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\title{An explicit nonideal rooted cluster morphism}
\author{Qiyue Tang}
\date{}
\begin{document}
\maketitle
\begingroup
\renewcommand{\thefootnote}{}
\footnotetext{\textit{2020 Mathematics Subject Classification.} Primary 13F60; Secondary 13B22.\par
\textit{Keywords.} Rooted cluster morphism, specialization, upper cluster algebra.}
\endgroup
\begin{abstract}
We construct an explicit, inducible rooted cluster morphism that is not ideal, answering a question of Chang and Zhu. Its image is a polynomial ring, whereas its image-seed algebra is an integrally closed, non-Noetherian subring. We also give a coefficient-free example in which specializing a mutable variable to one does not map into the cluster algebra of the deleted seed.
\end{abstract}

\section{Introduction}

A rooted cluster algebra is a cluster algebra with a specified initial seed.
Assem, Dupont and Schiffler introduced rooted cluster morphisms to study maps
that respect initial variables and mutation along biadmissible sequences
\cite{ADS}. Such a morphism determines an image seed, whose cluster algebra is
contained in its ring-theoretic image. The morphism is called ideal when these
two algebras agree. Chang and Zhu constructed nonideal morphisms whose values
on the initial variables do not determine the morphism. They then asked
whether every explicit rooted cluster morphism is ideal
\cite[\S2.2, Problem 2]{CZ}. Here explicit means uniquely determined, for
fixed source and target, by its values on the initial variables.

We answer this question negatively with three mutable variables in both
source and target and one frozen variable in the source. Throughout the paper,
the ground ring is $\Z$, and frozen variables are not inverted. This is the
coefficient convention of \cite{ADS,CZ}.

Here is the construction. Let $C=\A(\Sigma')$ have mutable variables
$(a,b,z)$, no frozen variables, and exchange matrix $B$. Let
$A=\A(\Sigma)$ have mutable variables $(a,b,u)$, frozen variable $t$, and
extended exchange matrix $\widetilde B$, where
\begin{equation}\label{eq:seeds}
B=\begin{pmatrix}0&2&-1\\-2&0&1\\1&-1&0\end{pmatrix},
\qquad
\widetilde B=\begin{pmatrix}0&2&-1\\-2&0&1\\1&-1&0\\1&-1&0\end{pmatrix}.
\end{equation}
Rows are ordered by the displayed mutable variables followed by the frozen
variable; columns are ordered by mutable variables. The source and target
have independent ambient coordinates; the repeated names $a,b$ specify
which coordinates the map identifies. In $\Q(a,b,z)$ put
\begin{equation}\label{eq:S}
s=\frac{a^2+b^2+z}{ab},\qquad P=\Z[a,b,s].
\end{equation}
Since $z=abs-a^2-b^2$, the elements $a,b,s$ are algebraically independent.
Thus $P$ is a polynomial ring in three variables. The notation $(a,b)P$
denotes its ideal generated by $a,b$.

\begin{theorem}\label{thm:main}
The assignment
\[
a\longmapsto a,\qquad b\longmapsto b,\qquad
u\longmapsto1,\qquad t\longmapsto z
\]
extends to an explicit, inducible rooted cluster morphism $f:A\to C$.
Its image seed has mutable variables $a,b$, frozen variable $z$, and matrix
\[
B_R=\begin{pmatrix}0&2\\-2&0\\1&-1\end{pmatrix}.
\]
Writing $R=\A(f(\Sigma))$, one has
\[
R=\Z+(a,b)P\subsetneq f(A)=P\subsetneq C.
\]
In particular, $f$ is not ideal.
\end{theorem}

The global definition of $f$ is the main issue: an assignment on initial
variables need not send all source cluster variables into the target cluster
algebra. We prove this containment by specializing Laurent expansions in
three source clusters and computing the intersection of the resulting rings.
The image-seed algebra is then determined by a rank-two recurrence.

The invariant $s$ has an established annulus interpretation. At $q=1$, the
loop formulas of Muller \cite[\S12.2, pp.~490--491]{Muller} give
\[
\ell x_i=x_{i+1}+x_{i-1},\qquad
\ell=\frac{x_i^2+x_{i+1}^2+r_0r_1}{x_ix_{i+1}},
\]
where $r_0,r_1$ are the two boundary variables. Setting $r_0=z,r_1=1$
recovers $s$. Muller's cluster algebra inverts these boundary variables.
Here the polynomial coefficient ring permits the strict inclusion in
Theorem~\ref{thm:main}. We compute its conductor and integral closure in
Section~\ref{sec:ring}.

Section~\ref{sec:deletion} gives a separate four-variable example for which
freezing fails to give a cluster localization. The same example answers the
specialization-at-one question of Assem--Dupont--Schiffler
\cite[Problem 6.16 in arXiv v1]{ADS} and the question following
Theorem~6.7 of Huang--Li--Yang \cite{HLY}.

\section{Mutation and polynomial coefficients}\label{sec:prelim}

We use finite seeds $\Sigma=(E,F,B)$. The disjoint sets $E$ and $F$ of
mutable and frozen variables are algebraically independent over $\Q$.
The integer matrix $B$ has rows indexed by $X=E\sqcup F$ and columns by
$E$. Its principal part is skew-symmetrizable: there are positive integers
$d_i$ for $i\in E$ with $d_i b_{ij}=-d_j b_{ji}$. The ambient field is
$\Q(X)$. For an integer $r$, write $[r]_+=\max(r,0)$.
Mutation at $k\in E$ replaces $x_k$ by $x_k'$ according to
\begin{equation}\label{eq:exchange}
x_kx_k'=\prod_{i\in X}x_i^{[b_{ik}]_+}
                  +\prod_{i\in X}x_i^{[-b_{ik}]_+},
\end{equation}
and replaces the matrix by
\begin{equation}\label{eq:matrix}
b'_{ij}=\begin{cases}
-b_{ij},&i=k\text{ or }j=k,\\
b_{ij}+[b_{ik}]_+[b_{kj}]_+-[-b_{ik}]_+[-b_{kj}]_+,&\text{otherwise}.
\end{cases}
\end{equation}
An empty product is $1$. The algebra $\A(\Sigma)$ is generated over $\Z$
by $F$ and all mutable variables obtained by finite mutation sequences.
For a set of independent variables $E$, the notation $\Z[E^{\pm1},F]$
means that only the variables in $E$ are inverted.

We use the Laurent phenomenon of Fomin--Zelevinsky
\cite[Theorem 3.1]{FZI} and its polynomial-coefficient form
\cite[Proposition 11.2]{FZII}. The latter occurs in a section explicitly
allowing infinite cluster type. The following argument explains why no
frozen inverses are needed.

\begin{lemma}\label{lem:laurent}
For every seed $(E_*,F,B_*)$ mutation equivalent to $\Sigma$,
\[
\A(\Sigma)\subseteq\Z[E_*^{\pm1},F].
\]
\end{lemma}
\begin{proof}
The ordinary Laurent phenomenon gives this inclusion with $F^{\pm1}$
in place of $F$. Fix a frozen variable $t\in F$ and use $E_*$ as base
coordinates. Regard rational functions as elements of $K(t)$, where
$K=\Q(E_*,F\setminus\{t\})$. The order at $t=0$ of a nonzero rational
function is the exponent of its first nonzero term in $t$; its leading
coefficient belongs to $K$.

Initially every mutable variable has order zero and a leading coefficient
given by a subtraction-free rational expression in the remaining
coordinates. Suppose these properties hold at a current seed. If the
entry of the $t$-row in the mutated column is $c$, the two terms on the
right of \eqref{eq:exchange} have orders $[c]_+$ and $[-c]_+$.
Their minimum is zero. Their leading coefficients, when both contribute,
cannot cancel: they are subtraction-free rational expressions and are
positive upon any positive specialization of the remaining coordinates.
Dividing by the old mutable variable preserves order zero and this
property of the leading coefficient. Induction along any mutation word
therefore gives order zero for every mutable cluster variable.

In the Laurent expansion already supplied by the ordinary theorem, order
zero excludes negative powers of $t$. Applying this to each frozen
variable proves the assertion for the generators and hence for the algebra.
\end{proof}

For clarity, we recall the morphism convention of
\cite[Definition 2.5]{CZ}. A unital ring map
$f:\A(\Sigma)\to\A(\Sigma')$ is a rooted cluster morphism if
$f(E)\subseteq E'\cup\Z$, $f(F)\subseteq E'\cup F'\cup\Z$, and it
commutes with mutation on every biadmissible sequence, on all initial
variables. A sequence is admissible if each term is mutable when used;
it is biadmissible if its termwise image is also admissible in the target.
Integer images remain unchanged. In particular, a frozen source variable
may map to a mutable target variable.

The image seed $f(\Sigma)$ retains from the target the mutable variables
$f(E)\cap E'$, all variables $f(X)\cap X'$, and the corresponding matrix
rows and mutable columns. Its other retained variables are frozen.
The morphism is ideal when $f(\A(\Sigma))=\A(f(\Sigma))$. It is
explicit when no other rooted cluster morphism with the same source,
target, and initial values differs from it. It is inducible when it
extends to a homomorphism between the ambient Laurent rings
$\Q[E^{\pm1},F]$ and $\Q[(E')^{\pm1},F']$
\cite[Definition 2.8]{CZ}.

\section{The morphism and its image}\label{sec:main}

We now use the two seeds in \eqref{eq:seeds} and the polynomial ring $P$
in \eqref{eq:S}. Denote the target mutations at $a$ and $b$, performed
separately from the initial seed, by
\begin{equation}\label{eq:neighbors}
a'=\frac{b^2+z}{a}=bs-a,\qquad
b'=\frac{a^2+z}{b}=as-b.
\end{equation}

\begin{lemma}\label{lem:intersection}
Inside $\Q(a,b,z)$ one has
\[
\Z[a^{\pm1},b^{\pm1},z]
\cap\Z[(a')^{\pm1},b^{\pm1},z]
\cap\Z[a^{\pm1},(b')^{\pm1},z]=P.
\]
\end{lemma}
\begin{proof}
The three rings are, respectively,
\begin{equation}\label{eq:charts}
P[(ab)^{-1}],\qquad P[(a'b)^{-1}],\qquad P[(ab')^{-1}].
\end{equation}
For the middle equality, use
\[
s=\frac{(a')^2+b^2+z}{a'b},\qquad a=bs-a'.
\]
The last equality follows by exchanging $a,b$, and the first follows
directly from \eqref{eq:S}.

The ring $P$ is a unique factorization domain. Any irreducible factor
remaining in a reduced denominator of an element of all three rings in
\eqref{eq:charts} must divide $ab$, $a'b$, and $ab'$. The only possible
factors of $ab$ are $a,b$, up to units. But modulo $a$ the element $a'b$
is $b^2s\ne0$, and modulo $b$ the element $ab'$ is $a^2s\ne0$.
Thus no denominator factor remains. Each ring contains $P$, proving equality.
\end{proof}

\begin{lemma}\label{lem:landing}
The Laurent-ring homomorphism
\[
\phi:\Z[a^{\pm1},b^{\pm1},u^{\pm1},t]
\longrightarrow\Z[a^{\pm1},b^{\pm1},z],\qquad
(a,b,u,t)\longmapsto(a,b,1,z)
\]
satisfies $\phi(A)\subseteq P\subseteq C$.
\end{lemma}
\begin{proof}
The map is defined because $a,b,1$ are units in its codomain.
The source one-step mutations at $a,b$ are
\[
\widetilde a'=\frac{b^2+ut}{a},\qquad
\widetilde b'=\frac{a^2+ut}{b}.
\]
Their images are $a',b'$. By Lemma~\ref{lem:laurent}, any $h\in A$
has a Laurent expansion in each of the source clusters
$(a,b,u)$, $(\widetilde a',b,u)$, and $(a,\widetilde b',u)$,
with coefficients in $\Z[t]$. To specialize a neighboring expansion,
multiply it by its mutable monomial denominator. The resulting identity
lies in the initial Laurent ring, so $\phi$ applies to it. The image of
that denominator is nonzero, since $a,b,a',b'$ are nonzero. Dividing in
$\Q(a,b,z)$ gives $\phi(h)$ in the corresponding target ring of
Lemma~\ref{lem:intersection}. Hence $\phi(h)\in P$.

For the other inclusion, mutate the target in positions $(3,1,2)$,
in that order. The first two new matrices are
\[
\begin{pmatrix}0&1&1\\-1&0&-1\\-1&1&0\end{pmatrix},
\qquad
\begin{pmatrix}0&-1&-1\\1&0&-1\\1&1&0\end{pmatrix}.
\]
The three new variables, denoted $p,q,v$, satisfy
\begin{equation}\label{eq:targetword}
p=\frac{a+b}{z},\qquad q=\frac{1+bp}{a},\qquad
v=\frac{q+p}{b}=\frac{(a+b)^2+z}{abz}.
\end{equation}
Thus $s=zv-2\in C$, and $a,b\in C$ imply $P\subseteq C$.
\end{proof}

\begin{proposition}\label{prop:morphism}
The restriction $f=\phi|_A:A\to C$ is an explicit, inducible rooted
cluster morphism, and $f(A)=P$.
\end{proposition}
\begin{proof}
Lemma~\ref{lem:landing} supplies the ring homomorphism. The initial
assignments satisfy the two conditions on mutable and frozen images.
A biadmissible sequence cannot have a first mutation in the third source
position: before that step the variable is still $u$, with image $1$.
Hence every biadmissible sequence uses only positions $1,2$.

Conversely, consider any word using those two positions. In the source,
the $u$- and $t$-rows remain equal by \eqref{eq:matrix}, because neither
row is mutated. Restricted to columns $1,2$, they agree with the target
$z$-row; the two mutable principal parts agree as well. In each exchange
monomial their contribution has the form $u^rt^r$, which specializes to
$z^r$. Induction proves that the current variables in positions $1,2$
map to the corresponding target variables. In particular the word is
biadmissible, and mutation commutes with $f$. The unchanged variables
$u,t$ and their images $1,z$ also satisfy the required identities.
This proves the mutation condition for every biadmissible sequence and
every initial variable.

Extending $\phi$ to rational coefficients and including its codomain in
$\Q[a^{\pm1},b^{\pm1},z^{\pm1}]$ proves inducibility. To prove
explicitness directly, let $g:A\to C$ be any unital ring map with the
same initial values. For $h\in A$, Lemma~\ref{lem:laurent} gives
nonnegative integers $i,j,k$ and a polynomial $Q\in\Z[a,b,u,t]$ with
$a^ib^ju^kh=Q$. Applying $f,g$ and cancelling their common nonzero
factor $a^ib^j$ in the domain $C$ gives $f(h)=g(h)$.

Finally, the source word $(3,1,2)$ produces
\[
\widetilde p=\frac{a+b}{u},\qquad
\widetilde q=\frac{t+b\widetilde p}{a},\qquad
\widetilde v=\frac{\widetilde q+\widetilde p}{b}
             =\frac{(a+b)^2+ut}{abu}.
\]
The frozen row is $(1,-1,0)$ after the first step and $(-1,0,1)$ after
the second, verifying the coefficients in these exchanges. Consequently
$f(\widetilde v)=s+2$. Since $a,b$ also belong to $f(A)$,
Lemma~\ref{lem:landing} now gives $f(A)=P$.
\end{proof}

\begin{proposition}\label{prop:image-seed}
The image-seed algebra is $R=\Z+(a,b)P$.
\end{proposition}
\begin{proof}
Its mutable variables are $a,b$, and $z$ is frozen, so its matrix is
$B_R$ in Theorem~\ref{thm:main}. Mutation at either mutable position
negates this entire matrix. Cancelling consecutive repetitions of a
position reduces every word to an alternating word. Thus all mutable
variables are the elements $x_i$, $i\in\Z$, defined in both directions by
\begin{equation}\label{eq:recurrence}
x_0=a,\quad x_1=b,\qquad x_{i-1}x_{i+1}=x_i^2+z.
\end{equation}
All clusters are adjacent pairs, up to their ordering. The identity
\[
\frac{x^2+y^2+z}{xy}
=\frac{x+x'}{y}
=\frac{(x')^2+y^2+z}{x'y},\qquad xx'=y^2+z,
\]
shows that the expression defining $s$ is invariant under mutation.
Therefore
\begin{equation}\label{eq:linear}
s=\frac{x_{i-1}^2+x_i^2+z}{x_{i-1}x_i},\qquad
x_{i+1}=sx_i-x_{i-1}\quad(i\in\Z).
\end{equation}
Every $x_i$ is a $\Z[s]$-linear combination of $a,b$.
Also $z=abs-a^2-b^2\in(a,b)P$. This proves
$R\subseteq\Z+(a,b)P$.

To prove the reverse inclusion, \eqref{eq:linear} gives, for every $n\geq0$,
\[
 x_{n+1}-bs^n,\ x_{-n}-as^n
 \ \in\ \sum_{j=0}^{n-1}\bigl(\Z as^j+\Z bs^j\bigr).
\]
For $n=0$ the sum is zero; the general formulas follow by induction from the two directions
of \eqref{eq:linear}. Simultaneous induction on $n$ now proves
$as^n,bs^n\in R$: subtract the lower-degree terms, already in $R$, from
the corresponding cluster variable. Multiplication by nonnegative powers
of $a,b$ then gives every monomial in $(a,b)P$. Thus $(a,b)P\subseteq R$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Propositions~\ref{prop:morphism} and \ref{prop:image-seed} give the
morphism and the two image descriptions. Modulo $(a,b)P$, the image of
$R$ is $\Z$, whereas $P/(a,b)P=\Z[s]$. Hence $s\notin R$ and
$R\subsetneq P$. Finally $P\subseteq\Z[a^{\pm1},b^{\pm1},z]$,
but the target variable $(a+b)/z$ in \eqref{eq:targetword} does not lie
in this ring. Therefore $P\subsetneq C$.
\end{proof}

\section{The image-seed ring}\label{sec:ring}

The exact image description permits a comparison with the upper algebra.
In the same polynomial coefficient convention, define
\[
\Upol(R)=\bigcap_{i\in\Z}\Z[z,x_i^{\pm1},x_{i+1}^{\pm1}]
\subseteq\Q(a,b,z),
\]
where the $x_i$ are defined in \eqref{eq:recurrence}.

\begin{proposition}\label{prop:upper}
One has
\[
\Upol(R)=P,\qquad R[z^{-1}]=P[z^{-1}]=C[z^{-1}].
\]
\end{proposition}
\begin{proof}
The invariant expression in \eqref{eq:linear} puts $s$ in every ring
defining $\Upol(R)$. Reversing the linear recurrence expresses $a,b$
as $\Z[s]$-linear combinations of any adjacent pair $x_i,x_{i+1}$.
Thus $P\subseteq\Upol(R)$. The initial pair and its two neighbors have
intersection $P$ by Lemma~\ref{lem:intersection}, giving equality.

Using \eqref{eq:neighbors} and $z=abs-a^2-b^2$, we obtain
\begin{equation}\label{eq:localize}
a'b'-ab=zs.
\end{equation}
All four variables on the left belong to $R$, so $s\in R[z^{-1}]$.
Since $R\subseteq P$ and $a,b\in R$, this proves $R[z^{-1}]=P[z^{-1}]$.
The target Laurent expansions in its initial cluster and its separate
neighbors at $a,b$ yield
\[
C\subseteq P[(abz)^{-1}]\cap P[(a'bz)^{-1}]
                  \cap P[(ab'z)^{-1}]=P[z^{-1}].
\]
Indeed, the denominator argument of Lemma~\ref{lem:intersection} allows
only prime factors dividing $z$ after these localizations. Together with
$P\subseteq C$, this proves $C[z^{-1}]=P[z^{-1}]$.
\end{proof}

Let $I=(a,b)P$, now also viewed as an ideal of $R=\Z+I$. The
\emph{conductor} of $R$ in $P$ is the ideal
$\{r\in R:rP\subseteq R\}$.

\begin{proposition}\label{prop:normal}
The conductor of $R$ in $P$ equals $I$. The ring $R$ is integrally closed
in its fraction field and is not Noetherian. The element $s$ is not integral
over $R$; in particular, $P$ is neither integral nor module-finite over $R$.
\end{proposition}
\begin{proof}
Since $I$ is a $P$-ideal contained in $R$, it lies in the conductor.
For $r=c+i$, with $c\in\Z$, $i\in I$, the image of $rs$ in
$P/I=\Z[s]$ is $cs$. This belongs to $R/I=\Z$ only if $c=0$.
The conductor is therefore exactly $I$.

The square of $I$ as an $R$-ideal is $(a^2,ab,b^2)P$: for example,
$a^2p=a(ap)$ with both factors in $I$, and the other generators are
handled identically. Consequently
\[
I/I^2\cong\Z[s]\overline a\oplus\Z[s]\overline b.
\]
Here $\overline a,\overline b$ denote the classes modulo $I^2$.
This is an infinite-rank free $\Z$-module, whereas the $R$-action
factors through $R/I=\Z$. A finite set of $R$-generators for $I$ would
give a finite set of $\Z$-generators for $I/I^2$, a contradiction.
Hence $R$ is not Noetherian, and in particular is not a finitely generated
$\Z$-algebra.

The relation $as\in R$ shows that
$\Frac(R)=\Frac(P)=\Q(a,b,s)$. If $w$ in this field is integral over
$R$, it is integral over the polynomial UFD $P$, and thus belongs to $P$.
Reducing a monic equation for $w$ modulo $I$ shows that its class in
$\Z[s]$ is integral over $\Z$. A nonconstant polynomial of degree $d>0$
cannot satisfy a monic equation of degree $m$ over $\Z$: its leading
term of degree $md$ cannot cancel with any lower power. The class of $w$
is therefore an integer, so $w\in\Z+I=R$.

The same reduction shows that $s$ is not integral over $R$. A
module-finite ring extension is integral, so $P$ is not module-finite
over $R$ either.
\end{proof}

\newpage
\section{Freezing and specialization at one}\label{sec:deletion}

We give a coefficient-free example where the difference persists after
the frozen variable is inverted. Let $C_4$ have independent mutable
variables $(a,b,c,z)$, no frozen variables, and matrix
\begin{equation}\label{eq:B4}
B_4=\begin{pmatrix}
0&2&-2&-1\\-2&0&2&1\\2&-2&0&0\\1&-1&0&0
\end{pmatrix}.
\end{equation}
Let $R_4$ be obtained by freezing $z$: its mutable variables are $a,b,c$,
its frozen variable is $z$, and its matrix consists of the first three
columns of $B_4$. Let $M$ be the coefficient-free Markov cluster algebra
obtained by deleting the row and column of $z$, with initial variables
$a,b,c$. Finally let
\[
\sigma:\Z[a^{\pm1},b^{\pm1},c^{\pm1},z^{\pm1}]
\longrightarrow\Z[a^{\pm1},b^{\pm1},c^{\pm1}]
\]
fix $a,b,c$ and send $z$ to $1$.
The freezing is called a cluster localization when
$R_4[z^{-1}]=C_4[z^{-1}]$.

\begin{theorem}\label{thm:deletion}
The inclusions
\[
R_4[z^{-1}]\subsetneq C_4[z^{-1}],\qquad
M\subsetneq\sigma(C_4)
\]
are strict. No unital ring homomorphism $C_4\to M$ sends
$(a,b,c,z)$ to $(a,b,c,1)$. The same assertions hold after extending
scalars from $\Z$ to $\Q$.
\end{theorem}
\begin{proof}
Mutate $B_4$ in positions $(4,1,2)$. The first two matrices are
\[
\begin{pmatrix}
0&1&-2&1\\-1&0&2&-1\\2&-2&0&0\\-1&1&0&0
\end{pmatrix},\qquad
\begin{pmatrix}
0&-1&2&-1\\1&0&0&-1\\-2&0&0&2\\1&1&-2&0
\end{pmatrix}.
\]
Denoting the three new variables by $p_4,q_4,v_4$, we obtain
\[
p_4=\frac{a+b}{z},\qquad q_4=\frac{c^2+bp_4}{a},\qquad
v_4=\frac{q_4+p_4}{b}=\frac{(a+b)^2+zc^2}{abz}\in C_4.
\]

Give the initial Laurent ring the integer grading determined by
$\deg a=\deg b=\deg c=1$ and $\deg z=0$; the degree of a Laurent
monomial is the sum of its exponents of $a,b,c$. Mutation of the
Markov principal matrix at any of its three positions negates that matrix.
Every exchange relation in $R_4$ consequently has the form
\[
y_ky_k'=z^{r_+}y_i^2+z^{r_-}y_j^2,
\qquad r_+,r_-\geq0,
\]
where $i,j,k$ are the three distinct mutable positions. Induction gives
degree one for every mutable cluster variable. Thus $R_4$ is
nonnegatively graded with degree-zero part $\Z[z]$, and
$(R_4[z^{-1}])_0=\Z[z^{\pm1}]$. But $v_4$ has degree zero and its
Laurent term $c^2/(ab)$ prevents it from belonging to $\Z[z^{\pm1}]$.
It is therefore outside $R_4[z^{-1}]$. Mutations avoiding $z$ identify
$R_4$ with a subalgebra of $C_4$, proving the first strict inclusion.

Lemma~\ref{lem:laurent} defines $\sigma$ on $C_4$. Every mutation
word of the deleted seed lifts to a word avoiding $z$, and setting
$z=1$ in its exchanges gives the corresponding Markov variables.
Therefore $M\subseteq\sigma(C_4)$. The same grading argument gives
$M_0=\Z$, whereas
\[
\sigma(v_4)=\frac{(a+b)^2+c^2}{ab}
\]
has degree zero and is not constant. This proves strictness. Moreover,
any ring homomorphism $h$ with the prescribed initial values would send
the identity $abzv_4=(a+b)^2+zc^2$ to
$ab\,h(v_4)=(a+b)^2+c^2$. Cancellation in $\Q(a,b,c)$ forces the
same forbidden value for $h(v_4)$.

Over $\Q$, the degree-zero parts become $\Q[z^{\pm1}]$ and $\Q$,
respectively. The identical witnesses prove all assertions there.
\end{proof}

This failure concerns the ordinary deleted algebra. For example, the
element $(a^2+b^2+c^2)/(abc)$ is invariant under Markov mutation, hence
Laurent in every Markov cluster. Its degree is $-1$, so it belongs to
the upper algebra but not to $M$. Thus the deletion in
Theorem~\ref{thm:deletion} does not satisfy the equality of ordinary and
upper cluster algebras required in the specialization theorem of
Keller--Plamondon--Qin \cite[Corollary 4.3]{KPQ}.

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I. Assem, G. Dupont and R. Schiffler,
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\end{thebibliography}

\bigskip
\noindent Qiyue Tang\\
Zhili College, Tsinghua University\\
Beijing 100084, China\\
\href{mailto:tangqy24@mails.tsinghua.edu.cn}{tangqy24@mails.tsinghua.edu.cn}
\end{document}