Gorenstein-projective: namely the module $M=M(1,-q,0)$ with $M^{*}=M^{\prime}(1,-q,0)$. Now we see: all the modules $M(1,-q,c)$ with $c\in k$ are modules which are semi-Gorenstein-projective and not torsionless, and that the $\Lambda$-duals $M(1,-q,c)^{*}\simeq M^{\prime}(1,-q^{-1},0)$ are semi-Gorenstein-projective. We should stress that this concerns a 1-parameter family $M(1,-q,c)$ (with $c\in k$) of semi-Gorenstein-projective left modules, and the single semi-Gorenstein-projective right module $M(1,-q^{-1},0)$. (9.6) Proposition. Let $b,c\in k.$

[TABLE]

Whereas we saw in (9.4) that all the right modules $M(1,b,c)^{*}$ are 3-dimensional and local, not all the modules $M^{\prime}(1,b,c)^{*}$ are 3-dimensional and local: the module $M^{\prime}(1,-1,0)^{*}=U(1,-q,0)+U(0,0,1)$ has dimension 4, whereas the modules $M^{\prime}(1,-1,c)^{*}=U(0,0,1)$ for $c\neq 0$ and, in case $q\neq 1$, the module $M^{\prime}(1,-q^{-1},0)^{*}=U(1,-1,0)$ are decomposable. Proof. According to (9.3), we only have to deal with the cases with $b\in\{-1,-q^{-1}\}.$ If $c=0$, then we can refer to Part I. For $b=-1$, the end of 7.1 in Part I shows that $M^{\prime}(1,-1,0)^{*}\simeq M(1,-q^{2},0)^{**}\simeq U(1,-q,0)+U(0,0,1)$. For $b=-q^{-1}\neq-1$, the end of 6.7 in Part I asserts that $M^{\prime}(1,-q^{-1},0)^{*}\simeq(M(1,-q,0)^{**}\simeq\Omega M(1,-1,0)\simeq U(1,-1,0).$ Now, we assume that $c\neq 0$. As in the proof of (9.4), we consider again 3 cases. Case 1. $b=-1$. The module $M^{\prime}(1,-1,c)$ with $c\neq 0$ is not torsionless, see (8.4). Since the factor module $M^{\prime}(1,-1,c)/M^{\prime}(1,-1,c)z$ is isomorphic to $(0,0,1)\Lambda$, it follows that $M^{\prime}(1,-1,c)^{*}\simeq((0,0,1)\Lambda)^{*}$ and an easy calculation yields $((0,0,1)\Lambda)^{*}\simeq U(0,0,1)$. Namely, the inclusion map $u\>z\Lambda\to\Lambda_{\Lambda}$ satisfies $yu=0$ and $zu=0$, thus a basis of $(z\Lambda)^{*}$ is given by $u,\ xu$ and the map $f\>z\Lambda\to\Lambda_{\Lambda}$ with $f(z)=yx$, so that $(z\Lambda)^{*}\simeq{}_{\Lambda}\Lambda/(\Lambda y+\Lambda z)\oplus k\simeq U(0,0,1).$ Case 2. $b=-q^{-1}$ and $o(q)=2.$ Thus, the characteristic of $k$ is different from $2$, $q=-1$ and $b=1$. The module $M^{\prime}(1,1,c)$ is torsionless: namely, by (8.1) we have $M^{\prime}(1,1,c)\simeq\Omega M^{\prime}(1,-1,-\frac{c}{2}),$ since $\omega^{\prime}(1,-1,-\frac{c}{2})=(1,1,c)$. Now, $\Omega M^{\prime}(1,-1,-\frac{c}{2})\simeq U^{\prime}(1,-1,\frac{c}{2})=(1,-1,\frac{c}{2})\Lambda$. Since $q\neq 1$, the right module $M^{\prime}(1,-1,-\frac{c}{2})$ is extensionless by (8.4), thus we can use (5.2) and (9.1) in order to see that $((1,-1,\frac{c}{2})\Lambda)^{*}\simeq\Lambda(1,-1,\frac{c}{2}).$ By (4.1) (2), we have $\Lambda(1,-1,\frac{c}{2})=U(1,-1,\frac{c}{2})\simeq\Omega M((1,-1,-\frac{c}{2}))\simeq M(0,0,1).$ Case 3. $b=-q^{-1}$ and $o(q)\geq 3.$ There is the $\Omega\mho$-sequence

[TABLE]

for $c^{\prime}=\lambda c$ with $\lambda\neq 0$ (here we use that $q^{2}\neq 1$). The $\Lambda$-dual is the exact sequence

[TABLE]

We assume that $q\neq 1$ and $q\neq 2$. Then by Proposition (9.2), we have $M^{\prime}(1,-q^{-2},c^{\prime})^{*}=M(1,-1,c^{\prime\prime})$ for some multiple $c^{\prime\prime}=\lambda^{\prime}c^{\prime}$ with $\lambda^{\prime}\neq 0$. It follows that $M^{\prime}(1,-q^{-1},c)^{*}=\Omega M(1,-1,c^{\prime\prime})$ and $c^{\prime\prime}=0$ if and only if $c=0.$ By (4.1), we have $\Omega M(1,-1,c^{\prime\prime})=M(0,0,1)$ in case $c\neq 0$, and $\Omega M(1,-1,0)=U(1,-1,0)$ in case $c=0.$ $\hfill\square$ (9.7) Corollary. Let $N$ be a right $\Lambda$-module of dimension at most $3$ which is semi-Gorenstein-projective, but not Gorenstein-projective. Then $N^{*}$ is not semi-Gorenstein-projective. Proof. According to (8.6), $N$ is isomorphic to a right module of the form $M^{\prime}(1,-q^{i},c)$ with $i\leq-1$ and $c\in k$ or of the form $M^{\prime}(1,-1,c)$ with $c\neq 0.$ We apply (9.6). If $i\leq-2$, then $N^{*}=M^{\prime}(1,-q^{i},c)^{*}=M(1,-q^{i+2},c^{\prime})$ for some $c^{\prime}$, and according to (1.5), $N^{*}$ is not semi-Gorenstein-projective, since $i+2\leq 0$. If $i=-1$, then $N^{*}$ is isomorphic to $M(0,0,1)$ or to $U(1,-1,0)$. If $N=M^{\prime}(1,-1,c)$ with $c\neq 0,$ then $N^{*}$ is isomorphic to $U(0,0,1).$ But by (1.5), $M(0,0,1),$ $U(1,-1,0)$ and $U(0,0,1)$ are not semi-Gorenstein-projective. $\hfill\square$

10. The general context. Our detailed study of the algebra $\Lambda(q)$ in Part I and Part II should be seen in the frame of looking at Gorenstein-projective (or, more general, semi-Gorenstein-projective and $\infty$-torsionfree modules) over short local algebras.

Let $A$ be a finite-dimensional local $k$-algebra with radical $J$ such that $A/J=k$. Such an algebra is said to be short provided $J^{3}=0.$ In commutative ring theory, the short local algebras have attracted a lot of interest, since some conjectures have been disproved by looking at modules over short algebras, see [AIŞ] for a corresponding account.

Let us assume now that $A$ is short, but not necessarily commutative. Let $e=\dim J/J^{2}$ and $a=\dim J^{2}$ (thus $0\leq a\leq e^{2}$). If there exists an indecomposable module which is semi-Gorenstein-projective or $\infty$-torsionfree, but not projective, then either $A$ is self-injective, so that $a\leq 1$ (and $e=1$ in case $a=0$), or else $a=e-1$ and $J^{2}=\operatorname{soc}{}_{A}A=\operatorname{soc}A_{A}$.

Of course, if $A$ is self-injective, then all modules are Gorenstein-projective, thus the interesting case is the case $a=e-1$. Our algebra $\Lambda(q)$ is of this kind (with $a=2$), as is the Jorgensen-Şega algebra [JŞ] (with $a=3$).

Not only the shape of the algebras is very restricted, also the modules themselves are very special: Let $A$ be a short local algebra which is not self-injective. Let $M$ be indecomposable and not projective. If $M$ is semi-Gorenstein-projective and torsionless, or if $M$ is $\infty$-torsionfree, (in particular, if $M$ is Gorenstein-projective), then $\operatorname{soc}M=\operatorname{rad}M$ and $\dim\operatorname{soc}M=a\cdot\dim\operatorname{top}M$ (by definition, $\operatorname{top}M=M/\operatorname{soc}M$). Also, if $M$ is semi-Gorenstein-projective and torsionless, then $\dim\Omega^{i}M=\dim M$ for all $i\in\mathbb{N}$, whereas if $M$ is $\infty$-torsionfree, then $\dim\mho^{i}M=\dim M$ for all $i\in\mathbb{N}$. These assertions have been shown by Christensen and Veliche in the case that $A$ is commutative, see [CV], but actually the proofs do not have to be modified in the general case. There is an essential difference between the commutative and the non-commutative algebras: If $A$ is commutative, then all local modules which are semi-Gorenstein-projective or $\infty$-torsionfree are Gorenstein-projective, whereas this is not true for $A$ non-commutative. For a general discussion, we refer to [RZ2] (and we have to thank D. Jorgensen for his advice concerning the present knowledge in the commutative case).

Thus, for our algebra $\Lambda(q)$, the non-projective indecomposable modules which are semi-Gorenstein-projective and torsionless, or which are $\infty$-torsionfree, are of dimension $3t$ with socle of dimension $2t$, where $t=\dim\operatorname{top}M$. For $t=1$, we deal with local modules with 2-dimensional socle: these are precisely the modules studied in the present paper.

Appendix. A diagrammatic description of the modules $M(a\,\mathpunct{:}b\,\mathpunct{:}c)$. If $M$ is a left $\Lambda$-module annihilated by $\operatorname{rad}^{2}\Lambda$, then it is a left $\overline{\Lambda}$-module. Since $\overline{\Lambda}$ is a commutative $k$-algebra, also $D(M)=\operatorname{Hom}(M,k)$ is a left $\overline{\Lambda}$-module, thus a left $\Lambda$-module. Proposition. Let $M$ be an indecomposable $3$-dimensional left $\Lambda$-module. Then $M$ or $D(M)$ is isomorphic to one of the following pairwise non-isomorphic $\overline{\Lambda}$-modules $M(a,b,c)$:

[TABLE]

The diagrams describe the modules $M=M(a,b,c)$ as follows: The elements $v,v_{1},v_{2}$ form a basis of $M$. Both elements $v_{1},v_{2}$ are annihilated by $x,y,z$. If there is drawn a solid arrow $v$ ............................................................ $v_{i}$ with $i\in\{1,2\}$ and with label $\alpha\in\{x,y,z\}$, then $\alpha v=v_{i}.$ If there is a dashed arrow $v$ ........................................ $v_{i}$ with label $\alpha$, then $\alpha v=c_{1}v_{1}+c_{2}v_{2}$ with $c_{i}\neq 0$ (and we provide the coefficients $c_{1},c_{2}$ below the diagram). Finally, $zv=0$ in case (1), $yv=0$ in case (2), $xv=0$ in case (3).

The last column provides a characterization of the corresponding modules $M(a,b,c)$: For example, a local 3-dimensional $\Lambda$-module $M$ is a case-(1)-module provided $zM=0,$ and so on. Remark. If $M$ is an indecomposable $3$-dimensional $\Lambda$-module, then its annihilator is equal to $U(a,b,c)$ for some $(a,b,c)\neq 0$ and $M$ considered as a $\Lambda/U(a,b,c)$-module is either the unique indecomposable projective $\Lambda/U(a,b,c)$-module (and then a local module, thus isomorphic to $M(a,b,c)$) or the unique indecomposable injective $\Lambda/U(a,b,c)$-module (and then a module with simple socle, thus isomorphic to $D(M(a,b,c))$).

Proof of the Proposition and the Remark. First, let us assume that $M$ is local. According to (2.6) and (1.4), we know that $M\simeq M(a\,\mathpunct{:}b\,\mathpunct{:}c)$ for some $(a\,\mathpunct{:}b\,\mathpunct{:}c)\in\mathbb{P}^{2}$ and that these modules are pairwise non-isomorphic. As representatives of the elements of $\mathbb{P}^{2}$, we choose (as usual) the triples $(c_{1},c_{2},c_{3})$ with $c_{i}=1$ for some $i$ and $c_{j}=0$ for $j<i$. Clearly, there are the seven cases (1) to (7) as listed above. It remains to choose in every case a basis $\Cal{B}(a,b,c)=\{v,v_{1},v_{2}\}$ of $M(a,b,c)$. Recall that $M(a,b,c)=\overline{\Lambda}/(a\,\mathpunct{:}b\,\mathpunct{:}c)$ is a factor module of $\Lambda$ and $\overline{\Lambda}$ has the basis $\{1,x,y,z\}$. We choose as elements of $\Cal{B}(a,b,c)$ the residue class $v=\overline{1}$ as well as two of the three residue classes $\overline{x},\overline{y},\overline{z}$, namely $v_{1}=\overline{x}$ if $a=0$ and $v_{1}=\overline{y}$ otherwise, and then $v_{2}=\overline{y}$ in case $(a,b,c)=(0,0,1)$ and $v_{2}=\overline{z}$ otherwise. (We should remark that the vertices and the arrows of the diagram are those of the coefficient quiver $\Gamma(M(a,b,c),\Cal{B}(a,b,c))$ as considered in [R], and the solid arrows focus the attention to a spanning tree.)

Second, assume that $M$ is not local. Since $M$ is an indecomposable module of length 3 and Loewy length 2, it follows that $M$ has simple socle, thus $D(M)$ is local and therefore of the form (1) to (7).

Finally, $M$ and $D(M)$ have the same annihilator, this is a 3-dimensional ideal, thus of the form $U(a,b,c)$. The 3-dimensional local algebra $\Lambda/U(a,b,c)$ has a unique 3-dimensional local module, this is the indecomposable projective $\Lambda/U(a,b,c)$-module, and dually, it has a unique 3-dimensional module with simple socle, this is the unique indecomposable injective $\Lambda/U(a,b,c)$-module. This completes the proof. $\hfill\square$

Reference.

[AIS] L. L. Avramov, S. B. Iyengar, L. M. Şega. Free resolutions over short local rings. J. London Math. Soc. (2) 78 (2008), 459–476.

[CV] L. W. Christensen, O. Veliche. Acyclicity over local rings with radical cube zero. Illinois J. Mathematics. 51 (2007), 1439–1454.

[JŞ] D. A. Jorgensen, L. M. Şega. Independence of the total reflexivity conditions for modules. Algebras and Representation Theory 9 (2006), 217–226.

[R] C. M. Ringel. Exceptional modules are tree modules. Lin. Alg. Appl. 275–276 (1998). 471–493.

[RZ1] C. M. Ringel, P. Zhang: Gorenstein-projective and semi-Gorenstein-projective modules. To appear. arXiv:1808.01809v3.

[RZ2] C. M. Ringel, P. Zhang. Gorenstein-projective modules over short local algebras. In preparation.

C. M. Ringel

Fakultät für Mathematik, Universität Bielefeld

POBox 100131, D-33501 Bielefeld, Germany

ringelmath.uni-bielefeld.de

P. Zhang

School of Mathematical Sciences, Shanghai Jiao Tong University

Shanghai 200240, P. R. China.

pzhangsjtu.edu.cn