# Algebraic spectral theory and Serre multiplicity formula

Anar Dosi (Dosiev)  Middle East Technical University Northern Cyprus Campus,
Guzelyurt, KKTC, Mersin 10, Turkey  (dosiev@yahoo.com<EMAIL_ADDRESS>
(Date: April 8, 2021)

###### Abstract.

The present paper is devoted to an algebraic treatment of the joint spectral
theory within the framework of Noetherian modules over an algebra finite
extension of an algebraically closed field. We prove the spectral mapping
theorem and analyze the index of tuples in purely algebraic case. The index
function over tuples from the coordinate ring of a variety is naturally
extended up to a numerical Tor-polynomial. Based on Serre’s multiplicity
formula, we deduce that Tor-polynomial is just the Samuel polynomial of the
local algebra.

###### Key words and phrases:

Taylor spectrum, Noetherian modules, integral extensions, Samuel polynomial,
Serre’s formula

###### 2000 Mathematics Subject Classification:

Primary 47A60, 13D03; Secondary 13D40, 46H30

##  1\. Introduction

The joint spectral theory plays a central role in operator theory and the
complex analytic geometry. Its origin goes back to Gelfand’s commutative
Banach algebras and their representations. The joint spectral theory was
mainly developed within the context of holomorphic functional calculus problem
by J. Taylor [25], [26]. An interesting link to the complex analytic geometry
was found by M. Putinar [23], [13] by constructing Taylor’s holomorphic
functional calculus within the context of Stein spaces. Taylor’s spectral
theory has a strong homological background developed independently by J. L.
Taylor in [26] and A. Ya. Helemskii in [17], that allows to have a further
generalization of the theory in the noncommutative setting. That played a
crucial role in foundations of noncommutative complex analytic geometry (see
[6], [8], [9] and [21]).

It is well known that many key results from the complex analytic geometry have
their analogs in schemes such as the fundamental theorem of Serre on vanishing
[16, 3.3.7]. More interesting sight is to have a scheme-theoretic analog of
the joint spectra found in [10] and [11] that indicates to the fundamental
nature of the joint spectral theory. Furthermore, Taylor’s multivariable
functional calculus has a scheme version suggested in [11]. Namely, Putinar
spectrum can be defined for a module over a scheme and it plays the same
central role in the functional calculus problem over Noetherian schemes as it
does in the complex analytic geometry. Actually, in the affine case the same
spectrum occurred in [18] and [19] based on the result of A. Neeman [20] that
there is a bijection between subsets of $\operatorname{Spec}\left(R\right)$
and localizing subcategories of the derived category of complexes of
$R$-modules. Taylor’s functional calculus through derived categories was
proposed in [22] by A. Yu. Pirkovskii, which is a complex analytic version of
Neeman’s result.

In the paper we intend to develop a spectral theory in purely algebraic case
with all its key properties and their links to algebraic geometry. As a basic
tool we are exploiting many results from commutative algebra. Actually, some
key results of commutative algebra can be retreated from point of view the
joint spectral theory. For example, associated primes of a Noetherian module
play the role of eigenvalues whereas support primes are spectral values. That
approach inherits Koszul homology groups of an algebraic variety and the
related index. In the case of a variety the index is reduced to the
multiplicity from the local theory [24], and we use the multiplicity formula
of Serre for calculation of Koszul homology groups.

Fix a commutative $k$-algebra $R/k$ with an $n$-tuple $x$ in $R$, and an
$R$-module $M$. It turns out that $x$ is a family of commuting linear
transformations acting on the $k$-vector space $M$, and we have their Taylor
spectrum $\sigma\left(x,M\right)$ to be the set of those $a\in\mathbb{A}^{n}$
such that the Koszul complex $\operatorname{Kos}\left(x-a,M\right)$ fails to
be exact. If the homology groups $H_{p}\left(x-a,M\right)$ of
$\operatorname{Kos}\left(x-a,M\right)$ are finite dimensional $k$-vector
spaces we have the index
$i\left(x-a\right)=\sum_{p=0}^{n}\left(-1\right)^{p}\dim_{k}\left(H_{p}\left(x-a,M\right)\right)$
of the tuple $x-a$. The last homology group $H_{n}\left(x-a,M\right)$ responds
to the submodule of all joint eigenvectors in $M$ whenever $a$ is a joint
eigenvalue of the tuple $x$. The set of all eigenvalues of $x$ is called the
point spectrum $\sigma_{\operatorname{p}}\left(x,M\right)$ of $x$. The central
result of Section 2 is the spectral mapping formula

(1.1)
$\sigma\left(p\left(x\right),M\right)=p\left(\sigma\left(x,M\right)\right)$

for all tuples $x$ and $p\left(x\right)$ (a polynomial tuple in $x$) from $R$
whenever $R/k$ is an algebra finite extension of an algebraically closed field
$k$ and $M$ is a Noetherian $R$-module. Moreover, $i\left(x\right)<\infty$ for
an $n$-tuple $x$ from $R$ whenever $R^{\prime}=k\left[x\right]\subseteq R$ is
an integral extension. In this case, $i\left(x\right)=i\left(y\right)$ for
every $n$-tuple $y$ from the subalgebra $R^{\prime}$ generating the same
maximal ideal $\left\langle x\right\rangle\subseteq R^{\prime}$. It is an
analog of the index stability result from analysis.

In Section 3, we analyze the general case of a scheme
$\left(\mathfrak{X},\mathcal{O}_{\mathfrak{X}}\right)$ and a module $M$ over
the global sections
$R=\Gamma\left(\mathfrak{X},\mathcal{O}_{\mathfrak{X}}\right)$. The spectrum
$\sigma\left(\mathfrak{X},M\right)$ of the $R$-module $M$ over the scheme
$\left(\mathfrak{X},\mathcal{O}_{\mathfrak{X}}\right)$ is defined as the
complement to the set of those $x\in\mathfrak{X}$ such that
$\operatorname{Tor}_{i}^{R}\left(\mathcal{O}_{\mathfrak{X}}\left(U\right),M\right)=0$,
$i\geq 0$ for an affine neighborhood $U$ of $x$. This spectrum was introduced
in [11] as a scheme-theoretic analog of Putinar spectrum of analytic sheaves
[23]. In the case of an affine scheme
$\mathfrak{X=}\operatorname{Spec}\left(R\right)$ the spectrum
$\sigma\left(\mathfrak{X},M\right)$ is reduced to the closure of the support
$\operatorname{Supp}_{R}\left(M\right)$ of the $R$-module $M$. If $M$ is a
finitely generated $R$-module then we come up with the support
$\operatorname{Supp}_{R}\left(M\right)$ rather than its closure. The point
spectrum $\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)$ is defined to
be the set $\operatorname{Ass}_{R}\left(M\right)$ of all associated primes of
$M$. The spectral mapping theorem for modules over schemes was proposed in
[11]. Its affine version states that

(1.2)
$\sigma\left(\mathfrak{Y},M\right)=f\left(\sigma\left(\mathfrak{X},M\right)\right)^{-}$

whenever $\mathfrak{X}=\operatorname{Spec}\left(R\right)$ and
$\mathfrak{Y}=\operatorname{Spec}\left(R^{\prime}\right)$ are affine schemes,
$f=\varphi^{\ast}:\mathfrak{X}\rightarrow\mathfrak{Y}$ is a morphism
responding to a ring map $\varphi:R^{\prime}\rightarrow R$, and $M\in
R$-$\operatorname{mod}$, which is $R^{\prime}$-module through $\varphi$
either. For a ring extension $\iota:R^{\prime}\subseteq R$ and a finitely
generated $R$-module $M$ we obtain that
$\iota^{\ast}\left(\operatorname{Supp}_{R}\left(M\right)\right)$ is dense in
$\operatorname{Supp}_{R^{\prime}}\left(M\right)$. In the case of the point
spectrum we prove more valuable equality without the related closure in (1.2)
(being so weak, the closure of Zariski topology covers up too much). If
$\iota:R^{\prime}\subseteq R$ is a ring extension with Noetherian $R$, and
$M\in R$-$\operatorname{mod}$, then (see Theorem 3.1)

$\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M\right)=\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\right).$

Notice that similar result for the spectrum (or support) is not true (see
below Remark 3.1). But that is the case if $R^{\prime}\subseteq R$ is integral
and we come with classics Krull-Cohen-Seidenberg Theory [1, Ch. 14], [2, 5.2].

The spectrum over a scheme and Taylor spectrum linked with each other when
$R=k\left[x\right]$ is an algebra finite extension. Namely,
$\mathfrak{X=}\operatorname{Spec}\left(R\right)\subseteq\mathbb{A}_{k}^{n}$ up
to a homeomorphism,
$\sigma\left(\mathfrak{X},M\right)=\sigma\left(\mathbb{A}_{k}^{n},M\right)$,
$\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)=\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)$,
and
$\sigma_{\operatorname{p}}\left(x,M\right)=\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)\cap\mathbb{A}^{n}$.
If $M$ is a Noetherian $R$-module then
$\sigma\left(x,M\right)=\sigma\left(\mathbb{A}_{k}^{n},M\right)\cap\mathbb{A}^{n}$
and it is a nonempty closed subset of $\mathbb{A}^{n}$ whose closure in the
scheme $\mathbb{A}_{k}^{n}$ is reduced to $\sigma\left(\mathfrak{X},M\right)$.
If $k\left[p\left(x\right)\right]\subseteq R$ is integral then

$\sigma_{\operatorname{p}}\left(p\left(x\right),M\right)=p\left(\sigma_{\operatorname{p}}\left(x,M\right)\right).$

Actually, (1.1) and (1.2) can be driven from the classics whenever
$k\left[p\left(x\right)\right]\subseteq R$ is integral and $M$ is Noetherian.
But the spectral mapping formula (1.1) is much stronger than classics, which
is true for all tuples. Thus Taylor spectrum with all its properties have an
independent value.

In Section 4 we consider the case of $M=R$ of the coordinate ring $R$ of a
variety $Y\subseteq\mathbb{A}^{n}$. In this case, the tuple $x\subseteq R$
consists of the coordinate functions and $\sigma\left(x,R\right)=Y$ whereas
$\sigma_{\operatorname{p}}\left(x,R\right)=\varnothing$. In the case of an
algebraic set $Y$ the point spectrum
$\sigma_{\operatorname{p}}\left(x,R\right)$ is the set of all isolated points
of $Y$. For every point $a\in\mathbb{A}^{n}$ we have $i\left(x-a\right)=0$ due
to Serre’s multiplicity formula from the local algebra. For a singular point
$a\in Y$ the dimension of $H_{1}\left(x-a,R\right)$ could get high integers
depending on the depth of singularity (see below Lemma 4.6). The link between
the index and the dimension of $Y$ is obtained through the numerical
polynomial $p:\mathbb{Z}\rightarrow\mathbb{Z}$,
$p\left(r\right)=\sum_{i=1}^{n}\left(-1\right)^{i}\dim_{k}\left(\operatorname{Tor}_{i}^{P}\left(P/\left\langle
X_{1},\ldots,X_{n}\right\rangle^{r},R\right)\right)$ called the
$\operatorname{Tor}$-polynomial, which is reduced to Hilbert-Samuel polynomial
of the localization $R_{\left\langle x-a\right\rangle}$.

Finally, I wish to thank G. G. Amosov, O. Yu. Aristov, B. Bilich and A. Yu.
Pirkovskii for their interest to the paper, useful discussions and to draw my
attention to the papers [18] and [19].

##  2\. The projection property of spectrum

In the present section we prove the projection property of spectrum based on
homology groups of the Koszul complex. The technical back up is the homology
of commutative rings related to Koszul homology groups.

###  2.1. Lemma Bourbaki

Let $R$ be a (unital) commutative ring, $K_{i}\in R$-$\operatorname{mod}$,
$i=0,1$, and let $\alpha:K_{1}\rightarrow K_{0}$ be an $R$-linear map (module
morphism). The chain complex $0\leftarrow
K_{0}\overset{\alpha}{\longleftarrow}K_{1}$ is denoted by $\mathcal{K}$. If
$\mathcal{C}$ is a chain complex $0\leftarrow
C_{0}\overset{d_{0}}{\longleftarrow}C_{1}\overset{d_{1}}{\longleftarrow}\cdots$
in $R$-$\operatorname{mod}$ then $\mathcal{K\otimes}_{R}\mathcal{C}$ is the
following complex

$\begin{array}[c]{ccccccc}&&\left(\mathcal{K\otimes}_{R}\mathcal{C}\right)_{p-1}&&\left(\mathcal{K\otimes}_{R}\mathcal{C}\right)_{p}&&\\\
&&\parallel&&\parallel&&\\\
\cdots&\overset{\partial_{p-2}}{\longleftarrow}&\left(K_{0}\mathcal{\otimes}_{R}C_{p-1}\right)\oplus\left(K_{1}\mathcal{\otimes}_{R}C_{p-2}\right)&\overset{\partial_{p-1}}{\longleftarrow}&\left(K_{0}\mathcal{\otimes}_{R}C_{p}\right)\oplus\left(K_{1}\mathcal{\otimes}_{R}C_{p-1}\right)&\overset{\partial_{p}}{\longleftarrow}&\cdots\end{array}$

with the differential

(2.1) $\partial_{p-1}\left(z_{0,p},z_{1,p-1}\right)=\left(\left(1\otimes
d_{p-1}\right)\left(z_{0,p}\right)+\left(-1\right)^{p-1}\left(\alpha\otimes
1\right)\left(z_{1,p-1}\right),\left(1\otimes
d_{p-2}\right)\left(z_{1,p-1}\right)\right),$

where $z_{i,j}\in K_{i}\mathcal{\otimes}_{R}C_{j}$. The canonical embedding
$i_{p}$ and projection $\pi_{p}$ define the exact sequence

$0\leftarrow
K_{1}\mathcal{\otimes}_{R}C_{p-1}\overset{\pi_{p}}{\longleftarrow}\left(K_{0}\mathcal{\otimes}_{R}C_{p}\right)\oplus\left(K_{1}\mathcal{\otimes}_{R}C_{p-1}\right)\overset{i_{p}}{\longleftarrow}K_{0}\mathcal{\otimes}_{R}C_{p}\leftarrow
0$

which splits. Since the diagram

$\begin{array}[c]{ccccccc}&&0&&0&&\\\ &&\downarrow&&\downarrow&&\\\
\cdots&\overset{1\otimes
d_{p-2}}{\longleftarrow}&K_{0}\mathcal{\otimes}_{R}C_{p-1}&\overset{1\otimes
d_{p-1}}{\longleftarrow}&K_{0}\mathcal{\otimes}_{R}C_{p}&\overset{1\otimes
d_{p}}{\longleftarrow}&\cdots\\\
&&\downarrow^{i_{p-1}}&&\downarrow^{i_{p}}&&\\\
\cdots&\overset{\partial_{p-2}}{\longleftarrow}&\left(\mathcal{K\otimes}_{R}\mathcal{C}\right)_{p-1}&\overset{\partial_{p-1}}{\longleftarrow}&\left(\mathcal{K\otimes}_{R}\mathcal{C}\right)_{p}&\overset{\partial_{p}}{\longleftarrow}&\cdots\\\
&&\downarrow^{\pi_{p-1}}&&\downarrow^{\pi_{p}}&&\\\ \cdots&\overset{1\otimes
d_{p-3}}{\longleftarrow}&K_{1}\mathcal{\otimes}_{R}C_{p-2}&\overset{1\otimes
d_{p-2}}{\longleftarrow}&K_{1}\mathcal{\otimes}_{R}C_{p-1}&\overset{1\otimes
d_{p-1}}{\longleftarrow}&\cdots\\\ &&\downarrow&&\downarrow&&\\\
&&0&&0&&\end{array}$

commutes, there is an exact sequence

$0\leftarrow
K_{1}\mathcal{\otimes}_{R}\mathcal{C}\overset{\pi}{\longleftarrow}\mathcal{K\otimes}_{R}\mathcal{C}\overset{i}{\longleftarrow}K_{0}\mathcal{\otimes}_{R}\mathcal{C}\leftarrow
0$

of complexes with $\deg\left(\pi\right)=-1$ and $\deg\left(i\right)=0$. The
sequence in turn generates a long exact sequence of homology groups

(2.2) $\cdots\longleftarrow
H_{p-1}\left(K_{0}\mathcal{\otimes}_{R}\mathcal{C}\right)\overset{\delta_{p-1}}{\longleftarrow}H_{p-1}\left(K_{1}\mathcal{\otimes}_{R}\mathcal{C}\right)\leftarrow
H_{p}\left(\mathcal{K\otimes}_{R}\mathcal{C}\right)\leftarrow
H_{p}\left(K_{0}\mathcal{\otimes}_{R}\mathcal{C}\right)\overset{\delta_{p}}{\longleftarrow}\cdots$

with the connecting morphisms $\delta_{p}$, $p\geq 0$. If $\mathcal{K}$ is
flat (that is, $K_{i}$ are flat modules) then
$H_{p}\left(K_{i}\mathcal{\otimes}_{R}\mathcal{C}\right)=K_{i}\otimes_{R}H_{p}\left(\mathcal{C}\right)$
for all $i$ and $p$.

######  Lemma 2.1.

The morphism
$\delta_{p-1}:H_{p-1}\left(K_{1}\mathcal{\otimes}_{R}\mathcal{C}\right)\rightarrow
H_{p-1}\left(K_{0}\mathcal{\otimes}_{R}\mathcal{C}\right)$ is acting by the
rule
$\delta_{p-1}\left(\omega^{\sim}\right)=\left(-1\right)^{p-1}\left(\left(\alpha\otimes
1\right)\omega\right)^{\sim}$ for all $\omega\in\ker\left(1\otimes
d_{p-2}\right)$, that is,
$\delta_{p-1}=\left(-1\right)^{p-1}\left(\alpha\otimes 1\right)^{\sim}$ for
all $p$. In the case of a flat complex $\mathcal{K}$ the morphism
$\delta_{p-1}$ is reduced to the following morphism
$\delta_{p-1}:K_{1}\otimes_{R}H_{p-1}\left(\mathcal{C}\right)\rightarrow
K_{0}\otimes_{R}H_{p-1}\left(\mathcal{C}\right)$,
$\delta_{p-1}=\left(-1\right)^{p-1}\alpha\otimes 1$.

###### Proof.

Take $\omega\in\ker\left(1\otimes d_{p-2}\right)$. Then
$\left(0,\omega\right)\in\left(K_{0}\mathcal{\otimes}_{R}C_{p}\right)\oplus\left(K_{1}\mathcal{\otimes}_{R}C_{p-1}\right)=\left(\mathcal{K\otimes}_{R}\mathcal{C}\right)_{p}$
and

$\partial_{p-1}\left(0,\omega\right)=\left(\left(-1\right)^{p-1}\left(\alpha\otimes
1\right)\left(\omega\right),0\right)=i_{p-1}\left(\left(-1\right)^{p-1}\left(\alpha\otimes
1\right)\left(\omega\right)\right)$

(see (2.1)). Hence $\left(-1\right)^{p-1}\left(\alpha\otimes
1\right)\left(\omega\right)\in\ker\left(1\otimes d_{p-2}\right)$ and
$\delta_{p-1}\left(\omega^{\sim}\right)=\left(-1\right)^{p-1}\left(\left(\alpha\otimes
1\right)\omega\right)^{\sim}$.

Finally, if $\mathcal{K}$ is flat and $\omega=x_{1}\otimes y_{p-1}$ with
$x_{1}\in K_{1}$, $y_{p-1}\in\ker\left(d_{p-2}\right)$ then

$\delta_{p-1}\left(x_{1}\otimes
y_{p-1}^{\sim}\right)=\delta_{p-1}\left(\omega^{\sim}\right)=\left(-1\right)^{p-1}\left(\alpha\left(x_{1}\right)\otimes
y_{p-1}\right)^{\sim}=\left(-1\right)^{p-1}\alpha\left(x_{1}\right)\otimes
y_{p-1}^{\sim},$

that is, $\delta_{p-1}=\left(-1\right)^{p-1}\alpha\otimes 1$. ∎

The following key assertion is taken from [3, 9.5, Lemma 3 ].

######  Lemma 2.2.

(N. Bourbaki) If $\alpha:K_{1}\rightarrow K_{0}$ is a morphism of flat
$R$-modules and $\mathcal{C}$ is a complex in $R$-$\operatorname{mod}$ then
the long homology sequence generates the following short exact sequences

$0\leftarrow\ker\left(\alpha\right)\otimes_{R}H_{p-1}\left(\mathcal{C}\right)\longleftarrow
H_{p}\left(\mathcal{K\otimes}_{R}\mathcal{C}\right)\longleftarrow\operatorname{coker}\left(\alpha\right)\otimes_{R}H_{p}\left(\mathcal{C}\right)\leftarrow
0$

of $R$-modules, $p\geq 1$.

###### Proof.

Using the long homology sequence (2.2) and Lemma 2.1, we obtain the following
exact sequence

$K_{0}\otimes_{R}H_{p-1}\left(\mathcal{C}\right)\overset{\left(-1\right)^{p-1}\alpha\otimes
1}{\longleftarrow}K_{1}\otimes_{R}H_{p-1}\left(\mathcal{C}\right)\leftarrow
H_{p}\left(\mathcal{K\otimes}_{R}\mathcal{C}\right)\leftarrow
K_{0}\otimes_{R}H_{p-1}\left(\mathcal{C}\right)\overset{\left(-1\right)^{p}\alpha\otimes
1}{\longleftarrow}K_{1}\otimes_{R}H_{p}\left(\mathcal{C}\right)$

of $R$-modules. Using again the flatness, we deduce that
$\ker\left(\left(-1\right)^{p-1}\alpha\otimes
1\right)=\ker\left(\alpha\right)\otimes_{R}H_{p-1}\left(\mathcal{C}\right)$
and $\operatorname{coker}\left(\left(-1\right)^{p}\alpha\otimes
1\right)=\operatorname{coker}\left(\alpha\right)\otimes_{R}H_{p}\left(\mathcal{C}\right)$.
The rest is clear. ∎

As a practical use of Lemma 2.2, let us consider the case of $K_{0}=K_{1}=R$
and $\alpha=x$ is an element of the ring $R$, which is acting over all
$R$-modules as a multiplication operator. Moreover,
$K_{i}\otimes_{R}H_{p}\left(\mathcal{C}\right)=H_{p}\left(\mathcal{C}\right)$
and $\alpha\otimes 1$ is the same action
$x:H_{p}\left(\mathcal{C}\right)\rightarrow H_{p}\left(\mathcal{C}\right)$ of
$x$ over the $R$-module $H_{p}\left(\mathcal{C}\right)$ denoted by
$x|H_{p}\left(\mathcal{C}\right)$. The complex
$\mathcal{K\otimes}_{R}\mathcal{C}$ is reduced to the cone
$\operatorname{Con}\left(x,\mathcal{C}\right)$ of the morphism
$x:\mathcal{C}\longrightarrow\mathcal{C}$. Namely,
$\operatorname{Con}\left(x,\mathcal{C}\right)$ is the following complex

$\cdots\longleftarrow C_{p-1}\oplus
C_{p-2}\overset{\partial_{p-1}}{\longleftarrow}C_{p}\oplus
C_{p-1}\leftarrow\cdots$

with the differential
$\partial_{p-1}\left(c_{p},c_{p-1}\right)=\left(d_{p-1}\left(c_{p}\right)+\left(-1\right)^{p-1}xc_{p-1},d_{p-2}\left(c_{p-1}\right)\right)$.

######  Corollary 2.1.

If $x\in R$ and $\mathcal{C}$ is a complex of $R$-modules then the following
canonical sequences

$0\leftarrow\ker\left(x|H_{p-1}\left(\mathcal{C}\right)\right)\longleftarrow
H_{p}\left(x,\mathcal{C}\right)\longleftarrow\operatorname{coker}\left(x|H_{p}\left(\mathcal{C}\right)\right)\leftarrow
0$

are exact, where $H_{p}\left(x,\mathcal{C}\right)$ are homology groups of the
cone $\operatorname{Con}\left(x,\mathcal{C}\right)$.

###### Proof.

Since
$\ker\left(\alpha\right)\otimes_{R}H_{p-1}\left(\mathcal{C}\right)=\ker\left(\alpha\otimes
1\right)=\ker\left(x|H_{p-1}\left(\mathcal{C}\right)\right)$ and
$\operatorname{coker}\left(\alpha\right)\otimes_{R}H_{p}\left(\mathcal{C}\right)=\operatorname{coker}\left(\alpha\otimes
1\right)=\operatorname{coker}\left(x|H_{p}\left(\mathcal{C}\right)\right)$,
the result follows from Lemma 2.2. ∎

In particular, $H_{p}\left(x,\mathcal{C}\right)=0$ iff
$x:H_{p-1}\left(\mathcal{C}\right)\rightarrow H_{p-1}\left(\mathcal{C}\right)$
is an injection and $x:H_{p}\left(\mathcal{C}\right)\rightarrow
H_{p}\left(\mathcal{C}\right)$ is a surjection.

###  2.2. Koszul homology groups

Fix a field $k$, $R/k$ a commutative $k$-algebra,
$x=\left(x_{1},\ldots,x_{n}\right)$ an $n$-tuple in $R$, and let $M$ be an
$R$-module, which in turn is a $k$-vector space with an $n$-tuple $x$ of
mutually commuting linear transformations acting on it. If it is necessary one
can replace $R$ by an algebra finite extension $k\left[x\right]$ of the field
$k$. For every point $a$ from the affine space $\mathbb{A}^{n}$ we have the
tuple $x-a=\left(x_{1}-a_{1},\ldots,x_{n}-a_{n}\right)$ on $M$, which is turn
defines the Koszul complex $\operatorname{Kos}\left(x-a,M\right)$:

$0\leftarrow
M\overset{\partial_{0}}{\longleftarrow}M\otimes_{k}k^{n}\overset{\partial_{1}}{\longleftarrow}\cdots\overset{\partial_{p-2}}{\longleftarrow}M\otimes_{k}\wedge^{p-1}k^{n}\overset{\partial_{p-1}}{\longleftarrow}M\otimes_{k}\wedge^{p}k^{n}\overset{\partial_{p}}{\longleftarrow}\cdots\overset{\partial_{n-1}}{\longleftarrow}M\leftarrow
0$

in $R$-$\operatorname{mod}$ with the differential

$\partial_{p-1}\left(m\otimes
v_{p}\right)=\sum_{s=1}^{p}\left(-1\right)^{s+1}\left(x_{i_{s}}-a_{i_{s}}\right)m\otimes
v_{p,s},$

where $m\in M$, $v_{p}=e_{i_{1}}\wedge\ldots\wedge e_{i_{p}}$,
$v_{p,s}=e_{i_{1}}\wedge\ldots\wedge\widehat{e_{i_{s}}}\wedge\ldots\wedge
e_{i_{p}}$ (the notation $\widehat{e_{i_{s}}}$ stands for skipping $e_{i_{s}}$
from the $p$-vector) and $\left(e_{1},\ldots,e_{n}\right)$ is the standard
basis for $k^{n}$. The homology groups of
$\operatorname{Kos}\left(x-a,M\right)$ are denoted by
$H_{p}\left(x-a,M\right)$, $p\geq 0$, which are $R$-modules. We put

$i\left(x\right)=\sum_{p=0}^{n}\left(-1\right)^{p+1}\dim_{k}\left(H_{p}\left(x,M\right)\right)$

to be the index of the tuple $x$ whenever
$\dim_{k}\left(H_{p}\left(x,M\right)\right)<\infty$ for all $p$. In the latter
case we write $i\left(x\right)<\infty$. Recall that the index
$i\left(t\right)$ of a single $k$-linear transformation $t:M\rightarrow M$ is
given by
$i\left(t\right)=\dim_{k}\ker\left(t\right)-\dim_{k}\operatorname{coker}\left(t\right)$
whenever both dimensions are finite. If $\dim_{k}\left(M\right)<\infty$ then
$i\left(t\right)=0$ for every $t$. The index of tuples of bounded linear
operators acting on a Banach space is a subject of Fredholm theory from
analysis [5], [14].

######  Lemma 2.3.

If $x^{\prime}=\left(x_{1},\ldots,x_{n-1}\right)$ then there is an exact
sequence

$0\leftarrow\ker\left(x_{n}|H_{p-1}\left(x^{\prime},M\right)\right)\longleftarrow
H_{p}\left(x,M\right)\longleftarrow\operatorname{coker}\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)\leftarrow
0$

of $R$-modules. In particular, if $H_{p}\left(x,M\right)\neq 0$ then
$H_{p}\left(x^{\prime},M\right)\neq 0$ or
$H_{p-1}\left(x^{\prime},M\right)\neq 0$. Moreover, if
$i\left(x\right)<\infty$ then
$i\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)<\infty$ for every $p$, and
$i\left(x\right)=\sum_{p=0}^{n-1}\left(-1\right)^{p}i\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)$.
If $i\left(x^{\prime}\right)<\infty$ then $i\left(x\right)=0$.

###### Proof.

Put $\mathcal{C=}\operatorname{Kos}\left(x^{\prime},M\right)$ to be a complex
of $R$-modules. Then $x_{n}$ defines an endomorphism of $\mathcal{C}$, which
in turn generates the cone $\operatorname{Con}\left(x_{n},\mathcal{C}\right)$.
The fact
$\operatorname{Con}\left(x_{n},\mathcal{C}\right)=\operatorname{Kos}\left(x,M\right)$
is know even in the noncommutative case (see [15, Lemma 1.5]). One needs to
use the vector space isomorphism
$\wedge^{p}k^{n-1}\oplus\wedge^{p-1}k^{n-1}\rightarrow\wedge^{p}k^{n}$,
$\left(v_{1},v_{2}\right)\mapsto v_{1}+v_{2}\wedge e_{n}$, which generates
$P$-module isomorphism $C_{p}\oplus C_{p-1}\rightarrow
M\otimes_{k}\wedge^{p}k^{n}$ for every $p$. If $m_{p}=m\otimes v_{p}\in
M\otimes_{k}\wedge^{p}k^{n}$ with $j_{p}\neq n$, then $m_{p}$ is identified
with an element $\left(c_{p},0\right)$ of $C_{p}\oplus C_{p-1}$ and
$\partial\left(c_{p},0\right)=\left(\partial^{\prime}\left(c_{p}\right),0\right)$,
where $\partial^{\prime}$ is the differential of $\mathcal{C}$. If $j_{p}=n$
then $m_{p}$ is identified with $\left(0,c_{p-1}\right)$, where
$c_{p-1}=m\otimes v_{p-1}\in C_{p-1}$, $v_{p-1}\wedge e_{n}=v_{p}$. Moreover,

$\displaystyle\partial\left(0,c_{p-1}\right)$
$\displaystyle=\partial\left(m_{p}\right)=\sum_{s=1}^{p-1}\left(-1\right)^{s+1}x_{i_{s}}m\otimes
v_{p-1,s}\wedge e_{n}+\left(-1\right)^{p+1}x_{n}m\otimes v_{p-1}$
$\displaystyle=\partial^{\prime}\left(c_{p-1}\right)\wedge
e_{n}+\left(-1\right)^{p+1}x_{n}c_{p-1}=\left(\left(-1\right)^{p+1}x_{n}c_{p-1},\partial^{\prime}\left(c_{p-1}\right)\right).$

Hence
$\partial\left(c_{p},c_{p-1}\right)=\left(\partial^{\prime}\left(c_{p}\right)+\left(-1\right)^{p-1}x_{n}c_{p-1},\partial^{\prime}\left(c_{p-1}\right)\right)$
for all $\left(c_{p},c_{p-1}\right)\in C_{p}\oplus C_{p-1}$, which means that
$\partial$ is the morphism of
$\operatorname{Con}\left(x_{n},\mathcal{C}\right)$. It remains to use
Corollary 2.1.

Finally, consider the case of $i\left(x\right)<\infty$. Since
$\dim_{k}\left(H_{p}\left(x,M\right)\right)<\infty$ for every $p$, it follows
that $\dim_{k}\ker\left(x_{n}|H_{p-1}\left(x^{\prime},M\right)\right)<\infty$
and
$\dim_{k}\operatorname{coker}\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)<\infty$
for every $p$, which means that
$i\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)<\infty$ for every $p$.
Then

$\displaystyle i\left(x\right)$
$\displaystyle=\sum_{p=0}^{n}\left(-1\right)^{p+1}\left(\dim_{k}\ker\left(x_{n}|H_{p-1}\left(x^{\prime},M\right)\right)+\dim_{k}\operatorname{coker}\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)\right)$
$\displaystyle=\sum_{p=0}^{n}\left(-1\right)^{p+1}\dim_{k}\operatorname{coker}\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)+\left(-1\right)^{p}\dim_{k}\ker\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)$
$\displaystyle=\sum_{p=0}^{n}\left(-1\right)^{p}i\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right),$

that is,
$i\left(x\right)=\sum_{p=0}^{n-1}\left(-1\right)^{p}i\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)$
(for $H_{n}\left(x^{\prime},M\right)=0$). In particular, if
$i\left(x^{\prime}\right)<\infty$ then
$\dim_{k}\left(H_{p}\left(x^{\prime},M\right)\right)<\infty$ and
$i\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)=0$ for all $p$. Therefore
$i\left(x\right)=0$. ∎

The Taylor spectrum $\sigma\left(x,M\right)$ of the operator tuple $x$ on $M$
is defined to be a subset of $\mathbb{A}^{n}$ of those $a$ such that
$\operatorname{Kos}\left(x-a,M\right)$ is not exact. The point spectrum
$\sigma_{\operatorname{p}}\left(x,M\right)$ consists of those
$a\in\mathbb{A}^{n}$ such that $H_{n}\left(x-a,M\right)\neq 0$. Taking into
account $H_{n}\left(x-a,M\right)=\ker\partial_{n-1}$, we conclude that
$a\in\sigma_{\operatorname{p}}\left(x,M\right)$ iff there is a nonzero $m\in
M$ such that $x_{i}m=a_{i}m$ for all $i$, that is, $a$ turns out to be a joint
eigenvalue and $m$ is the related joint eigenvector (see [4]).

######  Corollary 2.2.

Let $R/k$ be an algebra, $M\in R$-mod, $x$ an $n$-tuple from $R$. Then
$\pi\left(\sigma\left(x,M\right)\right)\subseteq\sigma\left(x^{\prime},M\right)$,
where $\pi:\mathbb{A}^{n}\rightarrow\mathbb{A}^{n-1}$ is the canonical
projection onto first $n-1$ coordinates.

###### Proof.

If $a\in\sigma\left(x,M\right)$ then $H_{p}\left(x-a,M\right)\neq 0$ for some
$p$. By Lemma 2.3, either $H_{p}\left(x^{\prime}-a^{\prime},M\right)\neq 0$ or
$H_{p-1}\left(x^{\prime}-a^{\prime},M\right)\neq 0$, where
$a^{\prime}=\pi\left(a\right)$. Whence
$a^{\prime}\in\sigma\left(x^{\prime},M\right)$. ∎

Put $M_{i}=M/\left\langle x_{1},\ldots,x_{i}\right\rangle M$, $i\geq 1$.
Recall [3, 9.6], [1, Ch. 23] that if $M_{n}\neq\left\\{0\right\\}$ and
$x_{i}\notin\operatorname{zdiv}\left(M_{i-1}\right)$ (zero-divisors) for all
$i$, then $x$ is called a regular sequence. In the case of a nonzero
Noetherian module $M$ and $x\subseteq\operatorname{rad}\left(M\right)$ the
condition $M_{n}\neq\left\\{0\right\\}$ is satisfied automatically. Indeed,
otherwise $\left\langle x\right\rangle M=M$ implies that
$\left(1+a\right)M=\left\\{0\right\\}$ for some $a\in\left\langle
x\right\rangle$ (Nakayama Lemma), which means that
$1+a\in\operatorname{Ann}\left(M\right)\subseteq\operatorname{rad}\left(M\right)$
or $1\in\operatorname{rad}\left(M\right)$, a contradiction.

######  Corollary 2.3.

Let $R/k$ be a $k$-algebra, $M\in R$-$\operatorname{mod}$, and let
$x=\left(x_{1},\ldots,x_{n}\right)\subseteq R$ be a regular sequence for $M$.
Then $H_{i}\left(x,M\right)=0$ for all $i>0$.

###### Proof.

We proceed by induction on $n$. If $n=1$ the result follows. In the general
case, we put $x^{\prime}=\left(x_{1},\ldots,x_{n-1}\right)$, which is regular
either. By induction hypothesis, $H_{i}\left(x^{\prime},M\right)=0$ for all
$i>0$. By Lemma 2.3, $H_{p}\left(x,M\right)=0$ for all $p\geq 2$. Moreover,
$H_{1}\left(x,M\right)=\ker\left(x_{n}|H_{0}\left(x^{\prime},M\right)\right)$
and $H_{0}\left(x^{\prime},M\right)=M/\left\langle x^{\prime}\right\rangle M$.
But $x_{n}:M/\left\langle x^{\prime}\right\rangle M\rightarrow M/\left\langle
x^{\prime}\right\rangle M$ is injective by assumption, therefore
$H_{1}\left(x,M\right)=0$. Whence $H_{i}\left(x,M\right)=0$ for all $i>0$. ∎

In the case of a local ring $R$ and a Noetherian module $M$ the statement of
Corollary 2.3 turns out to be a criteria for regularity of a tuple (see [7]
for regularity in the noncommutative setting). One can easily seen that the
tuple $X=\left(X_{1},\ldots,X_{n}\right)$ is a regular sequence for the
polynomial algebra $P=k\left[X\right]$ to be a $P$-module. By Corollary 2.3,
$H_{i}\left(X,P\right)=0$ for all $i>0$, and
$H_{0}\left(X,P\right)=P/\mathfrak{t}=k$, where $\mathfrak{t=}\left\langle
X\right\rangle$ is the maximal ideal of $P$. Thus $0\leftarrow
k\longleftarrow\operatorname{Kos}\left(X,P\right)$ is exact or
$\operatorname{Kos}\left(X,P\right)$ provides a free $P$-module resolution for
the $P$-module $k$. In particular, for every $P$-module $M$ we obtain that
$\operatorname{Kos}\left(X,P\right)\otimes_{P}M=\operatorname{Kos}\left(x,M\right)$
and
$\operatorname{Tor}_{i}^{P}\left(k,M\right)=H_{i}\left(\operatorname{Kos}\left(X,P\right)\otimes_{P}M\right)=H_{i}\left(x,M\right)$,
where $x$ is the $n$-tuple of linear transformations on $M$ given by
$X$-action on $M$ (see also below Subsection 3.4).

Let $R/k$ be a $k$-algebra, $R^{\prime}=k\left[x\right]\subseteq R$ a
subalgebra with an $n$-tuple $x$. An $m$-tuple $y$ from $R^{\prime}$ is said
to be related to $x$ if $x$ and $y$ generate the same (maximal) ideal
$\left\langle x\right\rangle=\left\langle y\right\rangle$ in $R^{\prime}$. In
particular, $x$ is always related to $x$ itself.

The following assertion is well know [25]. Below we provide its modified (a
bit) version with the detailed proof.

######  Lemma 2.4.

If $R/k$ is an algebra, $x$ an $n$-tuple in $R$ and $y$ is an $m$-tuple
related to $x$, then the $k\left[x\right]$-module structure of
$H_{p}\left(y,M\right)$ is reduced to its $k$-vector space structure for every
$p\geq 0$. In particular, the action of a polynomial $p\left(x\right)\in R$ on
$H_{p}\left(y,M\right)$ is just the multiplication by $p\left(0\right)$
operator.

###### Proof.

For every $i$ let us define the $P$-linear map

$\gamma_{i}:M\otimes_{k}\wedge^{p}k^{m}\rightarrow
M\otimes_{k}\wedge^{p+1}k^{m},\quad\gamma_{i}\left(u_{p}\right)=u\otimes\left(e_{i}\wedge
v_{p}\right),$

where $u_{p}=u\otimes v_{p}$, $e_{i}\wedge
v_{p}=\left(-1\right)^{s-1}e_{i_{1}}\wedge\ldots\wedge e_{i}\wedge\ldots\wedge
e_{i_{p}}=\left(-1\right)^{s-1}v_{p+1}$ whenever
$v_{p}=e_{i_{1}}\wedge\ldots\wedge e_{i_{p}}$ and
$i_{1}<\cdots<i_{s-1}<i<i_{s}<\cdots<i_{p}$ for some $s$. Then

$\displaystyle\partial_{p}\gamma_{i}\left(u_{p}\right)$
$\displaystyle=\left(-1\right)^{s-1}\sum_{k<s}\left(-1\right)^{k+1}y_{i_{k}}u\otimes
v_{p+1,k}+x_{i}u\otimes v_{p}+\left(-1\right)^{s-1}\sum_{k\geq
s}\left(-1\right)^{k}y_{i_{k}}u\otimes v_{p+1,k}$
$\displaystyle=-\sum_{k<s}\left(-1\right)^{k+1}y_{i_{k}}u\otimes e_{i}\wedge
v_{p,k}+\sum_{k\geq s}\left(-1\right)^{k}y_{i_{k}}u\otimes e_{i}\wedge
v_{p,k}+x_{i}m\otimes v_{p}$
$\displaystyle=-\gamma_{i}\partial_{p-1}\left(u_{p}\right)+y_{i}u_{p}.$

If $u_{p}\in\ker\left(\partial_{p-1}\right)$ then
$y_{i}m_{p}=\partial_{p}\gamma_{i}\left(u_{p}\right)\in\operatorname{im}\left(\partial_{p}\right)$,
which means that the action of $y_{i}$ over $H_{p}\left(y,M\right)$ is
trivial, that is, $yH_{p}\left(y,M\right)=\left\\{0\right\\}$. But
$H_{p}\left(y,M\right)$ is an $R$-module, therefore $\left\langle
y\right\rangle H_{p}\left(y,M\right)=\left\\{0\right\\}$. Taking into account
that $\left\langle x\right\rangle=\left\langle y\right\rangle$ in
$R^{\prime}$, we deduce that $\left\langle x\right\rangle
H_{p}\left(y,M\right)=\left\\{0\right\\}$.

Finally, take $p\left(x\right)\in R^{\prime}$. Then
$p\left(x\right)=p\left(0\right)+g$, $g\in\left\langle x\right\rangle$ and
$p\left(x\right)H_{p}\left(y,M\right)=p\left(0\right)H_{p}\left(y,M\right)$. ∎

Since $R^{\prime}=k\left[x\right]=k\left[x-a\right]$ and $\left\langle
x-a\right\rangle\subseteq R^{\prime}$ is a maximal ideal, we conclude that
$k\left[x\right]$-module structure of $H_{p}\left(x-a,M\right)$ is reduced to
its $k$-vector space structure for every $p\geq 0$. Actually, the action of a
polynomial $p\left(x\right)\in R$ on $H_{p}\left(x-a,M\right)$ is just the
multiplication by $p\left(a\right)$ operator.

Again consider $R^{\prime}=k\left[x\right]\subseteq R$ a subalgebra with an
$n$-tuple $x$, and an $m$-tuple $y$ from $R^{\prime}$ related to $x$. If
$p\left(x\right)\in R^{\prime}$ then $z=\left(y,p\left(x\right)\right)$ is an
$m+1$-tuple in $R$.

######  Proposition 2.1.

If $b=\left(0,\lambda\right)\in\sigma\left(z,M\right)$ for some $\lambda\in
k$, then $0\in\sigma\left(y,M\right)$ and $\lambda=p\left(0\right)$. Moreover,
$\dim_{k}\left(H_{p}\left(z-b,M\right)\right)=\dim_{k}H_{p-1}\left(y,M\right)+\dim_{k}H_{p}\left(y,M\right)$
for all $p$, and $i\left(z-b\right)=0$ whenever $i\left(y\right)<\infty$.

###### Proof.

Suppose $b\in\sigma\left(z,M\right)$ and $H_{j}\left(z-b,M\right)\neq 0$ for
some $j$. Using Corollary 2.2, we derive that $0\in\sigma\left(y,M\right)$. By
Lemma 2.3, the following sequence

$0\leftarrow\ker\left(p\left(x\right)-\lambda|H_{j-1}\left(y,M\right)\right)\longleftarrow
H_{j}\left(z-b,M\right)\longleftarrow\operatorname{coker}\left(p\left(x\right)-\lambda|H_{j}\left(y,M\right)\right)\leftarrow
0$

turns out to be exact. By Lemma 2.4,
$p\left(x\right)H_{p}\left(y,M\right)=p\left(0\right)H_{p}\left(y,M\right)$
for every $p\left(x\right)\in R^{\prime}$. Thus the actions of
$p\left(x\right)-\lambda$ over the homology groups $H_{i}\left(y,M\right)$ are
reduced to the constant multiplication operators by $p\left(0\right)-\lambda$.
If $\lambda\neq p\left(0\right)$ then $p\left(x\right)-\lambda$ defines an
invertible action over homology groups. Therefore the corners of the exact
sequence are vanishing, which means that $H_{j}\left(z-b,M\right)=0$, a
contradiction. Hence $\lambda=p\left(0\right)$. Using again Lemma 2.3, for
every $p$ we obtain the following

$0\leftarrow H_{p-1}\left(y,M\right)\longleftarrow
H_{p}\left(z-b,M\right)\longleftarrow H_{p}\left(y,M\right)\leftarrow 0$

exact sequence. In particular,
$\dim_{k}\left(H_{p}\left(z-b,M\right)\right)=\dim_{k}H_{p-1}\left(y,M\right)+\dim_{k}H_{p}\left(y,M\right)$
whenever $i\left(y\right)<\infty$. Finally,

$\displaystyle i\left(z-b\right)$
$\displaystyle=\sum_{p=0}^{n+1}\left(-1\right)^{p+1}\dim_{k}\left(H_{p}\left(z-b,M\right)\right)$
$\displaystyle=\sum_{p=0}^{n+1}\left(-1\right)^{p+1}\dim_{k}H_{p-1}\left(y,M\right)+\left(-1\right)^{p+1}\dim_{k}H_{p}\left(y,M\right)=0,$

that is, $i\left(z-b\right)=0$. ∎

######  Remark 2.1.

The assertion just proven is equally true for the tuple
$z=\left(p\left(x\right),y\right)$ swapped. Namely, if
$b=\left(\lambda,0\right)\in\sigma\left(z,M\right)$ for some $\lambda\in k$,
then $0\in\sigma\left(y,M\right)$ and $\lambda=p\left(0\right)$.

######  Corollary 2.4.

If $w=\left(x,p\left(x\right)\right)$ is an $n+1$-tuple in $R$ and
$c=\left(a,\lambda\right)\in\sigma\left(w,M\right)$ for some
$a\in\mathbb{A}^{n}$ and $\lambda\in k$, then $a\in\sigma\left(x,M\right)$ and
$\lambda=p\left(a\right)$.

###### Proof.

Put $y=x-a$, $R^{\prime}=k\left[y\right]$, $z=\left(y,q\left(y\right)\right)$
and $b=\left(0,\lambda\right)$, where $q\left(y\right)=p\left(y+a\right)$.
Then $z-b=\left(y,q\left(y\right)-\lambda\right)=w-c$, which means that
$b\in\sigma\left(z,M\right)$. It remains to use Proposition 2.1 for $y$
instead of $x$. Then $0\in\sigma\left(y,M\right)$ (or
$a\in\sigma\left(x,M\right)$) and $\lambda=q\left(0\right)=p\left(a\right)$. ∎

Now let us define the following one-to-one function

$f_{n}:\mathbb{Z}_{+}^{n+1}\rightarrow\mathbb{Z}_{+}^{n+2},\tau_{n}\left(d_{0},\ldots,d_{n}\right)=\left(d_{0},d_{0}+d_{1},d_{1}+d_{2},\ldots,d_{n-1}+d_{n},d_{n}\right)$

that generates Fibonacci numbers by increasing $n+1$-tuples of nonnegative
integers. If $x$ is an $n$-tuple from $R$ with $i\left(x\right)<\infty$ then
$d\left(x\right)=\left(d_{0},\ldots,d_{n}\right)\in\mathbb{Z}_{+}^{n+1}$ is a
tuple of dimensions $d_{p}=\dim_{k}H_{p}\left(x,M\right)$, $0\leq p\leq n$.

######  Corollary 2.5.

Let $R/k$ be an algebra, $R^{\prime}=k\left[x\right]\subseteq R$ a subalgebra
with an $n$-tuple $x$, and related to $x$ an $m$-tuple $y$. If
$\left(0,\lambda\right)\in\sigma\left(\left(x,y\right),M\right)$ and both
$i\left(x\right)$ and $i\left(y\right)$ are finite, then $\lambda=0$,
$d\left(x\right)=f_{n-m-1}\ldots f_{m}\left(d\left(y\right)\right)$ and
$i\left(x\right)=\delta_{nm}i\left(y\right)$ whenever $m\leq n$.

###### Proof.

Put $z=\left(x,y\right)$ to be $n+m$-tuple from $R$. Note that $y$ is a tuple
of polynomials from $R^{\prime}$ such that $y\subseteq\left\langle
x\right\rangle\subseteq R^{\prime}$, that is, $y\left(0\right)=0$. By
Corollary 2.4, $0\in\sigma\left(x,M\right)$ and
$\lambda=y\left(0\right)=0\in\sigma\left(y,M\right)$ (see Corollary 2.2). Thus
$0\in\sigma\left(z,M\right)$ and $d\left(z\right)=f_{n+m-1}\ldots
f_{n}\left(d\left(x\right)\right)$ thanks Proposition 2.1. By symmetry (see
Remark 2.1), we deduce $d\left(z\right)=f_{n+m-1}\ldots
f_{m}\left(d\left(y\right)\right)$ either. If $m\leq n$ then
$d\left(x\right)=f_{n-m-1}\ldots f_{m}\left(d\left(y\right)\right)$ due to the
property to be a one-to-one function. If $n>m$ then $i\left(x\right)=0$ (see
Proposition 2.1). But if $m=n$ then $d\left(x\right)=d\left(y\right)$ and
$i\left(x\right)=i\left(y\right)$. ∎

######  Remark 2.2.

If $R^{\prime}=k\left[x\right]=k\left[y\right]$ for an $m$-tuple
$y\subseteq\left\langle x\right\rangle$ then $y$ is a tuple related to $x$.
Indeed, $x_{i}=q_{i}\left(y\right)\in k\left[y\right]=R^{\prime}$ for every
$i$. It follows that $x_{i}=q_{i}\left(0\right)+h_{i}\left(y\right)$,
$h_{i}\left(y\right)\in\left\langle y\right\rangle\subseteq\left\langle
x\right\rangle$. Hence $q_{i}\left(0\right)=0$, that is, $x_{i}\in\left\langle
y\right\rangle$. Thus $\left\langle x\right\rangle=\left\langle
y\right\rangle$ in $R^{\prime}$.

###  2.3. Noetherian modules and triangular actions

As above we have an algebra $R/k$ and an $R$-module $M$ with an $n$-tuple $x$
from $R$. If $R$ is Noetherian and $M$ is a finitely generated $R$-module then
it turns out to be Noetherian automatically. Conversely, a Noetherian
$R$-module has a finitely many generators. For brevity we say that $M$ is a
Noetherian $R$-module with an acting tuple $x$ on $M$. Since
$M\otimes_{k}\wedge^{i}k^{n}$ is a Noetherian $R$-module, it follows that so
is $H_{i}\left(x,M\right)$ for every $i$.

######  Lemma 2.5.

Let $R=k\left[x\right]$ be an algebra finite extension with an $n$-tuple $x$,
$\mathfrak{m=}\left\langle x\right\rangle\subseteq R$ a maximal ideal, and let
$M$ be an $R$-module. Then
$H_{i}\left(x,M\right)_{\mathfrak{m}}=H_{i}\left(x,M\right)$ for all $i\geq
0$. If $M$ is Noetherian and $H_{p}\left(x,M\right)=0$ for some $p$, then
$H_{j}\left(x,M\right)=0$ for all $j\geq p$.

###### Proof.

By Lemma 2.4, the action of every $p\left(x\right)\notin\mathfrak{m}$ on
$H_{i}\left(x,M\right)$ is reduced to the diagonal operator $p\left(0\right)$,
which is invertible. Therefore
$H_{i}\left(x,M\right)_{\mathfrak{m}}=H_{i}\left(x,M\right)$ (see [1, 12.1,
12.4]).

Now assume that $M$ is a Noetherian $R$-module and prove that
$H_{p}\left(x,M\right)=0$ implies that $H_{j}\left(x,M\right)=0$ for all
$j\geq p$. If $R$ is a local $k$-algebra with its maximal ideal $\mathfrak{m}$
and $x\subseteq\mathfrak{m}$ then the result follows thanks to Lemma 2.3.
Namely, we proceed by induction on $n$ for a nonzero Noetherian $R$-module
$M$. Since $M/x_{1}M\neq 0$ (Nakayama lemma), the result follows in the case
of $n=1$. Put $x^{\prime}=\left(x_{1},\ldots,x_{n-1}\right)$ and suppose
$H_{p}\left(x,M\right)=0$. By Lemma 2.3, we obtain that
$\operatorname{coker}\left(x_{n}|H_{p}\left(x^{\prime},M\right)\right)=0$.
Since $H_{p}\left(x^{\prime},M\right)$ is a Noetherian $R$-module and
$x_{n}\in\operatorname{rad}R$, it follows that
$H_{p}\left(x^{\prime},M\right)/x_{n}H_{p}\left(x^{\prime},M\right)\neq 0$
whenever $H_{p}\left(x^{\prime},M\right)\neq 0$ (Nakayama lemma). Hence
$H_{p}\left(x^{\prime},M\right)=0$, and $H_{j}\left(x^{\prime},M\right)=0$,
$j\geq p$ by induction hypothesis. Using again Lemma 2.3, we obtain the exact
sequence

$0\leftarrow\ker\left(x_{n}|H_{j}\left(x^{\prime},M\right)\right)\longleftarrow
H_{j+1}\left(x,M\right)\longleftarrow\operatorname{coker}\left(x_{n}|H_{j+1}\left(x^{\prime},M\right)\right)\leftarrow
0$

for every $j\geq p$. It follows that $H_{j+1}\left(x,M\right)=0$ for all
$j\geq p$. Optionally one can use [12, Theorem 17.6].

In the general case, $M_{\mathfrak{m}}$ is a Noetherian
$R_{\mathfrak{m}}$-module and
$H_{i}\left(x,M\right)=H_{i}\left(x,M\right)_{\mathfrak{m}}=H_{i}\left(x/1,M_{\mathfrak{m}}\right)$
(the localization is an exact functor) $i\geq 0$, where $x/1$ is the tuple
representing
$\left(x_{1}/1,\ldots,x_{n}/1\right)\subseteq\mathfrak{m}R_{\mathfrak{m}}=\operatorname{rad}R_{\mathfrak{m}}$.
If $H_{p}\left(x,M\right)=0$ then $H_{p}\left(x/1,M_{\mathfrak{m}}\right)=0$
and we come up with the local $k$-algebra case $R_{\mathfrak{m}}$. Based on
the fact just proven, we conclude that
$H_{j}\left(x/1,M_{\mathfrak{m}}\right)=0$ for all $j\geq p$. Hence
$H_{j}\left(x,M\right)=0$ for all $j\geq p$. ∎

Now we analyze tuples $y$ from an algebra finite extension $R/k$ which admit
finite index. Everywhere below we assume that $k$ is an algebraically closed
field. A tuple $y$ whose extension $k\left[y\right]\subseteq k\left[x\right]$
is integral plays a key role in this manner.

######  Theorem 2.1.

Let $k\subseteq R^{\prime}\subseteq R$ be a ring extensions of the field $k$
such that $R/k$ is an algebra finite extension and $R/R^{\prime}$ is integral.
If $M$ is a Noetherian $R$-module then $i\left(y\right)<\infty$ whenever $y$
is a tuple in $R$ generating a maximal ideal in $R^{\prime}$. In this case,
the $R$-module structure on every $H_{p}\left(y,M\right)$ is triangularizable
(or it is a semilocal $R$-module) whereas its $R^{\prime}$-module structure is
diagonalizable.

###### Proof.

Since $R/R^{\prime}$ is algebra finite and integral, it follows that $R$ is
module finite over $R^{\prime}$ [1, 10.18]. But $M$ is module finite over $R$,
therefore $M$ is module finite over $R^{\prime}$ (see [1, 10.16]). Moreover,
$R^{\prime}/k$ is an algebra finite extension by Artin-Tate [1, 16.17], and it
is Noetherian (Hilbert Basis). In particular, $M$ is a Noetherian
$R^{\prime}$-module either, and so are all homology groups
$H_{p}\left(y,M\right)$ for all tuples $y$ in $R^{\prime}$. Assume that $y$ is
an $m$-tuple generating a maximal ideal $\left\langle y\right\rangle$ in
$R^{\prime}$. But $R^{\prime}=k\left[y^{\prime}\right]$ for a certain
$s$-tuple $y^{\prime}\subseteq R^{\prime}$, and the maximal ideals of
$R^{\prime}$ respond to points from $\mathbb{A}^{s}$ by Zariski
Nullstellensatz [1, 15.4]. In particular, $\left\langle
y\right\rangle=\left\langle y^{\prime}-a^{\prime}\right\rangle$ for some
$a^{\prime}\in\mathbb{A}^{s}$. Thus we can assume that
$R^{\prime}=k\left[y^{\prime}\right]$ and $\left\langle
y\right\rangle=\left\langle y^{\prime}\right\rangle$, which means that $y$ is
a tuple related to $y^{\prime}$. By Lemma 2.4, the $R^{\prime}$-module
structure of $H_{p}\left(y,M\right)$ is diagonalizable, and it is just the
$k$-vector space structure. In particular, every ascending chain of vector
subspaces in $H_{p}\left(y,M\right)$ turns out to be a chain of
$R^{\prime}$-submodules which has to stabilize being a Noetherian
$R^{\prime}$-module. Hence $H_{p}\left(y,M\right)$ is a finite dimensional
$k$-vector space, and $i\left(y\right)<\infty$.

But $H_{p}\left(y,M\right)$ has also Noetherian $R$-module structure on. In
particular, $H_{p}\left(y,M\right)$ has the finite length as an $R$-module,
which means that there is a Jordan-Hölder series with the quotients
$R/\mathfrak{m}_{j}$ for some maximal ideals $\mathfrak{m}_{j}\subseteq R$.
Thus $H_{p}\left(y,M\right)$ is a semilocal $R$-module, that is,
$\operatorname{Supp}_{R}\left(H_{p}\left(y,M\right)\right)$ is finite. As
above $R=k\left[x\right]$ for some $n$-tuple $x$, and maximal ideals respond
to points from $\mathbb{A}^{n}$. Hence there is a chain $0=V_{s}\varsubsetneq
V_{s-1}\varsubsetneq\cdots\varsubsetneq V_{0}=H_{p}\left(y,M\right)$ of
$R$-submodules such that $V_{j}/V_{j+1}=R/\mathfrak{m}_{j}=k$ with
$\mathfrak{m}_{j}=\left\langle x-b^{\left(j\right)}\right\rangle$,
$b^{\left(j\right)}\in\mathbb{A}^{n}$, $1\leq j\leq s$. Thus
$s=\dim_{k}\left(H_{p}\left(x,M\right)\right)$ and

$\operatorname{Supp}_{R}\left(H_{p}\left(y,M\right)\right)=\operatorname{Ass}_{R}\left(H_{p}\left(y,M\right)\right)=\operatorname{Max}_{R}\left(H_{p}\left(y,M\right)\right)=\left\\{\mathfrak{m}_{1},\ldots,\mathfrak{m}_{s}\right\\}$

with their multiplicities (see [1, 19.4]). Based on the chain we can easily
construct a basis $\omega=\left(\omega_{1},\ldots,\omega_{s}\right)$ for
$H_{p}\left(y,M\right)$ such that

$x_{i}|H_{p}\left(y,M\right)=\left[\begin{array}[c]{ccc}b_{i}^{\left(1\right)}&&0\\\
&\ddots&\\\ \ast&&b_{i}^{\left(s\right)}\end{array}\right]$

for all $i$, $1\leq i\leq n$ with respect to $\omega$. In particular,

$p\left(x\right)|H_{p}\left(y,M\right)=\left[\begin{array}[c]{ccc}p\left(b^{\left(1\right)}\right)&&0\\\
&\ddots&\\\ \ast&&p\left(b^{\left(s\right)}\right)\end{array}\right]$

for every $p\left(x\right)\in R$, which means that $R$-action on
$H_{p}\left(y,M\right)$ is triangularizable. ∎

######  Corollary 2.6.

Let $R/k$ be an algebra finite extension, and let $\mathfrak{a\varsubsetneq}R$
be an ideal. The following assertions are equivalent: $\left(i\right)$ there
is a tuple $y$ generating $\mathfrak{a}$ such that $k\left[y\right]\subseteq
R$ is integral; $\left(ii\right)$ there is a tuple $y$ generating
$\mathfrak{a}$ such that $i\left(y\right)<\infty$ for every Noetherian
$R$-module $M$ and the $R$-module structure on every $H_{p}\left(y,M\right)$
is triangularizable; $\left(iii\right)$ the radical $\sqrt{\mathfrak{a}}$ is a
finite intersection of maximal ideals of $R$.

###### Proof.

The implication $\left(i\right)\Rightarrow\left(ii\right)$ is due to Theorem
2.1. Let prove the implication $\left(ii\right)\Rightarrow\left(iii\right)$.
Suppose that $\mathfrak{a=}\left\langle y\right\rangle$ for some tuple $y$
such that $i\left(y\right)<\infty$ for every Noetherian $R$-module $M$, and
the $R$-module structure on every $H_{p}\left(y,M\right)$ is triangularizable.
By Lemma 2.4, $k\left[y\right]$-module structure on $H_{p}\left(y,M\right)$ is
diagonalizable. In particular, $\mathfrak{a=}\left\langle
y\right\rangle\subseteq\operatorname{Ann}_{R}\left(H_{p}\left(y,M\right)\right)$,
where $\operatorname{Ann}_{R}\left(H_{p}\left(y,M\right)\right)$ is the
annihilator of the $R$-module $H_{p}\left(y,M\right)$. Taking into account
that $R$-module $H_{p}\left(y,M\right)$ is semilocal (or triangularizable), we
deduce that

$\displaystyle\sqrt{\mathfrak{a}}$
$\displaystyle\subseteq\sqrt{\operatorname{Ann}_{R}\left(H_{p}\left(y,M\right)\right)}=\operatorname{nil}_{R}\left(H_{p}\left(y,M\right)\right)=\cap\operatorname{Supp}_{R}\left(H_{p}\left(y,M\right)\right)=\cap\operatorname{Ass}_{R}\left(H_{p}\left(y,M\right)\right)$
$\displaystyle=\cap_{j=1}^{s}\mathfrak{m}_{j},$

that is, $\sqrt{\mathfrak{a}}\subseteq\cap_{j=1}^{s}\mathfrak{m}_{j}$ (see to
the proof of Theorem 2.1, and [1, 13.6]). If $p=0$ and $M=R$, then we obtain
that $H_{0}\left(y,R\right)=R/\left\langle y\right\rangle=R/\mathfrak{a}$ and
$\cap_{j=1}^{s}\mathfrak{m}_{j}=\operatorname{nil}_{R}\left(R/\mathfrak{a}\right)=\sqrt{\mathfrak{a}}$.

Finally prove that $\left(iii\right)\Rightarrow\left(i\right)$. Suppose that
$\sqrt{\mathfrak{a}}=\cap_{j=1}^{s}\mathfrak{m}_{j}$ for some maximal ideals
$\left\\{\mathfrak{m}_{j}\right\\}$. It is a primary decomposition and
$\operatorname{Ass}\left(R/\sqrt{\mathfrak{a}}\right)\subseteq\left\\{\mathfrak{m}_{j}\right\\}\subseteq\operatorname{Max}\left(R/\sqrt{\mathfrak{a}}\right)$
(see [1, 18.17]). Since $R/\sqrt{\mathfrak{a}}$ is Noetherian, we deduce that
$\operatorname{Spec}\left(R/\sqrt{\mathfrak{a}}\right)=\operatorname{Max}\left(R/\sqrt{\mathfrak{a}}\right)$
(see [1, 17.14]) or the Krull dimension
$\dim\left(R/\sqrt{\mathfrak{a}}\right)=0$. By Akizuki-Hopkins Theorem [1,
19.8], we conclude that $R/\sqrt{\mathfrak{a}}$ is an Artinian ring. Using
again Zariski Nullstellensatz, we deduce that
$\dim_{k}\left(R/\sqrt{\mathfrak{a}}\right)<\infty$ (the gaps of a Jordan-
Hölder series are $R/\mathfrak{m}_{j}=k$). But $R=k\left[x\right]$ for some
$n$-tuple $x$, and every $x_{i}$ defines a linear transformation on
$R/\sqrt{\mathfrak{a}}$. By Cayley-Hamilton Theorem,
$q_{i}\left(x_{i}\right)\in\sqrt{\mathfrak{a}}$ for a monic polynomial
$q_{i}\in k\left[t\right]$. It follows that
$y_{i}=p_{i}\left(x_{i}\right)\in\mathfrak{a}$ for a monic polynomial
$p_{i}\in k\left[t\right]$, $1\leq i\leq n$. Put
$y=\left(y_{1},\ldots,y_{n}\right)$ and consider the subalgebra
$R^{\prime}=k\left[y\right]\subseteq R$. For every $i$ we have
$\left(p_{i}-y_{i}\right)\left(x_{i}\right)=0$, that is, $x_{i}$ is integral
over $R^{\prime}$ and $R=R^{\prime}\left[x\right]$. Hence $R$ is module finite
over $R^{\prime}$ or $R^{\prime}\subseteq R$ is an integral extension [1,
10.18]. But $y$ can easily be extended up to generators of the ideal
$\mathfrak{a}$. Just consider $\mathfrak{b=}\left\langle
y\right\rangle\subseteq\mathfrak{a}$ and pick up generators from
$\mathfrak{b/a}$, or just add up another generators of $\mathfrak{a}$ to $y$
being a Noetherian ideal. Whence $\mathfrak{a=}\left\langle y\right\rangle$
and $k\left[y\right]\subseteq R$ is an integral extension. ∎

Recall that if $M$ is a module over a $k$-algebra $R$ and $t\in R$, which
defines a linear transformation $t|M$ over the $k$-vector space $M$, then its
spectrum $\sigma\left(t|M\right)$ is defined to be a subset of those
$\lambda\in k$ such that $t-\lambda$ is not an invertible linear
transformation over $M$. Actually, it is just the Taylor spectrum
$\sigma\left(t,M\right)$ of the single tuple $t$ from the algebra $R$. Note
also that it is not expected that $t^{-1}\in R$ in the case of an invertible
linear transformation $t$ over $M$, that is, $0\notin\sigma\left(t\right)$
(see below Example 2.2).

######  Proposition 2.2.

(Non voidness) Let $R/k$ be an algebra finite extension and $M$ a Noetherian
$R$-module. Then $\sigma\left(t|M\right)\neq\varnothing$ for every $t\in R$.
For every chain $0=M_{0}\varsubsetneq M_{1}\varsubsetneq\cdots\varsubsetneq
M_{n-1}\varsubsetneq M_{n}=M$ of submodules with
$M_{i}/M_{i-1}=R/\mathfrak{p}_{i}$ and
$\operatorname{Ass}\left(M\right)\subseteq\left\\{\mathfrak{p}_{1},\ldots,\mathfrak{p}_{n}\right\\}\subseteq\operatorname{Supp}\left(M\right)$,
we have
$\sigma\left(t|M\right)=\cup_{i}\sigma\left(t|R/\mathfrak{p}_{i}\right)$. In
particular, $\sigma\left(t|M\right)$ is either finite set or it is a dense
subset of $\mathbb{A}^{1}$.

###### Proof.

We can assume that $M\neq 0$ and $R=k\left[x\right]$ for an $n$-tuple $x$.
Then $\operatorname{Supp}\left(M\right)\neq\varnothing$. Actually,
$\operatorname{Supp}\left(M\right)=V\left(\operatorname{Ann}\left(M\right)\right)$
(see [1, 13.4]) to be the set of all primes containing the annihilator
$\operatorname{Ann}\left(M\right)$ of $M$. In particular, there is a maximal
ideal $\mathfrak{m\subseteq}R$ from $\operatorname{Supp}\left(M\right)$, that
is, $M_{\mathfrak{m}}\neq 0$. By Zariski Nullstellensatz,
$\mathfrak{m=}\left\langle x-a\right\rangle$ for some $a\in\mathbb{A}^{n}$.
Take $t=t\left(x\right)\in R$. Then $t-t\left(a\right)\in\mathfrak{m}$ and
$\left\langle\left(t/1\right)-t\left(a\right)\right\rangle\subseteq\mathfrak{m}R_{\mathfrak{m}}=\operatorname{rad}\left(R_{\mathfrak{m}}\right)$.
By Nakayama lemma, $\left(M/\left\langle t-t\left(a\right)\right\rangle
M\right)_{\mathfrak{m}}=M_{\mathfrak{m}}/\left\langle\left(t/1\right)-t\left(a\right)\right\rangle
M_{\mathfrak{m}}\neq 0$. It follows that $M/\left\langle
t-t\left(a\right)\right\rangle M\neq 0$ or
$\operatorname{im}\left(t-t\left(a\right)\right)\neq M$, that is,
$t\left(a\right)\in\sigma\left(t|M\right)$.

Further, consider a chain $0=M_{0}\varsubsetneq
M_{1}\varsubsetneq\cdots\varsubsetneq M_{n-1}\varsubsetneq M_{n}=M$ of
submodules with $M_{i}/M_{i-1}=R/\mathfrak{p}_{i}$ and
$\operatorname{Ass}\left(M\right)\subseteq\left\\{\mathfrak{p}_{1},\ldots,\mathfrak{p}_{n}\right\\}\subseteq\operatorname{Supp}\left(M\right)$
(see [1, 17.16]). If $t-\lambda$ is invertible on $M$ then
$\left(t-\lambda\right)^{-1}$ is an $R$-linear map too. Hence $t-\lambda$ is
invertible iff so are all $\left(t-\lambda\right)|M_{i}/M_{i-1}$, that is,
$\sigma\left(t|M\right)=\cup_{i}\sigma\left(t|R/\mathfrak{p}_{i}\right)$.

Finally, consider the case of $M=R/\mathfrak{p}$ for some prime
$\mathfrak{p}$, which is the coordinate ring of the variety
$Y=V\left(\mathfrak{p}\right)\cap\mathbb{A}^{n}\subseteq\mathbb{A}^{n}$. Then
$\operatorname{zdiv}\left(M\right)=\left\\{0\right\\}$, and
$\lambda\in\sigma\left(t|M\right)$ iff $t-\lambda\in\mathfrak{m}$ for some
maximal ideal $\mathfrak{m}$ containing $\mathfrak{p}$. By Zariski
Nullstellensatz, $\mathfrak{m=}\left\langle x-a\right\rangle$ for some $a\in
Y$, that is, $\lambda=t\left(a\right)$. Hence
$\sigma\left(t|M\right)=t\left(Y\right)$ is an irreducible subset of
$\mathbb{A}^{1}$ (see [2, 2.4.4]). But $\sigma\left(t|M\right)$ is irreducible
iff so is its closure $\sigma\left(t|M\right)^{-}$ in $\mathbb{A}^{1}$ (see
[2, 2.4.2]). So $\sigma\left(t|M\right)^{-}$ is either singleton or
$\sigma\left(t|M\right)^{-}=\mathbb{A}^{1}$. But
$\sigma\left(t|M\right)\neq\varnothing$, therefore $\sigma\left(t|M\right)$ is
either singleton or it is a dense irreducible subset of $\mathbb{A}^{1}$. ∎

######  Example 2.1.

As an example of the case $\sigma\left(t|M\right)=\mathbb{A}^{1}$ consider the
action of $X$ on the PID $k\left[X\right]$. If $\ell_{R}\left(M\right)<\infty$
then $\operatorname{Ass}\left(M\right)=\operatorname{Max}\left(M\right)$ and
$M$ is a finite dimensional $k$-vector space (see [1, 19.4]). In this case,
$t$ is a triangularizable whose diagonal entries consist of its finite
spectrum $\sigma\left(t|M\right)$.

######  Example 2.2.

The spectrum can be empty set unless $M$ is Noetherian. Put
$R=k\left[X\right]$ and $M=k\left(X\right)=\operatorname{Frac}\left(R\right)$.
Then $M$ is an $R$-module and $X$ is acting as a multiplication operator over
$M$. For every $\lambda\in k$ we have $X-\lambda$ is invertible and
$\left(X-\lambda\right)^{-1}$ is the multiplication operator on $M$ by
$1/\left(X-\lambda\right)$. Note also that $\left(X-\lambda\right)^{-1}\notin
R$.

######  Example 2.3.

The spectrum $\sigma\left(t|M\right)$ can be an open subset of
$\mathbb{A}^{1}$. Let $M$ be the coordinate ring of the hyperbola
$Y=\left\\{xy=1\right\\}$ in $\mathbb{A}^{2}$, and $t=x$. Then
$\sigma\left(t|M\right)=t\left(Y\right)=\mathbb{A}^{1}-\left\\{0\right\\}$ is
an open dense subset of $\mathbb{A}^{1}$.

Now we can prove a key result of the present section.

######  Theorem 2.2.

(The projection property) Let $R/k$ be an algebra finite extension of the
field $k$, $M$ a Noetherian $R$-module, $y=\left(y_{1},\ldots,y_{m}\right)$ an
$m$-tuple in $R$, and let $y^{\prime}=\left(y_{1},\ldots,y_{m-1}\right)$. Then
$\sigma\left(y^{\prime},M\right)=\pi\left(\sigma\left(y,M\right)\right)$,
where $\pi:\mathbb{A}^{m}\rightarrow\mathbb{A}^{m-1}$ is the canonical
projection onto first $m-1$ coordinates.

###### Proof.

The inclusion
$\pi\left(\sigma\left(y,M\right)\right)\subseteq\sigma\left(y^{\prime},M\right)$
holds in the general case thanks to Corollary 2.2. Conversely, take
$a^{\prime}\in\sigma\left(y^{\prime},M\right)$. Then
$H_{p}\left(y^{\prime}-a^{\prime},M\right)\neq 0$ for some $p$. But
$H_{p}\left(y^{\prime}-a^{\prime},M\right)$ is a Noetherian $R$-module, and
$y_{m}$ defines a linear transformation on it. By Proposition 2.2,
$\sigma\left(y_{m}|H_{p}\left(y^{\prime}-a^{\prime},M\right)\right)\neq\varnothing$,
that is,

$y_{m}-a_{m}:H_{p}\left(y^{\prime}-a^{\prime},M\right)\rightarrow
H_{p}\left(y^{\prime}-a^{\prime},M\right)$

is not invertible for some $a_{m}\in k$. If
$\operatorname{coker}\left(y_{m}-a_{m}|H_{p}\left(y^{\prime}-a^{\prime},M\right)\right)\neq
0$ then using the following exact (Lemma 2.3) sequence

$0\leftarrow\ker\left(y_{m}-a_{m}|H_{p-1}\left(y^{\prime}-a^{\prime},M\right)\right)\longleftarrow
H_{p}\left(y-a,M\right)\longleftarrow\operatorname{coker}\left(y_{m}-a_{m}|H_{p}\left(y^{\prime}-a^{\prime},M\right)\right)\leftarrow
0,$

we conclude that $H_{p}\left(y-a,M\right)\neq 0$. If
$\ker\left(y_{m}-a_{m}|H_{p}\left(y^{\prime}-a^{\prime},M\right)\right)\neq 0$
then using the exact sequence

$0\leftarrow\ker\left(y_{m}-a_{m}|H_{p}\left(y^{\prime}-a^{\prime},M\right)\right)\longleftarrow
H_{p+1}\left(y-a,M\right)\longleftarrow\operatorname{coker}\left(y_{m}-a_{m}|H_{p+1}\left(y^{\prime}-a^{\prime},M\right)\right)\leftarrow
0,$

we conclude that $H_{p+1}\left(y-a,M\right)\neq 0$. Hence either
$H_{p}\left(y-a,M\right)\neq 0$ or $H_{p+1}\left(y-a,M\right)\neq 0$. Anyway
$a\in\sigma\left(y,M\right)$ and $\pi\left(a\right)=a^{\prime}$. ∎

######  Corollary 2.7.

Let $R/k$ be an algebra finite extension of the field $k$ and let $M$ be a
Noetherian $R$-module. Then $\sigma\left(y,M\right)\neq\varnothing$ for every
tuple $y$ in $R$.

###### Proof.

Indeed, using Theorem 2.2, we conclude that
$\pi_{1}\left(\sigma\left(y,M\right)\right)=\sigma\left(y_{1},M\right)=\sigma\left(y_{1}|M\right)$,
where $\pi_{1}:\mathbb{A}^{m}\rightarrow\mathbb{A}^{1}$ is the canonical
projection onto the first coordinate. By Proposition 2.2,
$\sigma\left(y_{1}|M\right)\neq\varnothing$. Therefore
$\sigma\left(y,M\right)\neq\varnothing$. ∎

Nonetheless the point spectrum $\sigma_{\operatorname{p}}\left(y,M\right)$
could be an empty set, which detects only zero-divisors in the module $M$.

###  2.4. The spectral mapping property

Finally, let us prove the main result for the present section. As above we fix
an algebra finite extension $R/k$, an $n$-tuple with
$x=\left(x_{1},\ldots,x_{n}\right)$ from $R$, and an $m$-tuple
$p\left(x\right)=\left(p_{1}\left(x\right),\ldots,p_{m}\left(x\right)\right)$
from the subalgebra $k\left[x\right]$. If $M$ is an $R$-module then
$k\left[p\left(x\right)\right]\subseteq R$ is a subalgebra and $M$ has the
natural $k\left[p\left(x\right)\right]$-module structure. Moreover, the tuple
$p\left(x\right)$ defines a morphism mapping
$p:\mathbb{A}^{n}\rightarrow\mathbb{A}^{m}$, $a\mapsto
p\left(a\right)=\left(p_{1}\left(a\right),\ldots,p_{m}\left(a\right)\right)$.

######  Theorem 2.3.

If $R/k$ is an algebra finite extension of the field $k$, $M$ a Noetherian
$R$-module, then
$\sigma\left(p\left(x\right),M\right)=p\left(\sigma\left(x,M\right)\right)$
for all tuples $x$ and $p\left(x\right)$ from $R$.

###### Proof.

Consider the $n+m$-tuple $y=\left(x,p\left(x\right)\right)$ in $R$. By Theorem
2.2, $\sigma\left(x,M\right)=\pi_{n}\left(\sigma\left(y,M\right)\right)$,
where $\pi_{n}:\mathbb{A}^{n+m}\rightarrow\mathbb{A}^{n}$ is the canonical
projection onto the first $n$ coordinates. Take
$\left(a,b\right)\in\sigma\left(y,M\right)$. By Corollary 2.4,
$b_{j}=p_{j}\left(a\right)$ for all $j$, $1\leq j\leq m$, that is,
$b=p\left(a\right)$. Thus
$\sigma\left(y,M\right)=\left\\{\left(a,p\left(a\right)\right):a\in\sigma\left(x,M\right)\right\\}$.
Using again Theorem 2.2, we conclude that
$\sigma\left(p\left(x\right),M\right)=\pi_{m}\left(\sigma\left(y,M\right)\right)=p\left(\sigma\left(x,M\right)\right)$,
where $\pi_{m}:\mathbb{A}^{n+m}\rightarrow\mathbb{A}^{m}$ is the canonical
projection onto the last $m$ coordinates. ∎

######  Corollary 2.8.

If $R/k$ is an algebra finite extension of the field $k$, $M$ a Noetherian
$R$-module, then
$\sigma\left(p\left(x\right)|M\right)=p\left(\sigma\left(x,M\right)\right)$
for every polynomial $p\left(x\right)\in R$.

######  Corollary 2.9.

Let $R/k$ be an algebra, $R^{\prime}=k\left[x\right]\subseteq R$ a
$k$-subalgebra with an $n$-tuple $x$ such that $R/R^{\prime}$ is integral, and
let $M$ be a Noetherian $R$-module. If $y$ is an $m$-tuple in $R^{\prime}$
related to $x$ with $m\leq n$, and $0\in\sigma\left(x,M\right)$, then
$d\left(x\right)=f_{n-m-1}\ldots f_{m}\left(d\left(y\right)\right)$ and
$i\left(x\right)=\delta_{nm}i\left(y\right)$.

###### Proof.

By Theorem 2.3,
$\left(0,\lambda\right)\in\sigma\left(\left(x,y\right),M\right)$ for a certain
$\lambda\in\mathbb{A}^{m}$. Actually, $\lambda=y\left(0\right)=0$, and
$0\in\sigma\left(\left(x,y\right),M\right)$. Since
$R^{\prime}=k\left[x\right]$ and $y$ generates the maximal ideal $\left\langle
x\right\rangle$ in $R^{\prime}$, we conclude that $i\left(x\right)<\infty$ and
$i\left(y\right)<\infty$ thanks to Theorem 2.1. It remains to use Corollary
2.5. ∎

##  3\. Spectra of a module over a scheme and extension property

In the present section we review some key facts on spectrum of a module over a
scheme from [11], and use them regarding our framework of spectra. We also
introduce the point spectrum in the general setting and investigate its
invariance under integral extensions. Thereafter we switch to the case of the
affine scheme $\mathbb{A}_{k}^{n}$ over a field $k$.

###  3.1. Spectra of a module over scheme

Let $\left(\mathfrak{X},\mathcal{O}_{\mathfrak{X}}\right)$ be a scheme with
the ring $R=\Gamma\left(\mathfrak{X},\mathcal{O}_{\mathfrak{X}}\right)$ of
global sections, and let $M\in R$-$\operatorname{mod}$. We say that a point
$x\in\mathfrak{X}$ belongs to a resolvent set
$\operatorname{res}\left(\mathfrak{X},M\right)$ if there is an affine
neighborhood $U$ of $x$ such that
$\mathcal{O}_{\mathfrak{X}}\left(U\right)\perp_{R}M$, that is,
$\operatorname{Tor}_{i}^{R}\left(\mathcal{O}_{\mathfrak{X}}\left(U\right),M\right)=0$
for all $i\geq 0$. Since $U$ is an affine neighborhood for all points close to
$x$, it follows that $\operatorname{res}\left(\mathfrak{X},M\right)$ is an
open set, whose complement set $\sigma\left(\mathfrak{X},M\right)$ is called
the spectrum of $M$ over the scheme. The property
$\mathcal{O}_{\mathfrak{X}}\left(U\right)\perp_{R}M$ implies that
$\mathcal{O}_{\mathfrak{X}}\left(V\right)\perp_{R}M$ for every open affine
$V\subseteq U$. Indeed, suppose $U=\operatorname{Spec}\left(B\right)$ and take
a free resolution

$\mathcal{P=}R\otimes G_{\circ},\quad 0\leftarrow R\otimes
G_{0}\overset{\partial_{0}}{\longleftarrow}R\otimes
G_{1}\overset{\partial_{1}}{\longleftarrow}\cdots$

of the module $M$, where $G_{j}$ are free abelian groups. Since
$\mathcal{O}_{\mathfrak{X}}$ is a quasi-coherent sheaf, it follows that
$\mathcal{O}_{\mathfrak{X}}|_{U}=\widetilde{B}$ with
$\mathcal{O}_{\mathfrak{X}}\left(U\right)=B$, and
$B\otimes_{R}\mathcal{P}=B\otimes G_{\circ}$, which is exact iff so are
$B_{x}\otimes G_{\circ}$, $x\in\operatorname{Spec}\left(B\right)$ (see [11,
Lemma 2.1]). But $B_{x}=\mathcal{O}_{\mathfrak{X},x}$ for all $x\in U$. Thus
$\mathcal{O}_{\mathfrak{X}}\left(U\right)\perp_{R}M$ is equivalent to
$\mathcal{O}_{\mathfrak{X},x}\perp_{R}M$ for all $x\in U$. In particular,
$\mathcal{O}_{\mathfrak{X},x}\perp_{R}M$ for all $x\in V$ imply that
$\mathcal{O}_{\mathfrak{X}}\left(V\right)\perp_{R}M$.

If $\mathfrak{X}=\operatorname{Spec}\left(R\right)$ is an affine scheme, then
$\mathcal{O}_{\mathfrak{X},x}\perp_{R}M$ means that
$R_{x}\otimes_{R}\mathcal{P}$ is exact. But $R_{x}$ is a flat $R$-module,
therefore the sequence $0\leftarrow R_{x}\otimes_{R}M\leftarrow
R_{x}\otimes_{R}\mathcal{P}$ remains exact. Hence
$\mathcal{O}_{\mathfrak{X},x}\perp_{R}M$ iff $M_{x}=0$. Thus
$x\in\operatorname{res}\left(\mathfrak{X},M\right)$ iff $M_{y}=0$ for all $y$
from a neighborhood of $x$. Since
$\sigma\left(\mathfrak{X},M\right)=\mathfrak{X}-\operatorname{res}\left(\mathfrak{X},M\right)$,
we derive that $x\in\sigma\left(\mathfrak{X},M\right)$ such that property
$M_{y}\neq 0$ holds for some $y$ in every neighborhood of $x$, that is,
$\sigma\left(\mathfrak{X},M\right)$ is the closure of
$\operatorname{Supp}\left(M\right)$ in $\mathfrak{X}$. Thus

(3.1)
$\sigma\left(\mathfrak{X},M\right)=\operatorname{Supp}\left(M\right)^{-}\subseteq
V\left(\operatorname{Ann}\left(M\right)\right).$

If $M$ is a finitely generated $R$-module then
$\operatorname{Supp}\left(M\right)$ is closed,
$\operatorname{res}\left(\mathfrak{X},M\right)=\left\\{x\in\mathfrak{X}:R_{x}\perp_{R}M\right\\}=\left\\{x\in\mathfrak{X}:M_{x}=0\right\\}$,
that is,

(3.2)
$\sigma\left(\mathfrak{X},M\right)=\operatorname{Supp}\left(M\right)=V\left(\operatorname{Ann}\left(M\right)\right).$

In the case of a general scheme the spectrum can be localized using an affine
covering. In particular,
$\sigma\left(\mathbb{P}_{k}^{r},M\right)=\mathbb{P}_{k}^{r}$ whenever $M$ is a
nonzero vector space over a field $k$ (see [11]). The related spectral mapping
formula was mentioned in (1.2).

###  3.2. The point spectrum

Let $\mathfrak{X}=\operatorname{Spec}\left(R\right)$ be an affine scheme and
let $M$ be an $R$-module. The set $\operatorname{Ass}_{R}\left(M\right)$ of
all primes associated to $M$ is called the point spectrum
$\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)$ of $M$. If $R$ is
Noetherian then
$\operatorname{Supp}\left(M\right)=\cup\left\\{V\left(\mathfrak{p}\right):\mathfrak{p}\in\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\right\\}$
and the minimal primes $\operatorname{Min}\left(M\right)$ from
$\operatorname{Supp}\left(M\right)$ belong to
$\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)$ [1, 17.14]. In
particular, $\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)$ is dense in
$\operatorname{Supp}\left(M\right)$, which in turn implies that
$\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)^{-}=\sigma\left(\mathfrak{X},M\right)$
in $\mathfrak{X}$ (see (3.1)).

Now let $R^{\prime}\subseteq R$ be a ring extension,
$\iota:R^{\prime}\rightarrow R$ is the realted inclusion map, and
$\mathfrak{X}^{\prime}=\operatorname{Spec}\left(R^{\prime}\right)$ with the
canonical mapping $\iota^{\ast}:\mathfrak{X}\rightarrow\mathfrak{X}^{\prime}$,
$\iota^{\ast}\left(\mathfrak{p}\right)=\iota^{-1}\left(\mathfrak{p}\right)=\mathfrak{p\cap}R$.
If $\mathfrak{p}^{\prime}\mathfrak{=p}\cap
R=\iota^{\ast}\left(\mathfrak{p}\right)$ we say that $\mathfrak{p}$ lies over
the prime $\mathfrak{p}^{\prime}$. Note also that $M$ has the canonical
$R^{\prime}$-module structure along the embedding $\iota$. If
$\mathfrak{a}=\operatorname{Ann}\left(m\right)$ is the annihilator of $m\in M$
in $R$ then $\mathfrak{a}^{\prime}\mathfrak{=a}\cap
R^{\prime}=\operatorname{Ann}^{\prime}\left(m\right)$ is the annihilator of
$m$ in $R^{\prime}$. It follows that
$\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\right)\subseteq\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M\right)$.

Notice also that
$\iota^{\ast}\left(\operatorname{Supp}\left(M\right)\right)\subseteq\operatorname{Supp}^{\prime}\left(M\right)$,
where
$\operatorname{Supp}^{\prime}\left(M\right)=\operatorname{Supp}_{R^{\prime}}\left(M\right)$.
Indeed, if $M_{\mathfrak{p}}\neq\left\\{0\right\\}$ for some
$\mathfrak{p}\in\mathfrak{X}$, then $sm\neq 0$ for some $m\in M$ and all $s\in
R-\mathfrak{p}$. But $\mathfrak{p}^{\prime}\mathfrak{=p\cap}R^{\prime}$, and
for $s^{\prime}\in R^{\prime}$ we have $s^{\prime}\in
R^{\prime}-\mathfrak{p}^{\prime}$ iff $s^{\prime}\in R-\mathfrak{p}$. Hence
$s^{\prime}m\neq 0$ for all $s^{\prime}\in R^{\prime}-\mathfrak{p}^{\prime}$,
which means that $m/1\neq 0$ in $M_{\mathfrak{p}^{\prime}}$ or
$\mathfrak{p}^{\prime}\in\operatorname{Supp}^{\prime}\left(M\right)$.

Using (3.1) and continuity of the map $\iota^{\ast}$, we deduce that
$\iota^{\ast}\left(\sigma\left(\mathfrak{X},M\right)\right)\subseteq\sigma\left(\mathfrak{X}^{\prime},M\right)$.
Actually $\iota^{\ast}\left(\sigma\left(\mathfrak{X},M\right)\right)$ is dense
in $\sigma\left(\mathfrak{X}^{\prime},M\right)$ due to the spectral mapping
property (1.2).

######  Theorem 3.1.

If $R^{\prime}\subseteq R$ is a ring extension and $M\in
R$-$\operatorname{mod}$ then
$\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M\right)=\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\right)$
whenever $R$ is Noetherian.

###### Proof.

First note that $M\neq\left\\{0\right\\}$ iff
$\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\neq\varnothing$ (see [1,
17.10]), and the result follows in the case of
$\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)=\varnothing$. Thus we
can assume that $M\neq\left\\{0\right\\}$, and take
$\mathfrak{q}\in\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M\right)$.
Then $\iota_{\mathfrak{p}}:R_{\mathfrak{q}}^{\prime}\rightarrow
R_{\mathfrak{q}}$ is a ring extension and
$M_{\mathfrak{q}}=R_{\mathfrak{q}}^{\prime}\otimes_{R^{\prime}}M=R_{\mathfrak{q}}^{\prime}\otimes_{R^{\prime}}R\otimes_{R}M=R_{\mathfrak{q}}\otimes_{R}M$
is an $R_{\mathfrak{q}}$-module either, where
$R_{\mathfrak{q}}=\iota\left(R^{\prime}-\mathfrak{q}\right)^{-1}R$. Moreover,
$\mathfrak{q}R_{\mathfrak{q}}^{\prime}\in\operatorname{Ass}_{R_{\mathfrak{q}}^{\prime}}\left(M_{\mathfrak{q}}\right)$
[1, 17.8], say
$\mathfrak{q}R_{\mathfrak{q}}^{\prime}=\operatorname{Ann}^{\prime}\left(m/1\right)$
for some $m/1\in M_{\mathfrak{q}}\backslash\left\\{0\right\\}$. If
$\operatorname{Ann}\left(m/1\right)$ is the annihilator of $m/1$ in
$R_{\mathfrak{q}}$ then $\operatorname{Ann}\left(m/1\right)\cap
R_{\mathfrak{q}}^{\prime}=\operatorname{Ann}^{\prime}\left(m/1\right)=\mathfrak{q}R_{\mathfrak{q}}^{\prime}$.
Pick

$\mathfrak{S=}\left\\{\operatorname{Ann}\left(n\right):n\in
M_{\mathfrak{q}},\operatorname{Ann}\left(n\right)\cap
R_{\mathfrak{q}}^{\prime}=\mathfrak{q}R_{\mathfrak{q}}^{\prime}\right\\},$

which is a nonempty set of ideals of the ring $R_{\mathfrak{q}}$. Since $R$ is
Noetherian, so is $R_{\mathfrak{q}}$ and $\mathfrak{S}$ has a maximal element
$\mathfrak{l}=\operatorname{Ann}\left(n\right)$ for some $n\in
M_{\mathfrak{q}}$. If $x\in R_{\mathfrak{q}}-\mathfrak{l}$ then $xn\neq 0$ (or
$1\notin\operatorname{Ann}\left(xn\right)$),
$\mathfrak{l}\subseteq\operatorname{Ann}\left(xn\right)\neq R_{\mathfrak{q}}$,
which in turn implies that

$\mathfrak{q}R_{\mathfrak{q}}^{\prime}=\operatorname{Ann}\left(n\right)\cap
R_{\mathfrak{q}}^{\prime}\subseteq\operatorname{Ann}\left(xn\right)\cap
R_{\mathfrak{q}}^{\prime}\subsetneqq R_{\mathfrak{q}}^{\prime}.$

But $R_{\mathfrak{q}}^{\prime}$ is local with its unique maximal ideal
$\mathfrak{q}R_{\mathfrak{q}}^{\prime}$, therefore
$\mathfrak{q}R_{\mathfrak{q}}^{\prime}=\operatorname{Ann}\left(xn\right)\cap
R_{\mathfrak{q}}^{\prime}$ or
$\operatorname{Ann}\left(xn\right)\in\mathfrak{S}$. It follows that
$\mathfrak{l=}\operatorname{Ann}\left(xn\right)$ whenever $x\in
R_{\mathfrak{q}}-\mathfrak{l}$. In particular, if $xy\in\mathfrak{l}$ with
$x\in R_{\mathfrak{q}}-\mathfrak{l}$ then $yxn=0$ or
$y\in\operatorname{Ann}\left(xn\right)=\mathfrak{l}$, which means that
$\mathfrak{l}$ is a prime or
$\mathfrak{l}\in\operatorname{Ass}_{R_{\mathfrak{q}}}\left(M_{\mathfrak{q}}\right)$,
and it is lying over $\mathfrak{q}R_{\mathfrak{q}}^{\prime}$. But
$\mathfrak{l}=\mathfrak{p}R_{\mathfrak{q}}$ for some
$\mathfrak{p}\in\operatorname{Ass}_{R}\left(M\right)$ [1, 17.8], [2, 4.1.5]
with $\mathfrak{p}\cap\left(R^{\prime}-\mathfrak{q}\right)=\varnothing$ (or
$\mathfrak{p}\cap R^{\prime}\subseteq\mathfrak{q}$). Prove that
$\mathfrak{p}\cap R^{\prime}=\mathfrak{q}$. Take $x^{\prime}\in\mathfrak{q}$.
Then
$x^{\prime}/1\in\mathfrak{q}R_{\mathfrak{q}}^{\prime}\subseteq\operatorname{Ann}\left(n\right)=\mathfrak{p}R_{\mathfrak{q}}$
or $x^{\prime}/1=x/t^{\prime}$ for some $x\in\mathfrak{p}$ and $t^{\prime}\in
R^{\prime}-\mathfrak{q}$. It follows that $sx^{\prime}\in\mathfrak{p}$ for
some $s\in R^{\prime}-\mathfrak{q}$. Taking into account that $s\in
R-\mathfrak{p}$, we conclude that $x^{\prime}\in\mathfrak{p}$ or
$x^{\prime}\in\mathfrak{p}\cap R^{\prime}$. Hence $\mathfrak{q=p}\cap
R^{\prime}=\iota^{\ast}\left(\mathfrak{p}\right)$ and
$\mathfrak{p}\in\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)$. ∎

######  Remark 3.1.

Note that $\operatorname{Supp}^{\prime}\left(M\right)$ can be much larger than
$\iota^{\ast}\left(\operatorname{Supp}\left(M\right)\right)$. For example, put
$R^{\prime}=\mathbb{Z\subset Q=}R$ and $M=\mathbb{Q}$. Then
$\sigma\left(\mathfrak{X},M\right)=\operatorname{Supp}\left(M\right)=\left\\{0\right\\}$
whereas
$\sigma\left(\mathfrak{X}^{\prime},M\right)=\operatorname{Supp}^{\prime}\left(M\right)^{-}=\operatorname{Spec}\left(\mathbb{Z}\right)=\mathfrak{X}^{\prime}$.
But
$\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M\right)=\operatorname{Ass}_{\mathbb{Z}}\left(\mathbb{Q}\right)=\left\\{0\right\\}=\iota^{\ast}\left(\left\\{0\right\\}\right)=\iota^{\ast}\left(\operatorname{Ass}_{\mathbb{Q}}\left(\mathbb{Q}\right)\right)=\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\right)$,
and it is dense in $\mathfrak{X}^{\prime}$.

######  Corollary 3.1.

Let $k$ be a field and let $k\subseteq R^{\prime}\subseteq R$ be $k$-algebra
extensions such that $R/k$ is algebra-finite. Then
$\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M\right)=\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\right)$
for every $R$-module $M$.

###### Proof.

The ring $R$ is Noetherian by Hilbert Basis Theorem. It remains to use Theorem
3.1. ∎

######  Corollary 3.2.

Let $R^{\prime}\subseteq R$ be a ring extension with Noetherian $R$, $M\in
R$-$\operatorname{mod}$ and let $Q\subseteq M$ be an $R$-submodule. If $Q$ is
$\mathfrak{p}$-primary then $Q$ is $\mathfrak{p}^{\prime}$-primary
$R^{\prime}$-submodule of $M$, where $\mathfrak{p}\in\mathfrak{X}$ and
$\mathfrak{p}^{\prime}=\iota^{\ast}\left(\mathfrak{p}\right)$. If $M$ is a
Noetherian $R$-module with its submodule $Q$ then $Q$ is
$\mathfrak{p}^{\prime}$-primary $R^{\prime}$-submodule of $M$ iff
$Q=Q_{1}\cap\cdots\cap Q_{r}$ admits an irredundant primary decomposition in
$R$-$\operatorname{mod}$ such that every $Q_{j}$ is
$\mathfrak{p}^{\prime}$-primary $R^{\prime}$-module.

###### Proof.

Suppose that $Q$ is a $\mathfrak{p}$-primary submodule, that is,
$\operatorname{Ass}_{R}\left(M/Q\right)=\left\\{\mathfrak{p}\right\\}$ or
$\sigma_{\operatorname{p}}\left(\mathfrak{X},M/Q\right)=\left\\{\mathfrak{p}\right\\}$.
Since
$\mathfrak{p}^{\prime}=\iota^{\ast}\left(\mathfrak{p}\right)\in\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M/Q\right)\right)\subseteq\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M/Q\right)$,
it follows that
$\mathfrak{p}^{\prime}\in\operatorname{Ass}_{R^{\prime}}\left(M/Q\right)$.
Conversely, if
$\mathfrak{q\in}\operatorname{Ass}_{R^{\prime}}\left(M/Q\right)$ then

$\mathfrak{q\in}\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M/Q\right)=\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M/Q\right)\right)=\iota^{\ast}\left(\operatorname{Ass}_{R}\left(M/Q\right)\right)=\iota^{\ast}\left(\left\\{\mathfrak{p}\right\\}\right)=\left\\{\mathfrak{p}^{\prime}\right\\}$

thanks to Theorem 3.1, that is, $\mathfrak{q=p}^{\prime}$. Hence
$\operatorname{Ass}_{R^{\prime}}\left(M/Q\right)=\left\\{\mathfrak{p}^{\prime}\right\\}$,
which means that $Q$ is a $\mathfrak{p}^{\prime}$-primary
$R^{\prime}$-submodule of $M$.

Finally, assume that $M$ is a finitely generated $R$-module. By Lasker-Noether
Theorem [1, 18.19], $Q=Q_{1}\cap\cdots\cap Q_{r}$ admits an irredundant
primary decomposition in $R$-$\operatorname{mod}$, where $Q_{j}$ is
$\mathfrak{p}_{j}$-primary. In this case,
$\left\\{\mathfrak{p}_{1},\ldots,\mathfrak{p}_{r}\right\\}$ are uniquely
defined, in fact they are all distinct primes of
$\operatorname{Ass}_{R}\left(M/Q\right)$ (First Uniqueness [1, 18.18]). In
particular,

$\operatorname{Ass}_{R^{\prime}}\left(M/Q\right)=\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M/Q\right)=\iota^{\ast}\left(\operatorname{Ass}_{R}\left(M/Q\right)\right)=\iota^{\ast}\left(\left\\{\mathfrak{p}_{1},\ldots,\mathfrak{p}_{r}\right\\}\right)$

by virtue of Theorem 3.1. If $Q$ is $\mathfrak{p}^{\prime}$-primary
$R^{\prime}$-submodule then
$\operatorname{Ass}_{R^{\prime}}\left(M/Q\right)=\left\\{\mathfrak{p}^{\prime}\right\\}$
and

$\operatorname{Ass}_{R^{\prime}}\left(M/Q_{j}\right)=\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M/Q_{j}\right)=\iota^{\ast}\left(\operatorname{Ass}_{R}\left(M/Q_{j}\right)\right)=\iota^{\ast}\left(\left\\{\mathfrak{p}_{j}\right\\}\right)=\left\\{\mathfrak{p}^{\prime}\right\\}$

for every $j$. Thus every $Q_{j}$ is $\mathfrak{p}^{\prime}$-primary
$R^{\prime}$-module.

Conversely, suppose that $Q=Q_{1}\cap\cdots\cap Q_{r}$ admits an irredundant
primary decomposition in $R$-$\operatorname{mod}$ such that every $Q_{j}$ is
$\mathfrak{p}^{\prime}$-primary $R^{\prime}$-module. Then
$\iota^{\ast}\left(\left\\{\mathfrak{p}_{j}\right\\}\right)=\left\\{\mathfrak{p}^{\prime}\right\\}$
for every $j$, and
$\operatorname{Ass}_{R^{\prime}}\left(M/Q\right)=\iota^{\ast}\left(\left\\{\mathfrak{p}_{1},\ldots,\mathfrak{p}_{r}\right\\}\right)=\left\\{\mathfrak{p}^{\prime}\right\\}$,
which means that $Q$ is a $\mathfrak{p}^{\prime}$-primary $R^{\prime}$-module.
∎

###  3.3. Integral extensions

Now assume that $R^{\prime}\subseteq R$ is an integral extension of rings and
$M\in R$-$\operatorname{mod}$. In this case, the mapping
$\iota^{\ast}:\mathfrak{X}\rightarrow\mathfrak{X}^{\prime}$ is surjective due
to Krull-Cohen-Seidenberg Theory [1, Ch. 14], [2, 5.2].

######  Proposition 3.1.

If $R^{\prime}\subseteq R$ is integral and $M\in R$-$\operatorname{mod}$ then
$\operatorname{Supp}^{\prime}\left(M\right)=\iota^{\ast}\left(\operatorname{Supp}\left(M\right)\right)$.

###### Proof.

The inclusion
$\iota^{\ast}\left(\operatorname{Supp}\left(M\right)\right)\subseteq\operatorname{Supp}^{\prime}\left(M\right)$
was proved above. Conversely, take
$\mathfrak{p}^{\prime}\in\operatorname{Supp}^{\prime}\left(M\right)$. Then
$M_{\mathfrak{p}^{\prime}}\neq\left\\{0\right\\}$, which means that
$s^{\prime}m\neq 0$ for all $s^{\prime}\in R^{\prime}-\mathfrak{p}^{\prime}$
and some $m\in M$. Put $\mathfrak{a=}\operatorname{Ann}\left(m\right)\subseteq
R$ and
$\mathfrak{a}^{\prime}=\operatorname{Ann}^{\prime}\left(m\right)\subseteq
R^{\prime}$ to be ideals with $\mathfrak{a}^{\prime}=\mathfrak{a}\cap
R^{\prime}$. Then $\mathfrak{a}^{\prime}\subseteq\mathfrak{p}^{\prime}$, and
using the key property Going Up [1, 14.3], we deduce that
$\mathfrak{p}^{\prime}=\iota^{\ast}\left(\mathfrak{p}\right)$ for some
$\mathfrak{p}\in\mathfrak{X}$ such that $\mathfrak{a}\subseteq\mathfrak{p}$.
The latter means that $sm\neq 0$ for all $s\in R-\mathfrak{p}$, that is,
$m/1\neq 0$ in $M_{\mathfrak{p}}$. Thus
$\mathfrak{p}\in\operatorname{Supp}\left(M\right)$ and
$\mathfrak{p}^{\prime}=\iota^{\ast}\left(\mathfrak{p}\right)\in\iota^{\ast}\left(\operatorname{Supp}\left(M\right)\right)$.
∎

######  Remark 3.2.

In the case of an integral algebra finite extension $R/R^{\prime}$ the set
$\left(\iota^{\ast}\right)^{-1}\left(\mathfrak{q}\right)$ is finite for every
$\mathfrak{q}\in\operatorname{Supp}^{\prime}\left(M\right)$. Indeed, since
$R/R^{\prime}$ is module finite, it follows that so is
$R_{\mathfrak{q}}/R_{\mathfrak{q}}^{\prime}$, which in turn implies that $A/k$
is module finite either, where
$A=R_{\mathfrak{q}}/\mathfrak{q}R_{\mathfrak{q}}$ and
$k=R_{\mathfrak{q}}^{\prime}/\mathfrak{q}R_{\mathfrak{q}}^{\prime}$ is the
residue field of the local ring $R_{\mathfrak{q}}^{\prime}$. In particular,
$\dim_{k}\left(A\right)<\infty$, and $A$ turns out to be an Artinian ring. By
Akizuki-Hopkins Theorem [1, 19.8], $\dim\left(A\right)=0$,
$\operatorname{Spec}\left(A\right)$ consists of maximal ideals and it is
finite. But $\left(\iota^{\ast}\right)^{-1}\left(\mathfrak{q}\right)$ is
canonically identified with a subset of $\operatorname{Spec}\left(A\right)$.

######  Corollary 3.3.

If $R^{\prime}\subseteq R$ is integral with Noetherian $R$ and $M\in
R$-$\operatorname{mod}$ with finite length
$\ell_{R^{\prime}}\left(M\right)<\infty$ as an $R^{\prime}$-module then
$\ell_{R}\left(M\right)<\infty$.

###### Proof.

Since $\ell_{R^{\prime}}\left(M\right)<\infty$, it follows that
$\operatorname{Supp}^{\prime}\left(M\right)$ consists of maximal ideals by
Jordan-Hőlder Theorem [1, 19.3]. By Proposition 3.1,
$\operatorname{Supp}^{\prime}\left(M\right)=\iota^{\ast}\left(\operatorname{Supp}\left(M\right)\right)$
and $\operatorname{Supp}\left(M\right)$ consists of primes lying over the
maximal ideals from $\operatorname{Supp}^{\prime}\left(M\right)$. By
Maximality [1, 14.3], $\operatorname{Supp}\left(M\right)$ consists of maximal
ideals of $R$ either. But $M$ is a finitely generated $R^{\prime}$-module
having finite length $\ell_{R^{\prime}}\left(M\right)$, therefore so is $M$ as
an $R$-module. It follows that $\ell_{R}\left(M\right)<\infty$ (see [1, 19.4
or 17.16]). ∎

######  Remark 3.3.

Let $k$ be a field, $k\subseteq R^{\prime}\subseteq R$ algebra extensions such
that $R/k$ is algebra-finite and $R/R^{\prime}$ is integral, and let $M\in
R$-$\operatorname{mod}$. Then $\ell_{R}\left(M\right)<\infty$ iff
$\ell_{R^{\prime}}\left(M\right)<\infty$. By Artin-Tate lemma [1, 16.21],
$R^{\prime}/k$ is algebra-finite too. In particular, $R^{\prime}$ is
Noetherian by Hilbert Basis. If $\left\\{M_{i}:0\leq i\leq n\right\\}$ is a
Jordan-Hőlder chain of submodules in $R^{\prime}$-$\operatorname{mod}$ (or
$R$-$\operatorname{mod}$) then
$M_{i}/M_{i-1}=R^{\prime}/\mathfrak{m}_{i}^{\prime}$ for some maximal ideals
$\mathfrak{m}_{i}^{\prime}$, $i\geq 1$. But $k\subseteq
R^{\prime}/\mathfrak{m}_{i}^{\prime}$ is an algebra finite extension, which is
finite by Zariski Nullstellensatz [1, 15.4]. It follows that
$\dim_{k}\left(M_{i}/M_{i-1}\right)<\infty$ for every $i$. Hence
$\dim_{k}\left(M\right)<\infty$. Thus $\ell_{R^{\prime}}\left(M\right)<\infty$
iff $\dim_{k}\left(M\right)<\infty$, which in turn is equivalent to
$\ell_{R}\left(M\right)<\infty$.

###  3.4. Algebraic varieties and spectra

Now let $P=k\left[X\right]$ be the algebra of all polynomials in $n$ variables
$X=\left(X_{1},\ldots,X_{n}\right)$ over the field $k$. The $P$-bimodule
$P\otimes_{k}P$ has the $n$-tuple $T=\left(T_{1},\ldots,T_{n}\right)$ of
mutually commuting operators $T_{i}=1\otimes X_{i}-X_{i}\otimes 1$ acting on,
and we have the Koszul complex
$\operatorname{Kos}\left(T,P\otimes_{k}P\right)$ augmented with the
multiplication (bi)operator $\pi:P\otimes_{k}P\rightarrow P$ provides a free
$P$-bimodule resolution of the algebra $P$. Actually,
$\operatorname{Kos}\left(T,P\otimes_{k}P\right)$ splits as a complex of
$P$-bimodules. If $M\in P$-$\operatorname{mod}$ with the actions
$x_{i}\left(m\right)=X_{i}\cdot m$, $1\leq i\leq n$, then derive that the
complex $0\leftarrow
P\otimes_{P}M\overset{1\otimes\pi}{\longleftarrow}\operatorname{Kos}\left(P\otimes_{k}P,T\right)\otimes_{P}M$
remains exact. But $P\otimes_{P}M=M$ and
$\operatorname{Kos}\left(T,P\otimes_{k}P\right)\otimes_{P}M=\operatorname{Kos}\left(t,P\otimes_{k}M\right)$
with $t_{i}=1\otimes x_{i}-X_{i}\otimes 1$, and $1\otimes\pi$ is reduced to
the homomorphism $\pi_{M}:P\otimes_{k}M\rightarrow M$,
$\pi_{M}\left(r\otimes_{k}m\right)=rm$. Thus
$\mathcal{P=}\operatorname{Kos}\left(t,P\otimes_{k}M\right)$ provides a finite
free resolution of the module $M$. If $R/k$ is an algebra finite extension
with $R=k\left[x\right]$ for an $n$-tuple $x$, then $R$ is a quotient of $P$
and $\operatorname{Kos}\left(t,P\otimes_{k}R\right)$ turns out to be a free
$R$-module resolution of $R$ either. Therefore it splits in
$R$-$\operatorname{mod}$.

Consider the affine space $\mathfrak{X}=\mathbb{A}_{k}^{n}$ over $k$. Then
$\mathfrak{p}\in\operatorname{res}\left(\mathfrak{X},M\right)$ iff there is an
open affine neighborhood $U=\operatorname{Spec}\left(B\right)$ of
$\mathfrak{p}$ such that $B\perp_{P}M$. It means that

$0\leftarrow B\otimes_{P}M\longleftarrow
B\otimes_{P}\mathcal{P=}\operatorname{Kos}\left(t^{B},B\otimes_{k}M\right)$

with $t_{i}^{B}=1\otimes x_{i}-\left(X_{i}|_{U}\right)\otimes 1$ is exact. Put
$k\left(\mathfrak{p}\right)=P_{\mathfrak{p}}/\mathfrak{p}P_{\mathfrak{p}}=\left(P/\mathfrak{p}\right)_{\mathfrak{p}}=\operatorname{Frac}\left(P/\mathfrak{p}\right)$
to be residue field of $\mathfrak{p}$. Then
$k\left(\mathfrak{p}\right)\otimes_{P}\mathcal{P=}\operatorname{Kos}\left(t^{\mathfrak{p}},k\left(\mathfrak{p}\right)\otimes_{k}M\right)$
with $t_{i}^{\mathfrak{p}}=1\otimes
x_{i}-X_{i}\left(\mathfrak{p}\right)\otimes 1$, $1\leq i\leq n$.

The following key result was proved in [11].

######  Theorem 3.2.

Let $\mathfrak{X}=\mathbb{A}_{k}^{n}$ be the affine space over $k$,
$P=k\left[X_{1},\ldots,X_{n}\right]$ and let $M\in P$-$\operatorname{mod}$. If
$\mathfrak{p}\in\operatorname{res}\left(\mathfrak{X},M\right)$ then
$k\left(\mathfrak{p}\right)\perp_{P}M$, that is, the complex
$\operatorname{Kos}\left(t^{\mathfrak{p}},k\left(\mathfrak{p}\right)\otimes_{k}M\right)$
is exact. Thus

$\operatorname{res}\left(\mathfrak{X},M\right)\subseteq\left\\{\mathfrak{p}\in\mathfrak{X}:k\left(\mathfrak{p}\right)\perp_{P}M\right\\}\subseteq\left\\{\mathfrak{p}\in\mathfrak{X}:M_{\mathfrak{p}}=\mathfrak{m}_{\mathfrak{p}}M_{\mathfrak{p}}\right\\},$

where
$\mathfrak{m}_{\mathfrak{p}}=\operatorname{rad}\left(P_{\mathfrak{p}}\right)$.
If $M$ is a Noetherian $P$-module then
$\operatorname{res}\left(\mathfrak{X},M\right)=\left\\{\mathfrak{p}\in\mathfrak{X}:k\left(\mathfrak{p}\right)\perp_{P}M\right\\}$.

If $\mathfrak{p=}\left\langle X-a\right\rangle$ is a maximal ideal
corresponding to a closed point $a\in\mathfrak{X}$ then
$k\left(\mathfrak{p}\right)=\operatorname{Frac}\left(P/\mathfrak{p}\right)=k\mathfrak{\
}$and
$k\left(\mathfrak{p}\right)\otimes_{P}\mathcal{P=}\operatorname{Kos}\left(t^{\mathfrak{p}},M\right)$
with $t_{i}^{\mathfrak{p}}=x_{i}-a_{i}$, $1\leq i\leq n$, that is,
$t^{\mathfrak{p}}=x-a$. Thus if $M$ is a finitely generated $P$-module, then
$\sigma\left(\mathfrak{X},M\right)=\operatorname{Supp}\left(M\right)$ (see
(3.2)), and $a\in\operatorname{res}\left(\mathfrak{X},M\right)$ iff
$\operatorname{Kos}\left(x-a,M\right)$ is exact thanks to Theorem 3.2.

######  Lemma 3.1.

Let $R/k$ be an algebra finite extension of the field $k$,
$\mathfrak{X=}\operatorname{Spec}\left(R\right)$, $M$ an $R$-module, and let
$R=k\left[x\right]$ for an $n$-tuple $x$. Then $M$ is a $P$-module,
$\mathfrak{X}\subseteq\mathbb{A}_{k}^{n}$ up to a homeomorphism,
$\operatorname{Supp}_{P}\left(M\right)=\operatorname{Supp}_{R}\left(M\right)$,
$\sigma\left(\mathfrak{X},M\right)=\sigma\left(\mathbb{A}_{k}^{n},M\right)$,
$\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)=\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)$,
and

$\sigma_{\operatorname{p}}\left(x,M\right)=\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)\cap\mathbb{A}^{n}.$

If $M$ is a Noetherian module then

$\sigma\left(x,M\right)=\sigma\left(\mathbb{A}_{k}^{n},M\right)\cap\mathbb{A}^{n}$

and it is a nonempty closed subset of $\mathbb{A}^{n}$ whose closure in the
scheme $\mathbb{A}_{k}^{n}$ is reduced to $\sigma\left(\mathfrak{X},M\right)$.

###### Proof.

Put $R=P/\mathfrak{a}$ for some ideal $\mathfrak{a\subseteq}P$. Then $M$ turns
out to be a $P$-module with the actions $X_{i}u=x_{i}u$, $u\in M$, $1\leq
i\leq n$ and $\mathfrak{a\subseteq}\operatorname{Ann}_{P}\left(M\right)$. Note
that
$\mathfrak{X=}V\left(\mathfrak{a}\right)\subseteq\operatorname{Spec}\left(P\right)=\mathbb{A}_{k}^{n}$
up to a homeomorphism. Moreover, $M_{\mathfrak{p}}=M_{\mathfrak{p}^{\prime}}$
whenever $\mathfrak{p\in}V\left(\mathfrak{a}\right)$ and
$\mathfrak{p}^{\prime}=\mathfrak{p/a}$. Indeed,
$R_{\mathfrak{p}^{\prime}}=\left(R-\mathfrak{p}^{\prime}\right)^{-1}R=\left(P-\mathfrak{p}\right)^{-1}R=R_{\mathfrak{p}}=P_{\mathfrak{p}}\otimes_{P}R$
(see [1, (11.15.1)]) and

$M_{\mathfrak{p}^{\prime}}=R_{\mathfrak{p}^{\prime}}\otimes_{R}M=P_{\mathfrak{p}}\otimes_{P}R\otimes_{R}M=P_{\mathfrak{p}}\otimes_{P}M=M_{\mathfrak{p}}.$

In particular,
$\operatorname{Supp}_{P}\left(M\right)=\operatorname{Supp}_{R}\left(M\right)$.
But
$R/\mathfrak{p}^{\prime}=\left(P/\mathfrak{a}\right)/\left(\mathfrak{p/a}\right)=P/\mathfrak{p}$,
which means that $R/\mathfrak{p}^{\prime}\hookrightarrow M$ iff
$P/\mathfrak{p}\hookrightarrow M$. Thus
$\operatorname{Ass}_{P}\left(M\right)=\operatorname{Ass}_{R}\left(M\right)$
(see [1, 17.3]) or
$\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)=\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)$.
Using (3.1), we obtain that
$\sigma\left(\mathfrak{X},M\right)=\operatorname{Supp}_{R}\left(M\right)^{-}=\operatorname{Supp}_{P}\left(M\right)^{-}=\sigma\left(\mathbb{A}_{k}^{n},M\right)$.

The variety $\mathbb{A}^{n}$ is the set of all closed points in
$\mathbb{A}_{k}^{n}$ obtained by means of the natural homeomorphism
$\beta:\mathbb{A}^{n}\rightarrow\mathbb{A}_{k}^{n}$,
$\beta\left(a\right)=\left\langle X-a\right\rangle$ (see [16, 2.2.6]) onto the
closed points. Put $\sigma_{c}\left(\mathbb{A}_{k}^{n},M\right)$ to be
$\sigma\left(\mathbb{A}_{k}^{n},M\right)\cap\mathbb{A}^{n}$ or
$\beta^{-1}\left(\sigma\left(\mathbb{A}_{k}^{n},M\right)\right)$. Since
$\sigma\left(\mathbb{A}_{k}^{n},M\right)$ is closed, the set
$\sigma_{c}\left(\mathbb{A}_{k}^{n},M\right)$ turns out to be a closed subset
of $\mathbb{A}^{n}$. If
$\mathfrak{p}\in\sigma\left(\mathbb{A}_{k}^{n},M\right)$ then
$V\left(\mathfrak{p}\right)=\overline{\left\\{\mathfrak{p}\right\\}}\subseteq\sigma\left(\mathbb{A}_{k}^{n},M\right)$
and $\mathfrak{m}\in\sigma_{c}\left(\mathbb{A}_{k}^{n},M\right)$ whenever
$\mathfrak{m}\in V\left(\mathfrak{p}\right)\cap\mathbb{A}^{n}$ is a maximal
ideal. By Hilbert Nullstellensatz [1, 15.7],
$\sigma\left(\mathbb{A}_{k}^{n},M\right)$ is the closure of
$\sigma_{c}\left(\mathbb{A}_{k}^{n},M\right)$ in $\mathbb{A}_{k}^{n}$.

Notice that
$\mathfrak{m\in}\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)\cap\mathbb{A}^{n}$
iff $\mathfrak{m=}\left\langle
X-a\right\rangle=\operatorname{Ann}\left(u\right)$ for some point
$a\in\mathbb{A}^{n}$ and $u\neq 0$ from $M$. The latter means that
$x_{i}u=a_{i}u$ for all $i$, which means that
$a\in\sigma_{\operatorname{p}}\left(x,M\right)$ is a joint eigenvalue of $x$
with the related eigenvector $u$.

Finally, assume that $M$ is a Noetherian $R$-module (or $P$-module). Based on
Theorem 3.2, we conclude that
$\operatorname{res}\left(\mathbb{A}_{k}^{n},M\right)\cap\mathbb{A}^{n}=\left\\{a\in\mathbb{A}^{n}:k\left(\left\langle
X-a\right\rangle\right)\perp_{P}M\right\\}$ or

$\sigma_{c}\left(\mathbb{A}_{k}^{n},M\right)=\left\\{a\in\mathbb{A}^{n}:\operatorname{Kos}\left(x-a,M\right)\text{
is not exact}\right\\}=\sigma\left(x,M\right)$

is the Taylor spectrum of the tuple $x$ on $M$. Hence $\sigma\left(x,M\right)$
is a closed subset of $\mathbb{A}^{n}$ and
$\sigma\left(x,M\right)^{-}=\sigma\left(\mathbb{A}_{k}^{n},M\right)=\sigma\left(\mathfrak{X},M\right)$.
By Corollary 2.7, $\sigma\left(x,M\right)\neq\varnothing$. ∎

######  Corollary 3.4.

Let $R/k$ be an algebra finite extension of the field $k$, $M$ an $R$-module
with $\ell_{R}\left(M\right)<\infty$, and let $x$ be an $n$-tuple in $R$ such
that $k\left[x\right]\subseteq R$ is integral. Then
$\sigma_{\operatorname{p}}\left(x,M\right)=\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)=\sigma\left(\mathbb{A}_{k}^{n},M\right)=\sigma\left(x,M\right)$
and the $k\left[x\right]$-module structure on $M$ is triangularizable. In this
case, $i\left(x-a\right)=0$ for all $a\in\mathbb{A}^{n}$.

###### Proof.

Based on Corollary 3.3, we can assume that $R=k\left[x\right]$ (see also
Remark 3.3). In this case,
$\operatorname{Ass}\left(M\right)=\operatorname{Supp}\left(M\right)=\operatorname{Max}\left(M\right)\subseteq\mathbb{A}^{n}$
[1, 19.4]. It remains to use Lemma 3.1. Further, all the gaps of a Jordan-
Hölder chain is $k$ by Zariski Nullstellensatz. There is a $k$-basis
$\omega=\left(\omega_{1},\ldots,\omega_{s}\right)$ for $M$ such that
$\operatorname{Ass}\left(M\right)=\left\\{a^{\left(1\right)},\ldots,a^{\left(s\right)}\right\\}$
and

$x=\left[\begin{array}[c]{ccc}a^{\left(1\right)}&&0\\\ &\ddots&\\\
\ast&&a^{\left(s\right)}\end{array}\right],$

which means that the $k\left[x\right]$-module structure on $M$ is
triangularizable.

Finally, for every $a\in\mathbb{A}^{n}$ the complex
$\operatorname{Kos}\left(x-a,M\right)$ consists of finite dimensional
$k$-vector space. It follows that its index coincides with its Euler
characteristics. Hence

$\displaystyle i\left(x-a\right)$
$\displaystyle=\sum_{j=0}^{n}\left(-1\right)^{j+1}\dim_{k}\left(H_{j}\left(x-a,M\right)\right)=\sum_{j=0}^{n}\left(-1\right)^{j+1}\dim_{k}\left(M\otimes_{k}\wedge^{j}k^{n}\right)$
$\displaystyle=\dim_{k}\left(M\right)\sum_{j=0}^{n}\left(-1\right)^{j+1}\dbinom{n}{j}=0,$

that is, $i\left(x-a\right)=0$ for all $a\in\mathbb{A}^{n}$. ∎

The following reformulation of Theorem 3.1 results in spectral mapping theorem
for the joint point spectrum.

######  Theorem 3.3.

Let $R=k\left[x\right]$ be an algebra finite extension with an $n$-tuple $x$,
and let $M$ be an $R$-module. Then

$\sigma_{\operatorname{p}}\left(p\left(x\right),M\right)=p^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)\right)\cap\mathbb{A}^{m}$

for every $m$-tuple $p\left(x\right)$ from $R$. If
$k\left[p\left(x\right)\right]\subseteq R$ is integral then
$\sigma_{\operatorname{p}}\left(p\left(x\right),M\right)=p\left(\sigma_{\operatorname{p}}\left(x,M\right)\right)$
and $\sigma\left(p\left(x\right),M\right)$ is a closed set.

###### Proof.

Put $\mathfrak{X=}\operatorname{Spec}\left(R\right)$,
$R^{\prime}=k\left[p\left(x\right)\right]$ to be a subalgebra of $R$ generated
by an $m$-tuple $p\left(x\right)$ from $R$ with the inclusion map
$\iota:R^{\prime}\rightarrow R$, and
$\iota^{\ast}:\mathfrak{X}\rightarrow\mathfrak{X}^{\prime}$ is the natural map
with $\mathfrak{X}^{\prime}=\operatorname{Spec}\left(R^{\prime}\right)$. There
are canonical surjective maps $k\left[X\right]\rightarrow R$ and
$k\left[Y\right]\rightarrow R^{\prime}$, where $Y$ is an $m$-tuple. In
particular, $\mathfrak{X\subseteq}\mathbb{A}_{k}^{n}$ and
$\mathfrak{X}^{\prime}\subseteq\mathbb{A}_{k}^{m}$ up to a canonical
homeomorphisms. We have the polynomial ring map $p:k\left[Y\right]\rightarrow
k\left[X\right]$, $Y_{j}\mapsto p_{j}\left(X\right)$, $1\leq j\leq m$, which
in turn generates the map
$p^{\ast}:\mathbb{A}_{k}^{n}\rightarrow\mathbb{A}_{k}^{m}$ of affine spaces.
If $\mathfrak{m}_{a}=\left\langle X-a\right\rangle\in\mathbb{A}^{n}$ then
$p^{\ast}\left(\mathfrak{m}_{a}\right)=p^{-1}\left(\mathfrak{m}_{a}\right)\supseteq\left\langle
Y-p\left(a\right)\right\rangle=\mathfrak{m}_{p\left(a\right)}$, for
$p\left(\left\langle
Y-p\left(a\right)\right\rangle\right)\subseteq\left\langle
p\left(X\right)-p\left(a\right)\right\rangle\subseteq\mathfrak{m}_{a}$. Since
$\mathfrak{m}_{p\left(a\right)}\in\mathbb{A}^{m}$ is maximal, we conclude that
$p^{\ast}\left(\mathfrak{m}_{a}\right)=\mathfrak{m}_{p\left(a\right)}$. Thus
$p^{\ast}\left(\mathbb{A}^{n}\right)\subseteq\mathbb{A}^{m}$ and
$p^{\ast}:\mathbb{A}^{n}\rightarrow\mathbb{A}^{m}$, $a\mapsto p\left(a\right)$
is a polynomial morphism. Moreover,
$p^{\ast}\left(\mathbb{A}_{k}^{n}\right)\cap\mathbb{A}^{m}=p^{\ast}\left(\mathbb{A}^{n}\right)$.
Indeed, if $p^{\ast}\left(\mathfrak{q}\right)=\mathfrak{m}_{b}$ for some
$\mathfrak{q}\in\mathbb{A}_{k}^{n}$ and $b\in\mathbb{A}^{m}$, then
$\mathfrak{q\subseteq m}_{a}$ for some $a\in\mathbb{A}^{n}$ and
$\mathfrak{m}_{b}\subseteq
p^{\ast}\left(\mathfrak{m}_{a}\right)=\mathfrak{m}_{p\left(a\right)}$, which
implies that $\mathfrak{m}_{b}=\mathfrak{m}_{p\left(a\right)}$ or
$b=p\left(a\right)$.

Using the duality correspondence between affine schemes and rings, we obtain
the following commutative diagrams

$\begin{array}[c]{ccccccc}k\left[Y\right]&\overset{p}{\longrightarrow}&k\left[X\right]&&\mathbb{A}_{k}^{m}&\overset{p^{\ast}}{\longleftarrow}&\mathbb{A}_{k}^{n}\\\
\downarrow&&\downarrow&\Leftrightarrow&\uparrow&&\uparrow\\\
R^{\prime}&\overset{\iota}{\longrightarrow}&R&&\mathfrak{X}^{\prime}&\overset{\iota^{\ast}}{\longleftarrow}&\mathfrak{X}\end{array}$

supporting each other. Using Theorem 3.1 and Lemma 3.1, we deduce that

$\displaystyle\sigma_{\operatorname{p}}\left(p\left(x\right),M\right)$
$\displaystyle=\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{m},M\right)\cap\mathbb{A}^{m}=\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M\right)\cap\mathbb{A}^{m}=\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\right)\cap\mathbb{A}^{m}$
$\displaystyle=p^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)\right)\cap\mathbb{A}^{m}.$

If $R^{\prime}\subseteq R$ is integral, then
$\sigma_{\operatorname{p}}\left(\mathfrak{X}^{\prime},M\right)\cap\mathbb{A}^{m}=\iota^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathfrak{X},M\right)\cap\mathbb{A}^{n}\right)$
by Maximality [1, 14.3] and Theorem 3.1. Using again Lemma 3.1, we deduce that
$\sigma_{\operatorname{p}}\left(p\left(x\right),M\right)=p^{\ast}\left(\sigma_{\operatorname{p}}\left(\mathbb{A}_{k}^{n},M\right)\cap\mathbb{A}^{n}\right)=p\left(\sigma_{\operatorname{p}}\left(x,M\right)\right)$.

Finally, prove that Taylor spectrum $\sigma\left(p\left(x\right),M\right)$ is
a closed set. Since $R/R^{\prime}$ is integral, $M$ is module finite over
$R^{\prime}$. By Lemma 3.1,
$\sigma\left(p\left(x\right),M\right)=\sigma\left(\mathbb{A}_{k}^{m},M\right)\cap\mathbb{A}^{m}$
is a closed set. ∎

######  Remark 3.4.

The same argument from the proof of Theorem 3.3, and Proposition 3.1, result
in the spectral mapping property
$\sigma\left(p\left(x\right),M\right)=p\left(\sigma\left(x,M\right)\right)$
for Taylor spectrum whenever $k\left[p\left(x\right)\right]\subseteq R$ is
integral and $M$ is Noetherian. But as we have seen above in Theorem 2.3, the
formula holds in the general case of all tuples $p\left(x\right)$.

If $\mathfrak{a\subseteq}P$ is a radical ideal with the related algebraic set
$Y=Z\left(\mathfrak{a}\right)=V\left(\mathfrak{a}\right)\cap\mathbb{A}^{n}$
(that is $I\left(Y\right)=\mathfrak{a}$), then the coordinate ring $R$ of $Y$
is the reduced to the Noetherian ring $P/\mathfrak{a}$. Put
$\mathfrak{Y=}\operatorname{Spec}\left(R\right)$. By (3.2), we obtain that
$\sigma\left(\mathbb{A}_{k}^{n},R\right)=\operatorname{Supp}\left(R\right)=V\left(\mathfrak{a}\right)=\mathfrak{Y}$
is the set of all subvarieties of $Y$, and by Lemma 3.1,
$Y=\mathfrak{Y}\cap\mathbb{A}^{n}=\sigma\left(x,R\right)$ is the Taylor
spectrum of the operator tuple $x$ in $R$, which consists of all coordinate
functions. For a Noetherian $k\left[x\right]$-module $M$, we obtain that the
tuple $x$ on the module $M$ is triangularizable whose diagonal entries are
varieties (see [11, 4.4]).

Let $p=\left(p_{1},\ldots,p_{m}\right):Y\rightarrow\mathbb{A}^{m}$ be a
morphism given by means of a ring extension
$R^{\prime}=k\left[y\right]\subseteq R$ with $y=p\left(x\right)$ to be an
$m$-tuple. In particular, we have the spectrum
$\sigma\left(\mathbb{A}_{k}^{m},R\right)$ of the $R^{\prime}$-module $R$. If
$R/R^{\prime}$ is integral then $p^{\ast}$ turns out to be a finite morphism
(see [16, 2.3]). Using (1.2), we deduce that
$\sigma\left(\mathbb{A}_{k}^{m},R\right)\cap\mathbb{A}^{m}=p\left(Y\right)^{-}$
(see [11] for the details).

######  Corollary 3.5.

If $Y\subseteq\mathbb{A}^{n}$ is an algebraic set over $k$ with its coordinate
ring $R$, then $\sigma_{\operatorname{p}}\left(x,R\right)$ is the set $Y_{is}$
of all isolated points of $Y$. If $p:Y\rightarrow\mathbb{A}^{m}$ is a finite
morphism given by a ring extension $k\left[p\left(x\right)\right]\subseteq R$,
then
$\sigma\left(p\left(x\right),R\right)=\sigma\left(\mathbb{A}_{k}^{m},R\right)\cap\mathbb{A}^{m}=p\left(Y\right)$
and
$\sigma_{\operatorname{p}}\left(p\left(x\right),R\right)=p\left(Y_{is}\right)$.

###### Proof.

Note that $a\in\sigma_{\operatorname{p}}\left(x,R\right)$ iff $\left\langle
X-a\right\rangle g\left(X\right)\subseteq\mathfrak{a}$ for some
$g\left(X\right)\notin\mathfrak{a}$. It means that
$Y=Z\left(\mathfrak{a}\right)\subseteq\left\\{a\right\\}\cup Z\left(g\right)$
and $Y\nsubseteq Z\left(g\right)$. Since $a\in Y$, we conclude that $a\notin
Z\left(g\right)$ and
$\left\\{a\right\\}=Y\cap\left(\mathbb{A}^{n}-Z\left(g\right)\right)$ is open
in $Y$, that is, $\left\\{a\right\\}$ is an isolated point. Conversely, take
$a\in Y_{is}$. Since the complements to hypersurfaces from $\mathbb{A}^{n}$ is
a topology base in $\mathbb{A}^{n}$, it follows that
$\left\\{a\right\\}=Y\cap\left(\mathbb{A}^{n}-Z\left(g\right)\right)$ for a
certain hypersurface $Z\left(g\right)$. Then
$Y=Z\left(\mathfrak{a}\right)\subseteq\left\\{a\right\\}\cup Z\left(g\right)$,
$Y\nsubseteqq Z\left(g\right)$ and

$\displaystyle\left\langle X-a\right\rangle g$
$\displaystyle\subseteq\left\langle
X-a\right\rangle\cap\sqrt{g}=I\left(\left\\{a\right\\}\right)\cap
I\left(Z\left(g\right)\right)=I\left(\left\\{a\right\\}\cup
Z\left(g\right)\right)$ $\displaystyle\subseteq
I\left(Y\right)=\sqrt{\mathfrak{a}}=\mathfrak{a}\text{,}$

which means that $a\in\sigma_{\operatorname{p}}\left(x,R\right)$. Further, if
$p$ is a finite morphism then $R$ is a Noetherian
$k\left[p\left(x\right)\right]$-module and
$\sigma\left(p\left(x\right),R\right)=\sigma\left(\mathbb{A}_{k}^{m},R\right)\cap\mathbb{A}^{m}$
is closed (see Lemma 3.1). Using Lemma 3.1 and Theorem 2.3, we derive that
$\sigma\left(\mathbb{A}_{k}^{m},R\right)\cap\mathbb{A}^{m}=\sigma\left(p\left(x\right),R\right)=p\left(\sigma\left(x,R\right)\right)=p\left(Y\right)$,
and
$\sigma_{\operatorname{p}}\left(p\left(x\right),R\right)=p\left(\sigma_{\operatorname{p}}\left(x,R\right)\right)=p\left(Y_{is}\right)$
thanks to Theorem 3.3. ∎

##  4\. Koszul homology groups of a variety

In the present section we focus on the case of a module $M$ which is the
coordinate ring $R$ of a variety $Y$, and investigate its Koszul homology
groups. We are targeting to find out a link between the dimensions of Koszul
homology groups of the standard tuple in $R$ and the dimension
$d=\dim\left(Y\right)$.

###  4.1. Multiplicity formula of Serre

Let $P=k\left[X\right]$ be the polynomial algebra over the field $k$ with
$n$-tuple $X=\left(X_{1},\ldots,X_{n}\right)$, and put
$h\left(r\right)=\dim_{k}\left(\mathfrak{t}^{r}/\mathfrak{t}^{r+1}\right)=\dbinom{r+n-1}{n-1}$
to be the Hilbert polynomial ($\deg\left(h\right)=n-1$) of the graded
$P$-algebra $G^{\circ}P$ associated with the filtration
$\left\\{\mathfrak{t}^{r}\right\\}$, where $\mathfrak{t=}\left\langle
X\right\rangle\subseteq P$ is the maximal ideal generated by $X$, and
$\dbinom{z}{s}=\dfrac{1}{s!}z\left(z-1\right)\cdots\left(z-s+1\right)$ is the
binomial coefficient. If
$\mathfrak{s}\left(P,r\right)=\dim_{k}\left(P/\mathfrak{t}^{r}\right)$ is the
Samuel polynomial of the filtration then
$\mathfrak{s}\left(P,r\right)=\sum_{i=0}^{r-1}h\left(i\right)=\sum_{i=0}^{r-1}\dbinom{i+n-1}{n-1}=\dbinom{r+n-1}{n}$.
If $M$ is a $P$-module given by an $n$-tuple $x$ of $k$-linear maps on $M$,
then $P/\mathfrak{t}^{r}\otimes_{k}M=M^{\binom{r+n-1}{n}}$ and there are
nilpotent (shift) operators $X_{i}\otimes 1$ acting on $M^{\binom{r+n-1}{n}}$.
Put $x_{i}^{\left(r\right)}=\left(1\otimes x_{i}\right)-X_{i}\otimes 1$ and
$x^{\left(r\right)}=\left(x_{1}^{\left(r\right)},\ldots,x_{n}^{\left(r\right)}\right)$
to be an $n$-tuple of mutually commuting operators on the inflation
$M^{\binom{r+n-1}{n}}$, which defines its $P$-module structure. For every
$a\in\mathbb{A}^{n}$ the $P$-module structure on $M^{\binom{r+n-1}{n}}$ given
by $x^{\left(r\right)}-a$ is the same inflation of the $P$-module $M$ given by
the tuple $x-a$. Just notice that
$x_{i}^{\left(r\right)}-a_{i}=1\otimes\left(x_{i}-a_{i}\right)-X_{i}\otimes
1=\left(x_{i}-a_{i}\right)^{\left(r\right)}$ for all $i$, or
$x^{\left(r\right)}-a=\left(x-a\right)^{\left(r\right)}=1\otimes\left(x-a\right)-X\otimes
1$. In this case one can also replace $\mathfrak{t}$ by $\left\langle
X-a\right\rangle$.

######  Lemma 4.1.

If $M$ is a Noetherian $P$-module then so is $M^{\binom{r+n-1}{n}}$ and
$\sigma\left(x,M\right)=\sigma\left(x^{\left(r\right)},M^{\binom{r+n-1}{n}}\right)$
for all $r$. Thus $i\left(\left(x-a\right)^{\left(r\right)}\right)<\infty$ for
all $a\in\mathbb{A}^{n}$ and $r\geq 1$.

###### Proof.

For every $r\geq 1$ consider the following canonical exact sequence

(4.1) $0\leftarrow P/\mathfrak{t}^{r}\longleftarrow
P/\mathfrak{t}^{r+1}\longleftarrow\mathfrak{t}^{r}/\mathfrak{t}^{r+1}\leftarrow
0,$

in $P$-$\operatorname{mod}$. Notice that the $P$-module structure of
$\mathfrak{t}^{r}/\mathfrak{t}^{r+1}$ is reduced to its $k$-vector space one
with
$\dim_{k}\left(\mathfrak{t}^{r}/\mathfrak{t}^{r+1}\right)=h\left(r\right)$.
Using (4.1), we generate the following exact sequences

$0\leftarrow P/\mathfrak{t}^{r}\otimes_{k}M\longleftarrow
P/\mathfrak{t}^{r+1}\otimes_{k}M\longleftarrow\mathfrak{t}^{r}/\mathfrak{t}^{r+1}\otimes_{k}M\leftarrow
0$

such that the $P$-module structure of
$\mathfrak{t}^{r}/\mathfrak{t}^{r+1}\otimes_{k}M$ is diagonal and therefore it
is Noetherian. By induction on $r$ we deduce that
$P/\mathfrak{t}^{r}\otimes_{k}M$ is Noetherian. By Spectral Mapping Theorem
2.3, we have

$\displaystyle\sigma\left(x^{\left(r\right)},M^{\binom{r+n-1}{n}}\right)$
$\displaystyle=\sigma\left(1\otimes x-X\otimes 1,M^{\binom{r+n-1}{n}}\right)$
$\displaystyle=\left\\{a-b:\left(a,b\right)\in\sigma\left(\left(1\otimes
x,X\otimes 1\right),M^{\binom{r+n-1}{n}}\right)\right\\}$

and $b\in\sigma\left(X\otimes 1,M^{\binom{r+n-1}{n}}\right)$. For every
$a\in\sigma\left(1\otimes x,M^{\binom{r+n-1}{n}}\right)$ there is $b$ such
that $\left(a,b\right)\in\sigma\left(\left(1\otimes x,X\otimes
1\right),M^{\binom{r+n-1}{n}}\right)$ (see Theorem 2.2). Moreover,
$b_{i}\in\sigma\left(\left(X_{i}\otimes
1\right)|M^{\binom{r+n-1}{n}}\right)=\left\\{0\right\\}$, for $X_{i}\otimes 1$
is nilpotent. Hence
$\sigma\left(x^{\left(r\right)},M^{\binom{r+n-1}{n}}\right)=\sigma\left(1\otimes
x,M^{\binom{r+n-1}{n}}\right)$. Finally,

$\operatorname{Kos}\left(1\otimes\left(x-a\right),M^{\binom{r+n-1}{n}}\right)=P/\mathfrak{t}^{r}\otimes_{k}\operatorname{Kos}\left(x-a,M\right),$

which is exact iff so is $\operatorname{Kos}\left(x-a,M\right)$, that is,
$\sigma\left(1\otimes x,M^{\binom{r+n-1}{n}}\right)=\sigma\left(x,M\right)$.
It remains to use Theorem 2.1. ∎

Let $A/k$ be a local Noetherian algebra with an $n$-tuple $x$ contained in the
maximal ideal of $A$, and let $M$ be a Noetherian $A$-module such that
$\ell\left(M/xM\right)<\infty$, that is, $x$ is a system of parameters for
$M$. The filtration $\left\\{\left\langle x\right\rangle^{r}M\right\\}$ in $M$
defines the Samuel polynomial
$\mathfrak{s}\left(M,r\right)=\ell\left(M/\left\langle
x\right\rangle^{r}M\right)$ whose degree is at most $n$, and
$\mathfrak{s}\left(M,r\right)=e_{x,M}\left(n\right)\dfrac{r^{n}}{n!}+q\left(r\right)$
with $e_{x,M}\left(n\right)=\Delta^{n}\mathfrak{s}\left(M,\circ\right)$ ($n$th
difference operator) and $\deg\left(q\right)<n$. By Theorem 2.1,
$\ell\left(H_{p}\left(x,M\right)\right)=\dim_{k}\left(H_{p}\left(x,M\right)\right)$
for all $p$, and the Euler characteristic
$\chi\left(x,M\right)=\sum_{p=0}^{n}\left(-1\right)^{p}\ell\left(H_{p}\left(x,M\right)\right)$
from [24, 4.A.3] is reduced to the opposite index $-i\left(x\right)$ of the
tuple $x$. The multiplicity formula of Serre [24, 4.A.3. Theorem 1] is
expressed in the following way

(4.2) $i\left(x\right)=-e_{x,M}\left(n\right).$

If $\dim\left(M\right)=n$ then $i\left(x\right)<0$, whereas
$i\left(x\right)=0$ in the case of $\dim\left(M\right)<n$.

###  4.2. The numerical Tor-polynomial

Now let $\mathfrak{p}\subseteq P$ be a nonzero prime ideal of a variety
$Y\subseteq\mathbb{A}^{n}$ with its coordinate ring $R=P/\mathfrak{p}$. The
actions of $X_{i}$ on $R$ are denoted by $x_{i}$, and
$x=\left(x_{1},\ldots,x_{n}\right)$ is an operator tuple on $R$. Thus
$R=k\left[x\right]$ is an algebra finite extension of $k$ which is a domain.
In particular, $\sigma_{\operatorname{p}}\left(x,R\right)=\varnothing$ or
$H_{n}\left(x\right)=0$. For every $r$ we have the inflated $P$-module
$R^{\binom{r+n-1}{n}}$ given by means of the $n$-tuple $x^{\left(r\right)}$.
Using Lemma 3.1 and Lemma 4.1, we deduce that

$Y=\operatorname{Spec}\left(R\right)\cap\mathbb{A}^{n}=\sigma\left(\mathbb{A}_{k}^{n},R\right)\cap\mathbb{A}^{n}=\sigma\left(x,R\right)=\sigma\left(x^{\left(r\right)},R^{\binom{r+n-1}{n}}\right)$

for all $r$. For every $a\in Y$ the homology groups
$H_{i}\left(x^{\left(r\right)}-a,R^{\binom{r+n-1}{n}}\right)$ denoted by
$H_{i}\left(x^{\left(r\right)}-a\right)$ are finite dimensional $k$-vector
spaces (Lemma 4.1). Put
$d_{p}\left(a^{\left(r\right)}\right)=\dim_{k}\left(H_{p}\left(x^{\left(r\right)}-a\right)\right)$
and define the tuple
$d\left(a^{\left(r\right)}\right)=\left(d_{p}\left(a^{\left(r\right)}\right)\right)_{p}\in\mathbb{Z}_{+}^{n+1}$.
In particular,
$i\left(x^{\left(r\right)}-a\right)=\sum_{p=0}^{n}\left(-1\right)^{p+1}d_{p}\left(a^{\left(r\right)}\right)$
and it defines the function $i_{Y}^{\left(r\right)}:Y\rightarrow\mathbb{Z}$,
$a\mapsto i\left(x^{\left(r\right)}-a\right)$. We also put $i_{Y}$ instead of
$i_{Y}^{\left(1\right)}$.

######  Lemma 4.2.

If $Y\subseteq\mathbb{A}^{n}$ is a variety then $i_{Y}=-\delta_{nd}$, where
$d=\dim\left(Y\right)$.

###### Proof.

If $Y=\mathbb{A}^{n}$ then $\operatorname{Kos}\left(X-a,P\right)$ is a free
$P$-module resolution for the $P$-module $k$ (see Corollary 2.3), that is,
$H_{0}\left(X-a,P\right)=k$ and $H_{i}\left(X-a,P\right)=0$, $i\geq 1$. In
particular, $d_{0}\left(a\right)=1$, $d_{i}\left(a\right)=0$, $i\geq 1$, and
$i_{Y}\left(a\right)=-1$ for all $a\in Y$.

Now assume that $Y\subseteq\mathbb{A}^{n}$ is a variety of dimension $d<n$,
and $a=0\in Y$, which responds to the maximal ideal $\left\langle
x\right\rangle\subseteq R$. Then $R_{\left\langle x\right\rangle}$ is a local
Noetherian algebra with its maximal ideal $\mathfrak{m=}\left\langle
x\right\rangle R_{\left\langle x\right\rangle}$. By Lemma 2.5, we have
$H_{p}\left(x,R\right)=H_{p}\left(x,R\right)_{\left\langle
x\right\rangle}=H_{p}\left(x/1,R_{\left\langle x\right\rangle}\right)$ for all
$p\geq 0$, and $\mathfrak{m=}\left\langle x/1\right\rangle$ for the $n$-tuple
$x/1$. Thus $i\left(x\right)=i\left(x/1\right)$, $x/1$ is a system of
parameters for $R_{\left\langle x\right\rangle}$ and the Samuel polynomial
$\mathfrak{s}\left(R_{\left\langle x\right\rangle},r\right)$ has the degree
$\dim\left(R_{\left\langle x\right\rangle}\right)$ [1, Theorem 21.4]. But
$\dim\left(R_{\left\langle x\right\rangle}\right)=\dim\left(R\right)=d<n$.
Using Serre’s formula (4.2), we conclude that $i\left(x\right)=0$. ∎

Similar result for the inflated $P$-module $R^{\binom{r+n-1}{n}}$ is not
trivial (at least is not straightforward), for its $P$-module structure is not
diagonal. The related result is proved below in Proposition 4.1.

######  Remark 4.1.

In the case of a nonsingular point $a=0\in Y$ of a variety
$Y\subseteq\mathbb{A}^{n}$ with $d<n$, one can skip Serre’s formula. Namely,
$d=\dim_{k}\left(\mathfrak{m}/\mathfrak{m}^{2}\right)$ is the minimal number
of generators of the $R_{\left\langle x\right\rangle}$-module $\mathfrak{m}$
[1, 10.9]. As above $\mathfrak{m=}\left\langle x/1\right\rangle$ for the
$n$-tuple $x/1$, and there is a new $d$-tuple $y\subseteq\mathfrak{m}$
generating $\mathfrak{m}$ due to the regularity of $R_{\left\langle
x\right\rangle}$, that is, $y$ is a tuple related to $x/1$. Note that $0\in
Y=\sigma\left(x,R\right)$ or $H_{p}\left(x,R\right)\neq 0$ for some $p$, which
in turn implies that $0\in\sigma\left(x/1,R_{\left\langle
x\right\rangle}\right)$. Using Corollary 2.9, we deduce that
$i\left(x\right)=i\left(x/1\right)=\delta_{nd}i\left(y\right)=0$.

######  Remark 4.2.

A singular point $a$ of a variety $Y$ is said to be integral if $x_{i}-a_{i}$
is integral over the subalgebra $k\left[x^{\prime}-a^{\prime}\right]\subseteq
R$ with $i$-th term skipped tuple $x^{\prime}$. Suppose that $a=0$, and say
$x_{n}$ is integral over $R^{\prime}=k\left[x^{\prime}\right]$ with
$x^{\prime}=\left(x_{1},\ldots,x_{n-1}\right)$. Then $R^{\prime}\subseteq R$
is integral and $i\left(x^{\prime}\right)<\infty$ thanks to Theorem 2.1.
Moreover, $0\in\sigma\left(x^{\prime},R\right)$ thanks to Corollary 2.2. By
Lemma 2.3, $i\left(x\right)=0$. Again we can avoid (4.2).

######  Example 4.1.

In the case of a point $Y=\left\\{a\right\\}$, we have $R=k$ and $i=0$ thanks
to Corollary 3.4. Actually, $\operatorname{Kos}\left(x,k\right)$ is the
following complex

$0\leftarrow k\leftarrow
k^{n}\leftarrow\cdots\leftarrow\wedge^{p}k^{n}\leftarrow\cdots\leftarrow
k\leftarrow 0$

with trivial morphisms, $H_{j}\left(x,k\right)=\wedge^{j}k^{n}$,
$d_{j}\left(a\right)=\dbinom{n}{j}$, $j\geq 0$, and
$i_{Y}=\sum_{j=0}^{n}\left(-1\right)^{j+1}\dbinom{n}{j}=0$.

Now assume that $Y$ is a variety and $a=0\in Y$, that is,
$\mathfrak{p\subseteq}\left\langle X\right\rangle$. For every $r\geq 1$
consider again the sequence (4.1). Notice that
$\operatorname{Tor}_{i}^{P}\left(\mathfrak{t}^{r}/\mathfrak{t}^{r+1},R\right)=\operatorname{Tor}_{i}^{P}\left(k,R\right)^{h\left(r\right)}=H_{i}\left(x\right)^{h\left(r\right)}$
for all $i\geq 0$. The functor $\circ\otimes_{P}R$ generates the following
long homology exact sequence

$\displaystyle 0$ $\displaystyle\leftarrow
P/\mathfrak{t}^{r}\otimes_{P}R\leftarrow
P/\mathfrak{t}^{r+1}\otimes_{P}R\leftarrow
k^{h\left(r\right)}\leftarrow\operatorname{Tor}_{1}^{P}\left(P/\mathfrak{t}^{r},R\right)\leftarrow\operatorname{Tor}_{1}^{P}\left(P/\mathfrak{t}^{r+1},R\right)\leftarrow
H_{1}\left(x\right)^{h\left(r\right)}$ $\displaystyle\cdots$
$\displaystyle\leftarrow
H_{i-1}\left(x\right)^{h\left(r\right)}\longleftarrow\operatorname{Tor}_{i}^{P}\left(P/\mathfrak{t}^{r},R\right)\leftarrow\operatorname{Tor}_{i}^{P}\left(P/\mathfrak{t}^{r+1},R\right)\leftarrow
H_{i}\left(x\right)^{h\left(r\right)}\leftarrow\cdots$

But $P/\mathfrak{t}^{r}\otimes_{P}R=R/\mathfrak{t}^{r}R=R/\left\langle
x\right\rangle^{r}$ and $0\leftarrow R/\left\langle
x\right\rangle^{r}\longleftarrow R/\left\langle
x\right\rangle^{r+1}\longleftarrow\left\langle x\right\rangle^{r}/\left\langle
x\right\rangle^{r+1}\leftarrow 0$ is exact. Thus we come with the exact
sequence

$0\leftarrow\left\langle x\right\rangle^{r}/\left\langle
x\right\rangle^{r+1}\leftarrow
k^{h\left(r\right)}\leftarrow\operatorname{Tor}_{1}^{P}\left(P/\mathfrak{t}^{r},R\right)\leftarrow\operatorname{Tor}_{1}^{P}\left(P/\mathfrak{t}^{r+1},R\right)\leftarrow
H_{1}\left(x\right)^{h\left(r\right)}\leftarrow\cdots.$

Note that $\mathcal{P=}\operatorname{Kos}\left(t,P\otimes_{k}R\right)$ is a
free resolution of $R$, and
$P/\mathfrak{t}^{r}\otimes_{P}\mathcal{P=}\operatorname{Kos}\left(x^{\left(r\right)},P/\mathfrak{t}^{r}\otimes_{k}R\right)=\operatorname{Kos}\left(x^{\left(r\right)},R^{\binom{r+n-1}{n}}\right)$
with $x^{\left(r\right)}=1\otimes x-X\otimes 1$ (see Lemma 4.1). In
particular,
$\operatorname{Tor}_{i}^{P}\left(P/\mathfrak{t}^{r},R\right)=H_{i}\left(P/\mathfrak{t}^{r}\otimes_{P}\mathcal{P}\right)=H_{i}\left(x^{\left(r\right)}\right)$
for all $i$ and $r$. Now we apply the localization functor
$\circ\otimes_{R}R_{\left\langle x\right\rangle}$ (or
$\circ\otimes_{P}P_{\left\langle X\right\rangle}$) at $\left\langle
x\right\rangle$. By Lemma 2.5, we have

$H_{i}\left(x^{\left(r\right)}/1,R_{\left\langle
x\right\rangle}^{\binom{r+n-1}{n}}\right)=H_{i}\left(x^{\left(r\right)},R^{\binom{r+n-1}{n}}\right)_{\left\langle
x\right\rangle}=H_{i}\left(x^{\left(r\right)},R^{\binom{r+n-1}{n}}\right)=H_{i}\left(x^{\left(r\right)}\right)$

for all $i$ and $r$. We obtain the following exact sequence

(4.3) $\displaystyle 0$
$\displaystyle\leftarrow\mathfrak{m}^{r}/\mathfrak{m}^{r+1}\leftarrow
k^{h\left(r\right)}\leftarrow H_{1}\left(x^{\left(r\right)}\right)\leftarrow
H_{1}\left(x^{\left(r+1\right)}\right)\leftarrow
H_{1}\left(x\right)^{h\left(r\right)}\leftarrow\cdots$
$\displaystyle\leftarrow H_{n-1}\left(x\right)^{h\left(r\right)}\leftarrow
H_{n}\left(x^{\left(r\right)}\right)\leftarrow
H_{n}\left(x^{\left(r+1\right)}\right)\leftarrow
H_{n}\left(x\right)^{h\left(r\right)}=0,$

where $\mathfrak{m=}\left\langle x\right\rangle R_{\left\langle
x\right\rangle}$ is the maximal ideal of the local ring $R_{\left\langle
x\right\rangle}$.

######  Lemma 4.3.

If $H_{j+1}\left(x\right)=0$ for some $j$, then
$H_{i}\left(x^{\left(r\right)}\right)=0$ for all $i>j$ and $r\geq 1$, and the
sequence

$\displaystyle 0$
$\displaystyle\leftarrow\mathfrak{m}^{r}/\mathfrak{m}^{r+1}\leftarrow
k^{h\left(r\right)}\leftarrow\cdots\leftarrow
H_{i-1}\left(x\right)^{h\left(r\right)}\leftarrow
H_{i}\left(x^{\left(r\right)}\right)\leftarrow
H_{i}\left(x^{\left(r+1\right)}\right)\leftarrow
H_{i}\left(x\right)^{h\left(r\right)}\leftarrow\cdots$
$\displaystyle\leftarrow H_{j-1}\left(x^{\left(r\right)}\right)\leftarrow
H_{j}\left(x^{\left(r\right)}\right)\leftarrow
H_{j}\left(x\right)^{h\left(r\right)}\leftarrow 0$

remains exact. In particular, $H_{n}\left(x^{\left(r\right)}\right)=0$ for all
$r$, and the sequence

$0\leftarrow\mathfrak{m}^{r}/\mathfrak{m}^{r+1}\leftarrow
k^{h\left(r\right)}\leftarrow\cdots\leftarrow
H_{n-1}\left(x^{\left(r\right)}\right)\leftarrow
H_{n-1}\left(x^{\left(r+1\right)}\right)\leftarrow
H_{n-1}\left(x\right)^{h\left(r\right)}\leftarrow 0$

is exact.

###### Proof.

Suppose that $H_{j+1}\left(x\right)=0$ for some $j$. Prove that
$H_{i}\left(x^{\left(r\right)}\right)=0$ for all $i>j$, $r\geq 1$ by induction
on $r$. By Lemma 2.5, $H_{i}\left(x\right)=0$ for all $i>j$. If $r=1$, then we
have
$\operatorname{Tor}_{i}^{P}\left(P/\mathfrak{t},R\right)=\operatorname{Tor}_{i}^{P}\left(k,R\right)=H_{i}\left(x\right)$,
$i\geq 0$ and we come with the exact pieces

$\cdots\leftarrow H_{i}\left(x\right)\leftarrow
H_{i}\left(x^{\left(2\right)}\right)\leftarrow
H_{i}\left(x\right)^{h\left(r\right)}\leftarrow\cdots.$

of the sequence (4.3) for $r=1$. If $i>j$ then
$H_{i}\left(x^{\left(2\right)}\right)=0$. By induction hypothesis,
$H_{i}\left(x^{\left(r\right)}\right)=0$ for all $i>j$. The exact part

$\cdots\leftarrow H_{i-1}\left(x\right)^{h\left(r\right)}\leftarrow
H_{i}\left(x^{\left(r\right)}\right)\leftarrow
H_{i}\left(x^{\left(r+1\right)}\right)\leftarrow
H_{i}\left(x\right)^{h\left(r\right)}\leftarrow\cdots$

of the sequence (4.3) for $i>j$, implies that
$H_{i}\left(x^{\left(j+1\right)}\right)=0$, and the long homology sequence
terminates at $H_{j}\left(x\right)^{h\left(r\right)}$. Finally, since
$H_{n}\left(x\right)=0$ we conclude that
$H_{n}\left(x^{\left(r\right)}\right)=0$ for all $r\geq 1$. ∎

Based on Lemma 4.1, we have $i\left(x^{\left(r\right)}\right)<\infty$ for all
$r$, and we define the following function

$p:\mathbb{Z}\longrightarrow\mathbb{Z}\text{,\quad}p\left(r\right)=\sum_{i=1}^{n}\left(-1\right)^{i+1}\dim_{k}\left(\operatorname{Tor}_{i}^{P}\left(P/\mathfrak{t}^{r},R\right)\right)$

called as of $\operatorname{Tor}$-polynomial. Note that
$-\dim_{k}\left(H_{0}\left(x^{\left(r\right)}\right)\right)+p\left(r\right)=i\left(x^{\left(r\right)}\right)$
for all $r$. Recall that $R_{\left\langle x\right\rangle}$ is a local
Noetherian algebra, and $d=\dim\left(Y\right)=\dim\left(R_{\left\langle
x\right\rangle}\right)=\deg\left(h_{R}\right)+1=\deg\mathfrak{s}\left(R_{\left\langle
x\right\rangle},\circ\right)\leq n$, where
$h_{R}\left(r\right)=\dim_{k}\left(\mathfrak{m}^{r}/\mathfrak{m}^{r+1}\right)$
is the Hilbert polynomial of $R_{\left\langle x\right\rangle}$ associated with
$\left\\{\mathfrak{m}^{r}\right\\}$.

######  Proposition 4.1.

The equality
$p\left(r\right)=\left(1-\delta_{nd}\right)\mathfrak{s}\left(R_{\left\langle
x\right\rangle},r\right)$ holds for all $r$, that is, the function $p$ is a
numerical polynomial of degree at most $d$. In particular,
$i_{Y}^{\left(r\right)}=\dbinom{r+n-1}{n}i_{Y}$ for all $r$.

###### Proof.

Using the exact sequence from Lemma 4.3 and its Euler characteristics, we
deduce that

$\displaystyle 0$
$\displaystyle=-h_{R}\left(r\right)+h\left(r\right)+\sum_{i=1}^{n-1}\left(-1\right)^{i}\dim_{k}\left(H_{i}\left(x^{\left(r\right)}\right)\right)+\sum_{i=1}^{n-1}\left(-1\right)^{i+1}\dim_{k}\left(H_{i}\left(x^{\left(r+1\right)}\right)\right)$
$\displaystyle+h\left(r\right)\sum_{i=1}^{n-1}\left(-1\right)^{i}\dim_{k}\left(H_{i}\left(x\right)\right).$

Note that $H_{0}\left(x\right)=R/\left\langle x\right\rangle=k$,
$H_{n}\left(x^{\left(r\right)}\right)=0$, and
$\sum_{i=0}^{n}\left(-1\right)^{i}\dim_{k}\left(H_{i}\left(x\right)\right)=1-p\left(1\right)=-i\left(x\right)$.
If $d<n$ then $i\left(x\right)=0$ thanks to Lemma 4.2. Thus
$p:\mathbb{Z}\rightarrow\mathbb{Z}$ is a function with the property

$p\left(z+1\right)-p\left(z\right)=h_{R}\left(z\right),$

that is, $p\left(z+1\right)-p\left(z\right)$ is a numerical polynomial for
large $z$. That means $p\left(z\right)\in\mathbb{Q}\left[z\right]$ is a
numerical polynomial (see [16, 1.7.3]) and
$p\left(r\right)=\sum_{j=0}^{r-1}h_{R}\left(j\right)=\mathfrak{s}\left(R_{\left\langle
x\right\rangle},r\right)$ for all $r\geq 1$. If $Y=\mathbb{A}^{n}$ then
$h_{R}\left(r\right)=h\left(r\right)$, $H_{i}\left(x\right)=0$, $i\geq 1$, and
$i\left(x\right)=-1$ (see Lemma 4.2). It follows that
$p\left(r+1\right)=p\left(r\right)$ for all $r\geq 1$. But
$p\left(1\right)=i\left(x\right)+1=0$, therefore $p=0$.

Finally, taking into account that

$H_{0}\left(x^{\left(r\right)}\right)=H_{0}\left(x^{\left(r\right)}\right)_{\left\langle
x\right\rangle}=\operatorname{Tor}_{0}^{P}\left(P/\mathfrak{t}^{r},R\right)_{\left\langle
x\right\rangle}=\left(P/\mathfrak{t}^{r}\otimes_{P}R\right)_{\left\langle
x\right\rangle}=\left(R/\left\langle x\right\rangle^{r}\right)_{\left\langle
x\right\rangle}=R_{\left\langle x\right\rangle}/\mathfrak{m}^{r},$

we deduce that
$\dim_{k}\left(H_{0}\left(x^{\left(r\right)}\right)\right)=\dim_{k}\left(R_{\left\langle
x\right\rangle}/\mathfrak{m}^{r}\right)=\mathfrak{s}\left(R_{\left\langle
x\right\rangle},r\right)$. It follows that

$i\left(x^{\left(r\right)}\right)=-\dim_{k}\left(H_{0}\left(x^{\left(r\right)}\right)\right)+p\left(r\right)=-\mathfrak{s}\left(R_{\left\langle
x\right\rangle},r\right)+\left(1-\delta_{nd}\right)\mathfrak{s}\left(R_{\left\langle
x\right\rangle},r\right)=-\delta_{nd}\mathfrak{s}\left(R_{\left\langle
x\right\rangle},r\right).$

Thus $i_{Y}^{\left(r\right)}=0$ whenever $d<n$, and
$i_{Y}^{\left(r\right)}=-\mathfrak{s}\left(P_{\left\langle
X\right\rangle},r\right)=-\mathfrak{s}\left(P,r\right)=-\dbinom{r+n-1}{n}$ for
$Y=\mathbb{A}^{n}$. ∎

######  Example 4.2.

If $d<n$ and $H_{2}\left(x\right)=0$, then
$H_{i}\left(x^{\left(r\right)}\right)=0$ for all $i\geq 2$, $r\geq 1$ thanks
to Lemma 4.3. Then
$p\left(r\right)=\dim_{k}\left(H_{1}\left(x^{\left(r\right)}\right)\right)$,
$i\left(x\right)=-1+\dim_{k}\left(H_{1}\left(x\right)\right)=0$ and
$\dim_{k}\left(H_{1}\left(x^{\left(r\right)}\right)\right)=\mathfrak{s}\left(R_{\left\langle
x\right\rangle},r\right)$, that is,
$r\mapsto\dim_{k}\left(H_{1}\left(x^{\left(r\right)}\right)\right)$ is a
polynomial of degree $d$. If $Y\subseteq\mathbb{A}^{2}$ is an irreducible
curve and $a\in Y$, then $d_{0}\left(a\right)=1$ and $d_{2}\left(a\right)=0$.
By Lemma 4.2, $i_{Y}\left(a\right)=0$, therefore $d_{1}\left(a\right)=1$, and
$r\mapsto\dim_{k}\left(H_{1}\left(\left(x-a\right)^{\left(r\right)}\right)\right)$
is a linear polynomial.

######  Remark 4.3.

The presence of nonzero $H_{1}\left(x\right)$ is well know. If
$H_{1}\left(x\right)=0$ then $R\overset{\partial_{0}}{\longleftarrow}R\otimes
k^{n}\overset{\partial_{1}}{\longleftarrow}R\otimes\wedge^{2}k^{n}$ is exact
and so is its localization $R_{\left\langle
x\right\rangle}\overset{\partial_{0}}{\longleftarrow}R_{\left\langle
x\right\rangle}\otimes
k^{n}\overset{\partial_{1}}{\longleftarrow}R_{\left\langle
x\right\rangle}\otimes\wedge^{2}k^{n}$, that is,
$x/1\subseteq\operatorname{rad}\left(R_{\left\langle x\right\rangle}\right)$
with $H_{1}\left(x/1\right)=0$. Using [3, 9.7, Theorem 1], we conclude that
$x/1$ is a regular sequence in $R_{\left\langle x\right\rangle}$. In
particular, $\operatorname{depth}\left(R_{\left\langle
x\right\rangle}\right)\geq n$. But $\operatorname{depth}\left(R_{\left\langle
x\right\rangle}\right)\leq\dim\left(R_{\left\langle
x\right\rangle}\right)=d<n$, a contradiction.

The evaluation map $a:P\rightarrow k$, $f\mapsto f\left(a\right)$ can be
lifted to a ring homomorphism $a:R\rightarrow k$, which is the quotient
mapping $R\rightarrow R/\left\langle x\right\rangle$. In particular, there are
$k$-linear maps
$\wedge^{r}a:R\otimes_{k}\wedge^{r}k^{n}\rightarrow\wedge^{r}k^{n}$, $r\geq
1$, and $\left(\wedge^{r}a\right)\partial_{r-1}=0$ for all $r$, which in turn
define $k$-linear maps
$a^{\left(r\right)}:H_{r}\left(x\right)\rightarrow\wedge^{r}k^{n}$ on the
homology groups. Note that
$H_{0}\left(x\right)=R/\operatorname{im}\left(\partial_{0}\right)=k$,
$a^{\left(0\right)}=1$, and $a^{\left(1\right)}:H_{1}\left(x\right)\rightarrow
k^{n}$ is a $k$-linear map.

###  4.3. Upper-triangular matrices

Now let us analyze the first homology group $H_{1}\left(x\right)$. We assume
that $\operatorname{char}\left(k\right)\neq 2$. Every
$g=\sum_{i<j}g_{ij}e_{i}\wedge e_{j}\in R\otimes_{k}\wedge^{2}k^{n}$ can be
represented by means of an upper triangular matrix
$g=\left[g_{ij}\right]_{i,j}\in M_{n}\left(R\right)$ whose $i$-th column
$C_{i}$ and $i$-th row $R_{i}$ (they are $n$-tuples in $R$) have the shapes

$\begin{array}[c]{cccccc}&g_{1i}&&&&\\\ &g_{2i}&&&&\\\ &\vdots&&&&\\\
&g_{i-1i}&&&&\\\ \cdots&0&g_{ii+1}&\cdots&g_{in-1}&g_{in}\\\
&\vdots&&&&\end{array}$

for all $i>1$, and $C_{1}$, $R_{n}$ consist of zeros. For $n$-tuples $\alpha$
and $\beta$ from $R^{n}$ we write $\left\langle\alpha,\beta\right\rangle$
instead of the sum $\sum_{i=1}^{n}\alpha_{i}\beta_{i}$ in $R$. Note that
$\partial_{1}:R\otimes_{k}\wedge^{2}k^{n}\rightarrow R\otimes_{k}k^{n}$ is an
$R$-linear map acting by the rule

(4.4)
$\partial_{1}\left(g\right)=\sum_{i<j}x_{i}g_{ij}e_{j}-x_{j}g_{ij}e_{j}=\sum_{i=1}^{n}\left(\sum_{s<i}x_{s}g_{si}-\sum_{i<s}x_{s}g_{is}\right)e_{i}=\sum_{i=1}^{n}\left(\left\langle
C_{i},x\right\rangle-\left\langle R_{i},x\right\rangle\right)e_{i}.$

For example, if

$g=\left[\begin{array}[c]{ccccc}0&\cdots&g_{j}&\cdots&g_{n}\\\
0&\cdots&0&\cdots&0\\\ &&\vdots&&\\\ &&&&\\\ 0&&0&&0\end{array}\right]\text{
with }\sum_{k=j}^{n}x_{k}g_{k}=0$