$\displaystyle(\frac{1}{2}-\delta)\left(\sum_{i}\frac{1}{c_{i}}\right)+(1-\delta)\frac{n^{2}}{2c_{0}}+\frac{1}{2}(\delta-1)\left(\sum_{i}\frac{1}{c_{i}}\right)=0$
$\displaystyle\Rightarrow$
$\displaystyle-\frac{1}{2}\delta\left(\sum_{i}\frac{1}{c_{i}}\right)+(1-\delta)\frac{n^{2}}{2c_{0}}=0$
$\displaystyle\Rightarrow$
$\displaystyle\frac{1}{2}\delta\left(\sum_{i}\frac{1}{c_{i}}\right)+\delta\frac{n^{2}}{2c_{0}}=\frac{n^{2}}{2c_{0}}$
$\displaystyle\Rightarrow$
$\displaystyle\delta^{*}=\frac{\frac{n^{2}}{c_{0}}}{\left(\sum_{i}\frac{1}{c_{i}}\right)+\frac{n^{2}}{c_{0}}}$
$\displaystyle\Rightarrow$
$\displaystyle\delta^{*}=\frac{n^{2}}{{n^{2}}+\sum_{i}\frac{c_{0}}{c_{i}}}.$

The second partial derivative
$\frac{\partial^{2}}{\partial\delta^{2}}\left(U_{G}+\sum_{i}U_{D_{i}}\right)=-\frac{1}{2}\left(\sum_{i}\frac{1}{c_{i}}\right)-\frac{n^{2}}{2c_{0}}$
which is negative $\forall c_{0},c_{i}>0$. Thus, $\delta^{VM}=\delta^{*}$
maximizes the sum of utilities. ∎

####  10.3.4 Egalitarian Bargaining Solution

######  Proposition 10.5 (Egalitarian Solution).

The Egalitarian Bargaining Solution to the multi-specialist fine-tuning game
with quadratic costs is given by:

$\delta^{\textit{Egal.}}=\frac{\sqrt{c_{0}^{2}K^{2}-2c_{0}nK+\frac{c_{0}}{c_{\text{max}}}n^{2}+n^{2}}+n-\frac{c_{0}}{c_{\text{max}}}n-c_{0}K}{n^{2}+2n-\frac{c_{0}}{c_{\text{max}}}-2c_{0}K}.$

Where $K=\sum_{i}\frac{1}{c_{i}}$ and $c_{\text{max}}=\max_{i}c_{i}$.

###### Proof of Proposition 10.5.

The egalitarian solution is given by:

$\delta^{\text{Egal.}}=\arg\max_{\delta}\left[\min\left(U_{G},\min_{i}\left(U_{D_{i}}\right)\right)\right].$

Finding a closed-form solution for the egalitarian bargaining agreement proves
more challenging in the multi-specialist case because many players in the
fine-tuning game, could, potentially, be worst-off. Instead of simply
equalizing utilities, the multi-specialist case might have an egalitarian
solution where only the two worst-off players have equal utility.

The proof for quadratic costs starts off by observing that the specialist with
the lowest utility must be the specialist $j$ with the highest cost
coefficient $c_{j}$. This is because revenue function $r(\cdot)=I(\cdot)$ is
constant across all domains, as is the bargaining parameter $\delta$ and
coefficient $k_{i}=2$. Thus, the highest-cost-coefficient domain specialist is
given by $j=\arg\max_{i}c_{i}$. This observation leaves us with two
possibilities for the lowest-utility player: 1) Player $D_{j}$ 2) Player $G$.
Thus, the egalitarian solution is the bargaining agreement that equalizes the
utilities between the generalist and the costliest player:

$U_{G}(\delta^{\text{Egal.}})=U_{D_{j}}(\delta^{\text{Egal.}}).$

Plugging in, we get:

$\displaystyle\frac{n^{2}}{4c_{0}}\delta^{2}+\delta\sum_{i}\frac{1}{2c_{i}}(1-\delta)=\frac{1}{4c_{\text{max}}}+\left(\frac{n}{2c_{0}}-\frac{1}{2c_{\text{max}}}\right)\delta+\left(\frac{1}{4c_{\text{max}}}-\frac{n}{2c_{0}}\right)\delta^{2}$
$\displaystyle\Rightarrow$
$\displaystyle\left[\frac{n^{2}}{4c_{0}}-\sum_{i}\frac{1}{2c_{i}}+\frac{n}{2c_{0}}-\frac{1}{4c_{\text{max}}}\right]\delta^{2}+\left[\sum_{i}\frac{1}{2c_{i}}+\frac{1}{2c_{\text{max}}}-\frac{n}{2c_{0}}\right]\delta-\frac{1}{4c_{\text{max}}}=0.$

Plugging in to the quadratic formula, we get two candidate solutions:

$\delta^{\textit{Egal.}}\stackrel{{\scriptstyle?}}{{=}}\left\\{\frac{\pm\sqrt{c_{0}^{2}(\sum_{i}\frac{1}{c_{i}})^{2}-2c_{0}n(\sum_{i}\frac{1}{c_{i}})+\frac{c_{0}}{c_{\text{max}}}n^{2}+n^{2}}+n-\frac{c_{0}}{c_{\text{max}}}n-c_{0}(\sum_{i}\frac{1}{c_{i}})}{n^{2}+2n-\frac{c_{0}}{c_{\text{max}}}-2c_{0}(\sum_{i}\frac{1}{c_{i}})}\right\\}.$

As a last step, we select the candidate solution that yields a feasible
$\delta\in[0,1]$:

$\delta^{\textit{Egal.}}=\frac{\sqrt{c_{0}^{2}(\sum_{i}\frac{1}{c_{i}})^{2}-2c_{0}n(\sum_{i}\frac{1}{c_{i}})+\frac{c_{0}}{c_{\text{max}}}n^{2}+n^{2}}+n-\frac{c_{0}}{c_{\text{max}}}n-c_{0}(\sum_{i}\frac{1}{c_{i}})}{n^{2}+2n-\frac{c_{0}}{c_{\text{max}}}-2c_{0}(\sum_{i}\frac{1}{c_{i}})}.$

∎

###  10.4 Theorem on the Three Specialist Regimes

###### Proof of Theorem 2.

We prove this theorem in a sequence of Lemmas. The proof follows for any given
specialist $D_{i}$ and revenue-sharing parameter $\delta$.

######  Lemma 10.1.

If fixed costs are under control, meaning
$r_{i}(\alpha_{0})>\frac{1}{1-\delta}\phi_{i}(\alpha_{0})$, then $D_{i}$ will
not abstain – instead, $D_{i}$ would always prefer to free-ride.

If $r_{i}(\alpha_{0})>\frac{1}{1-\delta}\phi_{i}(\alpha_{0})$, then
$U_{D_{i}}\big{|}_{\alpha_{i}=\alpha_{0}}=r_{i}(\alpha_{0})-\frac{1}{1-\delta}\phi_{i}(\alpha_{0})$
is simply the RHS minus the LHS of the inequality. This means $U_{D_{i}}$ must
be positive at $\alpha_{i}=\alpha_{0}$. Thus, as long as fixed costs are under
control, the specialist prefers free-riding to abstaining.

######  Lemma 10.2.

If fixed costs are not under control, meaning
$r_{i}(\alpha_{0})<\frac{1}{1-\delta}\phi_{i}(\alpha_{0})$, then $D_{i}$ will
not free-ride – instead, $D_{i}$ would always prefer to abstain.

If $r_{i}(\alpha_{0})<\frac{1}{1-\delta}\phi_{i}(\alpha_{0})$, then
$U_{D_{i}}\big{|}_{\alpha_{i}=\alpha_{0}}=r_{i}(\alpha_{0})-\frac{1}{1-\delta}\phi_{i}(\alpha_{0})$
is simply the RHS minus the LHS of the inequality. This means $U_{D_{i}}$ must
be negative at $\alpha_{i}=\alpha_{0}$. Thus, as long as fixed costs are not
under control, the specialist prefers abstaining to free-riding.

######  Lemma 10.3.

If it is marginally profitable to invest in the technology, meaning
$r_{i}^{\prime}(\alpha_{0})>\frac{1}{1-\delta}\phi_{i}^{\prime}(\alpha_{0})$,
then $D_{i}$ will not free-ride – instead, $D_{i}$ would always prefer to
contribute.

If
$r_{i}^{\prime}(\alpha_{0})>\frac{1}{1-\delta}\phi_{i}^{\prime}(\alpha_{0})$,
then $\frac{\partial
U_{D_{i}}}{\partial\alpha_{i}}\big{|}_{\alpha_{i}=\alpha_{0}}=r_{i}^{\prime}(\alpha_{0})-\frac{1}{1-\delta}\phi_{i}^{\prime}(\alpha_{0})$
is simply the RHS minus the LHS of the inequality. This means $U_{D_{i}}$ is
increasing at $\alpha_{i}=\alpha_{0}$. Thus, as long as it is marginally
profitable to improve the technology, the specialist prefers contributing to
free-riding.

######  Lemma 10.4.

If it is marginally costly to invest in the technology, meaning
$r_{i}^{\prime}(\alpha_{0})<\frac{1}{1-\delta}\phi_{i}^{\prime}(\alpha_{0})$,
then $D_{i}$ will not contribute – instead, $D_{i}$ would always prefer to
free-ride.

If
$r_{i}^{\prime}(\alpha_{0})<\frac{1}{1-\delta}\phi_{i}^{\prime}(\alpha_{0})$,
then $\frac{\partial
U_{D_{i}}}{\partial\alpha_{i}}\big{|}_{\alpha_{i}=\alpha_{0}}=r_{i}^{\prime}(\alpha_{0})-\frac{1}{1-\delta}\phi_{i}^{\prime}(\alpha_{0})$
is simply the RHS minus the LHS of the inequality. This means $U_{D_{i}}$ is
decreasing at $\alpha_{i}=\alpha_{0}$. Thus, as long as it is marginally
costly to improve the technology, the specialist prefers free-riding to
contributing.

Taken together, we can conclude the following about combinations of
conditions:

  * •

Fixed costs under control, marginally profitable investment: A<F, F<C (Lemmas
10.1 and 10.3). Thus the specialist would contribute.

  * •

Fixed costs under control, marginally costly: A<F, C<F (Lemmas 10.1 and 10.4).
Thus the specialist would free-ride.

  * •

Fixed costs not under control, marginally profitable: F<A, F<C (Lemmas 10.2
and 10.3). Thus the specialist would either abstain or contribute.

  * •

Fixed costs not under control, marginally costly: F<A, C<F (Lemmas 10.2 and
10.4). Thus the specialist would abstain.

Above, the short-hand notation ‘A,’ ‘F,’ and ‘C’ refer to the strategies of
abstaining, free-riding, and contributing, respectively. The optimal
strategies follow from the two marginal conditions. This completes the proof.
∎


# Prime ideals in categories of representations of quivers of type $A$

Shunsuke Tada<EMAIL_ADDRESS>
###### Abstract.

We study prime tensor ideals in tensor abelian categories of quiver
representations. Specifically, we classify the prime tensor ideals in the
category of representations of zigzag quivers (with bounded path length) whose
vertex set is the set of integers. We show that prime tensor ideals in these
categories are in canonical bijection with prime ideals of a Boolean algebra,
the power set of integers.

###### Key words and phrases:

quiver, quiver representation, prime tensor ideal, tensor triangulated
geometry

###### 2010 Mathematics Subject Classification:

18D10

##  1\. Introduction

We will say a quiver (directed graph) is _of type $A$_ if its vertex set is
the set of integers, and there is a unique arrow between $i$ and $i+1$ for
each $i$ like below

$\dots\longleftrightarrow(-3)\longleftrightarrow(-2)\longleftrightarrow(-1)\longleftrightarrow
0\longleftrightarrow 1\longleftrightarrow 2\longleftrightarrow
3\longleftrightarrow\dots,$

where $\longleftrightarrow$ means that we can choose the direction of each
arrow. Such quivers’ representations are used to study persistent homology
which has topological information about the structure of data sets, [1]. The
representations help visualize the topological information via the “barcode”,
[2]. The theory of persistent homology is used in many areas. For example, see
[3] and [4] for applications to biology, [5] for an application to chemistry,
and [6] for an application to mathematics.

The category of representations of a quiver is a tensor (monoidal) category,
that is, an abelian category with a monoidal structure. In this category, we
can consider prime $\otimes$-ideals (see Definition 2.7) as defined by Tobias
J. Peter in [7]. This concept is related to the tensor triangulated geometry
developed by Paul Balmer [8].

In this paper, we classify the prime $\otimes$-ideals of the category
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ of point-wise
finite-dimensional representations (Definition 2.2) of any quiver
$A_{\mathbb{Z}}$ of type $A$ which has bounded path length. That is, we have
assumed that there exists $n\geq 1$ in $\mathbb{N}$ such that any path in
$A_{\mathbb{Z}}$ has length smaller than $n$. These prime $\otimes$-ideals are
related to the prime ideals of the ring
$\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}})$. We will obtain the
relationship between the prime $\otimes$-ideals of
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ and the prime ideals
of $\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}})$ using the power set
$\mathcal{P}(\mathbb{Z})$ of the set of integers $\mathbb{Z}$. We will obtain
following theorem.

######  Theorem 1.1.

Let $A_{\mathbb{Z}}$ be a quiver of type $A$ with bounded path length. Then,
there is a one-to-one correspondence between each of the three sets

  1. (1)

prime tensor ideals of the category
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$,

  2. (2)

prime ideals of the ring $\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}})$,
and

  3. (3)

prime ideals of the boolean algebra $\mathcal{P}(\mathbb{Z})$.

In this theorem, we need the condition

bounded path length

to have above relations. Without it, we could not have the classification of
the prime $\otimes$-ideals of the category
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$. For example, if the
quiver has an infinite length path like

$\dots\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\dots,$

then we will have an inclusion map from the prime ideals of
$\mathcal{P}(\mathbb{Z})$ to the prime $\otimes$-ideals of
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$, but the map is not
surjective. We will prove this in Corollary 3.22.

Outline. This paper is organized as follows. In Section 2, we will see the
basics of the quiver representations and lattice theory. In Section 3, we see
the classification of prime $\otimes$-ideals of
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ and some topological
properties of the spectrum of
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$.

Acknowledgements. I would like to thank Shane Kelly for suggesting me the
problem, discussing it.

##  2\. Preliminaries

###  2.1. Quiver and its representation

Here are some basics about quivers and their representations. See [9] for more
details. Throughout, $K$ will be a field.

######  Definition 2.1 (Quiver of type $A$).

A quiver $Q=(Q_{0},Q_{1},s,t)$ is a quadruple consisting of two sets: $Q_{0}$
called the set of vertices and $Q_{1}$ called the set of arrows, and two maps
$s$, $t$ : $Q_{1}\to Q_{0}$. For every arrow $\alpha\in Q_{1}$, $s(\alpha)\in
Q_{0}$ and $t(\alpha)\in Q_{0}$ are called $source$ and $target$ of $\alpha$,
respectively. We usually denote the arrow $\alpha$ by $\alpha:s(\alpha)\to
t(\alpha)$.

We will say a quiver is _of type $A$_ if its vertex set is the set of
integers, and there is a unique arrow between $i$ and $i+1$ for each $i$.

######  Definition 2.2 (Point-wise finite-dimensional representations).

Let $Q=(Q_{0},Q_{1},s,t)$ be a quiver. The category $\operatorname{Rep}(Q)$ of
representations of $Q$ is defined as follows. An object is pair consisting of
a family of $K$-vector spaces indexed by $Q_{0}$, and a family of
$K$-homomorphisms indexed by $Q_{1}$:

$V:=((V_{a})_{a\in Q_{0}},\ (V_{\alpha}:V_{s(\alpha)}\to
V_{t(\alpha)})_{\alpha\in Q_{1}}).$

Let $V=((V_{a})_{a\in Q_{0}},(V_{\alpha})_{\alpha\in Q_{1}})$ and
$W=((W_{a})_{a\in Q_{0}},(W_{\alpha})_{\alpha\in Q_{1}})$ be representations.
A morphism in $\operatorname{Rep}(Q)$ from $V$ to $W$ is a family of
$K$-homomorphisms $(f_{a}:V_{a}\to W_{a})_{a\in Q_{0}}$ satisfying

$f_{t(\alpha)}\circ V_{\alpha}=W_{\alpha}\circ f_{s(\alpha)}$

for every arrow $\alpha\in Q_{1}$.

A representation $V$ of $\operatorname{Rep}(Q)$ is said to be _point-wise
finite-dimensional_ if the dimension of vector space
$\operatorname{dim}_{K}(V_{a})$ is finite for every vertex $a$ in $Q$. We
write $\operatorname{Rep}^{\operatorname{pfd}}(Q)$ for the full subcategory of
$\operatorname{Rep}(Q)$ whose objects are point-wise finite-dimensional
representations.

######  Definition 2.3 (Interval representation).

Let $Q=(Q_{0},Q_{1},s,t)$ be a quiver, and let
$I=(I_{0},I_{1},s^{\prime},t^{\prime})$ be a subquiver of $Q$. Define
$K_{I}\in\operatorname{Rep}(Q)$ by

$(K_{I})_{a}:=\left\\{\begin{array}[]{cc}K&(a\in I_{0})\\\
0&(\text{otherwise}),\end{array}\right.$
$(K_{I})_{\alpha}:=\left\\{\begin{array}[]{cc}id_{K}&(\alpha\in I_{1})\\\
0&(\text{otherwise}).\end{array}\right.$

If $I$ is connected and convex (and therefore also full), then $K_{I}$ is
called an _interval representation_.

######  Theorem 2.4.

Let $A_{\mathbb{Z}}$ be a quiver of type $A$. Then every point-wise finite-
dimensional representation of $\operatorname{Rep}(A_{\mathbb{Z}})$ is direct
sum of interval representations.

###### Proof.

See [10] for the proof. ∎

###  2.2. Tensor category and its tensor ideal

Here are some basics about tensor categories and their tensor ideals. See
[11], [7] for more details.

######  Definition 2.5 (Tensor category [11]).

A _tensor category_ (or _monoidal category_) is
$\nu=(\nu_{0},\otimes,I,a,\ell,r)$ consisting of a category $\nu_{0}$, a
functor $\otimes:\nu_{0}\times\nu_{0}\to\nu_{0}$ , an object $I$ of $\nu_{0}$
and natural isomorphisms $a_{XYZ}:(X\otimes Y)\otimes Z\to X\otimes(Y\otimes
Z),\ \ell_{X}:I\otimes X\to X,\ r_{X}:X\otimes I\to X,$ subject to two
coherence axioms expressing the commutativity of the following diagrams:

---
$\textstyle{((W\otimes X)\otimes Y)\otimes
Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{a\otimes
1}$$\textstyle{(W\otimes X)\otimes(Y\otimes
Z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\textstyle{W\otimes(X\otimes(Y\otimes
Z))}$$\textstyle{(W\otimes(X\otimes Y))\otimes
Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\textstyle{W\otimes((X\otimes
Y)\otimes
Z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\otimes
a}$

and

---
$\textstyle{(X\otimes I)\otimes
Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{r\otimes
1}$$\textstyle{X\otimes(I\otimes
Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\otimes\ell}$$\textstyle{X\otimes
Y}$

######  Example 2.6.

The tuple ($\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}),\
\otimes,\ K_{A_{\mathbb{Z}}},\ a,\ \ell,\ r\ )$ is a tensor category. Let $V,\
V^{\prime},\ W$ and $W^{\prime}$ be objects of
$\operatorname{Rep}(A_{\mathbb{Z}})$. We define a functor

$\otimes:\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})\times\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})\to\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$

by

$\otimes(V,W):=((V_{n}\otimes_{K}W_{n})_{n\in(A_{\mathbb{Z}})_{0}},\
(V_{\alpha}\otimes_{K}W_{\alpha})_{\alpha\in(A_{\mathbb{Z}})_{1}})$

for objects and

$\displaystyle\otimes((f,g)$ $\displaystyle:(V,W)\to(V^{\prime},W^{\prime}))$
$\displaystyle:=(f_{n}\otimes
g_{n})_{n\in(A_{\mathbb{Z}})_{0}}:((V_{n}\otimes_{K}W_{n})_{n\in(A_{\mathbb{Z}})_{0}}\to((V^{\prime}_{n}\otimes_{K}W^{\prime}_{n})_{n\in(A_{\mathbb{Z}})_{0}}$

for arrows. Then, we have natural isomorphisms $a_{XYZ}:(X\otimes Y)\otimes
Z\to X\otimes(Y\otimes Z)$, $\ell_{X}:K_{A_{\mathbb{Z}}}\otimes X\to X$ and
$r_{X}:X\otimes K_{A_{\mathbb{Z}}}\to X$ for any $X,\ Y$ and $Z$ in
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$. These natural
isomorphisms are subject to two coherence axioms of tensor category.

######  Definition 2.7 (Prime tensor ideal [7]).

Let $\mathcal{A}$ be an abelian category. A full subcategory $\mathcal{M}$ of
$\mathcal{A}$ is called $coherent$ if

  1. (1)

it contains the zero object,

  2. (2)

it is closed under extensions and

  3. (3)

kernels and cokernels in $\mathcal{A}$ of morphisms between objects of
$\mathcal{M}$ are in $\mathcal{M}$.

Suppose now that $\mathcal{A}$ is a tensor abelian category. A _(coherent)
tensor ideal_ of $\mathcal{A}$ is a coherent subcategory
$\mathcal{M}\subset\mathcal{A}$ such that
$\mathcal{M}\otimes\mathcal{A}\subset\mathcal{M}$. A proper ideal
$\mathcal{M}\subsetneq\mathcal{A}$ is called a _(coherent) prime ideal_ if
$A\otimes B\in\mathcal{M}$ implies $A\in\mathcal{M}$ or $B\in\mathcal{M}$.

We write the set of all the prime tensor ideals by
$\operatorname{Spc}(\mathcal{A})$.

######  Example 2.8.

Let $Q$ be a quiver, and consider the tensor category
($\operatorname{Rep}^{\operatorname{pfd}}(Q),\ \otimes,\ K_{Q},\ a,\ \ell,\ r\
)$ and let $\mathcal{M}$ be the full subcategory whose objects are $\\{V\in
Q:V_{n}=0\\}$ for some vertex $n\in Q_{0}$. We will show that the full
subcategory $\mathcal{M}$ is a prime $\otimes$-ideal.

First, we show that $\mathcal{M}$ is closed under kernels and cokernels. Let
$V$ and $W$ be representations in $\mathcal{M}$ and $(f_{n}:V_{n}\to
W_{n})_{n\in Q}$ be a morphism from a representation $V$ to $W$. Since $V_{n}$
and $W_{n}$ are isomorphic to zero, the kernel and the cokernel of
$f_{n}:V_{n}\to W_{n}$ is isomorphic to zero. Thus,
$\operatorname{Ker}((f_{n})_{n\in A_{\mathbb{Z}}})$ and
$\operatorname{Coker}((f_{n})_{n\in A_{\mathbb{Z}}})$ is in $\mathcal{M}$. So
the full subcategory $\mathcal{M}$ is closed under kernels and cokernels.

Next, we show that $\mathcal{M}$ is closed under extension. Take an exact
sequence

$0\to V^{\prime}\to V\to V^{\prime\prime}\to 0$

where $V^{\prime}$ and $V^{\prime\prime}$ are objects of $\mathcal{M}$. Then,
there is an exact sequence of $K$-homomorphisms

$0\to V_{n}^{\prime}\to V_{n}\to V_{n}^{\prime\prime}\to 0.$

Since the representations $V^{\prime}$ and $V^{\prime\prime}$ are in
$\mathcal{M}$, we have $V^{\prime}_{n}=0=V^{\prime\prime}_{n}$, so $V_{n}$ is
isomorphic to $0$. Thus the representation $V$ is in $\mathcal{M}$. So the
full subcategory $\mathcal{M}$ is closed under extensions.

Then, for any representations $V\in\operatorname{Rep}^{\operatorname{pfd}}(Q)$
and $W\in\mathcal{M}$, we have $(V\otimes W)_{n}=V_{n}\otimes W_{n}=0$. Thus
$\mathcal{M}$ is a $\otimes$-ideal.

Finally, we show that $\mathcal{M}$ is a prime $\otimes$-ideal. Let $V$ and
$W$ be the representations of $\operatorname{Rep}^{\operatorname{pfd}}(Q)$. If
$V\otimes W$ is in $\mathcal{M}$, then we have $0=(V\otimes
W)_{n}=V_{n}\otimes W_{n}$. Thus $V_{n}$ or $W_{n}$ is isomorphic to $0$ and
therefore $V$ or $W$ is in $\mathcal{M}$.

###  2.3. Boolean rings and Boolean algebras

Here are some basics about Boolean rings and Boolean algebras. See [12] for
more details.

######  Definition 2.9 (Boolean ring).

A ring $B$ with an identity is called a $Boolean$ $ring$ if $x^{2}=x$ for all
${x\in B}$.

######  Proposition 2.10.

Any Boolean ring is a commutative ring, every prime ideal of a Boolean ring is
a maximal ideal, and every residue field is isomorphic to
$\mathbb{Z}/2\mathbb{Z}$.

###### Proof.

See exercise 57 on page 194 of [13] and exercise 11.(ii) on page 11 of [14]. ∎

######  Definition 2.11 (Boolean algebra).

Let $B$ be a poset. The _Boolean algebra_ is defined to be a tuple $(B;\lor,\
\land,\ ^{\prime},\ 0,\ 1)$ such that

  1. (i)

$x\lor y:=\text{sup}\\{x,y\\}$ and $x\land y:=\text{inf}\\{x,y\\}$ exist for
all $x,y\in B,$

  2. (ii)

$(B;\lor,\ \land)$ is distributive, i.e.,

$a\land(b\lor c)=(a\land b)\lor(a\land c),$

  3. (iii)

$0$ is the minimal element, and $1$ is the maximal element of $B$,

  4. (iv)

every element $x$ admits a unique complement, i.e., an element satisfying
$x\land x^{\prime}=0$ and $x\lor x^{\prime}=1$.

######  Definition 2.12 (Ideal/prime ideal of Boolean algebra).

Let $B$ be a Boolean algebra. A non-empty subset $J$ of $B$ is called an
_ideal_ if

  1. (i)

$a,b\in J$ implies $a\lor b\in J$,

  2. (ii)

$a\in B,b\in J$ and $a\leq b$ imply a $\in J$.

An ideal is _proper_ if $J\subsetneq B$. A proper ideal $J$ of $B$ is said to
be a prime ideal if $a,b\in B$ and $a\land b\in J$ imply $a\in J$ or $b\in J$.
We denote the sets of prime ideals by $\operatorname{Spc}(B).$

######  Example 2.13.

Let us write $\mathcal{P}(\mathbb{Z})$ for the power set of $\mathbb{Z}$. Then
$(\mathcal{P}(\mathbb{Z});\cap,\ \cup,\ ^{c},\ \emptyset,\ \mathbb{Z})$ is a
Boolean algebra where c is the operation of taking the complement of an
element. Then $\mathcal{P}_{a}:=\\{I\in\mathcal{P}(\mathbb{Z}):a\notin I\\}$
is the prime ideal for every $a\in\mathbb{Z}.$

######  Proposition 2.14.

  1. (i)

Let $B$ be a Boolean algebra and define + and $\cdot$ on $B$ by

$x+y:=(x\land y^{\prime})\lor(x^{\prime}\land y),\qquad x\cdot y:=x\land y.$

Then ($B$; +, ·) is a Boolean ring.

  2. (ii)

Let $B$ be a Boolean ring and define $\land$, $\lor$ and ′ on $B$ by

$x\lor y:=x+y+xy,\qquad x\land y:=xy,\qquad x^{\prime}:=1+x.$

Then $(B;\land,\lor,^{\prime},0,1)$ is a Boolean algebra.

  3. (iii)

The correspondence between Boolean algebras and Boolean rings established in
(i) and (ii) is a bijective one.

  4. (iv)

There is a one-to-one inclusion preserving correspondence between prime ideals
of a Boolean algebra and the prime ideals of the associated Boolean ring.

###### Proof.

For the proof of (i) $\sim$ (iii) see [12, Ex.4.27]. As for the proof of (iv),
by [12, Ex.4.29] and Proposition 2.14 (i), (ii), and (iii), a subset of a
Boolean algebra is an ideal of the Boolean algebra if and only if the subset
is an ideal of the corresponding Boolean ring. We can easily see that the
correspondence of their ideals gives the correspondence of their prime ideals.

∎

By this proposition, we can see Boolean algebras and Boolean rings are
equivalent.

######  Example 2.15.

Consider the Boolean algebra
$(\mathcal{P}(\mathbb{Z});\cap,\cup,^{c},\emptyset,\mathbb{Z})$ corresponding
to the power set of the set of integers. The Boolean ring corresponding to
$\mathcal{P}(\mathbb{Z})$ is isomorphic to
$\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}})$ via the map

$\phi:\mathcal{P}(\mathbb{Z})\to\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}}),\quad
I\mapsto(a_{i})_{i\in\mathbb{Z}}$

where $a_{i}=1+2\mathbb{Z}$ if $i\in I$ otherwise $a_{i}=0+2\mathbb{Z}$. Note
$\phi$ is a bijective ring homomorphism, i.e., a ring isomorphism. Here, for
each $I,\ J\in\mathcal{P}(\mathbb{Z})$, the sum $I+J:=(I\cup
J)\backslash(I\cap J)$ and the product $I\cdot J:=I\cap J$ on the left are
those defined in Proposition 2.14(i). By Proposition 2.14(iv), The map $\phi$
gives a bijection between $\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$ and
$\operatorname{Spc}(\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}}))$.

##  3\. Prime tensor ideals of
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$

###  3.1. Classification of the prime tensor ideals of
$\text{Rep}^{\text{pfd}}$($A_{\mathbb{Z}}$)

Here, we will classify the prime $\otimes$-ideals of
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ when the quiver
$A_{\mathbb{Z}}$ has bounded path length.

Denote the class of prime ideals of
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ and
$(\mathcal{P}(\mathbb{Z});\cap,\cup,^{c},\emptyset,\mathbb{Z})$ by
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
and $\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))$ respectively. We will
consider a subset $Q^{\prime}_{0}$ of the vertices of a quiver
$Q=(Q_{0},Q_{1},s,t)$ as a full subquiver of $Q$. A full subquiver is
determined by the set of vertices. For example, we will be able to see
$\mathbb{Z}=(A_{\mathbb{Z}})_{0}$ as $A_{\mathbb{Z}}$.

######  Proposition 3.1.

Assume that there is a natural number $n\geq 1$ such that the length of paths
in the quiver $A_{\mathbb{Z}}$ is less than $n$. Then, for any representation
$V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ and a prime
$\otimes$-ideal
$\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$,
we have $K_{\operatorname{supp}(V)}\in\mathcal{M}$ if and only if
$V\in\mathcal{M}$.

###### Proof.

$(\Rightarrow)$. Since $\mathcal{M}$ is a prime $\otimes$-ideal, we have
$V=V\otimes K_{\operatorname{supp}(V)}\in\mathcal{M}$.
$(\Leftarrow)$. Let the representation $V$ be in $\mathcal{M}$. We have
$0=K_{\operatorname{supp}(V)}\otimes
K_{\operatorname{supp}(V)^{c}}\in\mathcal{M}$. Since $\mathcal{M}$ is a prime
$\otimes$-ideal, either $K_{\operatorname{supp}(V)}$ or
$K_{\operatorname{supp}(V)^{c}}$ are in $\mathcal{M}$. We show that
$K_{\operatorname{supp}(V)}\in\mathcal{M}$ by contradiction. More precisely,
we will show that $K_{\operatorname{supp}(V)^{c}}\in\mathcal{M}$ implies
$K_{\mathbb{Z}}\in\mathcal{M}$, and this contradicts the assumption that
$\mathcal{M}$ is prime.

Since $\mathcal{M}$ is a prime $\otimes$-ideal, $\mathcal{M}$ is closed under
extension. Thus, for an exact sequence,

$0\to V\to V\oplus K_{\operatorname{supp}(V)^{c}}\to
K_{\operatorname{supp}(V)^{c}}\to 0$

the representation $V\oplus K_{\operatorname{supp}(V)^{c}}$ is in
$\mathcal{M}$. Then, we have $K^{\prime}_{\mathbb{Z}}\otimes(V\oplus
K_{supp(V)^{c}})\in\mathcal{M}$ where the representation
$K^{\prime}_{\mathbb{Z}}$ is defined by

$(K^{\prime}_{\mathbb{Z}})_{a}:=K\ (\ a\ \text{is any integer})$
$(K^{\prime}_{\mathbb{Z}})_{\alpha}:=0\ (\ \alpha\ \text{is any arrow})$

Then, the representation $K^{\prime}_{\mathbb{Z}}$ is in $\mathcal{M}$ because
there exists a morphism

$K^{\prime}_{\mathbb{Z}}\otimes(V\oplus K_{\operatorname{supp}(V)^{c}})\to
K^{\prime}_{\mathbb{Z}}\otimes(V\oplus K_{\operatorname{supp}(V)^{c}})$

such that its kernel is $K^{\prime}_{\mathbb{Z}}$. Here, note that support of
$K^{\prime}_{\mathbb{Z}}\otimes(V\oplus K_{\operatorname{supp}(V)^{c}})$ is
$\mathbb{Z}$.

Recall that we assume that all paths of the quiver $A_{\mathbb{Z}}$ have
length smaller than or equal to $n$. This implies that the quiver has
infinitely many sinks and sources. Take an arbitrary sink and let the sink be
$t_{0}$ ($t$ comes from “target”). And define a source $s_{1}$ as the smallest
source larger than $t_{0}$ . Define a sink $t_{2}$ as the smallest sink larger
than $s_{1}$. Inductively, we can define sinks $t_{2i}$ and sources $s_{2i+1}$
for all non negative integers $i$. Similarly, we can define sinks $t_{2i}$ and
sources $s_{2i-1}$ for all non positive integers $i$.

Let $B\subset\mathbb{Z}$ be $\bigcup_{i\in\mathbb{Z}}[s_{4i-1},s_{4i+1}]$, and
$D\subset\mathbb{Z}$ be $\bigcup_{i\in\mathbb{Z}}[s_{4i+1},s_{4i+3}]$ and $C$
be $\bigcup_{i\in\mathbb{Z}}\\{s_{4i-1},s_{4i+1}\\}$. Then we obtain

$K_{C}=K_{B}\otimes K_{D}$

If we write pictures of the sets $B$, $D$ and $C$, the sets will be the
integers marked by $\underbrace{}$ where B is like

$\dots\longleftarrow\underbrace{s_{-5}\longrightarrow t_{-4}\longleftarrow
s_{-3}}\longrightarrow t_{-2}\longleftarrow\underbrace{s_{-1}\longrightarrow
t_{0}\longleftarrow s_{1}}\longrightarrow
t_{2}\longleftarrow\underbrace{s_{3}\longrightarrow}\dots$

and $D$ is like

$\dots\underbrace{\longleftarrow s_{-5}}\longrightarrow
t_{-4}\longleftarrow\underbrace{s_{-3}\longrightarrow t_{-2}\longleftarrow
s_{-1}}\longrightarrow t_{0}\longleftarrow\underbrace{s_{1}\longrightarrow
t_{2}\longleftarrow s_{3}}\longrightarrow\dots$

and $C$ is like

$\dots\longleftarrow\underbrace{s_{-5}}\longrightarrow
t_{-4}\longleftarrow\underbrace{s_{-3}}\longrightarrow
t_{-2}\longleftarrow\underbrace{s_{-1}}\longrightarrow
t_{0}\longleftarrow\underbrace{s_{1}}\longrightarrow
t_{2}\longleftarrow\underbrace{s_{3}}\longrightarrow\dots$

Since $K_{B}\otimes K_{D}=K_{C}=K_{C}\otimes
K^{\prime}_{\mathbb{Z}}\in\mathcal{M}$, we have $K_{B}$ or $K_{D}$ are in
$\mathcal{M}$. We will assume that $K_{B}$ is in $\mathcal{M}$. (we can assume
$K_{D}\in\mathcal{M}$ instead of $K_{B}\in\mathcal{M}$. The proof is similar.
)

For this $B$, we will show the subsets $B^{\prime}$, $B^{\prime\prime}$ of
$\mathbb{Z}$ where
$B^{\prime}:=B\cup\bigcup_{i\in\mathbb{Z}}[s_{4i+1},t_{4i+2}-1]$ and
$B^{\prime\prime}:=B^{\prime}\cup\bigcup_{i\in\mathbb{Z}}[t_{4i-2}+1,s_{4i+2}-1]$
satisfy $K_{B^{\prime}}\in\mathcal{M}$ and
$K_{B^{\prime\prime}}\in\mathcal{M}$. We will show that
$K_{\mathbb{Z}}\in\mathcal{M}$ using $K_{B^{\prime\prime}}\in\mathcal{M}$.

(Proof of $K_{B^{\prime}}\in\mathcal{M}$)
For the $B$, we show $K_{B^{\prime}}\in\mathcal{M}$ where
$B^{\prime}:=B\cup\bigcup_{i\in\mathbb{Z}}[s_{4i+1},t_{4i+2}-1]$. Define a
subset $B_{1}:=B\cup\\{s_{4i+1}+1:i\in\mathbb{Z},\ s_{4i+1}+1\ \text{is not a
sink}\\}$. Then, we have an exact sequence

$0\to K_{B_{1}\setminus B}\to K_{B_{1}}\to K_{B}\to 0$

where $K_{B_{1}\setminus B}=K_{B_{1}\setminus B}\otimes
K^{\prime}_{\mathbb{Z}}\in\mathcal{M}$ and since $\mathcal{M}$ is closed under
extensions, the representation $K_{B_{1}}$ is in $\mathcal{M}$. Inductively,
we can define subsets of $\mathbb{Z}$ by

$B_{\ell+1}:=B_{\ell}\cup\\{s_{4i+1}+(\ell+1)\ :\ i\in\mathbb{Z},\
s_{4i+1}+1,\dots,s_{4i+1}+\ell\ \text{are not sinks}\\}$

for non negative integer $\ell$. Then we can show that
$K_{B_{\ell}}\in\mathcal{M}$ by mathematical induction. For $\ell=1$, we have
already shown that $K_{B_{1}}\in\mathcal{M}$. If we have
$K_{B_{\ell}}\in\mathcal{M}$, then we can show $K_{B_{\ell+1}}\in\mathcal{M}$
because for $K_{B_{\ell}}\in\mathcal{M}$, there is an exact sequence

$0\to K_{B_{\ell+1}\setminus B_{\ell}}\to K_{B_{\ell+1}}\to K_{B_{\ell}}\to 0$

We obtain $K_{B_{\ell+1}}$ in $\mathcal{M}$ because $\mathcal{M}$ is closed
under extensions with $K_{B_{\ell}}$ and $K_{B_{\ell+1}\setminus
B_{\ell}}\in\mathcal{M}$. Thus, by mathematical induction, we have
$K_{B_{\ell}}\in\mathcal{M}$ for any $\ell\geq 1$. Note that there exists
$m\leq n$ such that $B_{m}=B_{m+1}=\dots$ because the length of paths are
smaller than or equal to $n$. Then, the set $B_{m}$ contains $s_{4i+2}-1$ for
any $i\in\mathbb{Z}$ and we have $B^{\prime}=B_{m}$ and
$K_{B^{\prime}}\in\mathcal{M}$.

(Proof of $K_{B^{\prime\prime}}\in\mathcal{M}$)
For the $B^{\prime}$ defined above, we show
$K_{B^{\prime\prime}}\in\mathcal{M}$ where
$B^{\prime\prime}:=B^{\prime}\cup\bigcup_{i\in\mathbb{Z}}[s_{4i-2}+1,s_{4i+2}-1]$.

Define a subset $B^{\prime}_{1}:=B^{\prime}\cup\\{s_{4i-1}-1:i\in\mathbb{Z},\
s_{4i-1}-1\ \text{is not a sink}\\}$. Then, we have an exact sequence

$0\to K_{B^{\prime}_{1}\setminus B^{\prime}}\to K_{B^{\prime}_{1}}\to
K_{B^{\prime}}\to 0$

where $K_{B^{\prime 1}\setminus B^{\prime}}=K_{B^{\prime 1}\setminus
B^{\prime}}\otimes K^{\prime}_{\mathbb{Z}}\in\mathcal{M}$ and since
$\mathcal{M}$ is closed under extensions, $K_{B^{\prime}_{1}}$ is in
$\mathcal{M}$. Inductively, we can define subsets of $\mathbb{Z}$ as

$B^{\prime}_{(\ell+1)}:=B^{\prime}_{\ell}\cup\\{s_{4i-1}-(\ell+1):i\in\mathbb{Z},\
s_{4i-1}-1,\dots,s_{4i-1}-(\ell+1)\ \text{are not sinks}\\}$

for non negative integer $\ell$. Then we can show that
$K_{B_{\ell}}\in\mathcal{M}$ for any non-negative integers by mathematical
induction. For $\ell=1$, we have already shown that
$K_{B^{\prime}_{1}}\in\mathcal{M}$. If we have
$K_{B^{\prime}_{\ell}}\in\mathcal{M}$, then we can show
$K_{B^{\prime}_{(\ell+1)}}\in\mathcal{M}$ because for
$K_{B^{\prime}_{\ell}}\in\mathcal{M}$, there is an exact sequence

$0\to K_{B^{\prime}_{(\ell+1)}\setminus B^{\prime}_{\ell}}\to
K_{B^{\prime}_{(\ell+1)}}\to K_{B^{\prime}_{\ell}}\to 0$

We obtain $K_{B^{\prime}_{(\ell+1)}}$ in $\mathcal{M}$ because $\mathcal{M}$
is closed under extensions. Thus, by mathematical induction, we have
$K_{B^{\prime}_{\ell}}\in\mathcal{M}$ for any $\ell\geq 1$. Note that there
exists $m\leq n$ such that $B^{\prime}_{m}=B^{\prime}_{(m+1)}=\dots$ because
the length of paths are smaller than or equal to $n$. Then, the set
$B^{\prime}_{m}$ contains $t_{4i-2}+1$ for any $i\in\mathbb{Z}$ and we have
$B^{\prime\prime}=B^{\prime}_{m}$ and $K_{B^{\prime\prime}}\in\mathcal{M}$.

(Proof of $K_{\mathbb{Z}}\in\mathcal{M}$)
Define $E=\\{t_{4i+2}:i\in\mathbb{Z}\\}$ and note that
$B^{\prime\prime}=\mathbb{Z}\setminus\\{t_{4i+2}:i\in\mathbb{Z}\\}=\mathbb{Z}\setminus
E$ and $K_{E}=K_{E}\otimes K^{\prime}_{\mathbb{Z}}\in\mathcal{M}$. We have the
exact sequence

$0\to K_{E}\to K_{\mathbb{Z}}\to K_{B^{\prime\prime}}\to 0.$

Since $\mathcal{M}$ is closed under extensions and
$K_{B^{\prime\prime}},K_{E}\in\mathcal{M}$, the representation
$K_{\mathbb{Z}}\in\mathcal{M}$. Thus, $K_{\mathbb{Z}}\in\mathcal{M}$ and it
leads $\mathcal{M}=\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$.
This is a contradiction. Hence, we have
$K_{\operatorname{supp}(V)}\in\mathcal{M}$. ∎

######  Proposition 3.2.

There is a bijection between
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
and $\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$.

###### Proof.

Define two maps

$\displaystyle\phi:\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
$\displaystyle\to\operatorname{Spc}(\mathcal{P}({\mathbb{Z}})),$
$\displaystyle\mathcal{M}$
$\displaystyle\mapsto\\{\operatorname{supp}(V)\in\mathcal{P}(\mathbb{Z}):V\in\mathcal{M}\\},$
$\displaystyle\psi:\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))$
$\displaystyle\to\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})),$
$\displaystyle\mathcal{Q}$
$\displaystyle\mapsto\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)\in\mathcal{Q}\\}.$

First, we check
$\phi(\mathcal{M})\in\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))$ for
$\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
and $\psi(\mathcal{Q})\in\
\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
for $\mathcal{Q}\in\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))$. After that,
we will show $\phi$ and $\psi$ are inverse maps of each other.

The following (i), (ii) and (iii) show that $\phi(\mathcal{M})$ is an ideal of
$\mathcal{P}({\mathbb{Z}})$ and (iv) shows that $\phi(\mathcal{M})$ is a prime
ideal of $\mathcal{P}({\mathbb{Z}})$.

(i) $\phi(\mathcal{M})$ is not empty because
$\emptyset=\text{supp}(0)\in\phi(\mathcal{M})$ where $0$ is the zero
representation.

(ii) For any $\operatorname{supp}(V)$ and
$\operatorname{supp}(V^{\prime})\in\phi(\mathcal{M})$ (where $V$ and
$V^{\prime}$ $\in\mathcal{M}$), we will show
$\operatorname{supp}(V)\cup\operatorname{supp}(V^{\prime})\in\phi(\mathcal{M})$.
Since $\mathcal{M}$ is a prime $\otimes$-ideal of
$\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))$, for the exact sequence

$0\to V\to V\oplus V^{\prime}\to V^{\prime}\to 0,$

the direct sum $V\oplus V^{\prime}$ of $V$ and $V^{\prime}$ is in
$\mathcal{M}$. Thus, we have
$\operatorname{supp}(V)\cup\operatorname{supp}(V^{\prime})=\operatorname{supp}(V\oplus
V^{\prime})\in\phi(\mathcal{M}).$

(iii) For any $I\in\mathcal{P}(\mathbb{Z})$ and for any
$\operatorname{supp}(V)\in\phi(\mathcal{M})$, we will show that if
$I\subset\operatorname{supp}(V)$, then $I\in\phi(\mathcal{M})$. Assume that
$I\subset\operatorname{supp}(V)$. Then we have
$\operatorname{supp}(K_{I})=\operatorname{supp}(K_{I}\otimes V)$. Since
$K_{I}\otimes V\in\mathcal{M}$, we obtain $K_{I}\in\mathcal{M}$ and
$I\in\phi(\mathcal{M})$.

By (i), (ii) and (iii), $\phi(\mathcal{M})$ is an ideal of
$\mathcal{P}(\mathbb{Z})$.

(iv) If $I\cap J\in\phi(\mathcal{M})$ for $I$, $J$
$\in\mathcal{P}(\mathbb{Z})$, we will show that $I$ or $J$ are in
$\phi(\mathcal{M})$. Since $I\cap J\in\phi(\mathcal{M})$, there exists $W$
$\in\mathcal{M}$ such that $I\cap J=\operatorname{supp}(W)$. Define
$V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ by

$V_{a}:=\left\\{\begin{array}[]{ll}W_{a}&a\in\operatorname{supp}(W)\\\ K&a\in
I\setminus\operatorname{supp}(W)\\\
0&\operatorname{otherwise}\end{array}\right.$
$V_{\alpha}:=\left\\{\begin{array}[]{ll}W_{\alpha}&s(\alpha),t(\alpha)\in\operatorname{supp}(W)\\\
0&\operatorname{otherwise}\end{array}\right.$

where $a$ is a vertex and $\alpha$ is an arrow of $A_{\mathbb{Z}}$.

Then we have $V\otimes K_{\operatorname{supp}(W)}=W\in\mathcal{M}$. Since
$\mathcal{M}$ is a prime $\otimes$-ideal, the representations $V$ or
$K_{\operatorname{supp}(W)}$ are in $\mathcal{M}$. If $V\in\mathcal{M}$, then
we have $I=\operatorname{supp}(V)\in\phi(\mathcal{M})$. On the other hand, if
$K_{\operatorname{supp}(W)}\in\mathcal{Q}$, then $K_{I}\otimes
K_{J}=K_{\operatorname{supp}(W)}\in\mathcal{M}$. Therefore $K_{I}$ or $K_{J}$
are in $\mathcal{M}$ because $\mathcal{M}$ is a prime $\otimes$-ideal. This
implies $I$ or $J$ are in $\phi(\mathcal{M}).$

Next, the following (i), (ii) and (iii) show that $\psi(\mathcal{Q})$ is a
$\otimes$-ideal of $\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$
and (iv) shows that $\psi(\mathcal{Q})$ is a prime $\otimes$-ideal of
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$.

(i) We will show that $\psi(\mathcal{Q})$ is closed under kernels and
cokernels. For any $f:V\to W$ in $\psi(\mathcal{Q})$, the kernel and the
cokernel of $f$ satisfy
$\operatorname{supp}(\operatorname{Ker}(f))\subseteq\operatorname{supp}(V)$
and
$\operatorname{supp}(\operatorname{Coker}(f))\subseteq\operatorname{supp}(W)$
where $\operatorname{supp}(V)$ and $\operatorname{supp}(W)$ are in the prime
ideal $\mathcal{Q}$ of $\mathcal{P}(\mathbb{Z})$. Thus, we have
$\operatorname{supp}(\operatorname{Ker}(f))\in\mathcal{Q}$ and
$\operatorname{supp}(\operatorname{Coker}(f))\in\mathcal{Q}$. It means that
$\operatorname{Ker}(f)$ and $\operatorname{Coker}(f)$ are in
$\psi(\mathcal{Q})$. Hence, $\psi(\mathcal{Q})$ is closed under kernels and
cokernels.

(ii) We will show that $\psi(\mathcal{Q})$ is closed under extensions. For
$V^{\prime},V^{\prime\prime}\in\psi(\mathcal{Q})$, take an exact sequence

$0\to V^{\prime}\to V\to V^{\prime\prime}\to 0$

where $V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$. Then
$\operatorname{supp}(V)=\operatorname{supp}(V^{\prime})\cup\operatorname{supp}(V^{\prime\prime})\in\mathcal{Q}$
holds because $\operatorname{supp}(V^{\prime})$ and
$\operatorname{supp}(V^{\prime\prime})$ are in the prime ideal $\mathcal{Q}$.
we have $V\in\psi(\mathcal{Q})$ and therefore, $\psi(\mathcal{Q})$ is closed
under extensions.

(iii) We will show that for any
$M\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ and
$V\in\psi(\mathcal{Q})$, the representation $M\otimes V$ is in
$\psi(\mathcal{Q})$. Since $\operatorname{supp}(M\otimes
V)=\text{supp}(M)\cap\text{supp}(V)\in\mathcal{Q}$ holds, we have $M\otimes
V\in\psi(\mathcal{Q})$.

By (i), (ii), and (iii), we have shown that $\psi(\mathcal{Q})$ is a
$\otimes$-ideal of $\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$.
Now we show that it is prime.

(iv) If $V\otimes W\in\psi(\mathcal{Q})$ for $V$ and $W$
$\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$, then
supp$(V)\cap\operatorname{supp}(W)$=supp$(V\otimes W)\in\mathcal{Q}$. Since
$\mathcal{Q}$ is a prime ideal of $\mathcal{P}(\mathbb{Z})$, we have
$\operatorname{supp}(V)$ or $\operatorname{supp}(W)\in\mathcal{Q}$. Hence, $V$
or $W\in\psi(\mathcal{Q})$ and therefore $\psi(\mathcal{Q})$ is a prime
$\otimes$-ideal of $\mathcal{P}(\mathbb{Z})$.

Finally, we show $\phi$ and $\psi$ are inverse maps of each other. For any
$\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
and $\mathcal{Q}\in\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$, we have

$\displaystyle\phi\circ\psi(\mathcal{Q})$
$\displaystyle=\phi(\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(\mathbb{Z}):\operatorname{supp}(V)\in\mathcal{Q}\\})$
$\displaystyle=\\{\operatorname{supp}(V)\in\mathcal{P}(\mathbb{Z}):V\in\\{V^{\prime}\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V^{\prime})\in\mathcal{Q}\\}\\}$
$\displaystyle=\\{\operatorname{supp}(V)\in\mathcal{P}(\mathbb{Z}):V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}),\
\operatorname{supp}(V)\in\mathcal{Q}\\}$ $\displaystyle=\mathcal{Q}.$

The last equality holds because every $I\subseteq\mathbb{Z}$ is the support of
some representation (e.g., $K_{I}$).

$\displaystyle\psi\circ\phi(\mathcal{M})$
$\displaystyle=\psi(\\{\operatorname{supp}(V)\in\mathcal{P}(\mathbb{Z}):V\in\mathcal{M}\\})$
$\displaystyle=\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)\in\\{\operatorname{supp}(V^{\prime})\in\mathcal{P}(\mathbb{Z}):V^{\prime}\in\mathcal{M}\\}\\}$
$\displaystyle=\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)=\operatorname{supp}(V^{\prime})\
\exists\ V^{\prime}\in\mathcal{M}\\}$
$\displaystyle=\mathcal{M}.\qquad(\because\text{Proposition
\ref{lemma:KsuppViffV}}\ )$

Here, we use Proposition 3.1 to show the relation

$\mathcal{M}\supseteq\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)=\operatorname{supp}(V^{\prime})\
\exists\ V^{\prime}\in\mathcal{M}\\}.$

We will prove it. For any
$V\in\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)=\operatorname{supp}(V^{\prime})\
\exists\ V^{\prime}\in\mathcal{M}\\}$, there exists $V^{\prime}\in\mathcal{M}$
such that $\operatorname{supp}(V)=\operatorname{supp}(V^{\prime})$. By
Proposition 3.1, the interval representation
$K_{\operatorname{supp}(V^{\prime})}\in\mathcal{M}$. Thus, we obtain
$K_{\operatorname{supp}(V)}=\operatorname{supp}(V)\otimes
K_{\operatorname{supp}(V^{\prime})}\in\mathcal{M}$. Therefore, the
representation $V=V\otimes K_{\operatorname{supp}(V)}$ is in $\mathcal{M}$. We
have shown

$\mathcal{M}=\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)=\operatorname{supp}(V^{\prime})\
\exists\ V^{\prime}\in\mathcal{M}\\}.$

The two maps $\phi$ and $\psi$ are inverse maps of each other. Therefore there
is a bijection between
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
and $\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$. ∎

The next corollary is our main theorem.

######  Corollary 3.3.

There is a one-to-one correspondence between the following three sets,
$\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$,
$\operatorname{Spc}(\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}}))$ and
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$.

###### Proof.

There is a bijection between $\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$ and
$\operatorname{Spc}(\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}}))$ by
Example 2.15. There is a bijection between
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
and $\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$ by Proposition 3.2. ∎

###  3.2. Topological properties of
Spc($\text{Rep}^{\text{pfd}}$($A_{\mathbb{Z}}$)

In this subsection, we will see that
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
is a compact Hausdorff space homeomorphic to the topological space
$\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))\cong\operatorname{Spc}{(\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}}))}$
when the quiver $A_{\mathbb{Z}}$ has bounded length of paths. That is, we show
the following proposition.

######  Proposition 3.4.

Let $\mathcal{P}(\mathbb{Z})$ be the power set of $\mathbb{Z}$ and consider it
as a Boolean ring (isomorphic to
$(\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}}))$ by Example 2.15). Then we
have a homeomorphism between
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
and $\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$, where both sets are
equipped with their respective Zariski topologies (defined below).

We begin with the definition of the Zariski topology on
$\operatorname{Spc}(\mathcal{A})$ for a general tensor abelian category
$\mathcal{A}$.

######  Definition 3.5 (Zariski topology on tensor categories [7], [8]).

Let $\mathcal{A}$ be a tensor abelian category. For every family of objects
$\mathcal{S}\subset\mathcal{A}$ we denote by $Z(\mathcal{S})$ the following
subset of $\operatorname{Spc}(\mathcal{A})$:

$Z(\mathcal{S}):=\\{\mathcal{M}\in\operatorname{Spc}{\mathcal{A}}:\mathcal{S}\cap\mathcal{M}=\emptyset\\}.$

We will see the collection
$\\{Z(\mathcal{S}):\mathcal{S}\subset\mathcal{A}\\}$ defines closed subsets of
$\operatorname{Spc}(\mathcal{A})$ at Proposition 3.6. We call the topological
space _Zariski topology_. The subsets
$Z(\\{V\\})=\\{\mathcal{M}\in\operatorname{Spc}(\mathcal{A}):V\notin\mathcal{M}\\}$
for $V\in\mathcal{A}$ form a basis of closed subsets.

######  Proposition 3.6.

The collection of sets $Z(\mathcal{S})$ from Definition 3.5 are the closed
subsets of a topology.

###### Proof.

Let $(\mathcal{S}_{j})_{j\in J}$ be a collection of families of objects of
$\mathcal{A}$ and let $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ be the families
of objects of $\mathcal{A}$. Then we have $\bigcap_{j\in
J}Z(\mathcal{S}_{j})=Z(\bigcup_{j\in J}\mathcal{S}_{j})$ and
$Z(\mathcal{S}_{1})\cup Z(\mathcal{S}_{2})=Z(\\{V_{1}\oplus
V_{2}:V_{1}\in\mathcal{S}_{1},\ V_{2}\in\mathcal{S}_{2}\\})$. Moreover, we
have $Z(\mathcal{A})=\emptyset$ and $Z(\emptyset)=\mathcal{A}$. Thus the
collection $\\{Z(\mathcal{S}):\mathcal{S}\subset\mathcal{A}\\}$ are the closed
subsets of a topology on $\operatorname{Spc}(\mathcal{A})$. ∎

######  Lemma 3.7.

Consider the tensor category
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$, where
$A_{\mathbb{Z}}$ is a quiver of type $A$ with bounded path length. Then, for
any representation
$V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ the closed set
$Z(\\{V\\})=\\{\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})):V\notin\mathcal{M}\\}$
is open.

###### Proof.

For any prime $\otimes$-ideal $\mathcal{M}$ in
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$,
the representation $K_{\operatorname{supp}{V}}$ is in $\mathcal{M}$ if and
only if $K_{\mathbb{Z}\setminus\operatorname{supp}{V}}$ is not in
$\mathcal{M}$. This is because if $K_{\operatorname{supp}{V}}\in\mathcal{M}$
and $K_{\mathbb{Z}\setminus\operatorname{supp}{V}}\in\mathcal{M}$, then we
have $K_{\operatorname{supp}{V}}\oplus
K_{\mathbb{Z}\setminus\operatorname{supp}{V}}\in\mathcal{M}$. It means
$K_{\mathbb{Z}}=K_{\operatorname{supp}(K_{\operatorname{supp}{V}}\oplus
K_{\mathbb{Z}\setminus\operatorname{supp}{V}})}\in\mathcal{M}$ by Lemma 3.1.
This contradicts that $\mathcal{M}$ is a prime $\otimes$-ideal. Thus
$K_{\operatorname{supp}{V}}$ or
$K_{\mathbb{Z}\setminus\operatorname{supp}{V}}$ is not in $\mathcal{M}$. Also,
one of the representations $K_{\mathbb{Z}\setminus\operatorname{supp}{V}}$ and
$K_{\operatorname{supp}{V}}$ is included. This is because
$K_{\operatorname{supp}{V}}\otimes
K_{\mathbb{Z}\setminus\operatorname{supp}{V}}=0$ in $\mathcal{M}$ and since
$\mathcal{M}$ is the prime $\otimes$-ideal, either
$K_{\operatorname{supp}{V}}$ or
$K_{\mathbb{Z}\setminus\operatorname{supp}{V}}$ is in $\mathcal{M}$. Thus we
have shown that the representation $K_{\operatorname{supp}{V}}$ is in
$\mathcal{M}$ if and only if $K_{\mathbb{Z}\setminus\operatorname{supp}{V}}$
is not in $\mathcal{M}$.

Now by Lemma 3.1, we have the equation
$Z(\\{V\\}):=\\{\mathcal{M}:V\notin\mathcal{M}\\}=\\{\mathcal{M}:K_{\operatorname{supp}{V}}\notin\mathcal{M}\\}=:Z(\\{K_{\operatorname{supp}{V}}\\})$.
Thus we have

$\displaystyle Z(\\{V\\})^{c}$
$\displaystyle=Z(\\{K_{\operatorname{supp}{V}}\\})^{c}$
$\displaystyle:=\\{\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})):K_{\operatorname{supp}{V}}\notin\mathcal{M}\\}^{c}$
$\displaystyle=\\{\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})):K_{\operatorname{supp}{V}}\in\mathcal{M}\\}$
$\displaystyle=\\{\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})):K_{\mathbb{Z}\setminus\operatorname{supp}{V}}\notin\mathcal{M}\\}$
$\displaystyle\qquad\qquad(\because K_{\operatorname{supp}(V)}\in\mathcal{M}\
\text{iff}\ K_{\mathbb{Z}\setminus\operatorname{supp}(V)}\notin\mathcal{M})$
$\displaystyle=:Z(\\{K_{\mathbb{Z}\setminus\operatorname{supp}{V}}\\}).$

Therefore $Z(\\{V\\})$ is open. ∎

######  Corollary 3.8.

Consider the tensor category
$\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$, where
$A_{\mathbb{Z}}$ is a quiver of type $A$ with bounded path length. Then, the
topological space
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
is a Hausdorff space.

###### Proof.

Let $\mathcal{M}_{1}$ and $\mathcal{M}_{2}$ be different prime
$\otimes$-ideals. Then there exists a representation $V$ such that $V$ is in
$\mathcal{M}_{1}\setminus\mathcal{M}_{2}$. Then $\mathcal{M}_{2}\in
Z(\\{V\\})$ and $\mathcal{M}_{1}\in Z(\\{V\\})^{c}$. By Lemma 3.7, the closed
set $Z(\\{V\\})$ is open. Thus
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
is a Hausdorff space. ∎

######  Lemma 3.9.

Let $R$ be a commutative ring. Then $\operatorname{Spc}(R)$ is compact as the
Zariski topology.

###### Proof.

We can refer to [14] Chapter 2, Exercise 17 (v). ∎

We rewrite the statement of Proposition 3.4 again.

######  Proposition (3.4).

Let $\mathcal{P}(\mathbb{Z})$ be the power set of $\mathbb{Z}$ and consider it
as a Boolean ring (isomorphic to
$(\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}}))$ by Example 2.15). Then we
have a homeomorphism between
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
and $\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))$, where both sets are
equipped with their respective Zariski topologies.

###### Proof.

We will show that the bijective map

$\displaystyle\psi:\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))$
$\displaystyle\to\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})),$
$\displaystyle\mathcal{Q}$
$\displaystyle\mapsto\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)\in\mathcal{Q}\\}$

defined in Proposition 3.2 is a homeomorphism. Since we know that
$\operatorname{Spc}(\mathcal{P}(\mathbb{Z}))=\operatorname{Spc}(\prod_{i\in\mathbb{Z}}(\mathbb{Z}/{2\mathbb{Z}}))$
is compact (see Lemma 3.9) and
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
is a Hausdorff space (see Lemma 3.8), it is enough to show that the bijective
map
$\psi:\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))\to\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$
is continuous (a continuous bijective map from compact space to a Hausdorff
space is homeomorphism, for example, [15, Proposition 13.26] for the proof).

Let $Z(\mathcal{S})=\bigcap_{W\in\mathcal{S}}Z(\\{W\\})$ be a closed set of
$\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$.
Then we obtain
$\psi^{-1}(Z(\mathcal{S}))=\bigcap_{W\in\mathcal{S}}{\psi^{-1}(Z(\\{W\\}))}$.
For each $\psi^{-1}(Z(\\{W\\}))$, we have the equation

$\displaystyle\psi^{-1}(Z(\\{W\\}))$
$\displaystyle=\psi^{-1}(\\{\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})):W\notin\mathcal{M}\\})$
$\displaystyle=\psi^{-1}(\\{\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})):K_{\operatorname{supp}(W)}\notin\mathcal{M}\\})\qquad(\because\text{by
Proposition \ref{lemma:KsuppViffV})}$
$\displaystyle=\\{\psi^{-1}(\mathcal{M})\in\operatorname{Spc}(\mathcal{P(\mathbb{Z})}):K_{\operatorname{supp}(W)}\notin\mathcal{M}\\}$
$\displaystyle=\\{\psi^{-1}(\mathcal{M})\in\operatorname{Spc}(\mathcal{P(\mathbb{Z})}):\operatorname{supp}(W)\notin\psi^{-1}(\mathcal{M})\\}$
$\displaystyle=\\{\mathcal{Q}\in\operatorname{Spc}(\mathcal{P(\mathbb{Z})}):\operatorname{supp}(W)\notin\mathcal{Q}\\}\qquad\qquad\quad(\text{Replace
$\psi^{-1}(\mathcal{M})$ with $\mathcal{Q}$}.)$
$\displaystyle=\\{\mathcal{Q}\in\operatorname{Spc}(\mathcal{P(\mathbb{Z})}):\mathbb{Z}\setminus\operatorname{supp}(W)\in\mathcal{Q}\\}\qquad\qquad(\because\mathcal{Q}\
\text{is a prime ideal})$
$\displaystyle=\\{\mathcal{Q}\in\operatorname{Spc}(\mathcal{P(\mathbb{Z})}):\mathbb{Z}\setminus\operatorname{supp}(W)\notin\mathcal{Q}\\}^{c}$
$\displaystyle={D_{\mathbb{Z}\setminus\operatorname{supp}(W)}}^{c}.$

Since $D_{\mathbb{Z}\setminus\operatorname{supp}(W)}$ is open, the complement
${D_{\mathbb{Z}\setminus\operatorname{supp}(W)}}^{c}=\psi^{-1}(Z(\\{W\\}))$ is
a closed set. Thus we can conclude
$\psi^{-1}(Z(\mathcal{S}))=\bigcap_{W\in\mathcal{S}}{\psi^{-1}(Z(\\{W\\}))}$
is a closed set and therefore the map $\psi$ is a homeomorphism.

∎

###  3.3. About the representations of the quiver $\vec{A_{\mathbb{Z}}}$

In the proof of Proposition 3.1, we need the condition

the length of paths are bounded.

Without it, we can not prove $K_{\operatorname{supp}(V)}\in\mathcal{M}$ if and
only if $V\in\mathcal{M}$ for any
$V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ and
$\mathcal{M}\in\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}))$.
For example, there may be a representation such that its support is
$\mathbb{Z}$, but the representation is contained in a prime $\otimes$-ideal.
We will provide such a representation when $A_{\mathbb{Z}}$ is the quiver

$\dots\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\diamond\longrightarrow\dots.$

which will be denote by $\vec{A}_{\mathbb{Z}}$. The proof seems to generalise,
to any quiver of type $A$ containing an unbounded path, but we restrict our
attention to $\vec{A}_{\mathbb{Z}}$.

Using this, we show that the map

$\displaystyle\psi:\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))$
$\displaystyle\to\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})),$
$\displaystyle\mathcal{Q}$
$\displaystyle\mapsto\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)\in\mathcal{Q}\\}$

which appeared in the proof of Proposition 3.2 is injective but not
surjective.

We will prepare some lemmas and definitions to show the existence of the
representation whose support is $\mathbb{Z}$.

######  Definition 3.10.

(An ideal generated by a set)
Let $Q$ be a quiver and $S$ a subset of objects of
$\operatorname{Rep}^{\operatorname{pfd}}(Q)$ which contains the representation
$0$. Set $S_{-1}:=S$, and for $n\geq 0$ define full subcategories
$S_{n},S_{n}^{\prime},S_{n}^{\prime\prime}$ of
$\operatorname{Rep}^{\operatorname{pfd}}(Q)$ inductively as follows.

$S_{n}:=\left\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(Q):\begin{array}[]{c}\text{
there exists an exact sequence }\\\ 0\ \to V^{\prime}\to V\to
V^{\prime\prime}\to 0\\\ \text{ with }\ V^{\prime},\ V^{\prime\prime}\in
S_{n-1}\end{array}\right\\},$ $S^{\prime}_{n}:=\\{W\otimes
V:W\in\operatorname{Rep}^{\operatorname{pfd}}(Q),\ V\in S_{n}\\},$
$S^{\prime\prime}_{n}:=\bigcup_{V,W\in
S^{\prime}_{n}}\\{\operatorname{Ker}(f),\
\operatorname{Coker}(f):V\xrightarrow{f}W\\}.$

We then define an ideal $\langle S\rangle$ of
$\operatorname{Rep}^{\operatorname{pfd}}(Q)$ by

$\langle S\rangle:=\bigcup_{n\in\mathbb{Z}_{\geq 0}}S_{n}$

and call it the $\otimes$-$ideal$ $generated$ $by$ $S$. Here, we have

$S\subseteq S_{0}\subseteq S^{\prime}_{0}\subseteq
S^{\prime\prime}_{0}\subseteq S_{1}\subseteq\dots\subseteq S_{n}\subseteq
S^{\prime}_{n}\subseteq S^{\prime\prime}_{n}\subseteq
S_{n+1}\subseteq\dots\subseteq\langle S\rangle.$

######  Proposition 3.11.

The $\langle S\rangle:=\bigcup_{n\in\mathbb{Z}_{\geq 0}}S_{n}$ defined above
is a $\otimes$-ideal of $\operatorname{Rep}^{\operatorname{pfd}}(Q)$. In fact,
it is the smallest $\otimes$-ideal containing the set $S$.

###### Proof.

First we show that $\langle S\rangle$ is closed under extensions. Let
$V^{\prime}$ and $V^{\prime\prime}$ be representations in $\langle S\rangle$.
Then, there exists $n\in\mathbb{Z}_{\geq 0}$ such that $V^{\prime}$ and
$V^{\prime\prime}$ are in $S_{n}$. Take an arbitrary exact sequence

$0\to V^{\prime}\to V\to V^{\prime\prime}\to 0,$

and we have $V\in S_{n+1}\subseteq\langle S\rangle$. Thus $\langle S\rangle$
is closed under extensions.

Next, we show that $\langle S\rangle$ is closed under kernels and cokernels.
Let $f:V\to W$ be a morphism in $\langle S\rangle$. Then, there exists
$n\in\mathbb{Z}_{\geq 0}$ such that $V$ and $W$ are in $S_{n}$ and its kernels
and cokernels are in $S_{n+1}\subseteq\langle S\rangle$.

Finally, we show that $\langle S\rangle$ is closed under tensor products. Let
$V$ and $W$ be in $\operatorname{Rep}^{\operatorname{pfd}}(Q)$ and $\langle
S\rangle$ respectively. There exists $n\in\mathbb{Z}_{\geq 0}$ such that $W\in
S_{n}$. We have $V\otimes W\in S_{n+1}\subseteq\langle S\rangle$. Therefore
$\langle S\rangle$ is a $\otimes$-ideal of
$\operatorname{Rep}^{\operatorname{pfd}}(Q)$.

Next, we will show that the $\langle S\rangle$ is the smallest $\otimes$-ideal
containing $S$. Let $\mathcal{I}$ be a $\otimes$-ideal containing $S$. By the
construction of $S_{n}$ and the definition of $\otimes$-tensor ideal, $S_{n}$
is included by $\mathcal{I}$ for every $n\in\mathbb{Z}_{\geq 0}$. Thus we have
$\langle S\rangle=\bigcup_{n\in\mathbb{Z}_{\geq 0}}S_{n}\subseteq\mathcal{I}$
and therefore $\langle S\rangle$ is the smallest $\otimes$-tensor ideal
containing $S$.

∎

######  Lemma 3.12.

Let $[a,b]$ be an interval in $\mathbb{Z}$. If there exists a monomorphism

$\phi:K_{[a,b]}\to\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}},$

then there exists $\lambda\in\Lambda$ such that $[a,b]\subseteq I_{\lambda}$
where $K_{[a,b]}$ and $\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}$ are
representations of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$.

Similarly, if there exists an epimorphism

$\psi:\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}\to K_{[a,b]},$

then there exists $\lambda\in\Lambda$ such that $[a,b]\subseteq I_{\lambda}$
where $K_{[a,b]}$ and $\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}$ are
representations of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$.

###### Proof.

We will show the monomorphism case by contradiction. The other case can be
shown in the same way.

Assume that for any $\lambda\in\Lambda$, we have $[a,b]\nsubseteq
I_{\lambda}$. Then we obtain non-commutative diagram

$\textstyle{(K_{[a,b]})_{a}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{a}}$$\scriptstyle{id_{K}}$$\textstyle{(\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}})_{a}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0}$$\textstyle{(K_{[a,b]})_{b}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{b}}$$\textstyle{(\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}})_{b}\
,}$

and it is the contradiction. Therefore there must be $\lambda\in\Lambda$ such
that $[a,b]\subseteq I_{\lambda}$. ∎

######  Lemma 3.13.

Let $I$ be an interval in $\mathbb{Z}$. If there exists a monomorphism

$\phi:K_{I}\to\bigoplus_{\lambda\in\Lambda}K_{J_{\lambda}},$

then there exists $\lambda\in\Lambda$ such that $I\subseteq J_{\lambda}$ where
$K_{I}$ and $\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}$ are representations
of $\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$.

Similarly, if there exists an epimorphism

$\psi:\bigoplus_{\lambda\in\Lambda}K_{J_{\lambda}}\to K_{I},$

then there exists $\lambda\in\Lambda$ such that $I\subseteq J_{\lambda}$ where
$K_{I}$ and $\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}$ are the
representations of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$.

###### Proof.

We will show the monomorphism case. The other case can be shown in the same
way.

If $I$ is bounded, then we have already shown in Lemma 3.12. So we will assume
that $I$ is not bounded.

In case of $I=\mathbb{Z}$, let $I_{n}$ be the interval $[-n,n]$ for
$n\in\mathbb{Z}_{\geq 0}$. By restricting the interval $I=\mathbb{Z}$ to
$I_{n}$, we obtain the monomorphism

$\phi:K_{I_{n}}\to\bigoplus_{\lambda\in\Lambda}K_{J_{\lambda}\cap I_{n}}.$

By Lemma 3.12, there exists $\lambda\in\Lambda$ such that $I_{n}\subseteq
J_{\lambda}\cap I_{n}$. Thus for any $n\in\mathbb{Z}_{\geq 0}$ we obtain
$I_{n}\subseteq J_{{\lambda_{n}}}$ for some $\lambda_{n}$. Thus the set
$C_{n}:=\\{J_{\lambda}:I_{n}\subseteq J_{\lambda},\
\lambda\in\Lambda\\}\subseteq\\{J_{\lambda}:0\in J_{\lambda},\
\lambda\in\Lambda\\}$ is not empty and finite because the cardinarity of the
set $\\{J_{\lambda}:0\in J_{\lambda},\ \lambda\in\Lambda\\}$ is equal to or
smaller than the finite dimension
dim${}_{K}(\bigoplus_{\lambda\in\Lambda}K_{J_{\lambda}})_{0}$. Thus, for the
descending sequence

$C_{0}\supseteq C_{1}\supseteq C_{2}\supseteq C_{3}\supseteq\cdots,$

there exists $m\in\mathbb{Z}_{\geq 0}$ such that:

$C_{m}=C_{m+1}=\cdots.$

We can conclude that there exists $\lambda\in\Lambda$ such that
$I=\mathbb{Z}=J_{\lambda}\in\bigcap_{n\in\mathbb{Z}_{\geq 0}}C_{n}$.

If $I$ is bounded above and unbounded below, then there exists
$a\in\mathbb{Z}$ such that $I=(-\infty,a]$. Let $I^{\prime}_{n}$ be the
interval $[a-n,a]$ for $n\in\mathbb{Z}_{\geq 0}$. We can show that
$C^{\prime}_{n}:=\\{J_{\lambda}:I^{\prime}_{n}\subseteq J_{\lambda},\
\lambda\in\Lambda\\}$ is not empty because we have a monomorphism

$\phi:K_{I^{\prime}_{n}}\to\bigoplus_{\lambda\in\Lambda}K_{J_{\lambda}\cap
I^{\prime}_{n}},$

such that there exists $\lambda\in\Lambda$ which satisfies
$I^{\prime}_{n}\subset J_{\lambda}\cap I^{\prime}_{n}\subseteq J_{\lambda}$ by
Lemma 3.12. Also, we can show $C^{\prime}_{n}\subseteq\\{J_{\lambda}:a\in
J_{\lambda},\ \lambda\in\Lambda\\}$ is finite because $\\{J_{\lambda}:a\in
J_{\lambda},\ \lambda\in\Lambda\\}$ is equal to or smaller that the
dim${}_{K}(\bigoplus_{\lambda\in\Lambda}K_{J_{\lambda}})_{a}$. We will obtain
the descending sequence that stops for a $m\in\mathbb{Z}_{\geq 0}$

$C^{\prime}_{0}\supseteq C^{\prime}_{1}\supseteq C^{\prime}_{2}\supseteq
C^{\prime}_{3}\supseteq\cdots\cdots\supseteq
C^{\prime}_{m}=C^{\prime}_{m+1}=\cdots.$

We can conclude that there exists $\lambda\in\Lambda$ such that
$I^{\prime}=(-\infty,a]=J_{\lambda}\in\bigcap_{n\in\mathbb{Z}_{\geq
0}}C^{\prime}_{n}$.

In case of $I$ is bounded below and unbounded above, the proof is similar to
the case that $I$ is bounded above and unbounded below. ∎

######  Lemma 3.14.

Let $V$ and $W$ be representations of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$ whose interval
decomposition are $\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}$ and
$\bigoplus_{\lambda^{\prime}\in\Lambda^{\prime}}K_{J_{\lambda^{\prime}}}$. If
there exists a monomorphism

$\phi:V\to W,$

then for any interval $I_{\lambda}$, there exists
$\lambda^{\prime}\in\Lambda^{\prime}$ such that $I_{\lambda}\subseteq
J_{\lambda^{\prime}}$.

Similarly if there exists an epimorphism

$\psi:V\to W,$

then for any interval $I_{\lambda}$, there exists
$\lambda^{\prime}\in\Lambda^{\prime}$ such that $I_{\lambda}\subseteq
J_{\lambda^{\prime}}$.

###### Proof.

We will show the monomorphism case. The other case can be shown in the same
way.

Since we have the natural monomorphism

$K_{I_{\lambda}}\to\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}\cong
V\xrightarrow{\phi}W\cong\bigoplus_{\lambda^{\prime}\in\Lambda^{\prime}}K_{J_{\lambda^{\prime}}},$

by Lemma 3.13, there exists $\lambda^{\prime}\in\Lambda^{\prime}$ such that
$I_{\lambda}\subseteq J_{\lambda^{\prime}}$. ∎

######  Corollary 3.15.

Let $V$ and $W$ be representations of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$ whose interval
decomposition are $\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}$ and
$\bigoplus_{\lambda^{\prime}\in\Lambda^{\prime}}K_{J_{\lambda^{\prime}}}$. If
there exists a monomorphism (resp. epimorphism)

$\phi:V\to W,$

then $V$ (resp. $W$) is a direct summund of $V\otimes W$.

###### Proof.

Now we have $V\otimes
W\cong(\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}})\otimes(\bigoplus_{\lambda^{\prime}\in\Lambda^{\prime}}K_{J_{\lambda^{\prime}}})\cong\bigoplus_{\lambda\in\Lambda,\lambda^{\prime}\in\lambda^{\prime}}K_{I_{\lambda}}\otimes
K_{J_{\lambda^{\prime}}}=\bigoplus_{\lambda\in\Lambda,\lambda^{\prime}\in\lambda^{\prime}}K_{I_{\lambda}\cap
J_{\lambda^{\prime}}}$.

If the map $\phi$ is a monomorphism, for any $\lambda\in\Lambda$ there exists
$\lambda^{\prime}\in\Lambda^{\prime}$ such that $I_{\lambda}\subseteq
J_{\lambda^{\prime}}$ by Lemma 3.14. For these $\lambda$ and
$\lambda^{\prime}$, we have $K_{I_{\lambda}}\otimes
K_{J_{\lambda^{\prime}}}=K_{I_{\lambda}\cap
J_{\lambda^{\prime}}}=K_{I_{\lambda}}$. Therefore $V\otimes
W\cong\bigoplus_{\lambda\in\Lambda,\lambda^{\prime}\in\Lambda^{\prime}}K_{I_{\lambda}\cap
J_{\lambda^{\prime}}}$ has the direct summand
$\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}\cong V$.

In the similar way, we can show the case of $\phi$ is epi. ∎

######  Lemma 3.16.

Let $V^{\prime}$ and $V^{\prime\prime}$ be representations in
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A}_{\mathbb{Z}})$. If these are
decomposed into interval representations whose intervals are bounded, then for
any exact sequence

$0\to V^{\prime}\to V\to V^{\prime\prime}\to 0,$

the representation $V$ is decomposed into interval representations whose
intervals are bounded. That is, if $V^{\prime}$ and $V^{\prime\prime}$ are of
the form $\oplus_{j\in J}K_{I_{j}}$ for some collection $\\{I_{j}\\}_{j\in J}$
of bounded $I_{i}\subseteq\mathbb{Z}$, then $V$ is also of this form.

###### Proof.

We prove this lemma by a contradiction. Note that $V$ is of the form
$\oplus_{j\in J}K_{I_{j}}$ for some collection $\\{I_{j}\\}_{j\in J}$ of
$I_{i}\subseteq\mathbb{Z}$ by Proposition 2.4. We will show that if some
$I_{i}$ is unbounded, then there is a contradiction. Assume that there is an
unbounded interval $I$ in $\vec{A}_{\mathbb{Z}}$ such that the interval
representation $K_{I}$ is a direct summand of $V$ and choose an isomorphism
$V\cong W\oplus K_{I}$. Then we can rewrite the exact sequence by

$0\to V^{\prime}\xrightarrow{(\phi,\psi)}W\oplus
K_{I}\xrightarrow{(f,g)}V^{\prime\prime}\to 0$

Now, assume that $I$ is unbounded above. (We can also assume $I$ is unbounded
below, but the proof is similar.) Then, $\psi_{a}=0$ for every
$a\in\mathbb{Z}$. This is because if there exists $a\in\mathbb{Z}$ such that
$\psi_{a}\neq 0$, we obtain non-commutative square

$\textstyle{V^{\prime}_{a}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(\phi,\psi)_{a}}$$\textstyle{(W\oplus
K_{I})_{a}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V^{\prime}_{a{+}m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(\phi,\psi)_{a{+}m}}$$\textstyle{(W\oplus
K_{I})_{a+m}}$

where $m$ is large enough for $V^{\prime}_{a}\to V^{\prime}_{a+m}$ to be zero.
Thus $\psi_{a}=0$ for all $a\in\mathbb{Z}$. Then, since
$0=$Im$(\psi)=\operatorname{Ker}(g)$, we have that $g:K_{I}\to
V^{\prime\prime}$ is injective. This implies $V^{\prime\prime}$ has a
unbounded interval representation as a direct summand. However, the
representation $V^{\prime\prime}$ is decomposed into interval representations
whose intervals are bounded. It is a contradiction. Hence, we can conclude
that $V$ is decomposed into interval representations whose intervals are
bounded. ∎

######  Proposition 3.17.

Let $\mathcal{I}$ be the full subcategory of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$ whose
representations are decomposed into bounded interval representations
$\oplus_{j\in J}K_{I_{j}}$ for some collection $\\{I_{j}\\}_{j\in J}$ of
bounded $I_{i}\subseteq\mathbb{Z}$. Then $\mathcal{I}$ is a tensor ideal of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$.

###### Proof.

We have already shown that $\mathcal{I}$ is closed under extensions in Lemma
3.16. Moreover $\mathcal{I}$ is closed under kernels and cokernels by Lemma
3.14.

Next, we will show that the representation $V\otimes W$ is in $\mathcal{I}$
for any $V\in\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$
and $W\in\mathcal{I}$. Let
$V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$ and
$W\in\mathcal{I}$ be isomorphic to the representations
$\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}}$ and
$\bigoplus_{\lambda^{\prime}\in\Lambda^{\prime}}K_{J_{\lambda^{\prime}}}$
where $J_{\lambda^{\prime}}\subseteq\mathbb{Z}$ is bounded for every
$\lambda^{\prime}\in\Lambda^{\prime}.$ Then we obtain
$(\bigoplus_{\lambda\in\Lambda}K_{I_{\lambda}})\otimes(\bigoplus_{\lambda^{\prime}\in\Lambda^{\prime}}K_{J_{\lambda^{\prime}}})\cong\bigoplus_{\lambda\in\Lambda,\lambda^{\prime}\in\lambda^{\prime}}K_{I_{\lambda}\cap
J_{\lambda^{\prime}}}\in\mathcal{I}$.

We have shown that $\mathcal{I}$ is a tensor ideal. ∎

######  Definition 3.18.

Define the representation
$K^{\prime}_{\mathbb{Z}}\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})$
by

$(K^{\prime}_{\mathbb{Z}})_{a}:=K,\qquad(K^{\prime}_{\mathbb{Z}})_{\alpha}:=0,\
$

for every $a\in\mathbb{Z}$ and every arrow $\alpha$.

######  Lemma 3.19.

Let $S$ be the set $\\{0,K^{\prime}_{\mathbb{Z}}\\}$ of objects of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A}_{\mathbb{Z}})$ and let
$\langle S\rangle$ be the $\otimes$-ideal generated by $S$. Then we have
$K_{\mathbb{Z}}\notin\langle S\rangle$.

###### Proof.

We show $K_{\mathbb{Z}}\notin\langle S\rangle$ by a contradiction. Assume that
$K_{\mathbb{Z}}\in\langle S\rangle$. Then there exists $n\in\mathbb{Z}_{\geq
0}$ such that $K_{\mathbb{Z}}\in S_{n}$ where $S_{n}$ is defined in Definition
3.10. By Lemma 3.16, we can see that any representations of $S_{0}$ are
decomposed into bounded interval representations. Since any representations of
$S^{\prime}_{0}$ and $S^{\prime\prime}_{0}$ are obtained by tensoring and
taking kernels and cokernels, their representations are decomposed into
bounded interval representations. Therefore any representation in $S_{1}$ are
also decomposed into bounded interval representations. Inductively, every
representation of $S_{n}$ are decomposed into bounded interval
representations. Since $K_{\mathbb{Z}}\in S_{n}$, the interval representation
$K_{\mathbb{Z}}$ are decomposed into bounded interval representations.
However, the representation $K_{\mathbb{Z}}$ is the indecomposable interval
representation whose interval is infinite length. This is a contradiction.
Hence we have $K_{\mathbb{Z}}\notin\langle S\rangle$. ∎

######  Proposition 3.20.

There exists a maximal $\otimes$-ideal in
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A}_{\mathbb{Z}})$ that contains
a representation whose support is $\mathbb{Z}$.

###### Proof.

Let $S$ be the set $\\{0,K^{\prime}_{\mathbb{Z}}\\}$ and $\langle S\rangle$ be
the ideal generated by $S$. Define a partially ordered set
$(\Sigma,\subseteq):=\\{\text{ proper ideals containing $\langle S\rangle$
}\\}$. Then $\Sigma$ is not an empty set because $\langle S\rangle\in\Sigma$.
Let $\sigma\subseteq\Sigma$ be a totally ordered subset. Then
$\bigcup_{\mathcal{I}\in\sigma}\mathcal{I}$ is a proper ideal (it does not
contain $K_{\mathbb{Z}}$) and upper bound of $\sigma$. Thus by Zorn’s lemma,
there exists a maximal ideal containing $K^{\prime}_{\mathbb{Z}}\in\langle
S\rangle$ whose support are $\mathbb{Z}$. ∎

######  Lemma 3.21.

Let $\mathcal{M}$ be a maximal $\otimes$-ideal of
$\mathcal{A}=\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$.
Then, the maximal $\otimes$-ideal is a prime $\otimes$-ideal.

###### Proof.

Let $\mathcal{M}$ be a maximal $\otimes$-ideal and suppose that $W\otimes
W^{\prime}\in\mathcal{M}$. We will show that if $W\notin\mathcal{M}$ then
$W^{\prime}\in\mathcal{M}$. Consider the set

$\mathcal{I}:=\\{V\in\mathcal{A}:V\otimes W^{\prime}\in\mathcal{M}\\}.$

Then $W\in\mathcal{I}$ and $\mathcal{M}\subseteq\mathcal{I}$ hold. Since we
have assumed $W\notin\mathcal{M}$, the set $\mathcal{I}$ is strictly larger
than $\mathcal{M}$. If we show $\mathcal{I}$ is a $\otimes$-ideal, then the
tensor unit $I$ is in $\mathcal{A}=\mathcal{I}$ by the maximality of
$\mathcal{M}$, and therefore, we can show $W^{\prime}=I\otimes
W^{\prime}\in\mathcal{M}.$ So it is enough to show that $\mathcal{I}$ is a
$\otimes$-ideal to prove $\mathcal{M}$ is a prime $\otimes$-ideal.

First, we will show that $\mathcal{I}$ is closed under extensions. Let
$V^{\prime}$ and $V^{\prime\prime}$ be in $\mathcal{I}$. For any short exact
sequence

$0\to V^{\prime}\to V\to V^{\prime\prime}\to 0,$

there exists an exact sequence

$0\to V^{\prime}\otimes W^{\prime}\to V\otimes W^{\prime}\to
V^{\prime\prime}\otimes W^{\prime}\to 0$

such that $V^{\prime}\otimes W^{\prime}$ and $V^{\prime\prime}\otimes
W^{\prime}$ are in the $\otimes$-ideal $\mathcal{M}.$ Thus, the representation
$V\otimes W^{\prime}$ is in $\mathcal{M}$ and therefore $V$ is in
$\mathcal{I}$.

Next, we will show that $\mathcal{I}$ is closed under multiplication. Let $V$
be in $\mathcal{A}$ and $V^{\prime}$ be in $\mathcal{I}$. We have $(V\otimes
V^{\prime})\otimes W^{\prime}=V\otimes(V^{\prime}\otimes
W^{\prime})\in\mathcal{M}$ and therefore $(V\otimes
V^{\prime})\in\mathcal{I}.$

Then, we will show that $\mathcal{I}$ is closed under kernels and cokernels.
Let $f:V\to V^{\prime}$ be in $\mathcal{I}$. Then, we have the monomorphism
$ker(f)\otimes id_{W^{\prime}}:\operatorname{Ker}(f)\otimes
W^{\prime}\hookrightarrow V\otimes W^{\prime}$ such that $V\otimes W^{\prime}$
is in $\mathcal{M}$. Thus $(\operatorname{Ker}(f)\otimes
W^{\prime})\otimes(V\otimes W^{\prime})$ is in $\mathcal{M}$. By Lemma 3.15,
we have $\operatorname{Ker}(f)\otimes W^{\prime}$ as a direct summand of
$(\operatorname{Ker}(f)\otimes W^{\prime})\otimes(V\otimes W^{\prime})$ and
since $\mathcal{M}$ is a $\otimes$-ideal, the representation
$\operatorname{Ker}(f)\otimes W^{\prime}$ is in $\mathcal{M}$. Hence
$\operatorname{Ker}(f)$ is in $\mathcal{I}$ and therefore $\mathcal{I}$ is
closed under kernels. Similarly we can show that $\mathcal{I}$ is closed under
cokernels.

We have showed $\mathcal{I}$ is a $\otimes$-ideal. ∎

Next corollary is what we wanted.

######  Corollary 3.22.

There exists a prime $\otimes$-ideal of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$ which contains
a representation whose support is $\mathbb{Z}$.

Thus the map

$\displaystyle\psi:\operatorname{Spc}(\mathcal{P}({\mathbb{Z}}))$
$\displaystyle\to\operatorname{Spc}(\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}})),$
$\displaystyle\mathcal{Q}$
$\displaystyle\mapsto\\{V\in\operatorname{Rep}^{\operatorname{pfd}}(A_{\mathbb{Z}}):\operatorname{supp}(V)\in\mathcal{Q}\\}$

constructed in Proposition 3.2 is not surjective.

###### Proof.

By Proposition 3.20, there exists a maximal $\otimes$-ideal of
$\operatorname{Rep}^{\operatorname{pfd}}(\vec{A_{\mathbb{Z}}})$ that contains
a representation whose support is $\mathbb{Z}$ and the maximal $\otimes$-ideal
is prime $\otimes$-ideal by Corollary 3.21. ∎

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