Integrality over ideal semifiltrations

# Integrality over ideal semifiltrations

Darij Grinberg

( 111updated and improved version of undergraduate work from 2010)

###### Abstract

Abstract. We study integrality over rings (all commutative in this paper) and
over ideal semifiltrations (a generalization of integrality over ideals). We
begin by reproving classical results, such as a version of the “faithful
module” criterion for integrality over a ring, the transitivity of
integrality, and the theorem that sums and products of integral elements are
again integral. Then, we define the notion of integrality over an ideal
semifiltration (a sequence $\left(I_{0},I_{1},I_{2},\ldots\right)$ of ideals
satisfying $I_{0}=A$ and $I_{a}I_{b}\subseteq I_{a+b}$ for all
$a,b\in\mathbb{N}$), which generalizes both integrality over a ring and
integrality over an ideal (as considered, e.g., in Swanson/Huneke [5]). We
prove a criterion that reduces this general notion to integrality over a ring
using a variant of the Rees algebra. Using this criterion, we study this
notion further and obtain transitivity and closedness under sums and products
for it as well. Finally, we prove the curious fact that if $u$, $x$ and $y$
are three elements of a (commutative) $A$-algebra (for $A$ a ring) such that
$u$ is both integral over $A\left[x\right]$ and integral over
$A\left[y\right]$, then $u$ is integral over $A\left[xy\right]$. We generalize
this to integrality over ideal semifiltrations, too.

###### Contents

  1. 0 Definitions and notations
  2. 1 Integrality over rings
    1. 1.1 The fundamental equivalence
    2. 1.2 Transitivity of integrality
    3. 1.3 Integrality of sums and products
    4. 1.4 Some further consequences
  3. 2 Integrality over ideal semifiltrations
    1. 2.1 Definitions of ideal semifiltrations and integrality over them
    2. 2.2 Polynomial rings and Rees algebras
    3. 2.3 Reduction to integrality over rings
    4. 2.4 Sums and products again
    5. 2.5 Transitivity again
  4. 3 Generalizing to two ideal semifiltrations
    1. 3.1 The product of two ideal semifiltrations
    2. 3.2 Half-reduction
    3. 3.3 Integrality of products over the product semifiltration
  5. 4 Accelerating ideal semifiltrations
    1. 4.1 Definition of $\lambda$-acceleration
    2. 4.2 Half-reduction and reduction
  6. 5 On a lemma by Lombardi
    1. 5.1 A lemma on products of powers
    2. 5.2 Integrality over $A\left[x\right]$ and over $A\left[y\right]$ implies integrality over $A\left[xy\right]$
    3. 5.3 Generalization to ideal semifiltrations
    4. 5.4 Second proof of Corollary 1.20

## Introduction

The purpose of this paper is to state (and prove) some theorems and proofs
related to integrality in commutative algebra in somewhat greater generality
than is common in the literature. I claim no novelty, at least not for the
underlying ideas, but I hope that this paper will be useful as a reference (at
least for myself).

Section 1 (Integrality over rings) mainly consists of known facts (Theorem
1.1, Theorem 1.6, Theorem 1.10) and a generalized exercise from [4] (Corollary
1.20) with a few minor variations (Theorem 1.18 and Corollary 1.22).

Section 2 (Integrality over ideal semifiltrations) merges the concept of
integrality over rings (as considered in Section 1) and integrality over
ideals (a less popular but still highly useful notion; the book [5] is devoted
to it) into one general notion: that of integrality over ideal semifiltrations
(Definition 2.4). This notion is very general, yet it can be reduced to the
basic notion of integrality over rings by a suitable change of base ring
(Theorem 2.14). This reduction allows to extend some standard properties of
integrality over rings to the general case (Theorem 2.18, Theorem 2.20 and
Theorem 2.23).

Section 3 (Generalizing to two ideal semifiltrations) continues Section 2,
adding one more layer of generality. Its main results are a “relative” version
of Theorem 2.14 (Theorem 3.3) and a known fact generalized once more (Theorem
3.6).

Section 4 (Accelerating ideal semifiltrations) generalizes Theorem 3.3 (and
thus also Theorem 2.14) a bit further by considering accelerated ideal
semifiltrations (a generalization of powers of an ideal).

Section 5 (On a lemma by Lombardi) is about an auxiliary result Henri Lombardi
used in [6] to prove Kronecker’s Theorem222Kronecker’s Theorem. Let $B$ be a
ring (“ring” always means “commutative ring with unity” in this paper). Let
$g$ and $h$ be two elements of the polynomial ring $B\left[X\right]$. Let
$g_{\alpha}$ be any coefficient of the polynomial $g$. Let $h_{\beta}$ be any
coefficient of the polynomial $h$. Let $A$ be a subring of $B$ which contains
all coefficients of the polynomial $gh$. Then, the element
$g_{\alpha}h_{\beta}$ of $B$ is integral over the subring $A$.. Here we show a
variant of this result (generalized in one direction, less general in
another).

This paper is supposed to be self-contained (only linear algebra and basic
knowledge about rings, modules, ideals and polynomials is assumed).

All proofs given in this paper are constructive (with the exception of the
proof of Lemma 5.1, which proceeds by contradiction out of sheer convenience;
a constructive version of this proof can be found in [7]).

### Note on the level of detail

This is the short version of this paper. It constitutes an attempt to balance
clarity and brevity in the proofs. See [7] for a more detailed version.

### Note on an old preprint

This is an updated and somewhat generalized version of my preprint “A few
facts on integrality”, which is still available in its old form as well:

  * •

brief version:
https://www.cip.ifi.lmu.de/~grinberg/IntegralityBRIEF.pdf

  * •

long version:
https://www.cip.ifi.lmu.de/~grinberg/Integrality.pdf .

Be warned that said preprint has been written in 2009–2010 when I was an
undergraduate, and suffers from bad writing and formatting.

### Acknowledgments

I thank Irena Swanson and Marco Fontana for enlightening conversations, and
Irena Swanson in particular for making her book [5] freely available (which
helped me discover the subject as an undergraduate).

##  0 Definitions and notations

We begin our study of integrality with some classical definitions and
conventions from commutative algebra:

######  Definition 0.1.

In the following, “ring” will always mean “commutative ring with unity”.
Furthermore, if $A$ is a ring, then “$A$-algebra” shall always mean
“commutative $A$-algebra with unity”. The unity of a ring $A$ will be denoted
by $1_{A}$ or by $1$ if no confusion can arise.

We denote the set $\left\\{0,1,2,\ldots\right\\}$ by $\mathbb{N}$, and the set
$\left\\{1,2,3,\ldots\right\\}$ by $\mathbb{N}^{+}$.

######  Definition 0.2.

Let $A$ be a ring. Let $M$ be an $A$-module.

If $n\in\mathbb{N}$, and if $m_{1},m_{2},\ldots,m_{n}$ are $n$ elements of
$M$, then we define an $A$-submodule $\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}$ of $M$ by

$\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}=\left\\{\sum\limits\limits_{i=1}^{n}a_{i}m_{i}\
\mid\ \left(a_{1},a_{2},\ldots,a_{n}\right)\in A^{n}\right\\}.$

This $A$-submodule $\left\langle m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}$ is
known as the $A$-submodule of $M$ generated by $m_{1},m_{2},\ldots,m_{n}$ (or
as the $A$-linear span of $m_{1},m_{2},\ldots,m_{n}$). It consists of all
$A$-linear combinations of $m_{1},m_{2},\ldots,m_{n}$, and in particular
contains all $n$ elements $m_{1},m_{2},\ldots,m_{n}$. Thus, it satisfies
$\left\\{m_{1},m_{2},\ldots,m_{n}\right\\}\subseteq\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}$.

Also, if $S$ is a finite set, and $m_{s}$ is an element of $M$ for every $s\in
S$, then we define an $A$-submodule $\left\langle m_{s}\ \mid\ s\in
S\right\rangle_{A}$ of $M$ by

$\left\langle m_{s}\ \mid\ s\in
S\right\rangle_{A}=\left\\{\sum\limits\limits_{s\in S}a_{s}m_{s}\ \mid\
\left(a_{s}\right)_{s\in S}\in A^{S}\right\\}.$

This $A$-submodule $\left\langle m_{s}\ \mid\ s\in S\right\rangle_{A}$ is
known as the $A$-submodule of $M$ generated by the family
$\left(m_{s}\right)_{s\in S}$ (or as the $A$-linear span of
$\left(m_{s}\right)_{s\in S}$). It consists of all $A$-linear combinations of
the elements $m_{s}$ with $s\in S$, and in particular contains all these
elements themselves.

Of course, if $m_{1},m_{2},\ldots,m_{n}$ are $n$ elements of $M$, then

$\left\langle m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}=\left\langle m_{s}\
\mid\ s\in\left\\{1,2,\ldots,n\right\\}\right\rangle_{A}.$

Let us observe a trivial fact that we shall use (often tacitly):

######  Lemma 0.3.

Let $A$ be a ring. Let $M$ be an $A$-module. Let $N$ be an $A$-submodule of
$M$. Let $S$ be a finite set; let $m_{s}$ be an element of $N$ for every $s\in
S$. Then, $\left\langle m_{s}\ \mid\ s\in S\right\rangle_{A}\subseteq N$.

######  Definition 0.4.

Let $A$ be a ring, and let $n\in\mathbb{N}$. Let $M$ be an $A$-module. We say
that the $A$-module $M$ is $n$-generated if there exist $n$ elements
$m_{1},m_{2},\ldots,m_{n}$ of $M$ such that $M=\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}$. In other words, the $A$-module $M$
is $n$-generated if and only if there exists a set $S$ and an element $m_{s}$
of $M$ for every $s\in S$ such that $\left|S\right|=n$ and $M=\left\langle
m_{s}\ \mid\ s\in S\right\rangle_{A}$.

######  Definition 0.5.

Let $A$ be a ring. Let $B$ be an $A$-algebra. (Let us recall that both rings
and algebras are always understood to be commutative and unital in this
paper.)

If $u_{1},u_{2},\ldots,u_{n}$ are $n$ elements of $B$, then we define an
$A$-subalgebra $A\left[u_{1},u_{2},\ldots,u_{n}\right]$ of $B$ by

$A\left[u_{1},u_{2},\ldots,u_{n}\right]=\left\\{P\left(u_{1},u_{2},\ldots,u_{n}\right)\
\mid\ P\in A\left[X_{1},X_{2},\ldots,X_{n}\right]\right\\}$

(where $A\left[X_{1},X_{2},\ldots,X_{n}\right]$ denotes the polynomial ring in
$n$ indeterminates $X_{1},X_{2},\ldots,X_{n}$ over $A$).

In particular, if $u$ is an element of $B$, then the $A$-subalgebra
$A\left[u\right]$ of $B$ is defined by

$A\left[u\right]=\left\\{P\left(u\right)\ \mid\ P\in A\left[X\right]\right\\}$

(where $A\left[X\right]$ denotes the polynomial ring in a single indeterminate
$X$ over $A$). Since

$A\left[X\right]=\left\\{\sum\limits\limits_{i=0}^{m}a_{i}X^{i}\ \mid\
m\in\mathbb{N}\text{ and }\left(a_{0},a_{1},\ldots,a_{m}\right)\in
A^{m+1}\right\\},$

this becomes

$\displaystyle A\left[u\right]$
$\displaystyle=\left\\{\left(\sum\limits\limits_{i=0}^{m}a_{i}X^{i}\right)\left(u\right)\
\mid\ m\in\mathbb{N}\text{ and }\left(a_{0},a_{1},\ldots,a_{m}\right)\in
A^{m+1}\right\\}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
\left(\begin{array}[c]{c}\text{where
}\left(\sum\limits\limits_{i=0}^{m}a_{i}X^{i}\right)\left(u\right)\text{ means
the}\\\ \text{polynomial }\sum\limits\limits_{i=0}^{m}a_{i}X^{i}\text{
evaluated at }X=u\end{array}\right)$
$\displaystyle=\left\\{\sum\limits\limits_{i=0}^{m}a_{i}u^{i}\ \mid\
m\in\mathbb{N}\text{ and }\left(a_{0},a_{1},\ldots,a_{m}\right)\in
A^{m+1}\right\\}$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{because
}\left(\sum\limits\limits_{i=0}^{m}a_{i}X^{i}\right)\left(u\right)=\sum\limits\limits_{i=0}^{m}a_{i}u^{i}\right).$

Obviously, $uA\left[u\right]\subseteq A\left[u\right]$ (since
$A\left[u\right]$ is an $A$-algebra and $u\in A\left[u\right]$).

######  Definition 0.6.

Let $B$ be a ring, and let $A$ be a subring of $B$. Then, $B$ canonically
becomes an $A$-algebra. The $A$-module structure of this $A$-algebra $B$ is
given by multiplication inside $B$.

Definition 0.6 shows that theorems about $A$-algebras (for a ring $A$) are
always more general than theorems about rings that contain $A$ as a subring.
Hence, we shall study $A$-algebras in the following, even though most of the
applications of the results we shall see are found at the level of rings
containing $A$.

##  1 Integrality over rings

###  1.1 The fundamental equivalence

Most of the theory of integrality is based upon the following result:

######  Theorem 1.1.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Thus, $B$ is canonically an
$A$-module. Let $n\in\mathbb{N}$. Let $u\in B$. Then, the following four
assertions $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ and $\mathcal{D}$ are
equivalent:

  * •

Assertion $\mathcal{A}$: There exists a monic polynomial $P\in
A\left[X\right]$ with $\deg P=n$ and $P\left(u\right)=0$.

  * •

Assertion $\mathcal{B}$: There exist a $B$-module $C$ and an $n$-generated
$A$-submodule $U$ of $C$ such that $uU\subseteq U$ and such that every $v\in
B$ satisfying $vU=0$ satisfies $v=0$. (Here, $C$ is an $A$-module, since $C$
is a $B$-module and $B$ is an $A$-algebra.)

  * •

Assertion $\mathcal{C}$: There exists an $n$-generated $A$-submodule $U$ of
$B$ such that $1\in U$ and $uU\subseteq U$. (Here and in the following, “$1$”
means “$1_{B}$”, that is, the unity of the ring $B$.)

  * •

Assertion $\mathcal{D}$: We have $A\left[u\right]=\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$.

We shall soon prove this theorem; first, let us explain what it is for:

######  Definition 1.2.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $n\in\mathbb{N}$. Let $u\in
B$. We say that the element $u$ of $B$ is $n$-integral over $A$ if it
satisfies the four equivalent assertions $\mathcal{A}$, $\mathcal{B}$,
$\mathcal{C}$ and $\mathcal{D}$ of Theorem 1.1.

Hence, in particular, the element $u$ of $B$ is $n$-integral over $A$ if and
only if it satisfies the assertion $\mathcal{A}$ of Theorem 1.1. In other
words, $u$ is $n$-integral over $A$ if and only if there exists a monic
polynomial $P\in A\left[X\right]$ with $\deg P=n$ and $P\left(u\right)=0$.

The notion of “$n$-integral” elements that we have just defined is a
refinement of the classical notion of integrality of elements over rings (see,
e.g., [1, Definition (10.21)] or [2, Chapter V, §1.1, Definition 1] or [3,
Definition 8.1.1] for this classical notion, and [5, Definition 2.1.1] for its
particular case when $A$ is a subring of $B$). Indeed, the classical notion
defines an element $u$ of $B$ to be integral over $A$ if and only if (using
the language of our Definition 1.2) there exists some $n\in\mathbb{N}$ such
that $u$ is $n$-integral over $A$. Since I believe the concrete value of $n$
to be worth more than its mere existence, I prefer the specificity of the
“$n$-integral” concept to the slickness of “integral”.

Theorem 1.1 is one of several similar results providing equivalent criteria
for the integrality of an element of an $A$-algebra. See [1, Proposition
(10.23)], [2, Chapter V, Section 1.1, Theorem 1] or [3, Theorem 8.1.6] for
other such results (some very close to Theorem 1.1, and all proven in similar
ways).

Before we prove Theorem 1.1, let us recall a classical property of matrices:

######  Lemma 1.3.

Let $B$ be a ring. Let $n\in\mathbb{N}$. Let $M$ be an $n\times n$-matrix over
$B$. Then,

$\det M\cdot I_{n}=\operatorname*{adj}M\cdot M.$

(Here, $I_{n}$ means the $n\times n$ identity matrix and
$\operatorname*{adj}M$ denotes the adjugate of the matrix $M$. The expressions
“$\det M\cdot I_{n}$” and “$\operatorname*{adj}M\cdot M$” have to be
understood as “$\left(\det M\right)\cdot I_{n}$” and
“$\left(\operatorname*{adj}M\right)\cdot M$”, respectively.)

Lemma 1.3 is well-known (for example, it follows from [8, Theorem 6.100],
applied to $\mathbb{K}=B$ and $A=M$).

######  Proof 1.4 (Proof of Theorem 1.1.).

We will prove the implications $\mathcal{A}\Longrightarrow\mathcal{C}$,
$\mathcal{C}\Longrightarrow\mathcal{B}$,
$\mathcal{B}\Longrightarrow\mathcal{A}$,
$\mathcal{A}\Longrightarrow\mathcal{D}$ and
$\mathcal{D}\Longrightarrow\mathcal{C}$.

Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{C}$. Assume that
Assertion $\mathcal{A}$ holds. Then, there exists a monic polynomial $P\in
A\left[X\right]$ with $\deg P=n$ and $P\left(u\right)=0$. Consider this $P$.
Since $P\in A\left[X\right]$ is a monic polynomial with $\deg P=n$, there
exist elements $a_{0},a_{1},\ldots,a_{n-1}$ of $A$ such that
$P\left(X\right)=X^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}X^{k}$. Consider
these $a_{0},a_{1},\ldots,a_{n-1}$. Substituting $u$ for $X$ in the equality
$P\left(X\right)=X^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}X^{k}$, we find
$P\left(u\right)=u^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}u^{k}$. Hence, the
equality $P\left(u\right)=0$ (which holds by definition of $P$) rewrites as
$u^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}u^{k}=0$. Hence,
$u^{n}=-\sum\limits\limits_{k=0}^{n-1}a_{k}u^{k}$.

Let $U$ be the $A$-submodule $\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$ of $B$. Then, $U=\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$ and

$u^{n}=-\sum\limits\limits_{k=0}^{n-1}a_{k}u^{k}\in\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}=U.$

Moreover, the $n$ elements $u^{0},u^{1},\ldots,u^{n-1}$ belong to $U$ (since
$U=\left\langle u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$). In other words,

$u^{i}\in U\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\left\\{0,1,\ldots,n-1\right\\}.$ (1)

This relation also holds for $i=n$ (since $u^{n}\in U$); thus, it holds for
all $i\in\left\\{0,1,\ldots,n\right\\}$. In other words, we have

$u^{i}\in U\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\left\\{0,1,\ldots,n\right\\}.$ (2)

Applying this to $i=0$, we find $u^{0}\in U$. Thus, $1=u^{0}\in U$.

Recall that $U=\left\langle u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$.
Hence, $U$ is an $n$-generated $A$-module (since $u^{0},u^{1},\ldots,u^{n-1}$
are $n$ elements of $U$).

Now, for each $s\in\left\\{0,1,\ldots,n-1\right\\}$, we have
$s+1\in\left\\{1,2,\ldots,n\right\\}\subseteq\left\\{0,1,\ldots,n\right\\}$
and thus $u^{s+1}\in U$ (by (2), applied to $i=s+1$). Hence, Lemma 0.3
(applied to $M=B$, $N=U$, $S=\left\\{0,1,\ldots,n-1\right\\}$ and
$m_{s}=u^{s+1}$) yields

$\left\langle u^{s+1}\ \mid\
s\in\left\\{0,1,\ldots,n-1\right\\}\right\rangle_{A}\subseteq U.$

Now, from $U=\left\langle u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$, we
obtain

$\displaystyle uU$ $\displaystyle=u\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}=\left\langle u\cdot u^{0},u\cdot
u^{1},\ldots,u\cdot u^{n-1}\right\rangle_{A}$
$\displaystyle=\left\langle\underbrace{u\cdot u^{s}}_{=u^{s+1}}\ \mid\
s\in\left\\{0,1,\ldots,n-1\right\\}\right\rangle_{A}=\left\langle u^{s+1}\
\mid\ s\in\left\\{0,1,\ldots,n-1\right\\}\right\rangle_{A}\subseteq U.$

Thus, we have found an $n$-generated $A$-submodule $U$ of $B$ such that $1\in
U$ and $uU\subseteq U$. Hence, Assertion $\mathcal{C}$ holds. Hence, we have
proved that $\mathcal{A}\Longrightarrow\mathcal{C}$.

Proof of the implication $\mathcal{C}\Longrightarrow\mathcal{B}$. Assume that
Assertion $\mathcal{C}$ holds. Then, there exists an $n$-generated
$A$-submodule $U$ of $B$ such that $1\in U$ and $uU\subseteq U$. Consider this
$U$. Every $v\in B$ satisfying $vU=0$ satisfies $v=0$ (since $1\in U$ and
$vU=0$ yield $v\cdot\underbrace{1}_{\in U}\in vU=0$ and thus $v\cdot 1=0$, so
that $v=0$). Set $C=B$. Then, $C$ is a $B$-module, and $U$ is an $n$-generated
$A$-submodule of $C$ (since $U$ is an $n$-generated $A$-submodule of $B$, and
$C=B$) such that $uU\subseteq U$ and such that every $v\in B$ satisfying
$vU=0$ satisfies $v=0$. Thus, Assertion $\mathcal{B}$ holds. Hence, we have
proved that $\mathcal{C}\Longrightarrow\mathcal{B}$.

Proof of the implication $\mathcal{B}\Longrightarrow\mathcal{A}$. Assume that
Assertion $\mathcal{B}$ holds. Then, there exist a $B$-module $C$ and an
$n$-generated $A$-submodule333where $C$ is an $A$-module, since $C$ is a
$B$-module and $B$ is an $A$-algebra $U$ of $C$ such that $uU\subseteq U$, and
such that every $v\in B$ satisfying $vU=0$ satisfies $v=0$. Consider these $C$
and $U$.

The $A$-module $U$ is $n$-generated. In other words, there exist $n$ elements
$m_{1},m_{2},\ldots,m_{n}$ of $U$ such that $U=\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}$. Consider these
$m_{1},m_{2},\ldots,m_{n}$. For any $k\in\left\\{1,2,\ldots,n\right\\}$, we
have $m_{k}\in U$ (since $U=\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}$) and thus

$um_{k}\in uU\subseteq U=\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A},$

so that there exist $n$ elements $a_{k,1},a_{k,2},\ldots,a_{k,n}$ of $A$ such
that

$um_{k}=\sum\limits\limits_{i=1}^{n}a_{k,i}m_{i}.$ (3)

Consider these $a_{k,1},a_{k,2},\ldots,a_{k,n}$.

The $A$-algebra $B$ gives rise to a canonical ring homomorphism
$\iota:A\rightarrow B$ (sending each $a\in A$ to $a\cdot 1_{B}\in B$). This
ring homomorphism, in turn, induces a ring homomorphism $\iota^{n\times
n}:A^{n\times n}\rightarrow B^{n\times n}$ (which acts on an $n\times
n$-matrix by applying $\iota$ to each entry of the matrix).

We are also going to work with matrices over $U$ (that is, matrices whose
entries lie in $U$). This might sound somewhat strange, because $U$ is not a
ring; however, we can still define matrices over $U$ just as one defines
matrices over any ring. While we cannot multiply two matrices over $U$
(because $U$ is not a ring), we can define the product of a matrix over $A$
with a matrix over $U$ as follows: If $P\in A^{\alpha\times\beta}$ is a matrix
over $A$, and $Q\in U^{\beta\times\gamma}$ is a matrix over $U$ (where
$\alpha,\beta,\gamma\in\mathbb{N}$), then we define the product $PQ\in
U^{\alpha\times\gamma}$ by setting

$\left(PQ\right)_{x,y}=\sum\limits_{z=1}^{\beta}P_{x,z}Q_{z,y}\ \ \ \ \ \ \ \
\ \ \text{for all }x\in\left\\{1,2,\ldots,\alpha\right\\}\text{ and
}y\in\left\\{1,2,\ldots,\gamma\right\\}.$

(Here, for any matrix $T$ and any integers $x$ and $y$, we denote by $T_{x,y}$
the entry of the matrix $T$ in the $x$-th row and the $y$-th column.)

It is easy to see that whenever $P\in A^{\alpha\times\beta}$, $Q\in
A^{\beta\times\gamma}$ and $R\in U^{\gamma\times\delta}$ are three matrices,
then

$\left(PQ\right)R=P\left(QR\right).$ (4)

This is proven in the same way as the fact that the multiplication of matrices
over a ring is associative.

Now define a matrix $V\in U^{n\times 1}$ by setting

$V_{i,1}=m_{i}\ \ \ \ \ \ \ \ \ \ \text{for all
}i\in\left\\{1,2,\ldots,n\right\\}.$

Define another matrix $S\in A^{n\times n}$ by setting

$S_{k,i}=a_{k,i}\ \ \ \ \ \ \ \ \ \ \text{for all
}k\in\left\\{1,2,\ldots,n\right\\}\text{ and
}i\in\left\\{1,2,\ldots,n\right\\}.$

Then, for any $k\in\left\\{1,2,\ldots,n\right\\}$, we have

$\displaystyle\left(uV\right)_{k,1}$
$\displaystyle=u\underbrace{V_{k,1}}_{\begin{subarray}{c}=m_{k}\\\ \text{(by
the definition of }V\text{)}\end{subarray}}=um_{k}\ \ \ \ \ \ \ \ \ \
\text{and}$ $\displaystyle\left(SV\right)_{k,1}$
$\displaystyle=\sum\limits\limits_{i=1}^{n}\underbrace{S_{k,i}}_{\begin{subarray}{c}=a_{k,i}\\\
\text{(by the definition of
}S\text{)}\end{subarray}}\underbrace{V_{i,1}}_{\begin{subarray}{c}=m_{i}\\\
\text{(by the definition of
}V\text{)}\end{subarray}}=\sum\limits\limits_{i=1}^{n}a_{k,i}m_{i}.$

Hence, the equality (3) rewrites as
$\left(uV\right)_{k,1}=\left(SV\right)_{k,1}$. Since this holds for every
$k\in\left\\{1,2,\ldots,n\right\\}$, we conclude that $uV=SV$. Thus,

$0=uV-SV=uI_{n}V-SV=\left(uI_{n}-S\right)V.$ (5)

Here, the “$S$” in “$uI_{n}-S$” means not the matrix $S\in A^{n\times n}$
itself, but rather its image under the ring homomorphism $\iota^{n\times
n}:A^{n\times n}\rightarrow B^{n\times n}$; thus, the matrix $uI_{n}-S$ is a
well-defined matrix in $B^{n\times n}$.

Now, let $P\in A\left[X\right]$ be the characteristic polynomial of the matrix
$S\in A^{n\times n}$. Then, $P$ is monic, and $\deg P=n$. Besides,
$P\left(X\right)=\det\left(XI_{n}-S\right)$, so that
$P\left(u\right)=\det\left(uI_{n}-S\right)$. Thus,

$\displaystyle P\left(u\right)\cdot V$
$\displaystyle=\det\left(uI_{n}-S\right)\cdot
V=\underbrace{\det\left(uI_{n}-S\right)I_{n}}_{\begin{subarray}{c}=\operatorname{adj}\left(uI_{n}-S\right)\cdot\left(uI_{n}-S\right)\\\
\text{(by Lemma \ref{lem.adj},}\\\ \text{applied to
}M=uI_{n}-S\text{)}\end{subarray}}\cdot V$
$\displaystyle=\left(\operatorname{adj}\left(uI_{n}-S\right)\cdot\left(uI_{n}-S\right)\right)\cdot
V$
$\displaystyle=\operatorname{adj}\left(uI_{n}-S\right)\cdot\underbrace{\left(\left(uI_{n}-S\right)V\right)}_{\begin{subarray}{c}=0\\\
\text{(by \eqref{pf.Theorem1.BA.short.matprod=0})}\end{subarray}}\ \ \ \ \ \ \
\ \ \ \left(\text{by \eqref{(PQ)R=P(QR)}}\right)$ $\displaystyle=0.$

Since the entries of the matrix $V$ are $m_{1},m_{2},\ldots,m_{n}$, this
yields $P\left(u\right)\cdot m_{k}=0$ for every
$k\in\left\\{1,2,\ldots,n\right\\}$. Now, from $U=\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}$, we obtain

$\displaystyle P\left(u\right)\cdot U$
$\displaystyle=P\left(u\right)\cdot\left\langle
m_{1},m_{2},\ldots,m_{n}\right\rangle_{A}=\left\langle P\left(u\right)\cdot
m_{1},P\left(u\right)\cdot m_{2},\ldots,P\left(u\right)\cdot
m_{n}\right\rangle_{A}$ $\displaystyle=\left\langle
0,0,\ldots,0\right\rangle_{A}\ \ \ \ \ \ \ \ \ \ \left(\text{since
}P\left(u\right)\cdot m_{k}=0\text{ for any
}k\in\left\\{1,2,\ldots,n\right\\}\right)$ $\displaystyle=0.$

This implies $P\left(u\right)=0$ (since every $v\in B$ satisfying $vU=0$
satisfies $v=0$). Thus, Assertion $\mathcal{A}$ holds. Hence, we have proved
that $\mathcal{B}\Longrightarrow\mathcal{A}$.

Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{D}$. Assume that
Assertion $\mathcal{A}$ holds. Then, there exists a monic polynomial $P\in
A\left[X\right]$ with $\deg P=n$ and $P\left(u\right)=0$. Consider this $P$.

Let $U$ be the $A$-submodule $\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$ of $B$. As in the Proof of the
implication $\mathcal{A}\Longrightarrow\mathcal{C}$, we can show that $U$ is
an $n$-generated $A$-module, and that $1\in U$ and $uU\subseteq U$.

Now, it is easy to show that

$u^{i}\in U\ \ \ \ \ \ \ \ \ \ \text{for any }i\in\mathbb{N}.$ (6)

[Indeed, this follows easily from $uU\subseteq U$ and $1\in U$ by induction on
$i$.]

Hence, for any $m\in\mathbb{N}$ and any
$\left(a_{0},a_{1},\ldots,a_{m}\right)\in A^{m+1}$, we have
$\sum\limits\limits_{i=0}^{m}a_{i}u^{i}\in U$ (since $U$ is an $A$-module, and
thus is closed under $A$-linear combination).

Now, the definition of $A\left[u\right]$ yields

$\displaystyle A\left[u\right]$
$\displaystyle=\left\\{\sum\limits\limits_{i=0}^{m}a_{i}u^{i}\ \mid\
m\in\mathbb{N}\text{ and }\left(a_{0},a_{1},\ldots,a_{m}\right)\in
A^{m+1}\right\\}$ $\displaystyle\subseteq U\ \ \ \ \ \ \ \ \ \
\left(\text{since }\sum\limits\limits_{i=0}^{m}a_{i}u^{i}\in U\text{ for any
}m\in\mathbb{N}\text{ and }\left(a_{0},a_{1},\ldots,a_{m}\right)\in
A^{m+1}\right)$ $\displaystyle=\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}.$

Combining this with the (obvious) relation $\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}\subseteq A\left[u\right]$, we find
$A\left[u\right]=\left\langle u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$.

Thus, Assertion $\mathcal{D}$ holds. Hence, we have proved that
$\mathcal{A}\Longrightarrow\mathcal{D}$.

Proof of the implication $\mathcal{D}\Longrightarrow\mathcal{C}$. Assume that
Assertion $\mathcal{D}$ holds. Then, $A\left[u\right]=\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$.

Let $U$ be the $A$-submodule $A\left[u\right]$ of $B$. Then, $U$ is an
$n$-generated $A$-module (since $U=A\left[u\right]=\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}$). Besides, $1=u^{0}\in
A\left[u\right]=U$. Finally, $U=A\left[u\right]$ yields $uU\subseteq U$. Thus,
Assertion $\mathcal{C}$ holds. Hence, we have proved that
$\mathcal{D}\Longrightarrow\mathcal{C}$.

Now, we have proved the implications $\mathcal{A}\Longrightarrow\mathcal{D}$,
$\mathcal{D}\Longrightarrow\mathcal{C}$,
$\mathcal{C}\Longrightarrow\mathcal{B}$ and
$\mathcal{B}\Longrightarrow\mathcal{A}$ above. Thus, all four assertions
$\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ and $\mathcal{D}$ are equivalent,
and Theorem 1.1 is proven.

For the sake of completeness (and as a very easy exercise), let us state a
basic property of integrality that we will not ever use:

######  Proposition 1.5.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $u\in B$. Let
$q\in\mathbb{N}$ and $p\in\mathbb{N}$ be such that $p\geq q$. Assume that $u$
is $q$-integral over $A$. Then, $u$ is $p$-integral over $A$.

###  1.2 Transitivity of integrality

Let us now prove the first and probably most important consequence of Theorem
1.1:

######  Theorem 1.6.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $v\in B$ and $u\in B$. Let
$m\in\mathbb{N}$ and $n\in\mathbb{N}$. Assume that $v$ is $m$-integral over
$A$, and that $u$ is $n$-integral over $A\left[v\right]$. Then, $u$ is
$nm$-integral over $A$.

(Here, we are using the fact that if $A$ is a ring, and if $v$ is an element
of an $A$-algebra $B$, then $A\left[v\right]$ is a subring of $B$, and
therefore $B$ is an $A\left[v\right]$-algebra.)

######  Proof 1.7 (Proof of Theorem 1.6.).

Since $v$ is $m$-integral over $A$, we have $A\left[v\right]=\left\langle
v^{0},v^{1},\ldots,v^{m-1}\right\rangle_{A}$ (this is the Assertion
$\mathcal{D}$ of Theorem 1.1, stated for $v$ and $m$ in lieu of $u$ and $n$).

Since $u$ is $n$-integral over $A\left[v\right]$, we have
$\left(A\left[v\right]\right)\left[u\right]=\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A\left[v\right]}$ (this is the
Assertion $\mathcal{D}$ of Theorem 1.1, stated for $A\left[v\right]$ in lieu
of $A$).

Let $S=\left\\{0,1,\ldots,n-1\right\\}\times\left\\{0,1,\ldots,m-1\right\\}$.
Then, $S$ is a finite set with size $\left|S\right|=nm$.

Let $x\in\left(A\left[v\right]\right)\left[u\right]$. Then, there exist $n$
elements $b_{0},b_{1},\ldots,b_{n-1}$ of $A\left[v\right]$ such that
$x=\sum\limits\limits_{i=0}^{n-1}b_{i}u^{i}$ (since
$x\in\left(A\left[v\right]\right)\left[u\right]=\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A\left[v\right]}$). Consider these
$b_{0},b_{1},\ldots,b_{n-1}$.

For each $i\in\left\\{0,1,\ldots,n-1\right\\}$, there exist $m$ elements
$a_{i,0},a_{i,1},\ldots,a_{i,m-1}$ of $A$ such that
$b_{i}=\sum\limits\limits_{j=0}^{m-1}a_{i,j}v^{j}$ (because $b_{i}\in
A\left[v\right]=\left\langle v^{0},v^{1},\ldots,v^{m-1}\right\rangle_{A}$).
Consider these $a_{i,0},a_{i,1},\ldots,a_{i,m-1}$. Thus,

$\displaystyle x$
$\displaystyle=\sum\limits\limits_{i=0}^{n-1}\underbrace{b_{i}}_{=\sum\limits\limits_{j=0}^{m-1}a_{i,j}v^{j}}u^{i}=\sum\limits\limits_{i=0}^{n-1}\sum\limits\limits_{j=0}^{m-1}a_{i,j}v^{j}u^{i}=\sum\limits\limits_{\left(i,j\right)\in\left\\{0,1,\ldots,n-1\right\\}\times\left\\{0,1,\ldots,m-1\right\\}}a_{i,j}v^{j}u^{i}$
$\displaystyle=\sum\limits\limits_{\left(i,j\right)\in S}a_{i,j}v^{j}u^{i}\ \
\ \ \ \ \ \ \ \ \left(\text{since
}\left\\{0,1,\ldots,n-1\right\\}\times\left\\{0,1,\ldots,m-1\right\\}=S\right)$
$\displaystyle\in\left\langle v^{j}u^{i}\ \mid\ \left(i,j\right)\in
S\right\rangle_{A}\ \ \ \ \ \ \ \ \ \ \left(\text{since }a_{i,j}\in A\text{
for every }\left(i,j\right)\in S\right).$

So we have proved that $x\in\left\langle v^{j}u^{i}\ \mid\ \left(i,j\right)\in
S\right\rangle_{A}$ for every
$x\in\left(A\left[v\right]\right)\left[u\right]$. In other words,
$\left(A\left[v\right]\right)\left[u\right]\subseteq\left\langle v^{j}u^{i}\
\mid\ \left(i,j\right)\in S\right\rangle_{A}$. Conversely, $\left\langle
v^{j}u^{i}\ \mid\ \left(i,j\right)\in
S\right\rangle_{A}\subseteq\left(A\left[v\right]\right)\left[u\right]$ (this
is trivial). Combining these two relations, we find
$\left(A\left[v\right]\right)\left[u\right]=\left\langle v^{j}u^{i}\ \mid\
\left(i,j\right)\in S\right\rangle_{A}$. Thus, the $A$-module
$\left(A\left[v\right]\right)\left[u\right]$ is $nm$-generated (since
$\left|S\right|=nm$).

Let $U=\left(A\left[v\right]\right)\left[u\right]$. Then, the $A$-module $U$
is $nm$-generated. Besides, $U$ is an $A$-submodule of $B$, and we have $1\in
U$ and $uU\subseteq U$.

Thus, the element $u$ of $B$ satisfies the Assertion $\mathcal{C}$ of Theorem
1.1 with $n$ replaced by $nm$. Hence, $u\in B$ satisfies the four equivalent
assertions $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ and $\mathcal{D}$ of
Theorem 1.1, all with $n$ replaced by $nm$. Thus, $u$ is $nm$-integral over
$A$. This proves Theorem 1.6.

###  1.3 Integrality of sums and products

Before the next significant consequence of Theorem 1.1, let us show an
essentially trivial fact:

######  Theorem 1.8.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $a\in A$. Then, $a\cdot
1_{B}\in B$ is $1$-integral over $A$.

######  Proof 1.9 (Proof of Theorem 1.8.).

The polynomial $X-a\in A\left[X\right]$ is monic and satisfies
$\deg\left(X-a\right)=1$; moreover, evaluating this polynomial at $a\cdot
1_{B}\in B$ yields $a\cdot 1_{B}-a\cdot 1_{B}=0$. Hence, there exists a monic
polynomial $P\in A\left[X\right]$ with $\deg P=1$ and $P\left(a\cdot
1_{B}\right)=0$ (namely, the polynomial $P\in A\left[X\right]$ defined by
$P\left(X\right)=X-a$). Thus, $a\cdot 1_{B}$ is $1$-integral over $A$. This
proves Theorem 1.8.

The following theorem is a standard result, generalizing (for example) the
classical fact that sums and products of algebraic integers are again
algebraic integers:

######  Theorem 1.10.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $x\in B$ and $y\in B$. Let
$m\in\mathbb{N}$ and $n\in\mathbb{N}$. Assume that $x$ is $m$-integral over
$A$, and that $y$ is $n$-integral over $A$.

(a) Then, $x+y$ is $nm$-integral over $A$.

(b) Furthermore, $xy$ is $nm$-integral over $A$.

Our proof of this theorem will rely on a simple lemma:

######  Lemma 1.11.

Let $A$ be a ring. Let $C$ be an $A$-algebra. Let $x\in C$.

Let $n\in\mathbb{N}$. Let $P\in A\left[X\right]$ be a monic polynomial with
$\deg P=n$. Define a polynomial $Q\in C\left[X\right]$ by
$Q\left(X\right)=P\left(X-x\right)$. Then, $Q$ is a monic polynomial with
$\deg Q=n$.

######  Proof 1.12 (Proof of Lemma 1.11.).

Recall that $P$ is a monic polynomial with $\deg P=n$; hence, we can write $P$
in the form

$P=X^{n}+\sum\limits_{i=0}^{n-1}a_{i}X^{i}$ (7)

for some $a_{0},a_{1},\ldots,a_{n-1}\in A$. Consider these
$a_{0},a_{1},\ldots,a_{n-1}$.

Now,

$\displaystyle Q\left(X\right)$
$\displaystyle=P\left(X-x\right)=\underbrace{\left(X-x\right)^{n}}_{=X^{n}+\left(\text{lower
order
terms}\right)}+\underbrace{\sum\limits_{i=0}^{n-1}a_{i}\left(X-x\right)^{i}}_{=\left(\text{lower
order terms}\right)}$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{here, we
have substituted }X-x\text{ for }X\text{ in \eqref{pf.Theorem5.b.1}}\right)$
$\displaystyle=X^{n}+\left(\text{lower order terms}\right),$

where “lower order terms” means a sum of terms of the form $bX^{i}$ with $b\in
C$ and $i<n$. Hence, $Q$ is a monic polynomial with $\deg Q=n$. This proves
Lemma 1.11.

######  Proof 1.13 (Proof of Theorem 1.10.).

Since $y$ is $n$-integral over $A$, there exists a monic polynomial $P\in
A\left[X\right]$ with $\deg P=n$ and $P\left(y\right)=0$. Consider this $P$.

(a) Let $C$ be the $A$-subalgebra $A\left[x\right]$ of $B$. Then,
$C=A\left[x\right]$, so that $x\in A\left[x\right]=C$.

Now, define a polynomial $Q\in C\left[X\right]$ by
$Q\left(X\right)=P\left(X-x\right)$. Then, Lemma 1.11 shows that $Q$ is a
monic polynomial with $\deg Q=n$. Also, substituting $x+y$ for $X$ in the
equality $Q\left(X\right)=P\left(X-x\right)$, we obtain
$Q\left(x+y\right)=P\left(\underbrace{\left(x+y\right)-x}_{=y}\right)=P\left(y\right)=0$.

Hence, there exists a monic polynomial $Q\in C\left[X\right]$ with $\deg Q=n$
and $Q\left(x+y\right)=0$. Thus, $x+y$ is $n$-integral over $C$. In other
words, $x+y$ is $n$-integral over $A\left[x\right]$ (since
$C=A\left[x\right]$). Thus, Theorem 1.6 (applied to $v=x$ and $u=x+y$) yields
that $x+y$ is $nm$-integral over $A$. This proves Theorem 1.10 (a).

(b) Recall that $P\in A\left[X\right]$ is a monic polynomial with $\deg P=n$.
Thus, there exist elements $a_{0},a_{1},\ldots,a_{n-1}$ of $A$ such that
$P\left(X\right)=X^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}X^{k}$. Consider
these $a_{0},a_{1},\ldots,a_{n-1}$. Substituting $y$ for $X$ in
$P\left(X\right)=X^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}X^{k}$, we find
$P\left(y\right)=y^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}y^{k}$. Thus,

$y^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}y^{k}=P\left(y\right)=0.$ (8)

Now, define a polynomial $Q\in\left(A\left[x\right]\right)\left[X\right]$ by
$Q\left(X\right)=X^{n}+\sum\limits\limits_{k=0}^{n-1}x^{n-k}a_{k}X^{k}$. Then,

$\displaystyle Q\left(xy\right)$
$\displaystyle=\underbrace{\left(xy\right)^{n}}_{=x^{n}y^{n}}+\sum\limits\limits_{k=0}^{n-1}x^{n-k}\underbrace{a_{k}\left(xy\right)^{k}}_{\begin{subarray}{c}=a_{k}x^{k}y^{k}\\\
=x^{k}a_{k}y^{k}\end{subarray}}=x^{n}y^{n}+\sum\limits\limits_{k=0}^{n-1}\underbrace{x^{n-k}x^{k}}_{=x^{n}}a_{k}y^{k}$
$\displaystyle=x^{n}y^{n}+\sum\limits\limits_{k=0}^{n-1}x^{n}a_{k}y^{k}=x^{n}\underbrace{\left(y^{n}+\sum\limits\limits_{k=0}^{n-1}a_{k}y^{k}\right)}_{\begin{subarray}{c}=0\\\
\text{(by \eqref{pf.Theorem5.c.1})}\end{subarray}}=0.$

Also, recall that
$Q\left(X\right)=X^{n}+\sum\limits\limits_{k=0}^{n-1}x^{n-k}a_{k}X^{k}$;
hence, the polynomial $Q\in\left(A\left[x\right]\right)\left[X\right]$ is
monic and $\deg Q=n$. Thus, there exists a monic polynomial
$Q\in\left(A\left[x\right]\right)\left[X\right]$ with $\deg Q=n$ and
$Q\left(xy\right)=0$. Thus, $xy$ is $n$-integral over $A\left[x\right]$.
Hence, Theorem 1.6 (applied to $v=x$ and $u=xy$) yields that $xy$ is
$nm$-integral over $A$. This proves Theorem 1.10 (b).

######  Corollary 1.14.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $x\in B$. Let
$m\in\mathbb{N}$. Assume that $x$ is $m$-integral over $A$. Then, $-x$ is
$m$-integral over $A$.

######  Proof 1.15 (Proof of Corollary 1.14.).

This is easy to prove directly (using Assertion $\mathcal{A}$ of Theorem 1.1),
but the slickest proof is using Theorem 1.10 (b): The element
$\left(-1\right)\cdot 1_{B}\in B$ is $1$-integral over $A$ (by Theorem 1.8,
applied to $a=-1$). Thus, $x\cdot\left(\left(-1\right)\cdot 1_{B}\right)$ is
$1m$-integral over $A$ (by Theorem 1.10 (b), applied to
$y=\left(-1\right)\cdot 1_{B}$ and $n=1$). In other words, $-x$ is
$m$-integral over $A$ (since $x\cdot\underbrace{\left(\left(-1\right)\cdot
1_{B}\right)}_{=-1_{B}}=x\cdot\left(-1_{B}\right)=-x\cdot 1_{B}=-x$ and
$1m=m$). This proves Corollary 1.14.

######  Corollary 1.16.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $x\in B$ and $y\in B$. Let
$m\in\mathbb{N}$ and $n\in\mathbb{N}$. Assume that $x$ is $m$-integral over
$A$, and that $y$ is $n$-integral over $A$. Then, $x-y$ is $nm$-integral over
$A$.

######  Proof 1.17 (Proof of Corollary 1.16.).

We know that $y$ is $n$-integral over $A$. Hence, Corollary 1.14 (applied to
$y$ and $n$ instead of $x$ and $m$) shows that $-y$ is $n$-integral over $A$.
Thus, Theorem 1.10 (a) (applied to $-y$ instead of $y$) shows that
$x+\left(-y\right)$ is $nm$-integral over $A$. In other words, $x-y$ is
$nm$-integral over $A$ (since $x+\left(-y\right)=x-y$). This proves Corollary
1.16.

###  1.4 Some further consequences

######  Theorem 1.18.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $n\in\mathbb{N}^{+}$. Let
$v\in B$. Let $a_{0},a_{1},\ldots,a_{n}$ be $n+1$ elements of $A$ such that
$\sum\limits\limits_{i=0}^{n}a_{i}v^{i}=0$. Let
$k\in\left\\{0,1,\ldots,n\right\\}$. Then,
$\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$ is $n$-integral over $A$.

######  Proof 1.19 (Proof of Theorem 1.18.).

Let $u=\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$. Then,

$\displaystyle 0$
$\displaystyle=\sum\limits\limits_{i=0}^{n}a_{i}v^{i}=\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i}+\sum\limits\limits_{i=k}^{n}a_{i}v^{i}=\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i}+\sum\limits\limits_{i=0}^{n-k}a_{i+k}\underbrace{v^{i+k}}_{=v^{i}v^{k}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{here, we substituted }i+k\text{
for }i\text{\ in the second sum}\right)$
$\displaystyle=\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i}+\underbrace{\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}v^{k}}_{=v^{k}\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}}=\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i}+v^{k}\underbrace{\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}}_{=u}=\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i}+v^{k}u,$

so that

$v^{k}u=-\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i}.$

Let $U$ be the $A$-submodule $\left\langle
v^{0},v^{1},\ldots,v^{n-1}\right\rangle_{A}$ of $B$. Thus, $U$ is an
$n$-generated $A$-module, and $1=v^{0}\in U$.

Now, we are going to show that

$uv^{s}\in U\ \ \ \ \ \ \ \ \ \ \text{for any
}s\in\left\\{0,1,\ldots,n-1\right\\}.$ (9)

[Proof of (9). Let $s\in\left\\{0,1,\ldots,n-1\right\\}$. Thus, we have either
$s<k$ or $s\geq k$. In the case $s<k$, the relation (9) follows from

$\displaystyle uv^{s}$
$\displaystyle=\sum\limits\limits_{i=0}^{n-k}a_{i+k}\underbrace{v^{i}\cdot
v^{s}}_{=v^{i+s}}\ \ \ \ \ \ \ \ \ \ \left(\text{since
}u=\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}\right)$
$\displaystyle=\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i+s}\in U$

(since every $i\in\left\\{0,1,\ldots,n-k\right\\}$ satisfies
$i+s\in\left\\{0,1,\ldots,n-1\right\\}$ 444Here we are using $s<k$., and thus
$\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i+s}\in\left\langle
v^{0},v^{1},\ldots,v^{n-1}\right\rangle_{A}=U$). In the case $s\geq k$, the
relation (9) follows from

$\displaystyle uv^{s}$
$\displaystyle=u\underbrace{v^{k+\left(s-k\right)}}_{=v^{k}v^{s-k}}=v^{k}u\cdot
v^{s-k}=-\sum\limits\limits_{i=0}^{k-1}a_{i}\underbrace{v^{i}\cdot
v^{s-k}}_{=v^{i+\left(s-k\right)}}\ \ \ \ \ \ \ \ \ \ \left(\text{since
}v^{k}u=-\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i}\right)$
$\displaystyle=-\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i+\left(s-k\right)}\in
U$

(since every $i\in\left\\{0,1,\ldots,k-1\right\\}$ satisfies
$i+\left(s-k\right)\in\left\\{0,1,\ldots,n-1\right\\}$ 555Here we are using
$s\geq k$ and $s\in\left\\{0,1,\ldots,n-1\right\\}$., and thus
$-\sum\limits\limits_{i=0}^{k-1}a_{i}v^{i+\left(s-k\right)}\in\left\langle
v^{0},v^{1},\ldots,v^{n-1}\right\rangle_{A}=U$). Hence, (9) is proven in both
possible cases, and thus the proof of (9) is complete.]

Now, from $U=\left\langle v^{0},v^{1},\ldots,v^{n-1}\right\rangle_{A}$, we
obtain

$\displaystyle uU$ $\displaystyle=u\left\langle
v^{0},v^{1},\ldots,v^{n-1}\right\rangle_{A}=\left\langle
uv^{0},uv^{1},\ldots,uv^{n-1}\right\rangle_{A}$ $\displaystyle=\left\langle
uv^{s}\ \mid\ s\in\left\\{0,1,\ldots,n-1\right\\}\right\rangle_{A}\subseteq U\
\ \ \ \ \ \ \ \ \ \left(\text{by \eqref{3} and Lemma \ref{lem.mil}}\right).$

Altogether, $U$ is an $n$-generated $A$-submodule of $B$ such that $1\in U$
and $uU\subseteq U$. Thus, $u\in B$ satisfies Assertion $\mathcal{C}$ of
Theorem 1.1. Hence, $u\in B$ satisfies the four equivalent assertions
$\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ and $\mathcal{D}$ of Theorem 1.1.
Consequently, $u$ is $n$-integral over $A$. Since
$u=\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$, this means that
$\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$ is $n$-integral over $A$. This
proves Theorem 1.18.

######  Corollary 1.20.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $\alpha\in\mathbb{N}$ and
$\beta\in\mathbb{N}$ be such that $\alpha+\beta\in\mathbb{N}^{+}$. Let $u\in
B$ and $v\in B$. Let $s_{0},s_{1},\ldots,s_{\alpha}$ be $\alpha+1$ elements of
$A$ such that $\sum\limits\limits_{i=0}^{\alpha}s_{i}v^{i}=u$. Let
$t_{0},t_{1},\ldots,t_{\beta}$ be $\beta+1$ elements of $A$ such that
$\sum\limits\limits_{i=0}^{\beta}t_{i}v^{\beta-i}=uv^{\beta}$. Then, $u$ is
$\left(\alpha+\beta\right)$-integral over $A$.

(This Corollary 1.20 generalizes [4, Exercise 2-5], which says that if $v$ is
an invertible element of an $A$-algebra $B$, then every element $u\in
A\left[v\right]\cap A\left[v^{-1}\right]$ is integral over $A$. To see how
this follows from Corollary 1.20, just pick $\alpha\in\mathbb{N}^{+}$ and
$\beta\in\mathbb{N}^{+}$ and $s_{0},s_{1},\ldots,s_{\alpha}\in A$ and
$t_{0},t_{1},\ldots,t_{\beta}\in A$ such that
$\sum\limits\limits_{i=0}^{\alpha}s_{i}v^{i}=u$ and
$\sum\limits\limits_{i=0}^{\beta}t_{i}\left(v^{-1}\right)^{i}=u$.)

######  Proof 1.21 (First proof of Corollary 1.20.).

Let $k=\beta$ and $n=\alpha+\beta$. Then, $k\in\left\\{0,1,\ldots,n\right\\}$
(since $\alpha\in\mathbb{N}$ and $\beta\in\mathbb{N}$) and
$n=\alpha+\beta\in\mathbb{N}^{+}$ and $n-\beta=\alpha$ (since
$n=\alpha+\beta$). Define $n+1$ elements $a_{0},a_{1},\ldots,a_{n}$ of $A$ by

$a_{i}=\begin{cases}t_{\beta-i},&\text{ if }i<\beta;\\\ t_{0}-s_{0},&\text{ if
}i=\beta;\\\ -s_{i-\beta},&\text{ if }i>\beta\end{cases}\ \ \ \ \ \ \ \ \ \
\text{for every }i\in\left\\{0,1,\ldots,n\right\\}.$

Then, from $n=\alpha+\beta$, we obtain

$\displaystyle\sum\limits\limits_{i=0}^{n}a_{i}v^{i}$
$\displaystyle=\sum\limits\limits_{i=0}^{\alpha+\beta}a_{i}v^{i}=\sum\limits\limits_{i=0}^{\beta-1}\underbrace{a_{i}}_{=t_{\beta-i}}v^{i}+\underbrace{a_{\beta}}_{=t_{0}-s_{0}}v^{\beta}+\sum\limits\limits_{i=\beta+1}^{\alpha+\beta}\underbrace{a_{i}}_{\begin{subarray}{c}=-s_{i-\beta}\end{subarray}}v^{i}$
$\displaystyle=\underbrace{\sum\limits\limits_{i=0}^{\beta-1}t_{\beta-i}v^{i}}_{\begin{subarray}{c}=\sum\limits\limits_{i=1}^{\beta}t_{i}v^{\beta-i}\\\
\text{(here, we have substituted }i\\\ \text{for }\beta-i\text{ in the
sum)}\end{subarray}}+\underbrace{\left(t_{0}-s_{0}\right)v^{\beta}}_{=t_{0}v^{\beta}-s_{0}v^{\beta}}+\underbrace{\sum\limits\limits_{i=\beta+1}^{\alpha+\beta}\left(-s_{i-\beta}\right)v^{i}}_{\begin{subarray}{c}=-\sum\limits\limits_{i=\beta+1}^{\alpha+\beta}s_{i-\beta}v^{i}=-\sum\limits\limits_{i=1}^{\alpha}s_{i}v^{i+\beta}\\\
\text{(here, we have substituted }i\\\ \text{for }i-\beta\text{ in the
sum)}\end{subarray}}$
$\displaystyle=\sum\limits\limits_{i=1}^{\beta}t_{i}v^{\beta-i}+t_{0}v^{\beta}-s_{0}v^{\beta}-\sum\limits\limits_{i=1}^{\alpha}s_{i}v^{i+\beta}$
$\displaystyle=\underbrace{\sum\limits\limits_{i=1}^{\beta}t_{i}v^{\beta-i}+t_{0}v^{\beta}}_{=\sum\limits\limits_{i=0}^{\beta}t_{i}v^{\beta-i}=uv^{\beta}}-\underbrace{\left(s_{0}v^{\beta}+\sum\limits\limits_{i=1}^{\alpha}s_{i}v^{i+\beta}\right)}_{\begin{subarray}{c}=\sum\limits\limits_{i=0}^{\alpha}s_{i}v^{i+\beta}=\left(\sum\limits\limits_{i=0}^{\alpha}s_{i}v^{i}\right)v^{\beta}=uv^{\beta}\\\
\text{(since
}\sum\limits\limits_{i=0}^{\alpha}s_{i}v^{i}=u\text{)}\end{subarray}}=uv^{\beta}-uv^{\beta}=0.$

Thus, Theorem 1.18 yields that $\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$ is
$n$-integral over $A$.

But $k=\beta$ and thus

$\displaystyle\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$
$\displaystyle=\sum\limits\limits_{i=0}^{n-\beta}a_{i+\beta}v^{i}=\sum\limits\limits_{i=\beta}^{n}a_{i}v^{i-\beta}\
\ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we have substituted
}i-\beta\\\ \text{for }i\text{ in the sum}\end{array}\right)$
$\displaystyle=\sum\limits\limits_{i=\beta}^{\beta}\underbrace{a_{i}}_{\begin{subarray}{c}=t_{0}-s_{0}\\\
\text{(by the}\\\ \text{definition of }a_{i}\text{,}\\\ \text{since
}i=\beta\text{)}\end{subarray}}v^{i-\beta}+\sum\limits\limits_{i=\beta+1}^{n}\underbrace{a_{i}}_{\begin{subarray}{c}=-s_{i-\beta}\\\
\text{(by the}\\\ \text{definition of }a_{i}\text{,}\\\ \text{since
}i>\beta\text{)}\end{subarray}}v^{i-\beta}$
$\displaystyle=\underbrace{\sum\limits\limits_{i=\beta}^{\beta}\left(t_{0}-s_{0}\right)v^{i-\beta}}_{\begin{subarray}{c}{}_{\begin{subarray}{c}=\left(t_{0}-s_{0}\right)v^{\beta-\beta}\end{subarray}}\\\
=\left(t_{0}-s_{0}\right)v^{0}\\\
=t_{0}v^{0}-s_{0}v^{0}\end{subarray}}+\underbrace{\sum\limits\limits_{i=\beta+1}^{n}\left(-s_{i-\beta}\right)v^{i-\beta}}_{\begin{subarray}{c}=-\sum\limits\limits_{i=\beta+1}^{n}s_{i-\beta}v^{i-\beta}=-\sum\limits\limits_{i=1}^{n-\beta}s_{i}v^{i}\\\
\text{(here, we have substituted }i\\\ \text{for }i-\beta\text{ in the
sum)}\end{subarray}}$
$\displaystyle=t_{0}v^{0}-s_{0}v^{0}-\sum\limits\limits_{i=1}^{n-\beta}s_{i}v^{i}=t_{0}v^{0}-s_{0}v^{0}-\sum\limits\limits_{i=1}^{\alpha}s_{i}v^{i}\
\ \ \ \ \ \ \ \ \ \left(\text{since }n-\beta=\alpha\right)$
$\displaystyle=t_{0}\underbrace{v^{0}}_{=1_{B}}-\underbrace{\left(s_{0}v^{0}+\sum\limits\limits_{i=1}^{\alpha}s_{i}v^{i}\right)}_{=\sum\limits\limits_{i=0}^{\alpha}s_{i}v^{i}=u}=t_{0}\cdot
1_{B}-u.$

Thus, $t_{0}\cdot 1_{B}-u$ is $n$-integral over $A$ (since
$\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$ is $n$-integral over $A$). Thus,
Corollary 1.14 (applied to $x=t_{0}\cdot 1_{B}-u$ and $m=n$) shows that
$-\left(t_{0}\cdot 1_{B}-u\right)$ is $n$-integral over $A$. In other words,
$u-t_{0}\cdot 1_{B}$ is $n$-integral over $A$ (since $-\left(t_{0}\cdot
1_{B}-u\right)=u-t_{0}\cdot 1_{B}$).

On the other hand, $t_{0}\cdot 1_{B}$ is $1$-integral over $A$ (by Theorem
1.8, applied to $a=t_{0}$). Thus, $t_{0}\cdot 1_{B}+\left(u-t_{0}\cdot
1_{B}\right)$ is $n\cdot 1$-integral over $A$ (by Theorem 1.10 (a), applied to
$x=t_{0}\cdot 1_{B}$, $y=u-t_{0}\cdot 1_{B}$ and $m=1$). In other words, $u$
is $\left(\alpha+\beta\right)$-integral over $A$ (since $t_{0}\cdot
1_{B}+\left(u-t_{0}\cdot 1_{B}\right)=u$ and $n\cdot 1=n=\alpha+\beta$). This
proves Corollary 1.20.

We will provide a second proof of Corollary 1.20 in Section 5.

######  Corollary 1.22.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $n\in\mathbb{N}^{+}$ and
$m\in\mathbb{N}$. Let $v\in B$. Let $b_{0},b_{1},\ldots,b_{n-1}$ be $n$
elements of $A$, and let $u=\sum\limits\limits_{i=0}^{n-1}b_{i}v^{i}$. Assume
that $vu$ is $m$-integral over $A$. Then, $u$ is $nm$-integral over $A$.

Corollary 1.22 generalizes a folklore fact about integrality, which states
that if $B$ is an $A$-algebra, and if an invertible $v\in B$ satisfies
$v^{-1}\in A\left[v\right]$, then $v$ is integral over $A$. (Indeed, this
latter fact follows from Corollary 1.22 by setting $u=v^{-1}$.)

######  Proof 1.23 (Proof of Corollary 1.22.).

Define $n+1$ elements $a_{0},a_{1},\ldots,a_{n}$ of $A\left[vu\right]$ by

$a_{i}=\begin{cases}-vu,&\text{ if }i=0;\\\ b_{i-1}\cdot 1_{B},&\text{ if
}i>0\end{cases}\ \ \ \ \ \ \ \ \ \ \text{for every
}i\in\left\\{0,1,\ldots,n\right\\}.$

(These are well-defined, since every positive
$i\in\left\\{0,1,\ldots,n\right\\}$ satisfies
$i\in\left\\{1,2,\ldots,n\right\\}$ and thus
$i-1\in\left\\{0,1,\ldots,n-1\right\\}$ and thus $b_{i-1}\in A$ and therefore
$b_{i-1}\cdot 1_{B}\in A\cdot 1_{B}\subseteq A\left[vu\right]$.)

The definition of $a_{0}$ yields $a_{0}=-vu$. Also,

$\displaystyle\sum\limits\limits_{i=0}^{n}a_{i}v^{i}$
$\displaystyle=\underbrace{a_{0}}_{=-vu}\underbrace{v^{0}}_{=1}+\sum\limits\limits_{i=1}^{n}\underbrace{a_{i}}_{\begin{subarray}{c}=b_{i-1}\cdot
1_{B}\\\ \text{(by the definition}\\\ \text{of }a_{i}\text{, since
}i>0\text{)}\end{subarray}}v^{i}=-vu+\sum\limits\limits_{i=1}^{n}b_{i-1}\cdot\underbrace{1_{B}v^{i}}_{=v^{i}=v^{i-1}v}$
$\displaystyle=-vu+\sum\limits\limits_{i=1}^{n}b_{i-1}v^{i-1}v=-vu+\underbrace{\sum\limits\limits_{i=0}^{n-1}b_{i}v^{i}}_{=u}v$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{here, we substituted }i\text{
for }i-1\text{\ in the sum}\right)$ $\displaystyle=-vu+uv=0.$

Let $k=1$. Then, $k=1\in\left\\{0,1,\ldots,n\right\\}$ (since
$n\in\mathbb{N}^{+}$).

Now, $A\left[vu\right]$ is a subring of $B$; hence, $B$ is an
$A\left[vu\right]$-algebra. The $n+1$ elements $a_{0},a_{1},\ldots,a_{n}$ of
$A\left[vu\right]$ satisfy $\sum\limits\limits_{i=0}^{n}a_{i}v^{i}=0$.

Hence, Theorem 1.18 (applied to the ring $A\left[vu\right]$ in lieu of $A$)
yields that $\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$ is $n$-integral over
$A\left[vu\right]$. But from $k=1$, we obtain

$\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}=\sum\limits\limits_{i=0}^{n-1}\underbrace{a_{i+1}}_{\begin{subarray}{c}=b_{\left(i+1\right)-1}\cdot
1_{B}\\\ \text{(by the definition}\\\ \text{of }a_{i+1}\text{, since
}i+1>0\text{)}\end{subarray}}v^{i}=\sum\limits\limits_{i=0}^{n-1}\underbrace{b_{\left(i+1\right)-1}}_{=b_{i}}\cdot\underbrace{1_{B}v^{i}}_{=v^{i}}=\sum\limits\limits_{i=0}^{n-1}b_{i}v^{i}=u.$

Hence, $u$ is $n$-integral over $A\left[vu\right]$ (since
$\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$ is $n$-integral over
$A\left[vu\right]$). But $vu$ is $m$-integral over $A$. Thus, Theorem 1.6
(applied to $vu$ in lieu of $v$) yields that $u$ is $nm$-integral over $A$.
This proves Corollary 1.22.

##  2 Integrality over ideal semifiltrations

###  2.1 Definitions of ideal semifiltrations and integrality over them

We now set our sights at a more general notion of integrality.

######  Definition 2.1.

Let $A$ be a ring, and let $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be a
sequence of ideals of $A$. Then, $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$
is called an ideal semifiltration of $A$ if and only if it satisfies the two
conditions

$\displaystyle I_{0}$ $\displaystyle=A;$ $\displaystyle I_{a}I_{b}$
$\displaystyle\subseteq I_{a+b}\ \ \ \ \ \ \ \ \ \ \text{for every
}a\in\mathbb{N}\text{ and }b\in\mathbb{N}.$

Two simple examples of ideal semifiltrations can easily be constructed from
any ideal:

######  Example 2.2.

Let $A$ be a ring. Let $I$ be an ideal of $A$. Then:

(a) The sequence $\left(I^{\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$. (Here, $I^{\rho}$ denotes the $\rho$-th power of $I$ in
the multiplicative monoid of ideals of $A$; in particular, $I^{0}=A$.)

(b) The sequence $\left(A,I,I,I,\ldots\right)=\left(\begin{cases}A,&\text{if
}\rho=0;\\\ I,&\text{if }\rho>0\end{cases}\right)_{\rho\in\mathbb{N}}$ is an
ideal semifiltration of $A$.

######  Proof 2.3 (Proof of Example 2.2.).

This is a straightforward exercise in checking axioms.

######  Definition 2.4.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Let $n\in\mathbb{N}$. Let $u\in B$.

We say that the element $u$ of $B$ is $n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$ if there exists
some $\left(a_{0},a_{1},\ldots,a_{n}\right)\in A^{n+1}$ such that

$\sum\limits\limits_{k=0}^{n}a_{k}u^{k}=0,\ \ \ \ \ \ \ \ \ \ a_{n}=1,\ \ \ \
\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ a_{i}\in I_{n-i}\text{ for every
}i\in\left\\{0,1,\ldots,n\right\\}.$

This definition generalizes [5, Definition 1.1.1] in multiple ways. Indeed, if
$I$ is an ideal of a ring $A$, and if $u\in A$ and $n\in\mathbb{N}$, then $u$
is $n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$ (here,
$\left(I^{\rho}\right)_{\rho\in\mathbb{N}}$ is the ideal semifiltration from
Example 2.2 (a)) if and only if there is an equation of integral dependence of
$u$ over $I$ (in the sense of [5, Definition 1.1.1]).

We further notice that integrality over an ideal semifiltration of a ring $A$
is a stronger claim than integrality over $A$:

######  Proposition 2.5.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Let $n\in\mathbb{N}$. Let $u\in B$ be such that $u$ is $n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. Then, $u$ is
$n$-integral over $A$.

######  Proof 2.6 (Proof of Proposition 2.5.).

We know that $u$ is $n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. Thus, by
Definition 2.4, there exists some $\left(a_{0},a_{1},\ldots,a_{n}\right)\in
A^{n+1}$ such that

$\sum\limits\limits_{k=0}^{n}a_{k}u^{k}=0,\ \ \ \ \ \ \ \ \ \ a_{n}=1,\ \ \ \
\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ a_{i}\in I_{n-i}\text{ for every
}i\in\left\\{0,1,\ldots,n\right\\}.$

Consider this $\left(a_{0},a_{1},\ldots,a_{n}\right)$.

Thus, there exists a monic polynomial $P\in A\left[X\right]$ with $\deg P=n$
and $P\left(u\right)=0$ (namely,
$P\left(X\right)=\sum\limits\limits_{k=0}^{n}a_{k}X^{k}$).

In other words, $u$ is $n$-integral over $A$. This proves Proposition 2.5.

We leave it to the reader to prove the following simple fact, which shows that
nilpotency is an instance of integrality over ideal semifiltrations:

######  Proposition 2.7.

Let $A$ be a ring. Let $0A$ be the zero ideal of $A$. Let $n\in\mathbb{N}$.
Let $u\in A$. Then, the element $u$ of $A$ is $n$-integral over
$\left(A,\left(\left(0A\right)^{\rho}\right)_{\rho\in\mathbb{N}}\right)$ if
and only if $u^{n}=0$.

###  2.2 Polynomial rings and Rees algebras

In order to study integrality over ideal semifiltrations, we shall now
introduce the concept of a Rees algebra – a subalgebra of a polynomial ring
that conveniently encodes an ideal semifiltration of the base ring. This,
again, generalizes a classical notion for ideals (namely, the Rees algebra of
an ideal – see [5, Definition 5.1.1]). First, we recall a basic fact:

######  Definition 2.8.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Then, there is a canonical ring
homomorphism $\iota:A\rightarrow B$ that sends each $a\in A$ to $a\cdot
1_{B}\in B$. This ring homomorphism $\iota$ induces a canonical ring
homomorphism $\iota\left[Y\right]:A\left[Y\right]\rightarrow B\left[Y\right]$
between the polynomial rings $A\left[Y\right]$ and $B\left[Y\right]$ that
sends each polynomial $\sum\limits\limits_{i=0}^{m}a_{i}Y^{i}\in
A\left[Y\right]$ (with $m\in\mathbb{N}$ and
$\left(a_{0},a_{1},\ldots,a_{m}\right)\in A^{m+1}$) to the polynomial
$\sum\limits\limits_{i=0}^{m}\iota\left(a_{i}\right)Y^{i}\in B\left[Y\right]$.
Thus, the polynomial ring $B\left[Y\right]$ becomes an
$A\left[Y\right]$-algebra.

######  Definition 2.9.

Let $A$ be a ring, and let $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an
ideal semifiltration of $A$. Thus, $I_{0},I_{1},I_{2},\ldots$ are ideals of
$A$, and we have $I_{0}=A$.

Consider the polynomial ring $A\left[Y\right]$. For each $i\in\mathbb{N}$, the
subset $I_{i}Y^{i}$ of $A\left[Y\right]$ is an $A$-submodule of the
$A$-algebra $A\left[Y\right]$ (since $I_{i}$ is an ideal of $A$). Hence, the
sum $\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}$ of these $A$-submodules
must also be an $A$-submodule of the $A$-algebra $A\left[Y\right]$.

Let $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ denote
this $A$-submodule $\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}$ of the
$A$-algebra $A\left[Y\right]$. Then,

$\displaystyle A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$
$\displaystyle=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}$
$\displaystyle=\left\\{\sum\limits_{i\in\mathbb{N}}a_{i}Y^{i}\ \mid\
\left(a_{i}\in I_{i}\text{ for all }i\in\mathbb{N}\right)\text{, }\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left.\text{and }\left(\text{only finitely
many }i\in\mathbb{N}\text{ satisfy }a_{i}\neq
0\right)\vphantom{\sum\limits_{i\in\mathbb{N}}}\right\\}$
$\displaystyle=\left\\{P\in A\left[Y\right]\ \mid\ \text{the }i\text{-th
coefficient of the polynomial }P\right.$ $\displaystyle\ \ \ \ \ \ \ \ \ \
\left.\text{lies in }I_{i}\text{ for every }i\in\mathbb{N}\right\\}.$

Clearly, $A\subseteq A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$, since

$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}\supseteq\underbrace{I_{0}}_{=A}\underbrace{Y^{0}}_{=1}=A\cdot
1=A.$

Hence, $1\in A\subseteq A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$. Also, the $A$-submodule
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ of
$A\left[Y\right]$ is an $A$-subalgebra of the $A$-algebra $A\left[Y\right]$
(by Lemma 2.10 below), and thus is a subring of $A\left[Y\right]$.

This $A$-subalgebra $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ is called the Rees algebra of the ideal semifiltration
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$.

######  Lemma 2.10.

Let $A$ be a ring, and let $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an
ideal semifiltration of $A$. Then, the $A$-submodule
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ of
$A\left[Y\right]$ is an $A$-subalgebra of the $A$-algebra $A\left[Y\right]$.

######  Proof 2.11 (Proof of Lemma 2.10.).

This is an easy exercise. (Use $I_{a}I_{b}\subseteq I_{a+b}$ to prove that
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ is closed
under multiplication.)

######  Remark 2.12.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ defined in
Definition 2.9.

The polynomial ring $B\left[Y\right]$ is an $A\left[Y\right]$-algebra (as
explained in Definition 2.8), and thus is an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra (since
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ is a subring
of $A\left[Y\right]$). Hence, if $p\in B\left[Y\right]$ is a polynomial and
$n\in\mathbb{N}$, then it makes sense to ask whether $p$ is $n$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. Questions of
this form will often appear in what follows.

We note in passing that the notion of a Rees algebra helps set up a bijection
between ideal semifiltrations of a ring $A$ and graded $A$-subalgebras of the
polynomial ring $A\left[Y\right]$:

######  Proposition 2.13.

Let $A$ be a ring. Consider the polynomial ring $A\left[Y\right]$ as a graded
$A$-algebra (with the usual degree of polynomials).

(a) If $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal semifiltration
of $A$, then its Rees algebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ is a graded
$A$-subalgebra of $A\left[Y\right]$.

(b) If $B$ is any graded $A$-subalgebra of $A\left[Y\right]$, and if
$\rho\in\mathbb{N}$, then we let $I_{B,\rho}$ denote the subset $\left\\{a\in
A\ \mid\ aY^{\rho}\in B\right\\}$ of $A$. Then, $I_{B,\rho}$ is an ideal of
$A$.

(c) The maps

$\displaystyle\left\\{\text{ideal semifiltrations of }A\right\\}$
$\displaystyle\rightarrow\left\\{\text{graded }A\text{-subalgebras of
}A\left[Y\right]\right\\},$
$\displaystyle\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ $\displaystyle\mapsto
A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$

and

$\displaystyle\left\\{\text{graded }A\text{-subalgebras of
}A\left[Y\right]\right\\}$ $\displaystyle\rightarrow\left\\{\text{ideal
semifiltrations of }A\right\\},$ $\displaystyle B$
$\displaystyle\mapsto\left(I_{B,\rho}\right)_{\rho\in\mathbb{N}}$

are mutually inverse bijections.

We shall not have any need for this proposition, so we omit its
(straightforward and easy) proof.

###  2.3 Reduction to integrality over rings

We start with a theorem which reduces the question of $n$-integrality over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$ to that of
$n$-integrality over a ring666Theorem 2.14 is inspired by [5, Proposition
5.2.1].:

######  Theorem 2.14.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Let $n\in\mathbb{N}$. Let $u\in B$.

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ defined in
Definition 2.9.

Then, the element $u$ of $B$ is $n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$ if and only if the
element $uY$ of the polynomial ring $B\left[Y\right]$ is $n$-integral over the
ring $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. (Here,
$B\left[Y\right]$ is an $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$-algebra, as explained in Remark 2.12.)

######  Proof 2.15 (Proof of Theorem 2.14.).

$\Longrightarrow:$ Assume that $u$ is $n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. Thus, by
Definition 2.4, there exists some $\left(a_{0},a_{1},\ldots,a_{n}\right)\in
A^{n+1}$ such that

$\sum\limits\limits_{k=0}^{n}a_{k}u^{k}=0,\ \ \ \ \ \ \ \ \ \ a_{n}=1,\ \ \ \
\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ a_{i}\in I_{n-i}\text{ for every
}i\in\left\\{0,1,\ldots,n\right\\}.$

Consider this $\left(a_{0},a_{1},\ldots,a_{n}\right)$.

Hence, there exists a monic polynomial
$P\in\left(A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)\left[X\right]$ with $\deg P=n$ and $P\left(uY\right)=0$ (viz.,
the polynomial
$P\left(X\right)=\sum\limits\limits_{k=0}^{n}a_{k}Y^{n-k}X^{k}$). Hence, $uY$
is $n$-integral over $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$. This proves the $\Longrightarrow$ direction of Theorem 2.14.

$\Longleftarrow:$ Assume that $uY$ is $n$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. Thus, there
exists a monic polynomial
$P\in\left(A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)\left[X\right]$ with $\deg P=n$ and $P\left(uY\right)=0$.
Consider this $P$. Since
$P\in\left(A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)\left[X\right]$ satisfies $\deg P=n$, there exists
$\left(p_{0},p_{1},\ldots,p_{n}\right)\in\left(A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)^{n+1}$ such that
$P\left(X\right)=\sum\limits\limits_{k=0}^{n}p_{k}X^{k}$. Consider this
$\left(p_{0},p_{1},\ldots,p_{n}\right)$. Note that $p_{n}=1$ (since $P$ is
monic and $\deg P=n$).

For every $k\in\left\\{0,1,\ldots,n\right\\}$, we have $p_{k}\in
A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}$, and thus there exists
a sequence $\left(p_{k,i}\right)_{i\in\mathbb{N}}\in A^{\mathbb{N}}$ such that
$p_{k}=\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}$, such that
$\left(p_{k,i}\in I_{i}\text{ for every }i\in\mathbb{N}\right)$, and such that
only finitely many $i\in\mathbb{N}$ satisfy $p_{k,i}\neq 0$. Consider this
sequence. Thus, $P\left(X\right)=\sum\limits\limits_{k=0}^{n}p_{k}X^{k}$
rewrites as
$P\left(X\right)=\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}X^{k}$
(since $p_{k}=\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}$). Hence,

$P\left(uY\right)=\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}\left(uY\right)^{k}=\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i+k}u^{k}.$

Therefore, $P\left(uY\right)=0$ rewrites as
$\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i+k}u^{k}=0$.
In other words, the polynomial
$\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i+k}u^{k}\in
B\left[Y\right]$ equals $0$. Hence, its coefficient before $Y^{n}$ equals $0$
as well. But its coefficient before $Y^{n}$ is
$\sum\limits\limits_{k=0}^{n}p_{k,n-k}u^{k}$, so we get
$\sum\limits\limits_{k=0}^{n}p_{k,n-k}u^{k}=0$.

Recall that $\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}=p_{k}$ for every
$k\in\left\\{0,1,\ldots,n\right\\}$ (by the definition of the $p_{k,i}$).
Thus, $\sum\limits\limits_{i\in\mathbb{N}}p_{n,i}Y^{i}=p_{n}=1$ in
$A\left[Y\right]$, and thus $p_{n,0}=1$ (by comparing coefficients before
$Y^{0}$).

Define an $\left(n+1\right)$-tuple $\left(a_{0},a_{1},\ldots,a_{n}\right)\in
A^{n+1}$ by $a_{k}=p_{k,n-k}$ for every $k\in\left\\{0,1,\ldots,n\right\\}$.
Then, $a_{n}=p_{n,0}=1$. Besides,
$\sum\limits\limits_{k=0}^{n}\underbrace{a_{k}}_{=p_{k,n-k}}u^{k}=\sum\limits\limits_{k=0}^{n}p_{k,n-k}u^{k}=0$.
Finally, $a_{k}=p_{k,n-k}\in I_{n-k}$ (since $p_{k,i}\in I_{i}$ for every
$i\in\mathbb{N}$) for every $k\in\left\\{0,1,\ldots,n\right\\}$. In other
words, $a_{i}\in I_{n-i}$ for every $i\in\left\\{0,1,\ldots,n\right\\}$.

Altogether, we now know that

$\sum\limits\limits_{k=0}^{n}a_{k}u^{k}=0,\ \ \ \ \ \ \ \ \ \ a_{n}=1,\ \ \ \
\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ a_{i}\in I_{n-i}\text{ for every
}i\in\left\\{0,1,\ldots,n\right\\}.$

Thus, by Definition 2.4, the element $u$ is $n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. This proves the
$\Longleftarrow$ direction of Theorem 2.14.

###  2.4 Sums and products again

Let us next state an analogue of Theorem 1.8 for integrality over ideal
semifiltrations:

######  Theorem 2.16.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Let $u\in A$. Then, $u\cdot 1_{B}$ is $1$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$ if and only if
$u\cdot 1_{B}\in I_{1}\cdot 1_{B}$.

######  Proof 2.17 (Proof of Theorem 2.16.).

Straightforward and left to the reader.

The next theorem is an analogue of Theorem 1.10 (a) for integrality over ideal
semifiltrations:

######  Theorem 2.18.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Let $x\in B$ and $y\in B$. Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$. Assume
that $x$ is $m$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$, and that $y$ is
$n$-integral over $\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$.

Then, $x+y$ is $nm$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$.

######  Proof 2.19 (Proof of Theorem 2.18.).

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. The
polynomial ring $B\left[Y\right]$ is an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra (as
explained in Remark 2.12).

Theorem 2.14 (applied to $x$ and $m$ instead of $u$ and $n$) yields that $xY$
is $m$-integral over $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ (since $x$ is $m$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$). Also, Theorem
2.14 (applied to $y$ instead of $u$) yields that $yY$ is $n$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ (since $y$ is
$n$-integral over $\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$).

Hence, Theorem 1.10 (a) (applied to
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$,
$B\left[Y\right]$, $xY$ and $yY$ instead of $A$, $B$, $x$ and $y$,
respectively) yields that $xY+yY$ is $nm$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. Since
$xY+yY=\left(x+y\right)Y$, this means that $\left(x+y\right)Y$ is
$nm$-integral over $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$. Hence, Theorem 2.14 (applied to $x+y$ and $nm$ instead of $u$ and
$n$) yields that $x+y$ is $nm$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. This proves
Theorem 2.18.

Our next theorem is a somewhat asymmetric analogue of Theorem 1.10 (b) for
integrality over ideal semifiltrations:

######  Theorem 2.20.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Let $x\in B$ and $y\in B$. Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$. Assume
that $x$ is $m$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$, and that $y$ is
$n$-integral over $A$.

Then, $xy$ is $nm$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$.

Before we prove this theorem, we require a trivial observation:

######  Lemma 2.21.

Let $A$ be a ring. Let $A^{\prime}$ be an $A$-algebra. Let $B^{\prime}$ be an
$A^{\prime}$-algebra. Let $v\in B^{\prime}$. Let $n\in\mathbb{N}$. Assume that
$v$ is $n$-integral over $A$. (Here, of course, we are using the fact that
$B^{\prime}$ is an $A$-algebra, since $B^{\prime}$ is an $A^{\prime}$-algebra
while $A^{\prime}$ is an $A$-algebra.)

Then, $v$ is $n$-integral over $A^{\prime}$.

######  Proof 2.22 (Proof of Theorem 2.20.).

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. The
polynomial ring $B\left[Y\right]$ is an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra (as
explained in Remark 2.12).

Theorem 2.14 (applied to $x$ and $m$ instead of $u$ and $n$) yields that $xY$
is $m$-integral over $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ (since $x$ is $m$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$). Also, we know
that $y$ is $n$-integral over $A$. Thus, Lemma 2.21 (applied to
$A^{\prime}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$,
$B^{\prime}=B\left[Y\right]$ and $v=y$) yields that $y$ is $n$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ (since
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ is an
$A$-algebra, and $B\left[Y\right]$ is an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra). On
the other hand, we know that $xY$ is $m$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. Hence,
Theorem 1.10 (b) (applied to
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$,
$B\left[Y\right]$ and $xY$ instead of $A$, $B$ and $x$, respectively) yields
that $xY\cdot y$ is $nm$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. Since
$xY\cdot y=xyY$, this means that $xyY$ is $nm$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. Hence,
Theorem 2.14 (applied to $xy$ and $nm$ instead of $u$ and $n$) yields that
$xy$ is $nm$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. This proves
Theorem 2.20.

It is easy to state analogues of Corollary 1.14 and Corollary 1.16 for ideal
semifiltrations. These analogues can be derived from Corollary 1.14 and
Corollary 1.16 in the same way as how we derived Theorem 2.18 from Theorem
1.10 (a).

###  2.5 Transitivity again

The next theorem imitates Theorem 1.6 for integrality over ideal
semifiltrations:

######  Theorem 2.23.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.

Let $v\in B$ and $u\in B$. Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$.

(a) Then, $\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}$ is an
ideal semifiltration of $A\left[v\right]$. (See Convention 2.1 below for the
meaning of “$I_{\rho}A\left[v\right]$”.)

(b) Assume that $v$ is $m$-integral over $A$, and that $u$ is $n$-integral
over
$\left(A\left[v\right],\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}\right)$.
Then, $u$ is $nm$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$.

Here and in the following, we are using the following convention:

######  Convention 2.1.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $v\in B$, and let $I$ be an
ideal of $A$. Then, you should read the expression “$IA\left[v\right]$” as
$I\cdot\left(A\left[v\right]\right)$, not as $\left(IA\right)\left[v\right]$.
For instance, you should read the term “$I_{\rho}A\left[v\right]$” (in Theorem
2.23 (a)) as $I_{\rho}\cdot\left(A\left[v\right]\right)$, not as
$\left(I_{\rho}A\right)\left[v\right]$.

Before we prove Theorem 2.23, let us state two lemmas. The first is a more
general (but still obvious) version of Theorem 2.23 (a):

######  Lemma 2.24.

Let $A$ be a ring. Let $A^{\prime}$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Then, $\left(I_{\rho}A^{\prime}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A^{\prime}$.

######  Proof 2.25 (Proof of Lemma 2.24.).

This is a straightforward verification of axioms.

######  Lemma 2.26.

Let $A$ be a ring. Let $A^{\prime}$ be an $A$-algebra. Let $B^{\prime}$ be an
$A^{\prime}$-algebra. Let $v\in B^{\prime}$. Then, $A^{\prime}\cdot
A\left[v\right]=A^{\prime}\left[v\right]$ (an equality between $A$-submodules
of $B^{\prime}$). (Here, we are using the fact that $B^{\prime}$ is an
$A$-algebra, because $B^{\prime}$ is an $A^{\prime}$-algebra while
$A^{\prime}$ is an $A$-algebra.)

Here, of course, the expression “$A^{\prime}\cdot A\left[v\right]$” means
“$A^{\prime}\cdot\left(A\left[v\right]\right)$”, not “$\left(A^{\prime}\cdot
A\right)\left[v\right]$”.

######  Proof 2.27 (Proof of Lemma 2.26.).

Left to the reader (see [7]).

We are now ready to prove Theorem 2.23:

######  Proof 2.28 (Proof of Theorem 2.23.).

(a) Lemma 2.24 (applied to $A^{\prime}=A\left[v\right]$) yields that
$\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A\left[v\right]$. This proves Theorem 2.23 (a).

(b) Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. Then,
$\left(A\left[v\right]\right)\left[Y\right]$ is an $A\left[Y\right]$-algebra
(since $A\left[v\right]$ is an $A$-algebra) and therefore also an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra (since
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ is a subring
of $A\left[Y\right]$). Hence,
$\left(A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)\left[v\right]$ is an $A$-subalgebra of
$\left(A\left[v\right]\right)\left[Y\right]$ (since $v\in
A\left[v\right]\subseteq\left(A\left[v\right]\right)\left[Y\right]$). On the
other hand,
$\left(A\left[v\right]\right)\left[\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ is an $A$-subalgebra of $\left(A\left[v\right]\right)\left[Y\right]$
(by its definition).

Note that $B$ is an $A\left[v\right]$-algebra (since $A\left[v\right]$ is a
subring of $B$). Hence, (as explained in Definition 2.8) the polynomial ring
$B\left[Y\right]$ is an $\left(A\left[v\right]\right)\left[Y\right]$-algebra.
Moreover, $B\left[Y\right]$ is an $A\left[Y\right]$-algebra (as explained in
Definition 2.8) and also an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra (as
explained in Remark 2.12).

Now, we will show that
$\left(A\left[v\right]\right)\left[\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}\ast
Y\right]=\left(A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)\left[v\right]$. (This is an equality between two subrings of
$\left(A\left[v\right]\right)\left[Y\right]$.)

In fact, Definition 2.9 yields
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}$. The same definition
(but applied to $A\left[v\right]$ and
$\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}$ instead of $A$ and
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$) yields

$\displaystyle\left(A\left[v\right]\right)\left[\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}\ast
Y\right]$
$\displaystyle=\sum\limits\limits_{i\in\mathbb{N}}I_{i}\underbrace{A\left[v\right]\cdot
Y^{i}}_{=Y^{i}\cdot
A\left[v\right]}=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}\cdot
A\left[v\right]$
$\displaystyle=\underbrace{\left(\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}\right)}_{=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]}\cdot
A\left[v\right]=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\cdot A\left[v\right]$
$\displaystyle=\left(A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)\left[v\right]$ (10)

(by Lemma 2.26, applied to
$A^{\prime}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ and
$B^{\prime}=\left(A\left[v\right]\right)\left[Y\right]$).

Recall that $B\left[Y\right]$ is an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra.
Hence, Lemma 2.21 (applied to
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$,
$B\left[Y\right]$ and $m$ instead of $A^{\prime}$, $B^{\prime}$ and $n$)
yields that $v$ is $m$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ (since $v$ is
$m$-integral over $A$).

Now, Theorem 2.14 (applied to $A\left[v\right]$ and
$\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}$ instead of $A$ and
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$) yields that the element $uY$ is
$n$-integral over
$\left(A\left[v\right]\right)\left[\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ (since $u$ is $n$-integral over
$\left(A\left[v\right],\left(I_{\rho}A\left[v\right]\right)_{\rho\in\mathbb{N}}\right)$).
In view of (10), this rewrites as follows: The element $uY$ is $n$-integral
over $\left(A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)\left[v\right]$. Hence, Theorem 1.6 (applied to
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$,
$B\left[Y\right]$ and $uY$ instead of $A$, $B$ and $u$) yields that $uY$ is
$nm$-integral over $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ (since $v$ is $m$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$). Thus,
Theorem 2.14 (applied to $nm$ instead of $n$) yields that $u$ is $nm$-integral
over $\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. This proves
Theorem 2.23 (b).

##  3 Generalizing to two ideal semifiltrations

Theorem 2.20 can be generalized: Instead of requiring $y$ to be integral over
the ring $A$, we can require $y$ to be integral over a further ideal
semifiltration $\left(J_{\rho}\right)_{\rho\in\mathbb{N}}$ of $A$. In that
case, $xy$ will be integral over the ideal semifiltration
$\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}$ (see Theorem 3.6 for the
precise statement). To get a grip on this, let us study two ideal
semifiltrations.

###  3.1 The product of two ideal semifiltrations

######  Theorem 3.1.

Let $A$ be a ring.

(a) Then, $\left(A\right)_{\rho\in\mathbb{N}}$ is an ideal semifiltration of
$A$.

(b) Let $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ and
$\left(J_{\rho}\right)_{\rho\in\mathbb{N}}$ be two ideal semifiltrations of
$A$. Then, $\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$.

######  Proof 3.2 (Proof of Theorem 3.1.).

The proof of this is just basic axiom checking (see [7] for details).

###  3.2 Half-reduction

Now let us generalize Theorem 2.14:

######  Theorem 3.3.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ and
$\left(J_{\rho}\right)_{\rho\in\mathbb{N}}$ be two ideal semifiltrations of
$A$. Let $n\in\mathbb{N}$. Let $u\in B$.

We know that $\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$ (according to Theorem 3.1 (b)).

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$.

We will abbreviate this $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ by
$A_{\left[I\right]}$.

(a) The sequence $\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$
is an ideal semifiltration of $A_{\left[I\right]}$.

(b) The element $u$ of $B$ is $n$-integral over
$\left(A,\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$ if and only
if the element $uY$ of the polynomial ring $B\left[Y\right]$ is $n$-integral
over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$.
(Here, we are using the fact that $B\left[Y\right]$ is an
$A_{\left[I\right]}$-algebra, because
$A_{\left[I\right]}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ is a subring of $A\left[Y\right]$ and because $B\left[Y\right]$ is
an $A\left[Y\right]$-algebra as explained in Definition 2.8.)

######  Proof 3.4 (Proof of Theorem 3.3.).

(a) We know that
$\left(J_{\tau}\right)_{\tau\in\mathbb{N}}=\left(J_{\rho}\right)_{\rho\in\mathbb{N}}$
is an ideal semifiltration of $A$. Thus, by Lemma 2.24 (applied to
$A_{\left[I\right]}$ and $\left(J_{\tau}\right)_{\tau\in\mathbb{N}}$ instead
of $A^{\prime}$ and $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$), the sequence
$\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$ is an ideal
semifiltration of $A_{\left[I\right]}$. This proves Theorem 3.3 (a).

(b) In order to verify Theorem 3.3 (b), we have to prove the $\Longrightarrow$
and $\Longleftarrow$ statements.

$\Longrightarrow:$ Assume that $u$ is $n$-integral over
$\left(A,\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. Thus, by
Definition 2.4 (applied to $\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}$
instead of $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$), there exists some
$\left(a_{0},a_{1},\ldots,a_{n}\right)\in A^{n+1}$ such that

$\sum\limits\limits_{k=0}^{n}a_{k}u^{k}=0,\ \ \ \ \ \ \ \ \ \ a_{n}=1,\ \ \ \
\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ a_{i}\in I_{n-i}J_{n-i}\text{ for
every }i\in\left\\{0,1,\ldots,n\right\\}.$

Consider this $\left(a_{0},a_{1},\ldots,a_{n}\right)$.

For each $k\in\left\\{0,1,\ldots,n\right\\}$, we have $a_{k}\in
I_{n-k}J_{n-k}\subseteq I_{n-k}$ (since $I_{n-k}$ is an ideal of $A$) and thus
$a_{k}Y^{n-k}\in
I_{n-k}Y^{n-k}\subseteq\sum\limits_{i\in\mathbb{N}}I_{i}Y^{i}=A_{\left[I\right]}$.
Thus, we can define an $\left(n+1\right)$-tuple
$\left(b_{0},b_{1},\ldots,b_{n}\right)\in\left(A_{\left[I\right]}\right)^{n+1}$
by $b_{k}=a_{k}Y^{n-k}$ for every $k\in\left\\{0,1,\ldots,n\right\\}$. This
$\left(n+1\right)$-tuple satisfies

$\sum\limits\limits_{k=0}^{n}b_{k}\cdot\left(uY\right)^{k}=0,\ \ \ \ \ \ \ \ \
\ b_{n}=1,\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ b_{i}\in
J_{n-i}A_{\left[I\right]}\text{ for every }i\in\left\\{0,1,\ldots,n\right\\}$

(as can be easily checked). Hence, by Definition 2.4 (applied to
$A_{\left[I\right]}$, $B\left[Y\right]$,
$\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$, $uY$ and
$\left(b_{0},b_{1},\ldots,b_{n}\right)$ instead of $A$, $B$,
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$, $u$ and
$\left(a_{0},a_{1},\ldots,a_{n}\right)$), the element $uY$ is $n$-integral
over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$.
This proves the $\Longrightarrow$ direction of Theorem 3.3 (b).

$\Longleftarrow:$ Assume that $uY$ is $n$-integral over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$.
Thus, by Definition 2.4 (applied to $A_{\left[I\right]}$, $B\left[Y\right]$,
$\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$, $uY$ and
$\left(p_{0},p_{1},\ldots,p_{n}\right)$ instead of $A$, $B$,
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$, $u$ and
$\left(a_{0},a_{1},\ldots,a_{n}\right)$), there exists some
$\left(p_{0},p_{1},\ldots,p_{n}\right)\in\left(A_{\left[I\right]}\right)^{n+1}$
such that

$\sum\limits\limits_{k=0}^{n}p_{k}\cdot\left(uY\right)^{k}=0,\ \ \ \ \ \ \ \ \
\ p_{n}=1,\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ p_{i}\in
J_{n-i}A_{\left[I\right]}\text{ for every }i\in\left\\{0,1,\ldots,n\right\\}.$

Consider this $\left(p_{0},p_{1},\ldots,p_{n}\right)$. For every
$k\in\left\\{0,1,\ldots,n\right\\}$, we have

$\displaystyle p_{k}$ $\displaystyle\in J_{n-k}A_{\left[I\right]}\ \ \ \ \ \ \
\ \ \ \left(\text{since }p_{i}\in J_{n-i}A_{\left[I\right]}\text{ for every
}i\in\left\\{0,1,\ldots,n\right\\}\right)$
$\displaystyle=J_{n-k}\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}\ \ \ \ \ \
\ \ \ \ \left(\text{since
}A_{\left[I\right]}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}\right)$
$\displaystyle=\sum\limits\limits_{i\in\mathbb{N}}J_{n-k}I_{i}Y^{i}=\sum\limits\limits_{i\in\mathbb{N}}I_{i}J_{n-k}Y^{i},$

and thus there exists a sequence $\left(p_{k,i}\right)_{i\in\mathbb{N}}\in
A^{\mathbb{N}}$ such that
$p_{k}=\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}$, such that
$\left(p_{k,i}\in I_{i}J_{n-k}\text{ for every }i\in\mathbb{N}\right)$, and
such that only finitely many $i\in\mathbb{N}$ satisfy $p_{k,i}\neq 0$.
Consider this sequence. Thus,

$\sum\limits\limits_{k=0}^{n}\underbrace{p_{k}}_{=\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}}\cdot\underbrace{\left(uY\right)^{k}}_{\begin{subarray}{c}=u^{k}Y^{k}\\\
=Y^{k}u^{k}\end{subarray}}=\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}\underbrace{Y^{i}\cdot
Y^{k}}_{=Y^{i+k}}u^{k}=\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i+k}u^{k}.$

Hence, $\sum\limits\limits_{k=0}^{n}p_{k}\cdot\left(uY\right)^{k}=0$ rewrites
as
$\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i+k}u^{k}=0$.
In other words, the polynomial
$\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i+k}u^{k}\in
B\left[Y\right]$ equals $0$. Hence, its coefficient before $Y^{n}$ equals $0$
as well. But its coefficient before $Y^{n}$ is
$\sum\limits\limits_{k=0}^{n}p_{k,n-k}u^{k}$. Hence, we obtain
$\sum\limits\limits_{k=0}^{n}p_{k,n-k}u^{k}=0$.

Recall that $\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}=p_{k}$ for every
$k\in\left\\{0,1,\ldots,n\right\\}$ (by the definition of the $p_{k,i}$).
Thus, $\sum\limits\limits_{i\in\mathbb{N}}p_{n,i}Y^{i}=p_{n}=1$ in
$A\left[Y\right]$, and thus $p_{n,0}=1$ (by comparing coefficients before
$Y^{0}$).

Define an $\left(n+1\right)$-tuple $\left(a_{0},a_{1},\ldots,a_{n}\right)\in
A^{n+1}$ by $a_{k}=p_{k,n-k}$ for every $k\in\left\\{0,1,\ldots,n\right\\}$.
Then, $a_{n}=p_{n,0}=1$. Besides,

$\sum\limits\limits_{k=0}^{n}\underbrace{a_{k}}_{=p_{k,n-k}}u^{k}=\sum\limits\limits_{k=0}^{n}p_{k,n-k}u^{k}=0.$

Finally, for every $k\in\left\\{0,1,\ldots,n\right\\}$, we have
$n-k\in\mathbb{N}$ and thus $a_{k}=p_{k,n-k}\in I_{n-k}J_{n-k}$ (since
$p_{k,i}\in I_{i}J_{n-k}$ for every $i\in\mathbb{N}$). Renaming the variable
$k$ as $i$ in this statement, we obtain the following: For every
$i\in\left\\{0,1,\ldots,n\right\\}$, we have $a_{i}\in I_{n-i}J_{n-i}$.

Altogether, we now know that the $\left(n+1\right)$-tuple
$\left(a_{0},a_{1},\ldots,a_{n}\right)\in A^{n+1}$ satisfies

$\sum\limits\limits_{k=0}^{n}a_{k}u^{k}=0,\ \ \ \ \ \ \ \ \ \ a_{n}=1,\ \ \ \
\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ a_{i}\in I_{n-i}J_{n-i}\text{ for
every }i\in\left\\{0,1,\ldots,n\right\\}.$

Thus, by Definition 2.4 (applied to
$\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}$ instead of
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$), the element $u$ is $n$-integral
over $\left(A,\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. This
proves the $\Longleftarrow$ direction of Theorem 3.3 (b), and thus Theorem 3.3
(b) is shown.

The reason why Theorem 3.3 (b) generalizes Theorem 2.14 (more precisely,
Theorem 2.14 is the particular case of Theorem 3.3 (b) for $J_{\rho}=A$) is
the following fact, which we mention here for the pure sake of completeness:

######  Theorem 3.5.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $n\in\mathbb{N}$. Let $u\in
B$.

We know that $\left(A\right)_{\rho\in\mathbb{N}}$ is an ideal semifiltration
of $A$ (according to Theorem 3.1 (a)).

Then, the element $u$ of $B$ is $n$-integral over
$\left(A,\left(A\right)_{\rho\in\mathbb{N}}\right)$ if and only if $u$ is
$n$-integral over $A$.

###  3.3 Integrality of products over the product semifiltration

Finally, let us generalize Theorem 2.20:

######  Theorem 3.6.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ and
$\left(J_{\rho}\right)_{\rho\in\mathbb{N}}$ be two ideal semifiltrations of
$A$.

Let $x\in B$ and $y\in B$. Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$. Assume
that $x$ is $m$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$, and that $y$ is
$n$-integral over $\left(A,\left(J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$.
Then, $xy$ is $nm$-integral over
$\left(A,\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$.

The proof of this theorem will require a generalization of Lemma 2.21:

######  Lemma 3.7.

Let $A$ be a ring. Let $A^{\prime}$ be an $A$-algebra. Let $B^{\prime}$ be an
$A^{\prime}$-algebra. Let $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an
ideal semifiltration of $A$. Let $v\in B^{\prime}$. Let $n\in\mathbb{N}$.
Assume that $v$ is $n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. (Here, of course,
we are using the fact that $B^{\prime}$ is an $A$-algebra, since $B^{\prime}$
is an $A^{\prime}$-algebra while $A^{\prime}$ is an $A$-algebra.)

Then, $v$ is $n$-integral over
$\left(A^{\prime},\left(I_{\rho}A^{\prime}\right)_{\rho\in\mathbb{N}}\right)$.
(Note that $\left(I_{\rho}A^{\prime}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A^{\prime}$, according to Lemma 2.24.)

######  Proof 3.8 (Proof of Lemma 3.7.).

This becomes obvious upon unraveling the definitions of “$n$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$” and of
“$n$-integral over
$\left(A^{\prime},\left(I_{\rho}A^{\prime}\right)_{\rho\in\mathbb{N}}\right)$”,
and by realizing that every $\rho\in\mathbb{N}$ and every $a\in I_{\rho}$
satisfy $a\cdot 1_{A^{\prime}}\in I_{\rho}A^{\prime}$. (See [7] for details.)

######  Proof 3.9 (Proof of Theorem 3.6.).

We have
$\left(J_{\rho}\right)_{\rho\in\mathbb{N}}=\left(J_{\tau}\right)_{\tau\in\mathbb{N}}$.
Hence, $y$ is $n$-integral over
$\left(A,\left(J_{\tau}\right)_{\tau\in\mathbb{N}}\right)$ (since $y$ is
$n$-integral over $\left(A,\left(J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$).
Also, $\left(J_{\tau}\right)_{\tau\in\mathbb{N}}$ is an ideal semifiltration
of $A$ (since $\left(J_{\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$, but we have
$\left(J_{\rho}\right)_{\rho\in\mathbb{N}}=\left(J_{\tau}\right)_{\tau\in\mathbb{N}}$).
Thus, $\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$ is an
ideal semifiltration of $A_{\left[I\right]}$ (by Lemma 2.24, applied to
$A_{\left[I\right]}$ and $\left(J_{\tau}\right)_{\tau\in\mathbb{N}}$ instead
of $A^{\prime}$ and $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$).

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. We will
abbreviate this $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ by
$A_{\left[I\right]}$. Thus,
$A_{\left[I\right]}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ is a subring of $A\left[Y\right]$. Hence, $B\left[Y\right]$ is an
$A_{\left[I\right]}$-algebra (since $B\left[Y\right]$ is an
$A\left[Y\right]$-algebra as explained in Definition 2.8).

Theorem 2.14 (applied to $x$ and $m$ instead of $u$ and $n$) yields that $xY$
is $m$-integral over $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ (since $x$ is $m$-integral over
$\left(A,\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\right)$). In other words,
$xY$ is $m$-integral over $A_{\left[I\right]}$ (since
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]=A_{\left[I\right]}$).

On the other hand, $A_{\left[I\right]}$ is an $A$-algebra, and
$B\left[Y\right]$ is an $A_{\left[I\right]}$-algebra. Hence, Lemma 3.7
(applied to $A_{\left[I\right]}$, $B\left[Y\right]$,
$\left(J_{\tau}\right)_{\tau\in\mathbb{N}}$ and $y$ instead of $A^{\prime}$,
$B^{\prime}$, $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ and $v$) yields that
$y$ is $n$-integral over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$
(since $y$ is $n$-integral over
$\left(A,\left(J_{\tau}\right)_{\tau\in\mathbb{N}}\right)$).

Hence, Theorem 2.20 (applied to $A_{\left[I\right]}$, $B\left[Y\right]$,
$\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$, $y$, $xY$, $n$
and $m$ instead of $A$, $B$, $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$, $x$,
$y$, $m$ and $n$, respectively) yields that $y\cdot xY$ is $mn$-integral over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$
(since $xY$ is $m$-integral over $A_{\left[I\right]}$).

Since $y\cdot xY=xyY$ and $mn=nm$, this means that $xyY$ is $nm$-integral over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$.
Hence, Theorem 3.3 (b) (applied to $xy$ and $nm$ instead of $u$ and $n$)
yields that $xy$ is $nm$-integral over
$\left(A,\left(I_{\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$. This
proves Theorem 3.6.

##  4 Accelerating ideal semifiltrations

###  4.1 Definition of $\lambda$-acceleration

We start this section with an obvious observation:

######  Theorem 4.1.

Let $A$ be a ring. Let $\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal
semifiltration of $A$. Let $\lambda\in\mathbb{N}$. Then,
$\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal semifiltration
of $A$.

I refer to the ideal semifiltration
$\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}$ in Theorem 4.1 as the
$\lambda$-acceleration of the ideal semifiltration
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$.

###  4.2 Half-reduction and reduction

Now, Theorem 3.3, itself a generalization of Theorem 2.14, can be generalized
once more:

######  Theorem 4.2.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ and
$\left(J_{\rho}\right)_{\rho\in\mathbb{N}}$ be two ideal semifiltrations of
$A$. Let $n\in\mathbb{N}$. Let $u\in B$. Let $\lambda\in\mathbb{N}$.

We know that $\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$ (according to Theorem 4.1).

Hence, $\left(I_{\lambda\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$ (according to Theorem 3.1 (b), applied to
$\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}$ instead of
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$).

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$.

We will abbreviate this $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ by
$A_{\left[I\right]}$.

(a) The sequence $\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$
is an ideal semifiltration of $A_{\left[I\right]}$.

(b) The element $u$ of $B$ is $n$-integral over
$\left(A,\left(I_{\lambda\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$ if
and only if the element $uY^{\lambda}$ of the polynomial ring
$B\left[Y\right]$ is $n$-integral over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$.
(Here, we are using the fact that $B\left[Y\right]$ is an
$A_{\left[I\right]}$-algebra, because
$A_{\left[I\right]}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$ is a subring of $A\left[Y\right]$ and because $B\left[Y\right]$ is
an $A\left[Y\right]$-algebra as explained in Definition 2.8.)

######  Proof 4.3 (Proof of Theorem 4.2.).

(a) This is precisely Theorem 3.3 (a).

(b) The definition of $A_{\left[I\right]}$ yields

$\displaystyle A_{\left[I\right]}$
$\displaystyle=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}\ \ \ \ \ \ \ \ \ \
\left(\text{by Definition~{}\ref{Definition8}}\right)$
$\displaystyle=\sum\limits\limits_{\ell\in\mathbb{N}}I_{\ell}Y^{\ell}\ \ \ \ \
\ \ \ \ \ \left(\text{here we renamed }i\text{ as }\ell\text{ in the
sum}\right).$

In order to verify Theorem 4.2 (b), we have to prove the $\Longrightarrow$ and
$\Longleftarrow$ statements.

$\Longrightarrow:$ Assume that $u$ is $n$-integral over
$\left(A,\left(I_{\lambda\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$.
Thus, by Definition 2.4 (applied to
$\left(I_{\lambda\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}$ instead of
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$), there exists some
$\left(a_{0},a_{1},\ldots,a_{n}\right)\in A^{n+1}$ such that

$\sum\limits\limits_{k=0}^{n}a_{k}u^{k}=0,\ \ \ \ \ \ \ \ \ \ a_{n}=1,\ \ \ \
\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ a_{i}\in
I_{\lambda\left(n-i\right)}J_{n-i}\text{ for every
}i\in\left\\{0,1,\ldots,n\right\\}.$

Consider this $\left(a_{0},a_{1},\ldots,a_{n}\right)$.

For each $k\in\left\\{0,1,\ldots,n\right\\}$, we have $a_{k}\in
I_{\lambda\left(n-k\right)}J_{n-k}\subseteq I_{\lambda\left(n-k\right)}$
(since $I_{\lambda\left(n-k\right)}$ is an ideal of $A$) and thus
$a_{k}Y^{\lambda\left(n-k\right)}\in
I_{\lambda\left(n-k\right)}Y^{\lambda\left(n-k\right)}\subseteq\sum\limits_{i\in\mathbb{N}}I_{i}Y^{i}=A_{\left[I\right]}$.
Thus, we can find an $\left(n+1\right)$-tuple
$\left(b_{0},b_{1},\ldots,b_{n}\right)\in\left(A_{\left[I\right]}\right)^{n+1}$
satisfying

$\sum\limits\limits_{k=0}^{n}b_{k}\cdot\left(uY^{\lambda}\right)^{k}=0,\ \ \ \
\ \ \ \ \ \ b_{n}=1,\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ b_{i}\in
J_{n-i}A_{\left[I\right]}\text{ for every }i\in\left\\{0,1,\ldots,n\right\\}.$
777Namely, the $\left(n+1\right)$-tuple
$\left(b_{0},b_{1},\ldots,b_{n}\right)\in\left(A_{\left[I\right]}\right)^{n+1}$
defined by $\left(b_{k}=a_{k}Y^{\lambda\left(n-k\right)}\text{ for every
}k\in\left\\{0,1,\ldots,n\right\\}\right)$ satisfies this. The proof is very
easy (see [7] for details).

Hence, by Definition 2.4 (applied to $A_{\left[I\right]}$, $B\left[Y\right]$,
$\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$, $uY^{\lambda}$
and $\left(b_{0},b_{1},\ldots,b_{n}\right)$ instead of $A$, $B$,
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$, $u$ and
$\left(a_{0},a_{1},\ldots,a_{n}\right)$), the element $uY^{\lambda}$ is
$n$-integral over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$.
This proves the $\Longrightarrow$ direction of Theorem 4.2 (b).

$\Longleftarrow:$ Assume that $uY^{\lambda}$ is $n$-integral over
$\left(A_{\left[I\right]},\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)$.
Thus, by Definition 2.4 (applied to $A_{\left[I\right]}$, $B\left[Y\right]$,
$\left(J_{\tau}A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}$, $uY^{\lambda}$
and $\left(p_{0},p_{1},\ldots,p_{n}\right)$ instead of $A$, $B$,
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$, $u$ and
$\left(a_{0},a_{1},\ldots,a_{n}\right)$), there exists some
$\left(p_{0},p_{1},\ldots,p_{n}\right)\in\left(A_{\left[I\right]}\right)^{n+1}$
such that

$\sum\limits\limits_{k=0}^{n}p_{k}\cdot\left(uY^{\lambda}\right)^{k}=0,\ \ \ \
\ \ \ \ \ \ p_{n}=1,\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ p_{i}\in
J_{n-i}A_{\left[I\right]}\text{ for every }i\in\left\\{0,1,\ldots,n\right\\}.$

Consider this $\left(p_{0},p_{1},\ldots,p_{n}\right)$. For every
$k\in\left\\{0,1,\ldots,n\right\\}$, we have

$\displaystyle p_{k}$ $\displaystyle\in
J_{n-k}A_{\left[I\right]}=J_{n-k}\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}\
\ \ \ \ \ \ \ \ \ \left(\text{since
}A_{\left[I\right]}=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}\right)$
$\displaystyle=\sum\limits\limits_{i\in\mathbb{N}}J_{n-k}I_{i}Y^{i}=\sum\limits\limits_{i\in\mathbb{N}}I_{i}J_{n-k}Y^{i},$

and thus there exists a sequence $\left(p_{k,i}\right)_{i\in\mathbb{N}}\in
A^{\mathbb{N}}$ such that
$p_{k}=\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}$, such that
$\left(p_{k,i}\in I_{i}J_{n-k}\text{ for every }i\in\mathbb{N}\right)$, and
such that only finitely many $i\in\mathbb{N}$ satisfy $p_{k,i}\neq 0$.
Consider this sequence. Thus,

$\sum\limits\limits_{k=0}^{n}\underbrace{p_{k}}_{=\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}}\cdot\left(uY^{\lambda}\right)^{k}=\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}\underbrace{Y^{i}\cdot\left(uY^{\lambda}\right)^{k}}_{=u^{k}Y^{i+\lambda
k}}=\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}u^{k}Y^{i+\lambda
k}.$

Compared with
$\sum\limits\limits_{k=0}^{n}p_{k}\cdot\left(uY^{\lambda}\right)^{k}=0$, this
yields
$\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}u^{k}Y^{i+\lambda
k}=0$. In other words, the polynomial
$\sum\limits\limits_{k=0}^{n}\sum\limits\limits_{i\in\mathbb{N}}\underbrace{p_{k,i}u^{k}}_{\in
B}Y^{i+\lambda k}\in B\left[Y\right]$ equals $0$. Hence, its coefficient
before $Y^{\lambda n}$ equals $0$ as well. But its coefficient before
$Y^{\lambda n}$ is
$\sum\limits\limits_{k=0}^{n}p_{k,\lambda\left(n-k\right)}u^{k}$ (since
$i+\lambda k=\lambda n$ holds if and only if $i=\lambda\left(n-k\right)$).
Hence, $\sum\limits\limits_{k=0}^{n}p_{k,\lambda\left(n-k\right)}u^{k}=0$.

Recall that $\sum\limits\limits_{i\in\mathbb{N}}p_{k,i}Y^{i}=p_{k}$ for every
$k\in\left\\{0,1,\ldots,n\right\\}$ (by the definition of the $p_{k,i}$).
Thus, $\sum\limits\limits_{i\in\mathbb{N}}p_{n,i}Y^{i}=p_{n}=1$ in
$A\left[Y\right]$, and thus $p_{n,0}=1$ (by comparing coefficients before
$Y^{0}$).

Define an $\left(n+1\right)$-tuple $\left(a_{0},a_{1},\ldots,a_{n}\right)\in
A^{n+1}$ by $a_{k}=p_{k,\lambda\left(n-k\right)}$ for every
$k\in\left\\{0,1,\ldots,n\right\\}$. Then, $a_{n}=p_{n,0}=1$. Besides,

$\sum\limits\limits_{k=0}^{n}\underbrace{a_{k}}_{=p_{k,\lambda\left(n-k\right)}}u^{k}=\sum\limits\limits_{k=0}^{n}p_{k,\lambda\left(n-k\right)}u^{k}=0.$

Finally, for every $k\in\left\\{0,1,\ldots,n\right\\}$, we have
$a_{k}=p_{k,\lambda\left(n-k\right)}\in I_{\lambda\left(n-k\right)}J_{n-k}$
(since $p_{k,i}\in I_{i}J_{n-k}$ for every $i\in\mathbb{N}$). Renaming the
variable $k$ as $i$ in this statement, we obtain the following: For every
$i\in\left\\{0,1,\ldots,n\right\\}$, we have $a_{i}\in
I_{\lambda\left(n-i\right)}J_{n-i}$.

Altogether, we now know that

$\sum\limits\limits_{k=0}^{n}a_{k}u^{k}=0,\ \ \ \ \ \ \ \ \ \ a_{n}=1,\ \ \ \
\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ a_{i}\in
I_{\lambda\left(n-i\right)}J_{n-i}\text{ for every
}i\in\left\\{0,1,\ldots,n\right\\}.$

Thus, by Definition 2.4 (applied to
$\left(I_{\lambda\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}$ instead of
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$), the element $u$ is $n$-integral
over
$\left(A,\left(I_{\lambda\rho}J_{\rho}\right)_{\rho\in\mathbb{N}}\right)$.
This proves the $\Longleftarrow$ direction of Theorem 4.2 (b), and thus
completes the proof.

A particular case of Theorem 4.2 (b) is the following fact:

######  Theorem 4.4.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Let $n\in\mathbb{N}$. Let $u\in B$. Let $\lambda\in\mathbb{N}$.

We know that $\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$ (according to Theorem 4.1).

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ defined in
Definition 2.9.

Then, the element $u$ of $B$ is $n$-integral over
$\left(A,\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}\right)$ if and only
if the element $uY^{\lambda}$ of the polynomial ring $B\left[Y\right]$ is
$n$-integral over the ring
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. (Here, we are
using the fact that $B\left[Y\right]$ is an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra,
because $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ is a
subring of $A\left[Y\right]$ and because $B\left[Y\right]$ is an
$A\left[Y\right]$-algebra as explained in Definition 2.8.)

######  Proof 4.5 (Proof of Theorem 4.4.).

Theorem 3.1 (a) states that $\left(A\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$.

Every $\rho\in\mathbb{N}$ satisfies $I_{\lambda\rho}=I_{\lambda\rho}A$ (since
$I_{\lambda\rho}$ is an ideal of $A$). Thus,
$\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}=\left(I_{\lambda\rho}A\right)_{\rho\in\mathbb{N}}$.

We will abbreviate the $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ of
$A\left[Y\right]$ by $A_{\left[I\right]}$. Thus, $B\left[Y\right]$ is an
$A_{\left[I\right]}$-algebra (since $B\left[Y\right]$ is an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra).

It is easy to see that $AA_{\left[I\right]}=A_{\left[I\right]}$ (since
$A_{\left[I\right]}$ is an $A$-algebra). Hence,
$\left(AA_{\left[I\right]}\right)_{\tau\in\mathbb{N}}=\left(A_{\left[I\right]}\right)_{\tau\in\mathbb{N}}=\left(A_{\left[I\right]}\right)_{\rho\in\mathbb{N}}$.

Now, we have the following chain of equivalences:

$\displaystyle\ \left(u\text{ is }n\text{-integral over
}\left(A,\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}\right)\right)$
$\displaystyle\Longleftrightarrow\ \left(u\text{ is }n\text{-integral over
}\left(A,\left(I_{\lambda\rho}A\right)_{\rho\in\mathbb{N}}\right)\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{since
}\left(I_{\lambda\rho}\right)_{\rho\in\mathbb{N}}=\left(I_{\lambda\rho}A\right)_{\rho\in\mathbb{N}}\right)$
$\displaystyle\Longleftrightarrow\ \left(uY^{\lambda}\text{ is
}n\text{-integral over
}\left(A_{\left[I\right]},\left(AA_{\left[I\right]}\right)_{\tau\in\mathbb{N}}\right)\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{by Theorem~{}\ref{Theorem15}
{(b)}, applied to
}\left(J_{\rho}\right)_{\rho\in\mathbb{N}}=\left(A\right)_{\rho\in\mathbb{N}}\right)$
$\displaystyle\Longleftrightarrow\ \left(uY^{\lambda}\text{ is
}n\text{-integral over
}\left(A_{\left[I\right]},\left(A_{\left[I\right]}\right)_{\rho\in\mathbb{N}}\right)\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{since
}\left(AA_{\left[I\right]}\right)_{\tau\in\mathbb{N}}=\left(A_{\left[I\right]}\right)_{\rho\in\mathbb{N}}\right)$
$\displaystyle\Longleftrightarrow\ \left(uY^{\lambda}\text{ is
}n\text{-integral over }A_{\left[I\right]}\right)$ $\displaystyle\ \ \ \ \ \ \
\ \ \ \left(\text{by Theorem~{}\ref{Theorem12}, applied to
}A_{\left[I\right]}\text{, }B\left[Y\right]\text{ and }uY^{\lambda}\text{
instead of }A\text{, }B\text{ and }u\right)$
$\displaystyle\Longleftrightarrow\ \left(uY^{\lambda}\text{ is
}n\text{-integral over }A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right)$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{since
}A_{\left[I\right]}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]\right).$

This proves Theorem 4.4.

Note that Theorem 2.14 is the particular case of Theorem 4.4 for $\lambda=1$.

Finally we can generalize even Theorem 1.18:

######  Theorem 4.6.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let
$\left(I_{\rho}\right)_{\rho\in\mathbb{N}}$ be an ideal semifiltration of $A$.
Let $n\in\mathbb{N}^{+}$. Let $v\in B$. Let $a_{0},a_{1},\ldots,a_{n}$ be
$n+1$ elements of $A$ such that $\sum\limits\limits_{i=0}^{n}a_{i}v^{i}=0$.
Assume further that $a_{i}\in I_{n-i}$ for every
$i\in\left\\{0,1,\ldots,n\right\\}$.

Let $k\in\left\\{0,1,\ldots,n\right\\}$. We know that
$\left(I_{\left(n-k\right)\rho}\right)_{\rho\in\mathbb{N}}$ is an ideal
semifiltration of $A$ (according to Theorem 4.1, applied to $\lambda=n-k$).

Then, $\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$ is $n$-integral over
$\left(A,\left(I_{\left(n-k\right)\rho}\right)_{\rho\in\mathbb{N}}\right)$.

######  Proof 4.7 (Proof of Theorem 4.6.).

Consider the polynomial ring $A\left[Y\right]$ and its $A$-subalgebra
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ defined in
Definition 2.9. Note that
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$ is a subring
of $A\left[Y\right]$; hence, $B\left[Y\right]$ is an
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$-algebra
(because $B\left[Y\right]$ is an $A\left[Y\right]$-algebra as explained in
Definition 2.8).

Definition 2.9 yields

$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]=\sum\limits\limits_{i\in\mathbb{N}}I_{i}Y^{i}=\sum\limits\limits_{\ell\in\mathbb{N}}I_{\ell}Y^{\ell}\
\ \ \ \ \ \ \ \ \ \left(\text{here we renamed }i\text{ as }\ell\text{ in the
sum}\right).$

Hence,
$\sum\limits\limits_{\ell\in\mathbb{N}}I_{\ell}Y^{\ell}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$.

Define $u\in B$ by

$u=\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}.$ (11)

In the ring $B\left[Y\right]$, we have

$\sum\limits_{i=0}^{n}a_{i}Y^{n-i}\underbrace{\left(vY\right)^{i}}_{=v^{i}Y^{i}=Y^{i}v^{i}}=\sum\limits_{i=0}^{n}a_{i}\underbrace{Y^{n-i}Y^{i}}_{=Y^{n}}v^{i}=Y^{n}\underbrace{\sum\limits_{i=0}^{n}a_{i}v^{i}}_{=0}=0.$

Besides, every $i\in\left\\{0,1,\ldots,n\right\\}$ satisfies

$\underbrace{a_{i}}_{\begin{subarray}{c}\in I_{n-i}\\\ \text{(by
assumption)}\end{subarray}}Y^{n-i}\in
I_{n-i}Y^{n-i}\subseteq\sum\limits\limits_{\ell\in\mathbb{N}}I_{\ell}Y^{\ell}=A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right].$

In other words, $a_{0}Y^{n-0},a_{1}Y^{n-1},\ldots,a_{n}Y^{n-n}$ are $n+1$
elements of $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$.
Hence, Theorem 1.18 (applied to
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$,
$B\left[Y\right]$, $vY$ and $a_{i}Y^{n-i}$ instead of $A$, $B$, $v$ and
$a_{i}$) yields that
$\sum\limits\limits_{i=0}^{n-k}a_{i+k}Y^{n-\left(i+k\right)}\left(vY\right)^{i}$
is $n$-integral over $A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast
Y\right]$. Since

$\sum\limits\limits_{i=0}^{n-k}a_{i+k}Y^{n-\left(i+k\right)}\underbrace{\left(vY\right)^{i}}_{=v^{i}Y^{i}=Y^{i}v^{i}}=\sum\limits\limits_{i=0}^{n-k}a_{i+k}\underbrace{Y^{n-\left(i+k\right)}Y^{i}}_{=Y^{\left(n-\left(i+k\right)\right)+i}=Y^{n-k}}v^{i}=\underbrace{\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}}_{\begin{subarray}{c}=u\\\
\text{(by (\ref{pf.Theorem17.u=}))}\end{subarray}}\cdot Y^{n-k}=uY^{n-k},$

this means that $uY^{n-k}$ is $n$-integral over
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$.

But Theorem 4.4 (applied to $\lambda=n-k$) yields that $u$ is $n$-integral
over
$\left(A,\left(I_{\left(n-k\right)\rho}\right)_{\rho\in\mathbb{N}}\right)$ if
and only if $uY^{n-k}$ is $n$-integral over the ring
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$. Since we know
that $uY^{n-k}$ is $n$-integral over the ring
$A\left[\left(I_{\rho}\right)_{\rho\in\mathbb{N}}\ast Y\right]$, this yields
that $u$ is $n$-integral over
$\left(A,\left(I_{\left(n-k\right)\rho}\right)_{\rho\in\mathbb{N}}\right)$. In
other words, $\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$ is $n$-integral over
$\left(A,\left(I_{\left(n-k\right)\rho}\right)_{\rho\in\mathbb{N}}\right)$
(since $u=\sum\limits\limits_{i=0}^{n-k}a_{i+k}v^{i}$). This proves Theorem
4.6.

##  5 On a lemma by Lombardi

###  5.1 A lemma on products of powers

Now, we shall show a rather technical lemma:

######  Lemma 5.1.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $x\in B$. Let
$m\in\mathbb{N}$ and $n\in\mathbb{N}$. Let $u\in B$. Let $\mu\in\mathbb{N}$
and $\nu\in\mathbb{N}$ be such that $\mu+\nu\in\mathbb{N}^{+}$. Assume that

$u^{n}\in\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\nu}\right\rangle_{A}$ (12)

and that

$\displaystyle u^{m}x^{\mu}$ $\displaystyle\in\left\langle
u^{0},u^{1},\ldots,u^{m-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu}\right\rangle_{A}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\left\langle u^{0},u^{1},\ldots,u^{m}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu-1}\right\rangle_{A}.$ (13)

Then, $u$ is $\left(n\mu+m\nu\right)$-integral over $A$.

This lemma can be seen as a variant of [6, Theorem 2]888Caveat: The notion
“integral over $\left(A,J\right)$” defined in [6] has nothing to do with our
notion “$n$-integral over
$\left(A,\left(I_{n}\right)_{n\in\mathbb{N}}\right)$”.. Indeed, the particular
case of [6, Theorem 2] when $J=0$ can easily be obtained from Lemma 5.1
(applied to $x$ and $\alpha$ instead of $u$ and $x$).

The proof of Lemma 5.1 is not difficult but rather elaborate. For a completely
detailed writeup of this proof, see [7]. Here let me give the skeleton of the
proof:

######  Proof 5.2 (Proof of Lemma 5.1 (sketched).).

Define the set

$\displaystyle S$
$\displaystyle=\left(\left\\{0,1,\ldots,n-1\right\\}\times\left\\{0,1,\ldots,\mu-1\right\\}\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \
\cup\left(\left\\{0,1,\ldots,m-1\right\\}\times\left\\{\mu,\mu+1,\ldots,\mu+\nu-1\right\\}\right).$

Clearly, $\left|S\right|=n\mu+m\nu$ and

$j<\mu+\nu\text{ for every }\left(i,j\right)\in S.$ (14)

Let $U$ be the $A$-submodule $\left\langle u^{i}x^{j}\ \mid\
\left(i,j\right)\in S\right\rangle_{A}$ of $B$. Then, $U$ is an
$\left(n\mu+m\nu\right)$-generated $A$-module (since
$\left|S\right|=n\mu+m\nu$). Besides, clearly,

$u^{i}x^{j}\in U\text{ for every }\left(i,j\right)\in S.$ (15)

Now, we will show that

$\text{every }i\in\mathbb{N}\text{ and }j\in\mathbb{N}\text{ satisfying
}j<\mu+\nu\text{ satisfy }u^{i}x^{j}\in U.$ (16)

[The proof of (16) can be done either by double induction (over $i$ and over
$j$) or by the well-ordering principle. The induction proof has the advantage
that it is completely constructive, but it is clumsy (I give this induction
proof in [7]). So, for the sake of brevity, the proof I am going to give here
is by the well-ordering principle:

For the sake of contradiction, we assume that (16) is not true. That is, there
exists some pair $\left(i,j\right)\in\mathbb{N}^{2}$ satisfying $j<\mu+\nu$
but not satisfying $u^{i}x^{j}\in U$. Let $\left(I,J\right)$ be the
lexicographically smallest999“Lexicographically smallest” means “smallest with
respect to the lexicographic order”. Here, the lexicographic order on
$\mathbb{N}^{2}$ is defined to be the total order on $\mathbb{N}^{2}$ in which
two pairs $\left(a_{1},b_{1}\right)\in\mathbb{N}^{2}$ and
$\left(a_{2},b_{2}\right)\in\mathbb{N}^{2}$ satisfy
$\left(a_{1},b_{1}\right)<\left(a_{2},b_{2}\right)$ if and only if either
$a_{1}<a_{2}$ or $\left(a_{1}=a_{2}\text{ and }b_{1}<b_{2}\right)$. It is
well-known that this total order is well-defined and turns $\mathbb{N}^{2}$
into a well-ordered set. such pair101010This is well-defined, since the
lexicographic order is a well-ordering on $\mathbb{N}^{2}$.. Then, $J<\mu+\nu$
but $u^{I}x^{J}\notin U$, and since $\left(I,J\right)$ is the
lexicographically smallest such pair, we have

$u^{I}x^{j}\in U\text{ for every }j\in\mathbb{N}\text{ such that }j<J$ (17)

and

$u^{i}x^{j}\in U\text{ for every }i\in\mathbb{N}\text{ and
}j\in\mathbb{N}\text{ such that }i<I\text{ and }j<\mu+\nu.$ (18)

Recall that $U$ is an $A$-module. Hence, (17) entails

$\left\langle u^{I}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}\subseteq U,$ (19)

and (18) entails

$\left\langle u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu+\nu-1}\right\rangle_{A}\subseteq U.$ (20)

Also, from $J<\mu+\nu$, we obtain $J\leq\mu+\nu-1$ (since $J$ and $\mu+\nu$
are integers).

We now want to prove that $u^{I}x^{J}\in U$.

We are in one of the following four cases:

Case 1: We have $I\geq m$ and $J\geq\mu$.

Case 2: We have $I<m$ and $J\geq\mu$.

Case 3: We have $I\geq n$ and $J<\mu$.

Case 4: We have $I<n$ and $J<\mu$.

In Case 1, we have $I-m\geq 0$ (since $I\geq m$) and $J-\mu\geq 0$ (since
$J\geq\mu$), thus

$\displaystyle\underbrace{u^{I}}_{=u^{I-m}u^{m}}\underbrace{x^{J}}_{=x^{\mu}x^{J-\mu}}$
$\displaystyle=u^{I-m}\underbrace{u^{m}x^{\mu}}_{\begin{subarray}{c}\in\left\langle
u^{0},u^{1},\ldots,u^{m-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu}\right\rangle_{A}+\left\langle
u^{0},u^{1},\ldots,u^{m}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu-1}\right\rangle_{A}\\\ \left(\text{by
(\ref{L18-2})}\right)\end{subarray}}x^{J-\mu}$ $\displaystyle\in
u^{I-m}\left(\left\langle
u^{0},u^{1},\ldots,u^{m-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu}\right\rangle_{A}\right.$ $\displaystyle\ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.+\left\langle
u^{0},u^{1},\ldots,u^{m}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu-1}\right\rangle_{A}\right)x^{J-\mu}$
$\displaystyle=\underbrace{u^{I-m}\left\langle
u^{0},u^{1},\ldots,u^{m-1}\right\rangle_{A}}_{\begin{subarray}{c}=\left\langle
u^{I-m},u^{I-m+1},\ldots,u^{I-1}\right\rangle_{A}\\\ \subseteq\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}\end{subarray}}\cdot\underbrace{\left\langle
x^{0},x^{1},\ldots,x^{\mu}\right\rangle_{A}x^{J-\mu}}_{\begin{subarray}{c}=\left\langle
x^{J-\mu},x^{J-\mu+1},\ldots,x^{J}\right\rangle_{A}\\\ \subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu+\nu-1}\right\rangle_{A}\\\ \text{(since }J-\mu\geq
0\text{ and }J\leq\mu+\nu-1\text{)}\end{subarray}}$ $\displaystyle\ \ \ \ \ \
\ \ \ \ +\underbrace{u^{I-m}\left\langle
u^{0},u^{1},\ldots,u^{m}\right\rangle_{A}}_{\begin{subarray}{c}=\left\langle
u^{I-m},u^{I-m+1},\ldots,u^{I}\right\rangle_{A}\\\ \subseteq\left\langle
u^{0},u^{1},\ldots,u^{I}\right\rangle_{A}\end{subarray}}\cdot\underbrace{\left\langle
x^{0},x^{1},\ldots,x^{\mu-1}\right\rangle_{A}x^{J-\mu}}_{\begin{subarray}{c}=\left\langle
x^{J-\mu},x^{J-\mu+1},\ldots,x^{J-1}\right\rangle_{A}\\\ \subseteq\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}\end{subarray}}$
$\displaystyle\subseteq\underbrace{\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu+\nu-1}\right\rangle_{A}}_{\begin{subarray}{c}\subseteq
U\\\ \text{(by (\ref{L18-indL2final}))}\end{subarray}}$ $\displaystyle\ \ \ \
\ \ \ \ \ \ +\underbrace{\left\langle
u^{0},u^{1},\ldots,u^{I}\right\rangle_{A}}_{=\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}+\left\langle
u^{I}\right\rangle_{A}}\cdot\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}$ $\displaystyle\subseteq
U+\underbrace{\left(\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}+\left\langle
u^{I}\right\rangle_{A}\right)\cdot\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}}_{=\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}+\left\langle
u^{I}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}}$ $\displaystyle=U+\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}\cdot\underbrace{\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}}_{\begin{subarray}{c}\subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu+\nu-1}\right\rangle_{A}\\\ \text{(since }J-1\leq
J\leq\mu+\nu-1\text{)}\end{subarray}}+\left\langle
u^{I}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}$ $\displaystyle\subseteq
U+\underbrace{\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu+\nu-1}\right\rangle_{A}}_{\begin{subarray}{c}\subseteq
U\\\ \text{(by
(\ref{L18-indL2final}))}\end{subarray}}+\underbrace{\left\langle
u^{I}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{J-1}\right\rangle_{A}}_{\begin{subarray}{c}\subseteq
U\\\ \text{(by (\ref{L18-indL1final}))}\end{subarray}}$
$\displaystyle\subseteq U+U+U\subseteq U\ \ \ \ \ \ \ \ \ \ \left(\text{since
}U\text{ is an }A\text{-module}\right).$

Thus, we have proved that $u^{I}x^{J}\in U$ holds in Case 1.

In Case 2, we have
$\left(I,J\right)\in\left\\{0,1,\ldots,m-1\right\\}\times\left\\{\mu,\mu+1,\ldots,\mu+\nu-1\right\\}\subseteq
S$ and thus $u^{I}x^{J}\in U$ (by (15), applied to $I$ and $J$ instead of $i$
and $j$). Thus, we have proved that $u^{I}x^{J}\in U$ holds in Case 2.

In Case 3, we have $I-n\geq 0$ (since $I\geq n$) and $J+\nu\leq\mu+\nu-1$
(since $J<\mu$ yields $J+\nu<\mu+\nu$, and since $J+\nu$ and $\mu+\nu$ are
integers), thus

$\displaystyle\underbrace{u^{I}}_{=u^{I-n}u^{n}}x^{J}$
$\displaystyle=u^{I-n}\underbrace{u^{n}}_{\begin{subarray}{c}\in\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\nu}\right\rangle_{A}\\\ \left(\text{by
(\ref{L18-1})}\right)\end{subarray}}x^{J}$
$\displaystyle\in\underbrace{u^{I-n}\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}}_{\begin{subarray}{c}=\left\langle
u^{I-n},u^{I-n+1},\ldots,u^{I-1}\right\rangle_{A}\\\ \subseteq\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}\end{subarray}}\cdot\underbrace{\left\langle
x^{0},x^{1},\ldots,x^{\nu}\right\rangle_{A}x^{J}}_{\begin{subarray}{c}=\left\langle
x^{J},x^{J+1},\ldots,x^{J+\nu}\right\rangle_{A}\\\ \subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu+\nu-1}\right\rangle_{A}\\\ \text{(since
}J+\nu\leq\mu+\nu-1\text{)}\end{subarray}}$
$\displaystyle\subseteq\left\langle
u^{0},u^{1},\ldots,u^{I-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\mu+\nu-1}\right\rangle_{A}\subseteq U\ \ \ \ \ \ \ \ \
\ \left(\text{by (\ref{L18-indL2final})}\right).$

Thus, we have proved that $u^{I}x^{J}\in U$ holds in Case 3.

In Case 4, we have
$\left(I,J\right)\in\left\\{0,1,\ldots,n-1\right\\}\times\left\\{0,1,\ldots,\mu-1\right\\}\subseteq
S$ and thus $u^{I}x^{J}\in U$ (by (15), applied to $I$ and $J$ instead of $i$
and $j$). Thus, we have proved that $u^{I}x^{J}\in U$ holds in Case 4.

By now, we have proved that $u^{I}x^{J}\in U$ holds in each of the four cases
1, 2, 3 and 4. Hence, $u^{I}x^{J}\in U$ always holds, contradicting
$u^{I}x^{J}\notin U$. This contradiction completes the proof of (16).]

Now that (16) is proven, we can easily conclude that $uU\subseteq U$ (since
every $\left(i,j\right)\in S$ satisfies $j<\mu+\nu$, and thus (16) shows that
$u^{i+1}x^{j}\in U$) and $1\in U$ (this follows by applying (16) to $i=0$ and
$j=0$). Altogether, $U$ is an $\left(n\mu+m\nu\right)$-generated $A$-submodule
of $B$ such that $1\in U$ and $uU\subseteq U$. Thus, $u\in B$ satisfies
Assertion $\mathcal{C}$ of Theorem 1.1 with $n$ replaced by $n\mu+m\nu$.
Hence, $u\in B$ satisfies the four equivalent assertions $\mathcal{A}$,
$\mathcal{B}$, $\mathcal{C}$ and $\mathcal{D}$ of Theorem 1.1 with $n$
replaced by $n\mu+m\nu$. Consequently, $u$ is
$\left(n\mu+m\nu\right)$-integral over $A$. This proves Lemma 5.1.

We record a weaker variant of Lemma 5.1:

######  Lemma 5.3.

Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $x\in B$ and $y\in B$ be
such that $xy\in A$. Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$. Let $u\in B$.
Let $\mu\in\mathbb{N}$ and $\nu\in\mathbb{N}$ be such that
$\mu+\nu\in\mathbb{N}^{+}$. Assume that

$u^{n}\in\left\langle
u^{0},u^{1},\ldots,u^{n-1}\right\rangle_{A}\cdot\left\langle
x^{0},x^{1},\ldots,x^{\nu}\right\rangle_{A}$ (21)

and that

$\displaystyle u^{m}$ $\displaystyle\in\left\langle
u^{0},u^{1},\ldots,u^{m-1}\right\rangle_{A}\cdot\left\langle
y^{0},y^{1},\ldots,y^{\mu}\right\rangle_{A}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\left\langle u^{0},u^{1},\ldots,u^{m}\right\rangle_{A}\cdot\left\langle
y^{1},y^{2},\ldots,y^{\mu}\right\rangle_{A}.$ (22)

Then, $u$ is $\left(n\mu+m\nu\right)$-integral over $A$.

######  Proof 5.4 (Proof of Lemma 5.3.).

(Again, this proof appears in greater detail in [7].) We have

$\left\langle
y^{0},y^{1},\ldots,y^{\mu}\right\rangle_{A}x^{\mu}\subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu}\right\rangle_{A},$ (23)

since every $i\in\left\\{0,1,\ldots,\mu\right\\}$ satisfies

$\displaystyle y^{i}\underbrace{x^{\mu}}_{=x^{\mu-i}x^{i}}$
$\displaystyle=y^{i}x^{\mu-i}x^{i}=\underbrace{x^{i}y^{i}}_{\begin{subarray}{c}=\left(xy\right)^{i}\in
A\\\ \text{(since }xy\in A\text{)}\end{subarray}}x^{\mu-i}\in
Ax^{\mu-i}=\left\langle x^{\mu-i}\right\rangle_{A}$ (24)
$\displaystyle\subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu}\right\rangle_{A}\ \ \ \ \ \ \ \ \ \
\left(\text{since }\mu-i\in\left\\{0,1,\ldots,\mu\right\\}\right).$

Besides,

$\left\langle
y^{1},y^{2},\ldots,y^{\mu}\right\rangle_{A}x^{\mu}\subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu-1}\right\rangle_{A},$ (25)

since every $i\in\left\\{1,2,\ldots,\mu\right\\}$ satisfies

$\displaystyle y^{i}x^{\mu}$ $\displaystyle\in\left\langle
x^{\mu-i}\right\rangle_{A}\ \ \ \ \ \ \ \ \ \ \left(\text{by
(\ref{L19-P1-short})}\right)$ $\displaystyle\subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu-1}\right\rangle_{A}\ \ \ \ \ \ \ \ \ \
\left(\text{since }\mu-i\in\left\\{0,1,\ldots,\mu-1\right\\}\right).$

Now, (22) yields

$\displaystyle u^{m}x^{\mu}$ $\displaystyle\in\left(\left\langle
u^{0},u^{1},\ldots,u^{m-1}\right\rangle_{A}\cdot\left\langle
y^{0},y^{1},\ldots,y^{\mu}\right\rangle_{A}\right.$ $\displaystyle\ \ \ \ \ \
\ \ \ \ \left.+\left\langle
u^{0},u^{1},\ldots,u^{m}\right\rangle_{A}\cdot\left\langle
y^{1},y^{2},\ldots,y^{\mu}\right\rangle_{A}\right)x^{\mu}$
$\displaystyle=\left\langle
u^{0},u^{1},\ldots,u^{m-1}\right\rangle_{A}\cdot\underbrace{\left\langle
y^{0},y^{1},\ldots,y^{\mu}\right\rangle_{A}x^{\mu}}_{\begin{subarray}{c}\subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu}\right\rangle_{A}\\\ \left(\text{by (\ref{L19-Pa-
short})}\right)\end{subarray}}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+\left\langle
u^{0},u^{1},\ldots,u^{m}\right\rangle_{A}\cdot\underbrace{\left\langle
y^{1},y^{2},\ldots,y^{\mu}\right\rangle_{A}x^{\mu}}_{\begin{subarray}{c}\subseteq\left\langle
x^{0},x^{1},\ldots,x^{\mu-1}\right\rangle_{A}\\\ \left(\text{by (\ref{L19-Pb-
short})}\right)\end{subarray}}$ $\displaystyle\subseteq\left\langle