# Projective modules over Rees-like algebras and its monoid extensions

Chandan Bhaumik, Md Abu Raihan and Husney Parvez Sarwar  (Chandan Bhaumik)
Department of Mathematics, Indian Institute of Technology Kharagpur, Kharagpur
721302, West Bengal, India<EMAIL_ADDRESS>(Md Abu Raihan)
Department of Mathematics, Indian Institute of Technology Kharagpur, Kharagpur
721302, West Bengal, India<EMAIL_ADDRESS>(H.P. Sarwar) Department of
Mathematics, Indian Institute of Technology Kharagpur, Kharagpur 721302, West
Bengal, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Key words and phrases:

Projective modules, unimodular element, elementary action, partially
cancellative monoids, Rees-like algebras, monoid algebras

###### 2020 Mathematics Subject Classification:

Primary 13C10; Secondary 19A13, 19A15

> _Abstract._ Let $A$ be a Rees-like algebra of dimension $d$ and $N$ a
> commutative partially cancellative torsion-free seminormal monoid. We prove
> the following results.
>
>   1. (1)
>
> Let $P$ be a finitely generated projective $A$-module of ${\rm rank}\geq d$.
> Then $(i)$ $P$ has a unimodular element; $(ii)$ The action of
> ${\operatorname{\rm E}}(A\oplus P)$ on ${\rm Um}(A\oplus P)$ is transitive.
>
>   2. (2)
>
> Let $P$ be a finitely generated projective $A[N]$-module of ${\rm
> rank}~{}r$. Then $(i)$ $P$ has a unimodular element for
> $r\geq{\operatorname{\rm max}}\\{3,d\\}$; $(ii)$ The action of
> ${\operatorname{\rm E}}(A[N]\oplus P)$ on ${\rm Um}(A[N]\oplus P)$ is
> transitive for $r\geq{\operatorname{\rm max}}\\{2,d\\}$.
>
>

>
> These improve the classical results of Serre [27] and Bass [1].

##  1\. Introduction

Let $R$ be a commutative Noetherian ring and $P$ a finitely generated
projective $R$-module of constant rank. We say that $P$ has a unimodular
element if $P$ splits off a free direct summand of rank one, i.e., $P\cong
Q\oplus R$ for some projective $R$-module $Q$, and $P$ is cancellative if
$P\oplus Q\cong P^{\prime}\oplus Q$ for some projective $R$-modules $Q$ and
$P^{\prime}$ implies $P\cong P^{\prime}$.

If ${\rm rank}(P)>\text{\rm dim}(R)$, the classical result of Serre [27] says
that $P$ has a unimodular element, and Bass [1] says that $P$ is cancellative.
It is well known that the results of Serre (2.6) and Bass (2.7) are the best
possible results in general, as it is evidenced by the tangent bundle over the
real algebraic sphere of dimension two. Therefore, when ${\rm
rank}(P)=\text{\rm dim}(R)$, the question of the existence of unimodular
elements and the cancellation problem is subtle. However, in this paper, we
prove the analogous results of Serre and Bass over Rees-like algebras
$R[\mathfrak{a}t,t^{2}]$ (see Definition 2.5) and its monoid extension
$R[\mathfrak{a}t,t^{2}][N]$, where $N$ is a finitely generated commutative
partially cancellative torsion-free seminormal monoid.

Before we state our results on the existence of a unimodular element, let us
recall the Serre dimension of a ring $R$, which is denoted by Serre dim $R$.
For a commutative ring $R$, we say Serre dim $R\leq n$ if every finitely
generated projective $R$-module $P$ of rank $>n$ has a unimodular element.

######  Theorem 1.1.

(Theorems 4.1, 4.2) Let $R$ be a commutative Noetherian ring of dimension $d$
and ${\mathfrak{a}}$ an ideal in $R$ such that ${\rm ht}({\mathfrak{a}})\geq
1$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a commutative partially
cancellative torsion-free seminormal monoid.

  1. (1)

Suppose $P$ is a finitely generated projective $A$-module of rank $\geq d+1$.
Then $P$ has a unimodular element. In other words, Serre dim $A\leq d$.

  2. (2)

Suppose $P$ is a finitely generated projective $A[N]$-module of rank
$\geq{\operatorname{\rm max}}\\{3,d+1\\}$. Assume Serre dim
$R[N\oplus{\mathbb{Z}}_{+}]\leq d$ and Serre dim $R[N\oplus
2{\mathbb{Z}}_{+}]\leq d$. Then $P$ has a unimodular element. In other words,
Serre dim $A[N]\leq{\operatorname{\rm max}}\\{2,d\\}$.

To establish the cancellation result of projective modules over Rees-like
algebra and its monoid extensions, we first show that the free module of rank
$d+1$ is cancellative. Using this result, we then prove that every finitely
generated projective module of rank $d+1$ is cancellative. More precisely, we
prove the following result.

######  Theorem 1.2.

(Theorems 5.2, 5.5) Let $R$ be a commutative Noetherian ring of dimension $d$
and ${\mathfrak{a}}$ an ideal in $R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a
commutative partially cancellative torsion-free seminormal monoid.

  1. (1)

Suppose $P$ is a finitely generated projective $A$-module of rank $\geq d+1$.
Then ${\operatorname{\rm E}}(A\oplus P)$ acts transitively on ${\rm
Um}(A\oplus P)$. In particular, $P$ is cancellative.

  2. (2)

Suppose $P$ is a finitely generated projective $A[N]$-module of rank $r$. Then
the action of ${\operatorname{\rm E}}(A[N]\oplus P)$ on ${\rm Um}(A[N]\oplus
P)$ is transitive for $r\geq{\operatorname{\rm max}}\\{2,d+1\\}$. In
particular, $P$ is cancellative for $r\geq d+1$.

The definition of Rees-like algebra was introduced by McCulloug and Peeva
[21], and it played a crucial role in providing a counterexample to the
Eisenbud-Goto regularity conjecture. They provide the counterexample to the
Eisenbud-Goto conjecture by constructing graded prime ideals in Rees-like
algebra with larger-than-expected regularity, which may then be homogenized.
In contrast to the usual Rees algebra, whose defining equations are difficult
to find in general, those of the Rees-like algebra are given explicitly. These
algebras naturally arise in algebraic geometry while blowing up a variety
along a subvariety. In the context of projective modules, the above two
theorems suggest that Rees-like algebra $R[\mathfrak{a}t,t^{2}]$ and the
polynomial ring $R[t]$ behave in a similar fashion.

Now we provide a historical background for the previous two theorems. A
similar type of above results are obtained for the polynomial rings of one
variable by Plumstead [22], for the Laurent polynomial rings of one variable
by Mandal [19], and for the overring of polynomial rings of one variable by
Rao [23]. The above results are extended to polynomial and Laurent polynomial
rings for several variables by Bhatwadekar–Roy [4], Bhatwadekar–Lindel–Rao [2]
and Lindel [18] respectively. For Rees algebras and symbolic Rees algebras
(need not be Noetherian), the analogous problems are considered in [24] and
[5] respectively.

For the monoid extensions over a commutative noetherian ring, the cancellation
of projective modules and the existence of a unimodular element in a
projective module were studied by Gubeladze [11, 12, 13], Sarwar [25, 26],
Dhorajia-Keshari [7], Keshari-Sarwar [15], Keshari-Mathew [14].

As an application of the above two theorems, we improve the bound for the
number of generators that are needed to generate a module as follows.

######  Theorem 1.3.

(Theorems 6.1, 6.2) Let $R$ be a commutative Noetherian ring of dimension $d$
and ${\mathfrak{a}}$ an ideal in $R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a
commutative partially cancellative torsion-free seminormal monoid.

  1. (1)

Let $M$ be a finitely generated $A$-module and $e(M):={\operatorname{\rm
max}}\\{\mu_{\mathfrak{p}}(M)+\text{\rm
dim}(A/\mathfrak{p}):\mathfrak{p}\in{\rm Spec}(A)\\}$. Then $M$ is generated
by $e(M)$ elements.

  2. (2)

Suppose $P$ is a finitely generated projective $A[N]$-module of rank
$r\geq{\operatorname{\rm max}}\\{3,d+1\\}$. Assume ${\rm
ht}({\mathfrak{a}})\geq 1$ and Serre dim $R[N\oplus{\mathbb{Z}}_{+}]\leq d$
and Serre dim $R[N\oplus 2{\mathbb{Z}}_{+}]\leq d$. Then $\mu(P)\leq
r+{\operatorname{\rm max}}\\{2,d\\}$.

For an arbitrary commutative Noetherian ring, the number of generators of a
module was studied by Eisenbud–Evans [8], Forster [9], and Swan [29]. For a
polynomial ring, the analogous problem was conjectured by Eisenbud and Evans,
and solved by Mohan Kumar [17]. A similar problem for Rees algebras is
considered in [24].

Finally, we prove the $K_{1}$-analog of Suslin’s results [28, Theorem 7.8] for
Rees-like algebras and its Laurent polynomial extensions.

######  Theorem 1.4.

(Theorems 6.4, 6.7) Let $R$ be a commutative Noetherian ring of dimension $d$
and ${\mathfrak{a}}$ an ideal in $R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and
$B=A[X_{1},\ldots,X_{n},Y_{1}^{\pm 1},\ldots,Y_{m}^{\pm 1}]$. Then, for
$r\geq{\operatorname{\rm max}}\\{3,d+2\\}$,

  1. (1)

the natural map $\phi:{\operatorname{\rm GL}}_{r}(A)/{\operatorname{\rm
E}}_{r}(A)\rightarrow K_{1}(A)$ is an isomorphism.

  2. (2)

the natural map $\phi:{\operatorname{\rm GL}}_{r}(B)/{\operatorname{\rm
E}}_{r}(B)\rightarrow K_{1}(B)$ is an isomorphism.

In view of the results in Section $9$ of [11], one can not expect that the
above result will hold for an arbitrary monoid extension.

The rest of the paper is structured as follows. Section 2 provides several
basic definitions and preliminary results. In Section 3, we studied some basic
computations of fiber product diagrams and some properties of Rees-like
algebra and its Krull dimension, which will be used in the main theorem. Our
main results concerning the existence of unimodular elements and cancellation
problems are proven in Section 4 and Section 5, respectively. Finally, in
Section 6, we provide some applications regarding the number of generators of
a module and $K_{1}$-analog of our results.

##  2\. Preliminaries

In this section, we fix some notations, recall key definitions, and summarize
pertinent results that will be referenced later in the paper. Throughout the
paper, we assume that the rings are commutative with a multiplicative identity
$1$, and the modules are finitely generated, while all projective modules are
of constant rank. All considered monoids are assumed to be finitely generated,
commutative, and unless specified otherwise, cancellative.

For a ring $R$, we denote its Krull dimension by $\text{\rm dim}(R)$ and its
nilradical by ${\rm nil}(R)$. The height of an ideal $I$ in $R$ is denoted by
${\rm ht}(I)$, and $R^{\times}$ represents the group of units in $R$. The set
of minimal prime ideals of $R$ is denoted by ${\rm Min}(R)$.

For an $R$-module $M$, $\mu(M)$ denotes the minimal number of generators of
$M$. For a prime ideal $\mathfrak{p}$ of $R$, let $\mu_{\mathfrak{p}}(M)$
denote the minimal number of elements needed to generate the
$R_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$.

###  2.1. Review of rings and modules

######  Definition 2.1.

(Unimodular Element) Let $R$ be a ring and $M$ a $R$-module. An element $m$ in
$M$ is called a unimodular element if there exists a surjective $R$-linear map
$\varphi\in M^{*}={\rm Hom}_{R}(M,R)$ such that $\varphi(m)=1$.

The set of all unimodular elements of $M$ is denoted by ${\rm Um}(M)$. For
$M=R^{n}$, we write ${\rm Um}_{n}(R)$ instead of ${\rm Um}(R^{n})$. An element
$(r_{1},\ldots,r_{n})\in R^{n}$ is called unimodular row if
$(r_{1},\ldots,r_{n})\in{\rm Um}_{n}(R)$. Observe that projective $R$-module
$P$ has a unimodular element if and only if P splits off a free summand of
rank one, i.e., $P\cong Q\oplus R$ for some projective $R$-module $Q$.

######  Definition 2.2.

(Cancellation) A projective $R$-module $P$ is said to be cancellative if
$P\oplus Q\cong P^{\prime}\oplus Q$ for some projective $R$-modules $Q$ and
$P^{\prime}$ implies $P\cong P^{\prime}$.

The group of all $R$-automorphisms of $M$ is denoted by ${\rm Aut}_{R}(M)$.
For $M=R^{n}$, we denote ${\rm Aut}_{R}(R^{n})$ by ${\operatorname{\rm
GL}}_{n}(R)$. The subgroup of ${\operatorname{\rm GL}}_{n}(R)$ consists of
element with determinant $1$, is denoted by ${\operatorname{\rm SL}}_{n}(R)$.
The elementary matrices $e_{ij}(\lambda),~{}i\neq j,~{}\lambda\in R$, is
defined by the matrices $I_{n}+\lambda e_{ij}$, where $e_{ij}$ are the matrix
with $1$ in the $(i,j)$-th position and zeros elsewhere. The subgroup of
${\operatorname{\rm SL}}_{n}(R)$ generated by $e_{ij}(\lambda)$, is denoted by
${\operatorname{\rm E}}_{n}(R)$. An element $\sigma\in{\operatorname{\rm
GL}}_{n}(R)$ is called stably elementary if $\begin{pmatrix}\sigma&0\\\
0&I_{k}\end{pmatrix}\in{\operatorname{\rm E}}_{n+k}(R)$ for some
$k\in{\mathbb{N}}$.

For an ideal $J$ of $R$, let ${\rm Um}(R\oplus M,J)$ denote the set of all
$(a,m)\in{\rm Um}(R\oplus M)$ with $a\in 1+J$ and $m\in JM$. Let E$(R\oplus
M,J)$ denote the subgroup of $\text{Aut}_{R}(R\oplus M)$ generated by the
automorphisms $\begin{pmatrix}1&a\varphi\\\ 0&id_{M}\end{pmatrix}\
\text{and}~{}\begin{pmatrix}1&0\\\ m&id_{M}\end{pmatrix}$, where $a\in
J,\varphi\in M^{*}$ and $m\in JM$. We shall write E$(R\oplus M)$ for
E$(R\oplus M,R)$.

######  Definition 2.3.

(Rees Algebra) Let $R$ be a ring and ${\mathfrak{a}}$ an ideal of $R$. The
Rees algebra of $R$ with respect to ${\mathfrak{a}}$ is a graded sub algebra
of $R[t]$ defined as

$R[{\mathfrak{a}}t]=R\oplus{\mathfrak{a}}t\oplus{\mathfrak{a}}^{2}t^{2}\oplus{\mathfrak{a}}^{3}t^{3}\oplus\cdots.$

The following theorem is for the Krull dimension of Rees algebra (see [5,
Theorem 5]).

######  Theorem 2.4.

Let $R$ be a Noetherian ring and ${\mathfrak{a}}$ an ideal of $R$. If
$\text{\rm dim}(R)$ is finite, then $\text{\rm dim}(R[{\mathfrak{a}}t])$ =
$\text{\rm dim}(R)+1$, if ${\mathfrak{a}}\nsubseteq\mathfrak{p}$ for some
$\mathfrak{p}\in{\rm Spec}(R)$ with $\text{\rm dim}(R/\mathfrak{p})=\text{\rm
dim}(R)$ and $\text{\rm dim}(R[{\mathfrak{a}}t])=\text{\rm dim}(R)$,
otherwise.

######  Definition 2.5.

(Rees-Like Algebra) Let $R$ be a ring and ${\mathfrak{a}}$ an ideal of $R$.
The Rees-like algebra of $R$ with respect to ${\mathfrak{a}}$ is a graded sub
algebra of $R[t]$ defined as

$R[{\mathfrak{a}}t,t^{2}]=R\oplus{\mathfrak{a}}t\oplus
Rt^{2}\oplus{\mathfrak{a}}t^{3}\oplus\cdots.$

Note that if $R$ is a Noetherian ring, then the Rees-like algebra
$R[{\mathfrak{a}}t,t^{2}]$ is a finitely generated $R$-algebra. We consider
the following conductor square, which will be used throughout the article (see
[20, Section 4, page 18]).

${R[{\mathfrak{a}}t,t^{2}]}$${R[t]}$${(R/{\mathfrak{a}})[t^{2}]}$${(R/{\mathfrak{a}})[t].}$$\scriptstyle{i_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\eta_{2}}$$\scriptstyle{i_{2}}$

where $i_{1}$, $i_{2}$ are inclusion maps and $\eta_{1}$, $\eta_{2}$ are
natural surjections.

We begin with two classical results due to Serre [27] and Bass [1]
respectively.

######  Theorem 2.6.

Let $R$ be a Noetherian ring of dimension $d$. Then any projective $R$-module
of rank $\geq d+1$ has a unimodular element. In other words, Serre dim $R\leq
d$.

######  Theorem 2.7.

Let $R$ be a Noetherian ring of dimension $d$ and $P$ a projective $R$-module
of rank $\geq d+1$. Then ${\operatorname{\rm E}}(R\oplus P)$ acts transitively
on ${\rm Um}(R\oplus P)$. In particular, $P$ is cancellative.

The following result is due to Lindel [18, Lemma 1.1].

######  Lemma 2.8.

Let $R$ be a Noetherian ring and $P$ a projective $R$-module of rank $r$. Then
there exists $s\in R$, $p_{i}\in P$ and $\phi_{i}\in P^{*}$ for $i=1,\ldots,r$
such that the following properties hold:

  1. (1)

$P_{s}$ is free,

  2. (2)

$(\phi_{i}(p_{j}))=\text{diagonal}(s,\ldots,s)$,

  3. (3)

$sP\subset p_{1}R+\cdots+p_{r}R$,

  4. (4)

the image of $s$ in $R_{red}$ is a non-zero divisor,

  5. (5)

$(0:sR)=(0:s^{2}R)$.

The following results are due to Dhorajia-Keshari [7, Lemma 2.4, Lemma 3.3,
Theorem 3.4]

######  Lemma 2.9.

Let $R$ be a Noetherian ring and $P$ a projective $R$-module. Let “bar” denote
reduction modulo the nil radical of $R$. If ${\operatorname{\rm
E}}(\overline{R}\oplus\overline{P})$ acts transitively on ${\rm
Um}(\overline{R}\oplus\overline{P})$, then ${\operatorname{\rm E}}(R\oplus P)$
acts transitively on ${\rm Um}(R\oplus P)$.

######  Lemma 2.10.

Let $R$ be a Noetherian ring and $P$ a projective $R$-module of rank $r$.
Choose $s\in R$ satisfying the properties of Lemma 2.8. Assume
${\operatorname{\rm E}}_{r+1}(A)$ acts transitively on ${\rm Um}_{r+1}(A)$,
where $A=R[X]/(X^{2}-s^{2}X)$. Then ${\operatorname{\rm E}}(R\oplus P)$ acts
transitively on ${\rm Um}(R\oplus P,s^{2}R)$.

######  Theorem 2.11.

Let $R$ be a Noetherian ring of dimension $d$ and $P$ a projective $R$-module
of rank $d$. Assume ${\operatorname{\rm E}}_{d+1}(A)$ acts transitively on
${\rm Um}_{d+1}(A)$, where $A$ is a finite extension of $R$. Then
${\operatorname{\rm E}}(R\oplus P)$ acts transitively on ${\rm Um}(R\oplus
P)$.

###  2.2. Review of monoids and monoid algebras

######  Definition 2.12.

(Monoids) A commutative monoid is a non-empty set $N$ with a binary operation
$\mu:N\times N\rightarrow N$ ( written $\mu(x,y)=x\cdot y$ ) which is
commutative (i.e. $x\cdot y=y\cdot x$), associative (i.e. $(x\cdot y)\cdot
z=x\cdot(y\cdot z)$ ), and has the neutral element $1$ (i.e. $x\cdot 1=x$).

A monoid $N$ is called cancellative if $xy=zy$ implies $x=z$ for all $x,y,z\in
N$ and torsion-free if for $x,y\in N$, whenever $x^{n}=y^{n}$ for some
$n\in\mathbb{N}$ implies $x=y$. A monoid $N$ is cancellative if and only if
the map $N\rightarrow{\rm gp}(N)$ is injective, where ${\rm gp}(N)$ is a group
of fractions of the monoid $N$. It is easy to check that a finitely generated
cancellative monoid $N$ is torsion-free if and only if ${\rm gp}(N)$ is
torsion-free.

The seminormalization of a cancellative monoid $N$ is the submonoid ${\rm
sn}(N)$ of ${\rm gp}(N)$ consisting of all elements $x\in{\rm gp}(N)$ such
that $x^{2}$, $x^{3}\in N$. We say that a cancellative monoid $N$ is
seminormal if ${\rm sn}(N)=N$. A cancellative monoid $N$ is called normal if
for any $x\in{\rm gp}(N)$ with $x^{m}\in N$ for some $m>0$ implies $x\in N$.

A subset $I$ of a monoid $N$ is called an ideal of $N$ if $IN\subset I$, where
$IN:=\\{xm\ |\ x\in I,\ m\in N\\}$. A monoid $N$ is called partially
cancellative if $N$ is isomorphic to $N^{\prime}/I$, where $N^{\prime}$ is a
cancellative monoid and $I$ is an ideal of $N^{\prime}$. A monoid $N$ is
partially cancellative torsion-free seminormal if $N$ is isomorphic to
$N^{\prime}/I$, where $N^{\prime}$ is a cancellative torsion-free seminormal
monoid and $I$ is an ideal of $N^{\prime}$.

The rank of the monoid $N$ is defined to be the dimension of
${\mathbb{Q}}\otimes_{{\mathbb{Z}}}{\rm gp}(N)$. An affine monoid $N$ of rank
$n$ is called $\Phi$-simplicial if the embedding
$N\hookrightarrow{\mathbb{Z}}_{+}^{n}$ is integral.

######  Definition 2.13.

(Monoid algebra) Let $R$ be a ring and $N$ a monoid. The monoid algebra $R[N]$
is generated as a free $R$-module with basis as elements of $N$ and
coefficients in $R$. Therefore, every element of $R[N]$ is of the form
$f=\sum_{x\in N}f_{x}x,\ f_{x}\in R,~{}f_{x}=0$ for all but finitely many
$x\in N$.

The following result is due to Gubeladze [10, Theorem 2.1].

######  Theorem 2.14.

Let $R$ be a principal ideal domain and $N$ a torsion-free seminormal monoid.
Then any projective $R[N]$-modules are free.

The following results are due to Sarwar [26, Lemma 3.3 and Theorem 3.4].

######  Lemma 2.15.

Let $R$ be a Noetherian ring of dimension $d$ and $N$ a partially cancellative
monoid. Then ${\operatorname{\rm E}}_{r}(R[N])$ acts transitively on ${\rm
Um}_{r}(R[N])$ for $r\geq{\operatorname{\rm max}}\\{3,d+2\\}$.

######  Theorem 2.16.

Let $R$ be a Noetherian ring of dimension $d$, $N$ a partially cancellative
torsion-free seminormal monoid, and $P$ a projective $R[N]$-module of rank
$r\geq{\operatorname{\rm max}}\\{2,d+1\\}$. Then ${\operatorname{\rm
E}}(R[N]\oplus P)$ acts transitively on ${\rm Um}(R[N]\oplus P)$.

##  3\. Some basic computations

In this section, we studied some basic computations of the Cartesian square
and explored some properties of Rees-like algebra and its Krull dimension.

###  3.1. Some results over Cartesian square

######  Definition 3.1.

(Fiber Product) Let $\mathcal{C}$ be any category. Let $X,Y$ and $Z$ be are
objects in the category $\mathcal{C}$ and $q_{1}:X\rightarrow Z$,
$q_{2}:Y\rightarrow Z$ are morphisms. The fiber product (also called a
pullback or Cartesian square) of objects $X$ and $Y$ over the object $Z$ is a
triple $(P,p_{1},p_{2})$, where $P$ is an object in $\mathcal{C}$,
$p_{1}:P\rightarrow X$ and $p_{2}:P\rightarrow Y$ are morphisms such that
$q_{1}\circ p_{1}=q_{2}\circ p_{2}$. The triple is universal in the sense that
given any other triple $(P^{\prime},p^{\prime}_{1},p^{\prime}_{2})$ of this
kind with $q_{1}\circ p^{\prime}_{1}=q_{2}\circ p^{\prime}_{2}$, there is a
unique morphism $h:P^{\prime}\rightarrow P$ such that $p_{1}\circ
h=p^{\prime}_{1}$ and $p_{2}\circ h=p^{\prime}_{2}$. In pictorially, the
following diagram

${P^{\prime}}$${P}$${X}$${Y}$${Z}$$\scriptstyle{p^{\prime}_{1}}$$\scriptstyle{p^{\prime}_{2}}$$\scriptstyle{\exists!h}$$\scriptstyle{p_{2}}$$\scriptstyle{p_{1}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$

is commutative.

######  Lemma 3.2.

Let $\mathcal{C}$ be any category. Let $X$, $Y$ and $Z$ be are objects in
$\mathcal{C}$. Consider the following commutative diagram

${P}$${X}$${Y}$${Z}$$\scriptstyle{p_{2}}$$\scriptstyle{p_{1}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$

in the category $\mathcal{C}$. This diagram is a fiber product diagram if and
only if for each pair of elements $x$ in $X$ and $y$ in $Y$ with
$q_{1}(x)=q_{2}(y)$, there is a unique element $u$ in $P$ such that
$p_{1}(u)=x$ and $p_{2}(u)=y$.

###### Proof.

The proof follows from the definitions. ∎

######  Lemma 3.3.

Consider the following fiber product diagram of commutative rings

${R}$${R_{1}}$${R_{2}}$${R_{3}.}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$

For each pair of elements $a=(a_{1},\ldots,a_{n})\in{\rm Um}_{n}(R_{1})$ and
$b=(b_{1},\ldots,b_{n})\in{\rm Um}_{n}(R_{2})$ such that $q_{1}(a)=q_{2}(b)$,
there is a unique element $c=(c_{1},\ldots,c_{n})\in{\rm Um}_{n}(R)$ with
$p_{1}(c)=a$ and $p_{2}(c)=b$.

###### Proof.

Let $a=(a_{1},\ldots,a_{n})\in{\rm Um}_{n}(R_{1})$ and
$b=(b_{1},\ldots,b_{n})\in{\rm Um}_{n}(R_{2})$ such that $q_{1}(a)=q_{2}(b)$.
So for each $i=1,\ldots,n$ the images of each pair of elements $a_{i}\in
R_{1}$ and $b_{i}\in R_{2}$ are the same, then by Lemma 3.2, we get a unique
element $c_{i}$ in $R$ with $p_{1}(c_{i})=a_{i}$ and $p_{2}(c_{i})=b_{i}$. Let
$c=(c_{1},\ldots,c_{n})$. Since $a$ and $b$ are unimodular rows, there exists
$a_{i}^{\prime}\in R_{1}$ and $b_{i}^{\prime}\in R_{2}$ such that
$\sum_{i=1}^{n}a_{i}a_{i}^{\prime}=1$ and
$\sum_{i=1}^{n}b_{i}b_{i}^{\prime}=1$. By similar argument, patching of
unimodular rows $(a_{1}^{\prime},\ldots,a_{n}^{\prime})$ and
$(b_{1}^{\prime},\ldots,b_{n}^{\prime})$, we obtain a row
$(c_{1}^{\prime},\ldots,c_{n}^{\prime})$ over $R$ such that
$p_{1}(c_{i}^{\prime})=a_{i}^{\prime}$ and
$p_{2}(c_{i}^{\prime})=b_{i}^{\prime}$. Let us consider the following fiber
product diagram

${R^{n}}$${R_{1}^{n}}$${R}$${R_{1}}$${R_{2}^{n}}$${R_{3}^{n}}$${R_{2}}$${R_{3}}$$\scriptstyle{\phi_{c}}$$\scriptstyle{\phi_{a}}$$\scriptstyle{\phi_{b}}$

where $\phi_{c}:R^{n}\rightarrow R$ is a $R$-linear homomorphism define by
$\phi_{c}((x_{1},\ldots,x_{n}))=\sum_{i=1}^{n}x_{i}c_{i}$. Observe that image
of $\sum_{i=1}^{n}c_{i}c_{i}^{\prime}$ under $p_{1}$ and $p_{2}$ are $1$. Also
note that patching the identity elements, we will get the identity. Then from
the above diagram, we have
$\phi_{c}((c_{1}^{\prime},\ldots,c_{n}^{\prime}))=1$. Hence $c\in{\rm
Um}_{n}(R)$. ∎

######  Lemma 3.4.

The fiber product diagram of commutative rings on the left will induce a fiber
product diagram of groups on the right

${R}$${R_{1}}$${R_{2}}$${R_{3}}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$${R^{\times}}$${R_{1}^{\times}}$${R_{2}^{\times}}$${R_{3}^{\times}.}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$

###### Proof.

Assume for each pair of invertible elements $a\in R_{1}^{\times}$ and $b\in
R_{2}^{\times}$, they have the same image in $R_{3}^{\times}$. Then, we will
get a unique element $c\in R$ with $p_{1}(c)=a$ and $p_{2}(c)=b$. Let
$a^{\prime}\in R_{1}^{\times}$ and $b^{\prime}\in R_{2}^{\times}$ be the
inverse of $a$ and $b$ respectively, i.e. $aa^{\prime}=1$ and $bb^{\prime}=1$.
By similar arguments the patching of $a^{\prime}$ and $b^{\prime}$, we get a
unique element $c^{\prime}\in R$ with $p_{1}(c^{\prime})=a^{\prime}$ and
$p_{2}(c^{\prime})=b^{\prime}$. Then from the fiber product diagram, we have
$cc^{\prime}=1$, which implies $c\in R^{\times}$. ∎

######  Proposition 3.5.

Consider the following fiber product diagram of commutative rings

${R}$${R_{1}}$${R_{2}}$${R_{3}.}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$

  1. (1)

For each pair of matrices $M_{1}\in{\operatorname{\rm GL}}_{n}(R_{1})$ and
$M_{2}\in{\operatorname{\rm GL}}_{n}(R_{2})$ such that
$q_{1}(M_{1})=q_{2}(M_{2})$, there is a unique $M\in{\operatorname{\rm
GL}}_{n}(R)$ with $p_{1}(M)=M_{1}$ and $p_{2}(M)=M_{2}$.

  2. (2)

For each pair of matrices $M_{1}\in{\operatorname{\rm SL}}_{n}(R_{1})$ and
$M_{2}\in{\operatorname{\rm SL}}_{n}(R_{2})$ such that
$q_{1}(M_{1})=q_{2}(M_{2})$, there is a unique $M\in{\operatorname{\rm
SL}}_{n}(R)$ with $p_{1}(M)=M_{1}$ and $p_{2}(M)=M_{2}$.

  3. (3)

Assume either $p_{1}$ or $p_{2}$ is surjective. For each pair of matrices
$M_{1}\in{\operatorname{\rm E}}_{n}(R_{1})$ and $M_{2}\in{\operatorname{\rm
E}}_{n}(R_{2})$ such that $q_{1}(M_{1})=q_{2}(M_{2})$, there is a unique
$M\in{\operatorname{\rm E}}_{n}(R)$ with $p_{1}(M)=M_{1}$ and
$p_{2}(M)=M_{2}$.

###### Proof.

$(1)$ Let $M_{1}=[a_{ij}]\in{\operatorname{\rm GL}}_{n}(R_{1})$ and
$M_{2}=[b_{ij}]\in{\operatorname{\rm GL}}_{n}(R_{2})$ such that for each
$i=1,\ldots,n$ the image of each pair of unimodular rows
$(a_{i1},\ldots,a_{in})$ in $R_{1}^{n}$ and $(b_{i1},\ldots,b_{in})$ in
$R_{2}^{n}$ are the same in $R_{3}^{n}$. By Lemma 3.3, for each $i=1,\ldots,n$
there is a unique row $(c_{i1},\ldots,c_{in})$ in $R^{n}$ with
$p_{1}((c_{i1},\ldots,c_{in}))=(a_{i1},\ldots,a_{in})$ and
$p_{2}((c_{i1},\ldots,c_{in}))=(b_{i1},\ldots,b_{in})$. Therefore, we can
construct a matrix $M=[c_{ij}]$ over $R$ with $p_{1}(M)=M_{1}$ and
$p_{2}(M)=M_{2}$.

It remains to prove that $M\in{\operatorname{\rm GL}}_{n}(R)$. Since $M_{1}$
and $M_{2}$ are invertible, there is a $M_{1}^{{}^{\prime}}$ and
$M_{2}^{{}^{\prime}}$ such that $M_{1}M_{1}^{{}^{\prime}}=I_{n}$ and
$M_{2}M_{2}^{{}^{\prime}}=I_{n}$. By similar arguments and patching of
$M_{1}^{{}^{\prime}}$ and $M_{2}^{{}^{\prime}}$, we obtain a matrix
$M^{{}^{\prime}}$ over $R$ such that
$p_{1}(M^{{}^{\prime}})=M_{1}^{{}^{\prime}}$ and
$p_{2}(M^{{}^{\prime}})=M_{2}^{{}^{\prime}}$. Consider the following fiber
product diagram of $R$-modules

${{\operatorname{\rm M}}_{n}(R)}$${{\operatorname{\rm
M}}_{n}(R_{1})}$${{\operatorname{\rm M}}_{n}(R_{2})}$${{\operatorname{\rm
M}}_{n}(R_{3}).}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$

Observe that image of $MM^{{}^{\prime}}$ under $p_{1}$ is $I_{n}$, and the
image of $MM^{{}^{\prime}}$ under $p_{2}$ is $I_{n}$. Also, by patching the
identity matrices, we obtain the identity matrix. Therefore, from the above
diagram, we have $MM^{{}^{\prime}}=I_{n}$. Thus $M$ is invertible.

$(2)$ Let $M_{1}=[a_{ij}]\in{\operatorname{\rm SL}}_{n}(R_{1})$ and
$M_{2}=[b_{ij}]\in{\operatorname{\rm SL}}_{n}(R_{2})$. From the first
paragraph of the proof of $(1)$, we obtain a matrix $M=[c_{ij}]$ over $R$ with
$p_{1}(M)=M_{1}$ and $p_{2}(M)=M_{2}$. It remains to prove that
$M\in{\operatorname{\rm SL}}_{n}(R)$. Using Lemma 3.4, we consider the
following fiber product diagram

${{\operatorname{\rm GL}}_{n}(R)}$${{\operatorname{\rm
GL}}_{n}(R_{1})}$${R^{\times}}$${R_{1}^{\times}}$${{\operatorname{\rm
GL}}_{n}(R_{2})}$${{\operatorname{\rm
GL}}_{n}(R_{3})}$${R_{2}^{\times}}$${R_{3}^{\times}}$$\scriptstyle{\varphi}$

of groups, where $\varphi:{\operatorname{\rm GL}}_{n}(R)\to R^{\times}$ is a
group homomorphism defined by $\varphi(M)=\text{det}(M)$. Also, by patching of
identity elements, we obtain the identity element in $R^{\times}$. Then from
the above cartesian square, we have $\varphi(M)=1$. Hence $\text{det}(M)=1$.

$(3)$ From $(2)$, the patching of the matrices $M_{1}\in{\operatorname{\rm
E}}_{n}(R_{1})$ and $M_{2}\in{\operatorname{\rm E}}_{n}(R_{2})$, we get
$M\in{\operatorname{\rm SL}}_{n}(R)$. Assume $p_{2}$ is surjective, so its
induced a surjection ${\operatorname{\rm E}}_{n}(R)\to{\operatorname{\rm
E}}_{n}(R_{2})$. Since $p_{2}(M)=M_{2}\in{\operatorname{\rm E}}_{n}(R_{2})$,
there exists $M^{\prime}\in{\operatorname{\rm E}}_{n}(R)$ such that
$p_{2}(M^{\prime})=p_{2}(M)$. Then $p_{2}(MM^{\prime-1})=I_{n}$, hence
$MM^{\prime-1}\in\ker(p_{2})\subset{\operatorname{\rm E}}_{n}(R)$. This
implies $M\in{\operatorname{\rm E}}_{n}(R)$. ∎

######  Remark 3.6.

In the view of Proposition 3.5, we observe that the following are fiber
product diagrams of groups (note that to get the third diagram, we assume
either $p_{1}$ or $p_{2}$ is surjective)

${{\operatorname{\rm GL}}_{n}(R)}$${{\operatorname{\rm
GL}}_{n}(R_{1})}$${{\operatorname{\rm GL}}_{n}(R_{2})}$${{\operatorname{\rm
GL}}_{n}(R_{3})}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$${{\operatorname{\rm
SL}}_{n}(R)}$${{\operatorname{\rm SL}}_{n}(R_{1})}$${{\operatorname{\rm
SL}}_{n}(R_{2})}$${{\operatorname{\rm
SL}}_{n}(R_{3})}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$${{\operatorname{\rm
E}}_{n}(R)}$${{\operatorname{\rm E}}_{n}(R_{1})}$${{\operatorname{\rm
E}}_{n}(R_{2})}$${{\operatorname{\rm
E}}_{n}(R_{3}).}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{q_{1}}$$\scriptstyle{q_{2}}$

######  Lemma 3.7.

Let $R$ be a ring, $I$ an ideal in $R$ and $\sigma\in{\operatorname{\rm
GL}}_{n}(R)$ such that $\bar{\sigma}\in{\operatorname{\rm E}}_{n}(R/I)$. Then
$\sigma\in{\operatorname{\rm E}}_{n}(R)$.

###### Proof.

The proof is a similar line as in the proof of Proposition
$\ref{patch_mat}(3)$. ∎

###  3.2. Some properties of Rees-like algebra

######  Proposition 3.8.

Let $R$ be a ring, ${\mathfrak{a}}$ an ideal in $R$, and $t$ a variable. Then
we have the following properties for the Rees-like algebra
$R[{\mathfrak{a}}t,t^{2}]$.

  1. (1)

$R[{\mathfrak{a}}t,t^{2}][X]=R[X][{\mathfrak{a}}[X]t,t^{2}]$, where $X$ is a
variable and ${\mathfrak{a}}[X]={\mathfrak{a}}R[X]$.

  2. (2)

$R[{\mathfrak{a}}t,t^{2}][N]=R[N][{\mathfrak{a}}[N]t,t^{2}]$, where $N$ is a
commutative monoid and ${\mathfrak{a}}[N]={\mathfrak{a}}R[N]$.

  3. (3)

$S^{-1}(R[{\mathfrak{a}}t,t^{2}])=S^{-1}R[(S^{-1}{\mathfrak{a}})t,t^{2}]$,
where $S\subset R$ is a multiplicative closed subset.

  4. (4)

$\frac{R[{\mathfrak{a}}t,t^{2}]}{{\mathfrak{b}}R[{\mathfrak{a}}t,t^{2}]}=\frac{R}{{\mathfrak{b}}}\left[\frac{{\mathfrak{a}}}{{\mathfrak{a}}{\mathfrak{b}}}t,t^{2}\right]$,
where ${\mathfrak{b}}$ is an ideal of $R$.

  5. (5)

${\rm nil}(R[{\mathfrak{a}}t,t^{2}])={\rm nil}(R)[{\rm
nil}({\mathfrak{a}})t,t^{2}]$, where ${\rm
nil}({\mathfrak{a}})={\mathfrak{a}}\cap{\rm nil}(R)$.

  6. (6)

$\frac{R[{\mathfrak{a}}t,t^{2}]}{{\rm
nil}(R[{\mathfrak{a}}t,t^{2}])}=\frac{R}{{\rm
nil}(R)}\left[\frac{{\mathfrak{a}}}{{\rm nil}({\mathfrak{a}})}t,t^{2}\right]$.

###### Proof.

$(1)$ and $(2)$ follows from the definition of $R[{\mathfrak{a}}t,t^{2}]$.

$(3)$ Since the localization commutes with direct sums, we have

$\begin{split}S^{-1}(R[{\mathfrak{a}}t,t^{2}])&=S^{-1}(R\oplus{\mathfrak{a}}\oplus
R\oplus{\mathfrak{a}}\oplus\cdots)\\\ &=S^{-1}R\oplus
S^{-1}{\mathfrak{a}}\oplus S^{-1}R\oplus S^{-1}{\mathfrak{a}}\oplus\cdots\\\
&=S^{-1}R[(S^{-1}{\mathfrak{a}})t,t^{2}].\end{split}$

$(4)$ Note that
${\mathfrak{b}}R[{\mathfrak{a}}t,t^{2}]={\mathfrak{b}}\oplus{\mathfrak{a}}{\mathfrak{b}}\oplus{\mathfrak{b}}\oplus{\mathfrak{a}}{\mathfrak{b}}\oplus\cdots$
is a homogeneous ideal of $R[{\mathfrak{a}}t,t^{2}]$. Then we have

$\begin{split}\frac{R[{\mathfrak{a}}t,t^{2}]}{{\mathfrak{b}}R[{\mathfrak{a}}t,t^{2}]}&=\frac{R}{{\mathfrak{b}}}\oplus\frac{{\mathfrak{a}}}{{\mathfrak{a}}{\mathfrak{b}}}\oplus\frac{R}{{\mathfrak{b}}}\oplus\frac{{\mathfrak{a}}}{{\mathfrak{a}}{\mathfrak{b}}}\oplus\cdots\\\
&=\frac{R}{{\mathfrak{b}}}\left[\frac{{\mathfrak{a}}}{{\mathfrak{a}}{\mathfrak{b}}}t,t^{2}\right].\end{split}$

$(5)$ Clearly, ${\rm nil}(R)[{\rm nil}({\mathfrak{a}})t,t^{2}]\subset{\rm
nil}(R[{\mathfrak{a}}t,t^{2}])$. Conversely, let
$f=a_{0}+a_{1}t+\cdots+a_{n}t^{n}\in R[{\mathfrak{a}}t,t^{2}]$ such that
$f^{m}=0$ for some $m\in{\mathbb{N}}$. Then $a_{n}^{m}t^{nm}=0$, hence $a_{n}$
is nilpotent. Since nilradical is closed under addition, $f-a_{n}t^{n}$ is
nilpotent. Apply the same argument to conclude that $a_{0},a_{1},\ldots,a_{n}$
are nilpotent, and hence $f\in{\rm nil}(R)[{\rm nil}({\mathfrak{a}})t,t^{2}]$.

$(6)$ From $(5)$, we have ${\rm nil}(R[{\mathfrak{a}}t,t^{2}])={\rm
nil}(R)\oplus{\rm nil}({\mathfrak{a}})\oplus{\rm nil}(R)\oplus{\rm
nil}({\mathfrak{a}})\oplus\cdots$, which is a homogeneous ideal of
$R[{\mathfrak{a}}t,t^{2}]$. Then

$\begin{split}\frac{R[{\mathfrak{a}}t,t^{2}]}{{\rm
nil}(R[{\mathfrak{a}}t,t^{2}])}&=\frac{R}{{\rm
nil}(R)}\oplus\frac{{\mathfrak{a}}}{{\rm
nil}({\mathfrak{a}})}\oplus\frac{R}{{\rm
nil}(R)}\oplus\frac{{\mathfrak{a}}}{{\rm nil}({\mathfrak{a}})}\oplus\cdots\\\
&=\frac{R}{{\rm nil}(R)}\left[\frac{{\mathfrak{a}}}{{\rm
nil}({\mathfrak{a}})}t,t^{2}\right].\qed\end{split}$

The following Proposition in the case of Rees algebra is due to Valla [30,
Poposition 1.1]. For the sake of completeness, we give the proof.

######  Proposition 3.9.

Let $R$ be a Noetherian ring and ${\mathfrak{a}}$ an ideal in $R$. We denote
${\mathfrak{b}}^{*}$ the homogeneous ideal ${\mathfrak{b}}R[t]\cap
R[{\mathfrak{a}}t,t^{2}]$, where ${\mathfrak{b}}$ is an ideal of $R$. Then the
following statements hold:

  1. (1)

If ${\mathfrak{b}}$ is an ideal of $R$, then
$\frac{R[{\mathfrak{a}}t,t^{2}]}{{\mathfrak{b}}R[t]\cap
R[{\mathfrak{a}}t,t^{2}]}=\frac{R}{{\mathfrak{b}}}\left[\frac{{\mathfrak{a}}}{{\mathfrak{a}}\cap{\mathfrak{b}}}t,t^{2}\right]$;

  2. (2)

If ${\mathfrak{b}}_{1}$ and ${\mathfrak{b}}_{2}$ are ideal of $R$, then
$({\mathfrak{b}}_{1}\cap{\mathfrak{b}}_{2})^{*}={\mathfrak{b}}_{1}^{*}\cap{\mathfrak{b}}_{2}^{*}$;

  3. (3)

If ${\mathfrak{b}}=\cap_{i=1}^{n}{\mathfrak{q}}_{i}$ is a minimal primary
decomposition of an ideal ${\mathfrak{b}}$ of $R$, where ${\mathfrak{q}}_{i}$
is ${\mathfrak{p}}_{i}$-primary, then ${\mathfrak{p}}_{i}^{*}$ is prime,
${\mathfrak{q}}_{i}^{*}$ is ${\mathfrak{p}}_{i}^{*}$-primary and
${\mathfrak{b}}^{*}=\cap_{i=1}^{n}{\mathfrak{q}}_{i}^{*}$ is a minimal primary
decomposition of an ideal ${\mathfrak{b}}^{*}$ of $R[{\mathfrak{a}}t,t^{2}]$.

###### Proof.

$(1)$ Since ${\mathfrak{b}}R[t]\cap
R[{\mathfrak{a}}t,t^{2}]={\mathfrak{b}}\oplus({\mathfrak{a}}\cap{\mathfrak{b}})\oplus{\mathfrak{b}}\oplus({\mathfrak{a}}\cap{\mathfrak{b}})\oplus\cdots$
is a homogeneous ideal of $R[{\mathfrak{a}}t,t^{2}]$, it follows that

$\begin{split}\frac{R[{\mathfrak{a}}t,t^{2}]}{{\mathfrak{b}}R[t]\cap
R[{\mathfrak{a}}t,t^{2}]}&=\frac{R}{{\mathfrak{b}}}\oplus\frac{{\mathfrak{a}}}{{\mathfrak{a}}\cap{\mathfrak{b}}}\oplus\frac{R}{{\mathfrak{b}}}\oplus\frac{{\mathfrak{a}}}{{\mathfrak{a}}\cap{\mathfrak{b}}}\oplus\cdots\\\
&=\frac{R}{{\mathfrak{b}}}\left[\frac{{\mathfrak{a}}}{{\mathfrak{a}}\cap{\mathfrak{b}}}t,t^{2}\right].\end{split}$

$(2)$ Observe that
$({\mathfrak{b}}_{1}\cap{\mathfrak{b}}_{2})R[t]={\mathfrak{b}}_{1}R[t]\cap{\mathfrak{b}}_{2}R[t]$.
Then we have

$\begin{split}({\mathfrak{b}}_{1}\cap{\mathfrak{b}}_{2})^{*}&=({\mathfrak{b}}_{1}\cap{\mathfrak{b}}_{2})R[t]\cap
R[{\mathfrak{a}}t,t^{2}]\\\ &=({\mathfrak{b}}_{1}R[t]\cap
R[{\mathfrak{a}}t,t^{2}])\cap({\mathfrak{b}}_{2}R[t]\cap
R[{\mathfrak{a}}t,t^{2}])\\\
&={\mathfrak{b}}_{1}^{*}\cap{\mathfrak{b}}_{2}^{*}.\end{split}$

$(3)$ It is easy to see that ${\mathfrak{p}}_{i}^{*}={\mathfrak{p}}R[t]\cap
R[{\mathfrak{a}}t,t^{2}]$ is prime. Since
$r({\mathfrak{p}}_{i})R[t]=r({\mathfrak{p}}_{i}R[t])$, it follows that
${\mathfrak{p}}_{i}^{*}=r({\mathfrak{q}}_{i}^{*})$, and hence
${\mathfrak{q}}_{i}^{*}$ is ${\mathfrak{p}}_{i}^{*}$-primary. Now the rest of
the proof follows from $(2)$. ∎

######  Proposition 3.10.

Let $R$ be a Noetherian ring and ${\mathfrak{a}}$ an ideal in $R$. If
$\text{\rm dim}(R)$ is finite, then $\text{\rm
dim}(R[{\mathfrak{a}}t,t^{2}])=\text{\rm dim}(R)+1$.

###### Proof.

By Proposition 3.8(6), we observe that $R[{\mathfrak{a}}t,t^{2}]/{\rm
nil}(R[{\mathfrak{a}}t,t^{2}])$ is a Rees-like algebra. Since $\text{\rm
dim}(R[{\mathfrak{a}}t,t^{2}]/{\rm nil}(R[{\mathfrak{a}}t,t^{2}]))=\text{\rm
dim}(R[{\mathfrak{a}}t,t^{2}])$, we may assume that $R[{\mathfrak{a}}t,t^{2}]$
is reduced, and hence $R$ is reduced. By Proposition $\ref{decom_rla}(3)$, the
minimal prime ideals of $R[{\mathfrak{a}}t,t^{2}]$ are the ideals
${\mathfrak{p}}^{*}$, where ${\mathfrak{p}}$ is a minimal prime ideal of $R$.
Thus, using Proposition $\ref{decom_rla}(1)$, we have

$\begin{split}\text{\rm dim}(R[{\mathfrak{a}}t,t^{2}])&={\operatorname{\rm
max}}\left\\{\text{\rm
dim}\left(\frac{R[{\mathfrak{a}}t,t^{2}]}{{\mathfrak{p}}^{*}}\right)~{}|~{}{\mathfrak{p}}^{*}\in{\rm
Min}(R[{\mathfrak{a}}t,t^{2}])\right\\}\\\ &={\operatorname{\rm
max}}\left\\{\text{\rm
dim}\left(\frac{R[{\mathfrak{a}}t,t^{2}]}{{\mathfrak{p}}R[t]\cap
R[{\mathfrak{a}}t,t^{2}]}\right)~{}|~{}{\mathfrak{p}}\in{\rm
Min}(R)\right\\}\\\ &={\operatorname{\rm max}}\left\\{\text{\rm
dim}\left(\frac{R}{{\mathfrak{p}}}\left[\frac{{\mathfrak{a}}}{{\mathfrak{a}}\cap{\mathfrak{p}}}t,t^{2}\right]\right)~{}|~{}{\mathfrak{p}}\in{\rm
Min}(R)\right\\}.\end{split}$

This implies that to determine $\text{\rm dim}(R[{\mathfrak{a}}t,t^{2}])$, we
may assume that $R$ is a domain.

If ${\mathfrak{a}}=0$, there is nothing to prove. So we assume
${\mathfrak{a}}\neq 0$, and let $a$ be a non-zero element of ${\mathfrak{a}}$.
Since $t=at/a\in{\rm Frac}(R[{\mathfrak{a}}t,t^{2}])$ and $t\in{\rm
Frac}(R[{\mathfrak{a}}t])$, it follows that ${\rm
Frac}(R[{\mathfrak{a}}t])={\rm Frac}(R[{\mathfrak{a}}t,t^{2}])={\rm
Frac}(R[t])=R(t)$. Consider $R[{\mathfrak{a}}t]\hookrightarrow
R[{\mathfrak{a}}t,t^{2}]\hookrightarrow R[t]\hookrightarrow R(t)$, then
combine [3, Proposition 3.2] and Theorem 2.4, we get $\text{\rm
dim}(R)+1=\text{\rm dim}(R[t])\leq\text{\rm
dim}(R[{\mathfrak{a}}t,t^{2}])\leq\text{\rm dim}(R[{\mathfrak{a}}t])=\text{\rm
dim}(R)+1$. This implies $\text{\rm dim}(R[{\mathfrak{a}}t,t^{2}])=\text{\rm
dim}(R)+1$. ∎

##  4\. Existence of unimodular element

######  Theorem 4.1.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$ such that ${\rm ht}({\mathfrak{a}})\geq 1$. Let $A=R[\mathfrak{a}t,t^{2}]$
and $P$ a projective $A$-module of rank $\geq d+1$. Then $P$ has a unimodular
element.

###### Proof.

By Proposition 3.10, $\text{\rm dim}(A)=d+1$ and if ${\rm rank}(P)\geq d+2$,
then the result follows from Theorem 2.6. So we can assume that ${\rm
rank}(P)=d+1$. If $d=0$, then we have ${\mathfrak{a}}=R$. In this case, the
theorem follows from [22, Corollary 2].

Assume $d\geq 1$. Note that $P\otimes_{A}R[t]$ is a projective $R[t]$-module
and $P\otimes_{A}(R/{\mathfrak{a}})[t^{2}]$ is a projective
$(R/{\mathfrak{a}})[t^{2}]$-module. Then by [22, Corollary 2],
$P\otimes_{A}R[t]$ has a unimodular element, say $p_{1}$ and
$P\otimes_{A}(R/{\mathfrak{a}})[t^{2}]$ has a unimodular element, say $p_{2}$.
Now consider the following conductor square of rings on the left, which
induces the Cartesian square of projective modules on the right

${R[{\mathfrak{a}}t,t^{2}]}$${R[t]}$${(R/{\mathfrak{a}})[t^{2}]}$${(R/{\mathfrak{a}})[t]}$$\scriptstyle{i_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\eta_{2}}$$\scriptstyle{i_{2}}$${P}$${P\otimes_{A}R[t]}$${P\otimes_{A}(R/{\mathfrak{a}})[t^{2}]}$${P\otimes_{A}(R/{\mathfrak{a}})[t].}$$\scriptstyle{i_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\eta_{2}}$$\scriptstyle{i_{2}}$

Let $\bar{u}$ and $\bar{v}$ be the images of $p_{1}$ and $p_{2}$ in
$P\otimes_{A}(R/{\mathfrak{a}})[t]$ respectively. Since $\bar{u}\in{\rm
Um}(P\otimes_{A}(R/{\mathfrak{a}})[t])$, it follows that
$P\otimes_{A}(R/{\mathfrak{a}})[t]\cong Q\oplus(R/{\mathfrak{a}})[t]$ for some
projective $(R/{\mathfrak{a}})[t]$-module $Q$ of rank $d$. Since ${\rm
ht}({\mathfrak{a}})\geq 1$, we must have $\text{\rm dim}(R/{\mathfrak{a}})\leq
d-1$.

If $d=1$, then $\text{\rm dim}(R/{\mathfrak{a}})=0$. To prove $Q$ is a free
$(R/{\mathfrak{a}})[t]$-module, we can assume $(R/{\mathfrak{a}})[t]$ is
reduced and hence $R/{\mathfrak{a}}$ is reduced. Then $R/{\mathfrak{a}}$ is a
finite product of fields. So without loss of generality, we can assume
$R/{\mathfrak{a}}$ is a field. Thus $Q$ is a free
$(R/{\mathfrak{a}})[t]$-module, this implies
$P\otimes_{A}(R/{\mathfrak{a}})[t]\cong(R/{\mathfrak{a}})[t]^{2}$. Then by
[23, Theorem 5.1(I)], there exists $\sigma\in{\operatorname{\rm
E}}(P\otimes_{A}(R/{\mathfrak{a}})[t])$ such that $\sigma(\bar{u})=\bar{v}$.
If $d\geq 2$, in this case also by [18, Theorem 2.6], there exists
$\sigma\in{\operatorname{\rm E}}(P\otimes_{A}(R/{\mathfrak{a}})[t])$ such that
$\sigma(\bar{u})=\bar{v}$. Let $\hat{\sigma}$ be the lift of $\sigma$ in
$P\otimes_{A}R[t]$. By changing the unimodular row $p_{1}$ to
$\hat{\sigma}(p_{1})$, we can assume that $\bar{u}=\bar{v}$. Consider the
following fiber product diagram of projective modules

${P}$${P\otimes_{A}R[t]}$${R[{\mathfrak{a}}t,t^{2}]}$${R[t]}$${P\otimes_{A}(R/{\mathfrak{a}})[t^{2}]}$${P\otimes_{A}(R/{\mathfrak{a}})[t]}$${(R/{\mathfrak{a}})[t^{2}]}$${(R/{\mathfrak{a}})[t]}$$\scriptstyle{\varphi}$$\scriptstyle{\varphi_{1}}$$\scriptstyle{\varphi_{2}}$

Since $\varphi_{1}$ and $\varphi_{2}$ are surjective homomorphisms as
indicated by the above diagram, we obtain a surjective homomorphism
$\varphi:P\twoheadrightarrow R[{\mathfrak{a}}t,t^{2}]$. Consequently, there
exists $p\in P$ such that $\varphi(p)=1$. Therefore, $P$ has a unimodular
element. ∎

######  Theorem 4.2.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a partially cancellative torsion-
free seminormal monoid. Assume ${\rm ht}({\mathfrak{a}})\geq 1$, Serre dim
$R[N\oplus{\mathbb{Z}}_{+}]\leq d$ and Serre dim $R[N\oplus
2{\mathbb{Z}}_{+}]\leq d$. Then any projective $A[N]$-module $P$ of rank
$r\geq{\operatorname{\rm max}}\\{3,d+1\\}$ has a unimodular element.

###### Proof.

Denote $N_{1}=N\oplus{\mathbb{Z}}_{+}$ and $N_{2}=N\oplus 2{\mathbb{Z}}_{+}$,
then we have $R[t][N]=R[N_{1}]$ and
$(R/{\mathfrak{a}})[t^{2}][N]=(R/{\mathfrak{a}})[N_{2}]$. Now consider the
following conductor square of rings on the left, which induces the Cartesian
square of monoid algebras on the right

${R[{\mathfrak{a}}t,t^{2}]}$${R[t]}$${(R/{\mathfrak{a}})[t^{2}]}$${(R/{\mathfrak{a}})[t]}$$\scriptstyle{i_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\eta_{2}}$$\scriptstyle{i_{2}}$${R[{\mathfrak{a}}t,t^{2}][N]}$${R[N_{1}]}$${(R/{\mathfrak{a}})[N_{2}]}$${(R/{\mathfrak{a}})[N_{1}].}$$\scriptstyle{i_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\eta_{2}}$$\scriptstyle{i_{2}}$

Note that $P\otimes_{A}R[N_{1}]$ is a projective $R[N_{1}]$-module and
$P\otimes_{A}(R/{\mathfrak{a}})[N_{2}]$ is a projective
$(R/{\mathfrak{a}})[N_{2}]$-module. By hypothesis, we have Serre dim
$R[N_{i}]\leq d$, where $i=1,2$. Let $p_{1}$ and $p_{2}$ be unimodular
elements of $P\otimes_{A}R[N_{1}]$ and $P\otimes_{A}(R/{\mathfrak{a}})[N_{2}]$
respectively. The above Cartesian square of rings induces the following
Cartesian square of projective modules

${P}$${P\otimes_{A}R[N_{1}]}$${P\otimes_{A}(R/{\mathfrak{a}})[N_{2}]}$${P\otimes_{A}(R/{\mathfrak{a}})[N_{1}].}$$\scriptstyle{i_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\eta_{2}}$$\scriptstyle{i_{2}}$

Consider $\bar{u}$ and $\bar{v}$ as the respective images of $p_{1}$ and
$p_{2}$ in $P\otimes_{A}(R/{\mathfrak{a}})[N_{1}]$. Since $\bar{u}\in{\rm
Um}(P\otimes_{A}(R/{\mathfrak{a}})[N_{1}])$, it follows that
$P\otimes_{A}(R/{\mathfrak{a}})[N_{1}]=Q\oplus(R/{\mathfrak{a}})[N_{1}]$ for
some projective $(R/{\mathfrak{a}})[N_{1}]$-module $Q$ of rank $r-1$. Since
${\rm ht}({\mathfrak{a}})\geq 1$, we have $\text{\rm
dim}(R/{\mathfrak{a}})\leq d-1$ and hence $r-1\geq{\operatorname{\rm
max}}\\{2,\text{\rm dim}(R/{\mathfrak{a}})+1\\}$. Note that $N_{1}$ is a
partially cancellation torsion-free seminormal monoid. Then by Theorem 2.16,
there exists $\sigma\in{\operatorname{\rm
E}}(P\otimes_{A}(R/{\mathfrak{a}})[N_{1}])$ such that
$\sigma(\bar{u})=\bar{v}$. Then using similar arguments as in the proof of
Theorem 4.1, we conclude that $P$ has a unimodular element. ∎

######  Remark 4.3.

One can remove the hypothesis ${\rm ht}({\mathfrak{a}})\geq 1$ in Theorems 4.1
and 4.2, by assuming $R$ is an integral domain. For this, if ${\rm
ht}({\mathfrak{a}})=0$ this implies ${\mathfrak{a}}=0$. Then the result
follows from [22, Corollary 2] and the assumption Serre dim $R[N\oplus
2{\mathbb{Z}}_{+}]\leq d$ respectively.

The definition of classes $\mathcal{C}(\Phi)$ and $\mathfrak{M}_{n}$ of
monoids introduced in [15] and [14] respectively.

######  Corollary 4.4.

Let $R$, ${\mathfrak{a}}$ and $A$ be as in Theorem 4.2. If
$N\in\mathfrak{M}_{n}$ is a $\Phi$-simplicial seminormal monoid of rank $n\geq
1$. Then Serre dim $A[N]\leq{\operatorname{\rm max}}\\{2,d\\}$.

###### Proof.

By [14, Corollary 4.3], monoids $N\oplus{\mathbb{Z}}_{+}$ and $N\oplus
2{\mathbb{Z}}_{+}$ are belongs to $\mathfrak{M}_{n+1}$ of rank $n+1$. Then
using [14, Theorem 4.1], we get Serre dim $R[N\oplus{\mathbb{Z}}_{+}]\leq d$
and Serre dim $R[N\oplus 2{\mathbb{Z}}_{+}]\leq d$. Therefore, the result
follows from Theorem 4.2. ∎

######  Example 4.5.

$(1)$ If $N={\mathbb{Z}}_{+}^{n}$ or $N\in\mathcal{C}(\Phi)$ of rank $n$, then
$N\in\mathfrak{M}_{n}$. Hence, by Corollary 4.4, Serre dim
$A[N]\leq{\operatorname{\rm max}}\\{2,d\\}$.

$(2)$ If $N\subset{\mathbb{Z}}^{2}_{+}$ is a normal monoid of rank $2$, then
by [15, Example 3.5(2)], $N\in\mathcal{C}(\Phi)$. Hence, Serre dim
$A[N]\leq{\operatorname{\rm max}}\\{2,d\\}$ by $(1)$.

$(3)$ If $N$ is a truncated or normal quasi-truncated monoid of rank $n\leq
3$, then by [15, Example 3.5(3)], $N\in\mathcal{C}(\Phi)$. Hence, Serre dim
$A[N]\leq{\operatorname{\rm max}}\\{2,d\\}$ by $(1)$.

######  Corollary 4.6.

Let $R$ be a one-dimensional Noetherian ring containing ${\mathbb{Q}}$ and
${\mathfrak{a}}$ an ideal in $R$ such that ${\rm ht}({\mathfrak{a}})\geq 1$.
Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a monoid generated by
$\\{x_{1}x_{3},x_{1}x_{4},x_{2}x_{3},x_{2}x_{4}\\}$. Then Serre dim $A[N]\leq
2$.

###### Proof.

Since $N$ is a normal monoid, it is easy to verify that
$N\oplus{\mathbb{Z}}_{+}$ and $N\oplus 2{\mathbb{Z}}_{+}$ are normal monoids.
Let $S$ be the normalization of $R$ and $C$ the conductor ideal of the
extension $R\subset S$. Note that $\text{\rm dim}(S/C)=0$, then $S/C$ is a
semilocal ring and hence $SK_{1}(S/C)=0$. Then by [16, Corollary 1.2], we have
$SK_{1}((S/C)[N\oplus{\mathbb{Z}}_{+}])=0$ and $SK_{1}((S/C)[N\oplus
2{\mathbb{Z}}_{+}])=0$. Therefore, using similar arguments as in the proof of
[16, Theorem 1.3], we conclude that Serre dim $R[N\oplus{\mathbb{Z}}_{+}]\leq
1$ and Serre dim $R[N\oplus 2{\mathbb{Z}}_{+}]\leq 1$. Then by Theorem 4.2,
Serre dim $A[N]\leq 2$. ∎

##  5\. Cancellation of Projective modules

######  Proposition 5.1.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Then ${\operatorname{\rm E}}_{r}(R[\mathfrak{a}t,t^{2}])$ acts
transitively on ${\rm Um}_{r}(R[\mathfrak{a}t,t^{2}])$ for $r\geq d+2$.

###### Proof.

Let $a=(a_{1},\ldots,a_{r})\in{\rm Um}_{r}(R[\mathfrak{a}t,t^{2}])$. By [23,
Theorem 5.1(I)], there exists $\sigma_{1}\in{\operatorname{\rm E}}_{r}(R[t])$
such that $(i_{1}(a))\sigma_{1}=(1,0,\ldots,0)$, and there exists
$\sigma_{2}\in{\operatorname{\rm E}}_{r}((R/{\mathfrak{a}})[t^{2}])$ such that
$(\eta_{1}(a))\sigma_{2}=(1,0,\ldots,0)$. Consider the following conductor
square of rings

${R[{\mathfrak{a}}t,t^{2}]}$${R[t]}$${(R/{\mathfrak{a}})[t^{2}]}$${(R/{\mathfrak{a}})[t].}$$\scriptstyle{i_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\eta_{2}}$$\scriptstyle{i_{2}}$

Let $\sigma_{1}^{\prime}$ (resp. $\sigma_{2}^{\prime}$) be the image of
$\sigma_{1}$ (resp. $\sigma_{2}$) in ${\operatorname{\rm
E}}_{r}(R/{\mathfrak{a}})[t])$. We define
$\phi^{\prime}=\sigma_{1}^{\prime-1}\sigma_{2}^{\prime}$. Since the image of
$i_{1}(a)$ under $\eta_{2}$ and the image of $\eta_{1}(a)$ under $i_{2}$ are
same in ${\rm Um}_{r}((R/{\mathfrak{a}})[t])$ as the above diagram is
commutative, it follows that $(1,0,\ldots,0)\phi^{\prime}=(1,0,\ldots,0)$.
Hence it is easy to see that

$\phi^{\prime}=(\begin{smallmatrix}1&0\\\ *&\eta^{\prime}\end{smallmatrix})$

Now we define

$\phi=(\begin{smallmatrix}1&0\\\ *&\eta\end{smallmatrix})$

such that the image of $\phi$ under $\eta_{2}$ is $\phi^{\prime}$. This is
possible because $\eta_{2}$ is surjective. Note that
$(1,0,\ldots,0)\phi=(1,0,\ldots,0)$, then
$(i_{1}(a))\sigma_{1}\phi=(1,0,\ldots,0)$. Since
$\phi^{\prime}\in{\operatorname{\rm E}}_{r}((R/{\mathfrak{a}})[t])$, by Lemma
3.7, $\phi\in{\operatorname{\rm E}}_{r}(R[t])$. Observe that the image of
$\sigma_{1}\phi$ is
$\sigma_{1}^{\prime}\phi^{\prime}=\sigma_{1}^{\prime}\sigma_{1}^{\prime-1}\sigma_{2}^{\prime}=\sigma_{2}^{\prime}$.
Therefore, replacing $\sigma_{1}$ by $\sigma_{1}\phi$, we can assume that
$\sigma_{1}^{\prime}=\sigma_{2}^{\prime}$. Hence by Proposition
$\ref{patch_mat}(3)$, there exists $\sigma\in{\operatorname{\rm
E}}_{r}(R[\mathfrak{a}t,t^{2}])$ such that $a\sigma=(1,0,\ldots,0)$. ∎

######  Theorem 5.2.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Suppose $P$ is a projective $R[\mathfrak{a}t,t^{2}]$-module of rank $\geq
d+1$. Then ${\operatorname{\rm E}}(R[\mathfrak{a}t,t^{2}]\oplus P)$ acts
transitively on ${\rm Um}(R[\mathfrak{a}t,t^{2}]\oplus P)$. In particular, $P$
is cancellative.

###### Proof.

Let $A=R[\mathfrak{a}t,t^{2}]$. We have already observed $\text{\rm
dim}(A)=d+1$ and if ${\rm rank}(P)\geq d+2$, then the result follows from
Theorem 2.7. Hence, it is enough to consider the case when ${\rm
rank}(P)=d+1$.

By Proposition 3.8, the ring $A/{\rm nil}(A)$ is a Rees-like algebra. In view
of Lemma 2.9, we can assume that $A$ is reduced, and hence $R$ is reduced. Let
$S$ be the set of all non-zero divisors of $R$. Since $\text{\rm
dim}(S^{-1}R)=0$, it follows that $S^{-1}R$ is a finite product of fields. To
prove $S^{-1}P$ is free, it is enough to prove for each connected component.
So without loss of generality, we may assume that $S^{-1}R$ is a field. Hence
by Proposition 3.8, $S^{-1}A=S^{-1}R[t]$ is a principal ideal domain.
Therefore $S^{-1}P$ is a free $S^{-1}A$-module. Since $P$ is finitely
generated, we can choose $s\in S$ that satisfies the properties of Lemma 2.8.
Utilizing Proposition 3.8, we have

$\frac{A[X]}{(X^{2}-s^{2}X)}=\frac{R[X][{\mathfrak{a}}[X]t,t^{2}]}{(X^{2}-s^{2}X)}=\frac{R[X]}{(X^{2}-s^{2}X)}\left[\frac{{\mathfrak{a}}[X]}{(X^{2}-s^{2}X){\mathfrak{a}}[X]}t,t^{2}\right],$

which is a Rees-like algebra. Since $\text{\rm
dim}(R[X]/(X^{2}-s^{2}X))=\text{\rm dim}(R)$, by Proposition 5.1,

${\operatorname{\rm
E}}_{d+2}\left(\frac{A[X]}{(X^{2}-s^{2}X)}\right)~{}\text{acts transitively
on}~{}{\rm Um}_{d+2}\left(\frac{A[X]}{(X^{2}-s^{2}X)}\right).$

Therefore, by Theorem 2.11, ${\operatorname{\rm E}}(A\oplus P)$ acts
transitively on ${\rm Um}(A\oplus P)$. ∎

######  Proposition 5.3.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a partially cancellative monoid.
Then ${\operatorname{\rm E}}_{r}(A[N])$ acts transitively on ${\rm
Um}_{r}(A[N])$ for $r\geq{\operatorname{\rm max}}\\{3,d+2\\}$.

###### Proof.

Let $a=(a_{1},\ldots,a_{r})\in{\rm Um}_{r}(A[N])$ and let
$e_{1}=(1,0,\ldots,0)$. Consider the following Cartesian square of monoid
algebras

${R[{\mathfrak{a}}t,t^{2}][N]}$${R[t][N]}$${(R/{\mathfrak{a}})[t^{2}][N]}$${(R/{\mathfrak{a}})[t][N].}$$\scriptstyle{i_{1}}$$\scriptstyle{\eta_{1}}$$\scriptstyle{\eta_{2}}$$\scriptstyle{i_{2}}$

It is easy to verify that $N\oplus{\mathbb{Z}}_{+}$ and $N\oplus
2{\mathbb{Z}}_{+}$ are partially cancellative monoid. Since
$R[t][N]=R[N\oplus{\mathbb{Z}}_{+}]$ and
$(R/{\mathfrak{a}})[t^{2}][N]=R[N\oplus 2{\mathbb{Z}}_{+}]$, by Lemma 2.15,
there exists $\sigma_{1}\in{\operatorname{\rm E}}_{r}(R[t][N])$ and
$\sigma_{2}\in{\operatorname{\rm E}}_{r}((R/{\mathfrak{a}})[t^{2}][N])$ such
that $(i_{1}(a))\sigma_{1}=e_{1}$ and $(\eta_{1}(a))\sigma_{2}=e_{1}$. Then
using similar arguments as in the proof of Proposition 5.1, we obtain
$\sigma\in{\operatorname{\rm E}}_{r}(A[N])$ such that $a\sigma=e_{1}$. ∎

######  Lemma 5.4.

Let $R$ be a reduced Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an
ideal in $R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a torsion-free seminormal
monoid. Suppose $P$ is a projective $A[N]$-module of rank $r$. Then there
exists $s\in R$ satisfying the properties of Lemma 2.8.

###### Proof.

Let $S$ be the set of all non-zero-divisors of $R$. From the second paragraph
of the proof of Theorem 5.2, we observe that $S^{-1}A$ is a principal ideal
domain. Then by Lemma 2.14, $S^{-1}P$ is a free $S^{-1}A[N]$-module of rank
$r$. Since $P$ is finitely generated, there exists $s\in S$ such that $P_{s}$
is free. We can choose $s\in R$ in such a way that it satisfies the properties
of Lemma 2.8. ∎

######  Theorem 5.5.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[{\mathfrak{a}}t,t^{2}]$ and $N$ a partially cancellative
torsion-free seminormal monoid. Suppose $P$ is a projective $A[N]$-module of
rank $r\geq{\operatorname{\rm max}}\\{2,d+1\\}$. Then ${\operatorname{\rm
E}}(A[N]\oplus P)$ acts transitively on ${\rm Um}(A[N]\oplus P)$.

###### Proof.

We observe that $A[N]/{\rm nil}(A[N])\cong A/{\rm nil}(A)[N]$, and by
Proposition 3.8, $A/{\rm nil}(A)$ is a Rees-like algebra. Then, in view of
Lemma 2.9, we can assume that $A[N]$ is reduced, and hence $R$ is reduced.

We shall use the induction on $d$. If $d=0$, then the argument given in the
proof of Lemma 5.4 implies that $P$ is free and hence the result follows from
Proposition 5.3.

Assume $d>0$, and let $(a,p)\in{\rm Um}(A[N]\oplus P)$. By Lemma 5.4, choose a
non-zero divisor $s\in R$ satisfying the properties of Lemma 2.8. Let “bar”
denote reduction modulo $s^{2}A[N]$. Since $\text{\rm dim}(R/s^{2}R)<\text{\rm
dim}(R)$, and $A[N]/s^{2}A[N]\cong(A/s^{2}A)[N]$, by induction on $d$, there
exists $\overline{\sigma}\in{\operatorname{\rm
E}}(\overline{A[N]}\oplus\overline{P})$ such that
$(\overline{a},\overline{p})\overline{\sigma}=(1,0)$. Let $\sigma$ be the lift
of $\overline{\sigma}$ in ${\operatorname{\rm E}}(A[N]\oplus P)$, then
$(a,p)\sigma\in{\rm Um}(A[N]\oplus P,s^{2}A[N])$. Utilizing Proposition 3.8,
we have

$\frac{A[N][X]}{(X^{2}-s^{2}X)}=\frac{R[{\mathfrak{a}}t,t^{2}][X]}{(X^{2}-s^{2}X)}[N]=\frac{R[X]}{(X^{2}-s^{2}X)}\left[\frac{{\mathfrak{a}}[X]}{(X^{2}-s^{2}X){\mathfrak{a}}[X]}t,t^{2}\right][N].$

Since $\text{\rm dim}(R[X]/(X^{2}-s^{2}X))=\text{\rm dim}(R)=d\geq 1$, by
Proposition 5.3,

${\operatorname{\rm
E}}_{r+1}\left(\frac{A[N][X]}{(X^{2}-s^{2}X)}\right)~{}\text{acts transitively
on}~{}{\rm Um}_{r+1}\left(\frac{A[N][X]}{(X^{2}-s^{2}X)}\right).$

Hence by Lemma 2.10, ${\operatorname{\rm E}}(A[N]\oplus P)$ acts transitively
on ${\rm Um}(A[N]\oplus P,s^{2}A[N])$. Then there exists
$\sigma_{1}\in{\operatorname{\rm E}}(A[N]\oplus P)$ such that
$(a,p)\sigma\sigma_{1}=(1,0)$. This completes the proof. ∎

######  Remark 5.6.

Note that rank-one projective $A[N]$-modules are always cancellative. To see
this, let $P$ be a projective $A[N]$-module of rank one and assume $P\oplus
Q\cong P^{\prime}\oplus Q$, where $Q$ is a projective $A[N]$-module. Then
there exists a $A[N]$-module $Q^{\prime}$ such that $Q\oplus Q^{\prime}\cong
A[N]^{n}$ for some positive integer $n$. Thus, we have $P\oplus A[N]^{n}\cong
P^{\prime}\oplus A[N]^{n}$. Hence, inductively we can assume $P\oplus
A[N]\cong P^{\prime}\oplus A[N]$. Then $P\cong\bigwedge^{2}(P\oplus
A[N])\cong\bigwedge^{2}(P^{\prime}\oplus A[N])\cong P^{\prime}$. Therefore,
the projective $A[N]$-module $P$ in Theorem 5.5 is cancellative for $r\geq
d+1$.

##  6\. Some Applications

In this section, we study the minimum number of generators of a module over
Rees-like algebra and its monoid extensions. Also, we study the $K_{1}$-analog
of our results over Rees-like algebra and its Laurent polynomial extensions.

###  6.1. Efficient generation of modules

######  Theorem 6.1.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[\mathfrak{a}t,t^{2}]$. If $M$ is an $A$-module and
$e(M):={\operatorname{\rm max}}\\{\mu_{\mathfrak{p}}(M)+\text{\rm
dim}(A/\mathfrak{p}):\mathfrak{p}\in{\rm Spec}(A)\\}$. Then $M$ is generated
by $e(M)$ elements.

###### Proof.

Let $\mu(M)=n$. Assume that $n>e(M)>d$. To prove that $M$ is generated by
$e(M)$ elements, it is enough to prove that $M$ is generated by $n-1$
elements. Let $K=\ker(\varphi)$, where $\varphi:A^{n}\twoheadrightarrow M$.
Then by [22, Theorem 2], there exists basic elements $u_{1}$ and $u_{2}$ of
the modules $K\otimes R[t]$ and $K\otimes(R/{\mathfrak{a}})[t^{2}]$
respectively. By patching of $u_{1}$ and $u_{2}$, we get a basic element $u\in
K$ which is a unimodular element in $A^{n}$. Thus we can write $A^{n}\cong
uA\oplus Q$, for some projective $A$-module $Q$ of rank $\geq n-1>d$. We are
now in the situation

${A^{n}}$${M}$${0}$${uA\oplus Q}$${M}$$\scriptstyle{\varphi}$

Since $\varphi(u)=0$, the map $\varphi$ induces the surjections
$\varphi\otimes R[t]:Q\otimes R[t]\twoheadrightarrow M\otimes R[t]$ and
$\varphi\otimes(R/{\mathfrak{a}})[t^{2}]:Q\otimes(R/{\mathfrak{a}})[t^{2}]\twoheadrightarrow
M\otimes(R/{\mathfrak{a}})[t^{2}]$. Then by patching, we get a surjection
$\varphi:Q\twoheadrightarrow M$. Moreover by Theorem 5.2, we have $Q\cong
A^{n-1}$. Hence $M$ is generated by $n-1$ elements. This completes the proof.
∎

######  Theorem 6.2.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a partially cancellative torsion-
free seminormal monoid. Assume ${\rm ht}({\mathfrak{a}})\geq 1$, Serre dim
$R[N\oplus{\mathbb{Z}}_{+}]\leq d$ and Serre dim $R[N\oplus
2{\mathbb{Z}}_{+}]\leq d$. Suppose $P$ is a projective $A[N]$-module of rank
$r\geq{\operatorname{\rm max}}\\{3,d+1\\}$. Then $\mu(P)\leq
r+{\operatorname{\rm max}}\\{2,d\\}$.

###### Proof.

The proof follows from [15, Theorem 4.1]. However, one needs to use the
Theorem 4.2 and Remark 5.6. ∎

The following result follows from Theorem 6.2 using Corollary 4.4 and
Corollary 4.6.

######  Corollary 6.3.

Let $R$, ${\mathfrak{a}}$ and $A$ be as in Theorem 6.2. Suppose $N$ is a
monoid and $P$ is a projective $A[N]$-module of rank $r\geq{\operatorname{\rm
max}}\\{3,d+1\\}$.

  1. (1)

If $N\in\mathfrak{M}_{n}$ is a $\Phi$-simplicial seminormal monoid of rank
$n\geq 1$, then $\mu(P)\leq r+{\operatorname{\rm max}}\\{2,d\\}$.

  2. (2)

If $R$ containing ${\mathbb{Q}}$, $d=1$ and $N$ is a monoid generated by
$\\{x_{1}x_{3},x_{1}x_{4},x_{2}x_{3},x_{2}x_{4}\\}$, then $\mu(P)\leq r+2$.

###  6.2. Stability theorems for ${\operatorname{\rm GL}}_{r}$

######  Theorem 6.4.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[\mathfrak{a}t,t^{2}]$. Then, for $r\geq{\operatorname{\rm
max}}\\{3,d+2\\}$, the natural map $\phi:{\operatorname{\rm
GL}}_{r}(A)/{\operatorname{\rm E}}_{r}(A)\rightarrow K_{1}(A)$ is an
isomorphism.

###### Proof.

To prove surjectivity, let us take an element
$\eta=[a_{ij}]\in{\operatorname{\rm GL}}_{n}(A)$ in $K_{1}(A)$ for some $n\geq
d+2$. By Proposition 5.1, there exists $\sigma_{1}\in{\operatorname{\rm
E}}_{n}(A)$ such that $(a_{n1},\ldots,a_{nn})\sigma_{1}=(0,\ldots,0,1)$.
Consider the elementary matrix $\sigma_{2}=e_{1n}(-b_{1n})\cdots
e_{(n-1)n}(-b_{(n-1)n})$ where $b_{ij}$ are $(i,j)$-th term of the matrix
$\eta\sigma_{1}$, i.e.,

$\sigma_{2}=\begin{pmatrix}1&0&\cdots&0&-b_{1n}\\\ 0&1&\cdots&0&-b_{2n}\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\ 0&0&\cdots&1&-b_{(n-1)n}\\\
0&0&\cdots&0&1\end{pmatrix},$

then we get $\sigma_{2}\eta\sigma_{1}=\begin{pmatrix}\eta^{\prime}&0\\\
0&1\end{pmatrix}$ where $\eta^{\prime}\in{\operatorname{\rm GL}}_{n-1}(A)$.
Therefore, the class of the element $\eta$ in $K_{1}(A)$ is represented by the
matrix of size $n-1$. In this way, we get an element in ${\operatorname{\rm
GL}}_{n-1}(A)$. If $n-1\leq r$, we are done. Otherwise, after the steps, we
obtain an element in ${\operatorname{\rm GL}}_{r}(A)/{\operatorname{\rm
E}}_{r}(A)$. Hence $[\eta]=[\eta^{\prime}]$ in $K_{1}(A)$, demonstrating
surjectivity.

To prove injectivity, it is enough to prove that any stably elementary matrix
is elementary over $A$. For this, let $\eta\in{\operatorname{\rm GL}}_{r}(A)$
be a stably elementary matrix. Let ${\mathfrak{p}}\in{\rm Spec}(R)$. Note that
$A_{{\mathfrak{p}}}=R_{{\mathfrak{p}}}[{\mathfrak{a}}_{{\mathfrak{p}}}t,t^{2}]$
has generalized dimension $\leq d$. Then by Eisenbud-Evans theorem (see, for
instance, [3, Section 2, page 118]), we can say $A_{{\mathfrak{p}}}$ has
stable range $\leq d+1$. So by [31, Theorem 3.2], any stably elementary matrix
in ${\operatorname{\rm GL}}_{r}(A_{{\mathfrak{p}}})$ is in ${\operatorname{\rm
E}}_{r}(A_{{\mathfrak{p}}})$ for $r\geq d+2$, this implies
$\eta_{{\mathfrak{p}}}\in{\operatorname{\rm E}}_{r}(A_{{\mathfrak{p}}})$. We
grade $A$ as $A=R\oplus A_{+}$, where $A_{+}={\mathfrak{a}}t\oplus
Rt^{2}\oplus{\mathfrak{a}}t^{3}\oplus\cdots$, then multiplying by an
elementary matrix, we can assume $\eta\in{\operatorname{\rm
GL}}_{r}(A,A_{+})$. Hence, by [11, Proposition 7.3], $\eta$ is elementary.
This completes the proof.

∎

######  Theorem 6.5.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $N$ a partially cancellative monoid.
Then, for $r\geq{\operatorname{\rm max}}\\{2,d+1\\}$, the natural map
$\phi:{\operatorname{\rm GL}}_{r}(A[N])/{\operatorname{\rm
E}}_{r}(A[N])\rightarrow K_{1}(A[N])$ is an epimorphism.

###### Proof.

The proof is a similar line as in the proof of Theorem 6.4. But, one needs to
use the Proposition 5.3. ∎

######  Theorem 6.6.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $B=A[X_{1},\ldots,X_{n}]$, where
$n\in{\mathbb{N}}\cup\\{0\\}$. Then, for $r\geq{\operatorname{\rm
max}}\\{3,d+2\\}$, the natural map $\phi:{\operatorname{\rm
GL}}_{r}(B)/{\operatorname{\rm E}}_{r}(B)\rightarrow K_{1}(B)$ is an
isomorphism.

###### Proof.

Surjectivity follows from the Theorem 6.5 by taking $N={\mathbb{Z}}_{+}^{n}$.
We prove injectivity by induction on $n$. For $n=0$, we are done by Theorem
6.4.

So we assume $n\geq 1$ and let $\sigma\in{\operatorname{\rm
GL}}_{r}(B)\cap{\operatorname{\rm E}}(B)$. Let $S$ be the multiplicatively
closed set consisting of all monic polynomials in $R[X_{n}]$. Then note that
we have the following

  1. (1)

$\text{\rm dim}(R[X_{n}]_{S})=d$;

  2. (2)

$(A[X_{n}])_{S}=R[X_{n}]_{S}[{\mathfrak{a}}[X_{n}]_{S}t,t^{2}]$ (by
Proposition 3.8);

  3. (3)

$B_{S}=(A[X_{n}])_{S}[X_{1},\ldots,X_{n-1}]$.

Since $\sigma_{S}\in{\operatorname{\rm GL}}_{r}(B_{S})\cap{\operatorname{\rm
E}}(B_{S})$, by induction hypothesis on $n$, $\sigma_{S}\in{\operatorname{\rm
E}}_{r}(B_{S})$. Then by [28, Corollary 5.7], $\sigma\in{\operatorname{\rm
E}}_{r}(B)$. ∎

######  Theorem 6.7.

Let $R$ be a Noetherian ring of dimension $d$ and ${\mathfrak{a}}$ an ideal in
$R$. Let $A=R[\mathfrak{a}t,t^{2}]$ and $B=A[X_{1},\ldots,X_{n},Y_{1}^{\pm
1},\ldots,Y_{m}^{\pm 1}]$, where $n,m\in{\mathbb{N}}\cup\\{0\\}$. Then, for
$r\geq{\operatorname{\rm max}}\\{3,d+2\\}$, the natural map
$\phi:{\operatorname{\rm GL}}_{r}(B)/{\operatorname{\rm E}}_{r}(B)\rightarrow
K_{1}(B)$ is an isomorphism.

###### Proof.

If we choose the monoid $N={\mathbb{Z}}_{+}^{n}\oplus{\mathbb{Z}}^{m}$, then
surjectivity follows from Theorem 6.5. We shall use induction on $m$ to prove
injectivity. For $m=0$, we are done by Theorem 6.6. So we assume $m\geq 1$ and
let $\sigma\in{\operatorname{\rm GL}}_{r}(B)\cap{\operatorname{\rm E}}(B)$.
Let $S=1+Y_{m}R[Y_{m}]$. Then, it is easy to verify the following

  1. (1)

$\text{\rm dim}(R[Y_{m},Y_{m}^{-1}]_{S})=d$;

  2. (2)

$(A[Y_{m}])_{Y_{m}S}=R[Y_{m}]_{Y_{m}S}[{\mathfrak{a}}[Y_{m}]_{Y_{m}S}t,t^{2}]$
(by Proposition 3.8);

  3. (3)

$B_{S}=(A[Y_{m}])_{Y_{m}S}[X_{1},\ldots,X_{n},Y_{1}^{\pm
1},\ldots,Y_{m-1}^{\pm 1}]$.

Since $\sigma_{S}\in{\operatorname{\rm GL}}_{r}(B_{S})\cap{\operatorname{\rm
E}}(B_{S})$, by induction hypothesis on $m$, $\sigma_{S}\in{\operatorname{\rm
E}}_{r}(B_{S})$. Hence there exists $f\in S$ such that
$\sigma_{f}\in{\operatorname{\rm E}}_{r}(B_{f})$. Let
$B^{\prime\prime}=A[X_{1},\ldots,X_{n},Y_{1}^{\pm 1},\ldots,Y_{m-1}^{\pm 1}]$
and $B^{\prime}=B^{\prime\prime}[Y_{m}]$, then $B^{\prime}_{Y_{m}}=B$ and
$B^{\prime}_{fY_{m}}=B_{f}$. Consider the following Cartesian square of rings

${B^{\prime}}$${B}$${B^{\prime}_{f}}$${B_{f}.}$

By [6, Lemma 3.2], there exists $\sigma_{1}\in{\operatorname{\rm E}}_{r}(B)$
and $\sigma_{2}\in{\operatorname{\rm SL}}_{r}(B^{\prime}_{f})$ such that
$\sigma_{f}=(\sigma_{2})_{Y_{m}}(\sigma_{1})_{f}$. Thus we have
$(\sigma\sigma_{1}^{-1})_{f}=(\sigma_{2})_{Y_{m}}$. Then by Proposition 3.5,
patching of $\sigma\sigma_{1}^{-1}$ and $\sigma_{2}$, we get
$\delta\in{\operatorname{\rm GL}}_{r}(B^{\prime})$ such that
$\delta_{Y_{m}}=\sigma\sigma_{1}^{-1}$ and $\delta_{f}=\sigma_{2}$.

Note that $\delta_{Y_{m}}=\sigma\sigma_{1}^{-1}$ is stably elementary matrix.
Since $\delta\in{\operatorname{\rm GL}}_{r}(B^{\prime\prime}[Y_{m}])$ and
$Y_{m}$ is a monic polynomial in $B^{\prime\prime}[Y_{m}]$, by [28, Corollary
5.7], $\delta$ is stably elementary matrix in $B^{\prime}$. By induction
hypothesis on $m$, $\delta$ is elementary, and hence
$\delta_{Y_{m}}=\sigma\sigma_{1}^{-1}\in{\operatorname{\rm E}}_{r}(B)$. This
implies $\sigma\in{\operatorname{\rm E}}_{r}(B)$. This completes the proof. ∎

Acknowledgements: Chandan Bhaumik would like to thank I.I.T. Kharagpur for the
financial support. Md Abu Raihan would like to thank C.S.I.R. India for the
financial support. Husney Parvez Sarwar would like to thank N.B.H.M., Govt. of
India for the grant No.02011/22/2023NBHM(R.P)/R&D II/,Dt.09-05-2023.

## References

  * [1] H. Bass, K-theory and stable algebra, Publ. Math. Inst. Hautes $\acute{E}$tudes Sci. 22 (1964), 5–60.
  * [2] S.M. Bhatwadekar, H. Lindel and R.A. Rao, The Bass-Murthy question: Serre dimension of Laurent polynomial extensions, Invent. Math. 81 (1985), 189–203.
  * [3] S.M. Bhatwadekar and A. Roy, Stability theorems for overrings of polynomial rings, Invent. math. 68 (1982), 117–127.
  * [4] S.M. Bhatwadekar and A. Roy, Some theorems about projective modules over polynomial rings, J. Algebra 86 (1984), 150–158.
  * [5] C. Bhaumik and H.P. Sarwar, Existence of unimodular element in a projective module over a symbolic Rees algebra, Proc. Indian Acad. Sci. Math. Sci. 133 (2023), no. 2, Paper No. 47.
  * [6] A. Dhorajia and M.K. Keshari, Projective modules over overrings of polynomial rings, J. Algebra 323 (2010), no. 2, 551–559.
  * [7] A. Dhorajia and M.K. Keshari, A note on cancellation of projective modules, J. Pure and Appl. Algebra 216 (2012), 126–129.
  * [8] D. Eisenbud and E.G. Evans, Jr., Generating modules efficiently: theorems from algebraic $K$-theory, J. of Algebra 27 (1973), 278–305.
  * [9] O. Forster, $\ddot{U}ber$ die Anzahl der Erzeugenden eines Ideals in einem Noetherschen ring, Math. Zeit. 84 (1964), 80–87.
  * [10] J. Gubeladze, Anderson’s conjecture and the maximal class of monoid over which projective modules are free, Math. USSR-Sb. 63 (1988), 165–188.
  * [11] J. Gubeladze, The elementary action on unimodular rows over a monoid ring, J. Algebra 148 (1992), 135–161.
  * [12] J. Gubeladze, The elementary action on unimodular rows over a monoid ring II, J. Algebra 155 (1993), 171–194.
  * [13] J. Gubeladze, Unimodular rows over monoid rings, Adv. Math. 337 (2018), 193–215.
  * [14] M.K. Keshari and M.A. Mathew, On Serre dimension of monoid algebras and Segre extensions, J. of Pure and Appl. Algebra 226 (2022), no. 9, Paper No. 107058, 16 pp.
  * [15] M.K. Keshari and H.P. Sarwar, Serre Dimension of Monoid Algebras, Proc. Indian Acad. Sci. Math. Sci. 127 (2017), no. 2, 269-–280.
  * [16] A. Krishna and H.P. Sarwar, $K$-theory of monoid algebras and a question of Gubeladze, J. Inst. Math. Jussieu, 18 (2019), no. 5, 1051–1085.
  * [17] N. Mohan Kumar, On two conjectures about polynomial rings. Invent. Math. 46 (1978), no. 3, 225–236.
  * [18] H. Lindel, Unimodular elements in projective modules, J. Algebra 172 (1995), no. 2, 301–319.
  * [19] S. Mandal, Basic elements and cancellation over Laurent polynomial rings, J. of Algebra 79 (1982), no. 2, 251–257.
  * [20] P. Mantero, J. McCullough and L.E. Miller, Canonical modules and class groups of Rees-like algebras, Michigan Math. J. 73 (2023), no. 3, 571–591.
  * [21] J. McCullough and I. Peeva, Counterexamples to the Eisenbud–Goto regularity conjecture, J. Amer. Math. Soc. 31 (2018), no. 2, 473–496.
  * [22] B. Plumstead. The conjectures of Eisenbud and Evans. Amer. J. of Math. 105 (1983), no. 6, 1417–1433.
  * [23] R.A. Rao, Stability theorems for overrings of polynomial rings. II, J. Algebra 78 (1982), no. 2, 437–444.
  * [24] R.A. Rao and H.P. Sarwar, Stability results for projective modules over Rees algebras. J. of Pure and Appl. Algebra 223 (2019), no. 1, 1–9.
  * [25] H.P. Sarwar, Some results about projective modules over monoid algebras, Comm. Algebra 44 (2016), no. 5, 2256–2263.
  * [26] H.P. Sarwar, $K_{0}$-stability over monoid algebras, Ann. of K-theory 6 (2021), no. 4, 629–649.
  * [27] J.P. Serre, Modules projectifs et spaces fibrés à fibre vectorielle, 1958 Séminaire P. Dubreil, M.-L. Dubreil-Jacotin et C. Pisot, 1957/58, Fasc. 2, Exposé 23 18 pp.
  * [28] A.A. Suslin, On the structure of the special linear group over polynomial rings, Math. USSR-Izv. 11 (1977), 221–238.
  * [29] R.G. Swan, The Number of Generators of a Module, Math. Zeit. 102 (1967), 318–322.
  * [30] G. Valla, Certain graded algebras are always Cohen-Macaulay, J. Algebra 42 (1976), 537–548.
  * [31] L.N. Vaserstein, On the stabilization of the general linear group over a ring, Math. USSR, Sb. 8 (1969), no. 3, 383–400.