# Bounds on Injective Dimension and
Exceptional Complete Intersection Maps

Hossein Faridian  Hossein Faridian, School of Mathematical and Statistical
Sciences, Clemson University, Clemson, SC 29634, USA<EMAIL_ADDRESS>
###### Abstract.

We prove that if $f:R\rightarrow S$ is a local homomorphism of noetherian
local rings, and $M$ is a non-zero finitely generated or artinian $S$-module
whose injective dimension over $R$ is bounded by the difference of the
embedding dimensions of $R$ and $S$, then $M$ is an injective $S$-module and
$f$ is an exceptional complete intersection map.

###### Key words and phrases:

injective dimension; exceptional complete intersection map; André-Quillen
homology

###### 2010 Mathematics Subject Classification:

13D05; 13C11; 13J99; 13D03.

##  1\. Introduction

In the past fifty years or so, a great body of research has been accomplished
by algebraists on the properties of rings in terms of those of their modules.
The trend has gradually shifted towards the relative situation in which one
studies the properties of ring homomorphisms in terms of those of the modules
along them. The focus of this work is on surjective homomorphisms of
noetherian local rings. Such a map is a complete intersection map if its
kernel is generated by a regular sequence. If in addition, the regular
sequence is part of a minimal generating set for the maximal ideal of the
source, then the homomorphism is called an exceptional complete intersection
map. This class of maps is of particular interest as they arise naturally in
various contexts; for example, the diagonal of a smooth map is locally
exceptional complete intersection.

It is proved in [BIK] that certain homological invariants of modules over ring
homomorphisms detect the exceptional complete intersection property. More
specifically, it is shown that if $f:R\rightarrow S$ is a local homomorphism
of noetherian local rings, and $M$ is a non-zero finitely generated $S$-module
with
$\operatorname{fd}_{R}(M)\leq\operatorname{edim}(R)-\operatorname{edim}(S)$,
then $M$ is a free $S$-module, $f$ is an exceptional complete intersection
map, and
$\operatorname{fd}_{R}(M)=\operatorname{edim}(R)-\operatorname{edim}(S)=\operatorname{fd}_{R}(S)$;
see [BIK, Theorem 3.1]. The goal of this article is to prove an analogue of
this theorem involving the injective dimension. More specifically, we prove
the following result; see Theorems 3.3 and 3.12.

######  Theorem.

Let $f:R\rightarrow S$ be a local homomorphism of noetherian local rings, and
$M$ a non-zero finitely generated or artinian $S$-module with
$\operatorname{id}_{R}(M)\leq\operatorname{edim}(R)-\operatorname{edim}(S)$.
Then the following assertions hold:

  1. (i)

$M$ is an injective $S$-module, i.e. $M\cong
E_{S}(l)^{\operatorname{type}_{S}(M)}$ where $l$ is the residue field of $S$.

  2. (ii)

$f$ is an exceptional complete intersection map.

  3. (iii)

$\operatorname{id}_{R}(M)=\operatorname{edim}(R)-\operatorname{edim}(S)=\operatorname{fd}_{R}(S)$.

In particular, if $M$ is injective as an $R$-module, then
$\operatorname{edim}(R)=\operatorname{edim}(S)$ and $f$ is flat.

We further prove an analogue of [BIK, Theorem 3.1] for artinian modules; see
Corollary 3.13. Along the way, we collect a few lemmas with suitable
references for the convenience of the reader, and also establish a handful of
lemmas which might be of independent interest.

##  2\. Basic Definitions and Observations

In what follow, all rings are assumed to be commutative with unity. Given a
local ring $(R,\mathfrak{m},k)$, $\mathfrak{m}$ denotes its unique maximal
ideal, $k\cong R/\mathfrak{m}$ is its residue field, and $\widehat{R}$
designates its $\mathfrak{m}$-adic completion. A ring homomorphism
$f:(R,\mathfrak{m})\rightarrow(S,\mathfrak{n})$ is local if
$f(\mathfrak{m})\subseteq\mathfrak{n}$. Given an $R$-module $M$, the notations
$\operatorname{pd}_{R}(M)$, $\operatorname{id}_{R}(M)$, and
$\operatorname{fd}_{R}(M)$ are used for the projective, injective, and flat
dimensions of $M$, respectively. Moreover, $\dim(R)$ and
$\operatorname{edim}(R)$ indicate the Krull and the embedding dimensions of
$R$, respectively.

We begin with a remark on André-Quillen homology.

######  Remark 2.1.

Let $f:R\rightarrow S$ be a ring homomorphism. By developing the machinery of
non-abelian homological algebra in the framework of Quillen’s model category
theory, one can construct the cotangent complex $\mathbb{L}^{S|R}$ associated
to $f$ as a well-defined object in the derived category $\mathcal{D}(S)$.
Given any $S$-module $M$, the cotangent complex $\mathbb{L}^{S|R}$ is used to
define the André-Quillen homology modules of $S$ over $R$ with coefficients in
$M$ as

$D_{i}(S|R;M):=\operatorname{Tor}_{i}^{S}\left(\mathbb{L}^{S|R},M\right)$

for every $i\geq 0$. For any $i\geq 0$, the André-Quillen homology
$D_{i}(S|R;M)$ is functorial in all three arguments. One further has

$D_{0}(S|R;M)\cong\Omega_{S|R}\otimes_{S}M$

where $\Omega_{S|R}$ is the module of Kähler differentials of $S$ over $R$. If
$f$ is surjective with $\operatorname{Ker}(f)=\mathfrak{a}$, then
$D_{0}(S|R;M)=0$ and

$D_{1}(S|R;M)\cong(\mathfrak{a}/\mathfrak{a}^{2})\otimes_{S}M.$

Given ring homomorphisms $R\rightarrow S\rightarrow T$ and a $T$-module $M$,
one has the following Jacobi-Zariski exact sequence of $T$-modules:

$\displaystyle\cdots\rightarrow D_{1}\left(S|R;M\right)\rightarrow
D_{1}\left(T|R;M\right)\rightarrow D_{1}\left(T|S;M\right)\rightarrow
D_{0}\left(S|R;M\right)\rightarrow D_{0}\left(T|R;M\right)\rightarrow
D_{0}\left(T|S;M\right)\rightarrow 0$

If $f:(R,\mathfrak{m},k)\rightarrow S$ is a local epimorphism of noetherian
local rings with $\operatorname{Ker}(f)=\mathfrak{a}$, then

$D_{1}(S|R;k)\cong(\mathfrak{a}/\mathfrak{a}^{2})\otimes_{S}k\cong\mathfrak{a}/\mathfrak{m}\mathfrak{a},$

so the minimum number of generators of $\mathfrak{a}$ can be computed as

$\mu(\mathfrak{a})=\operatorname{rank}_{k}(\mathfrak{a}/\mathfrak{m}\mathfrak{a})=\operatorname{rank}_{k}\left(D_{1}(S|R;k)\right),$

which in particular yields

$\operatorname{edim}(R)=\mu(\mathfrak{m})=\operatorname{rank}_{k}\left(D_{1}(k|R;k)\right).$

It is worth noting that the general theory of André-Quillen homology uses
simplicial resolutions; see [An], [GS], [Iy], [Qu1], [Qu2], and [Qu3].
However, the lower André-Quillen homologies $D_{i}(S|R;M)$ for $i=0,1,2$ can
be constructed without any appeal to simplicial resolutions; see [LS].

Using André-Quillen homology, one can prove the following lemma.

######  Lemma 2.2.

Let $f:(R,\mathfrak{m})\rightarrow S$ be a flat local homomorphism of
noetherian local rings whose closed fiber $S/\mathfrak{m}S$ is regular. Then
we have:

$\operatorname{edim}\left(S/\mathfrak{m}S\right)=\operatorname{edim}(S)-\operatorname{edim}(R)=\dim(S)-\dim(R)$

###### Proof.

See [BIK, 2.6]. ∎

We next recall the notion of an exceptional complete intersection map.

######  Remark 2.3.

Let $f:(R,\mathfrak{m},k)\rightarrow S$ be a local homomorphism of noetherian
local rings. Then $f$ admits a Cohen factorization, i.e. it fits into a
commutative diagram

${R}$${S}$${R^{\prime}}$${\widehat{S}}$$\scriptstyle{f}$$\scriptstyle{\dot{f}}$$\scriptstyle{f^{\prime}}$

of noetherian local rings in which $\dot{f}$ is flat with regular closed fiber
$R^{\prime}/\mathfrak{m}R^{\prime}$ and $f^{\prime}$ is surjective; see [AFH,
Theorem 1.1]. Accordingly, $f$ is said to be a complete intersection map if
there is a Cohen factorization of $f$ such that
$\operatorname{Ker}(f^{\prime})$ is generated by a regular sequence on
$R^{\prime}$. This property is independent of the choice of the Cohen
factorization; see [Av, Theorem 1.2]. In particular, $f$ is complete
intersection if and only if $f^{\prime}$ is complete intersection. Moreover,
the class of complete intersection maps is closed under composition and flat
base change. When $f$ is complete intersection, there is an inequality

$\operatorname{edim}(R)-\dim(R)\leq\operatorname{edim}(S)-\dim(S);$

see [BIK, Lemma 2.13]. Then $f$ is said to be an exceptional complete
intersection map if equality holds above. When $f$ is surjective, this
property is equivalent to the condition that $\operatorname{Ker}(f)$ can be
generated by a regular sequence whose image in $\mathfrak{m}/\mathfrak{m}^{2}$
is a linearly independent set over $k$; see [ILP, Lemma 3.2]. It follows from
Lemma 2.2 that $f$ is exceptional complete intersection if and only if
$f^{\prime}$ is exceptional complete intersection. For more information on
(exceptional) complete intersection maps, refer to [Av], [BIK], [ILP], and
[BILP].

The following lemma will be used later.

######  Lemma 2.4.

Let $f:R\rightarrow S$ be a surjective exceptional complete intersection map,
$k$ the residue field of $S$, and $\pi:S\rightarrow k$ the quotient map. Then
$D_{1}(\pi|R;k):D_{1}(S|R;k)\rightarrow D_{1}(k|R;k)$ is injective.

###### Proof.

Follows from the proof of [BIK, Lemma 2.16]. ∎

##  3\. Main Results

In this section, we prove our main result, i.e. the theorem outlined in the
introduction. We first handle the artinian case.

######  Lemma 3.1.

Let $(R,\mathfrak{m},k)$ be a noetherian local ring, and $M$ a non-zero
finitely generated or artinian $R$-module. Let $E_{R}(k)$ denote the injective
envelope of $k$, and
$\operatorname{type}_{R}(M)=\operatorname{rank}_{k}\left(\operatorname{Ext}_{R}^{0pt_{R}(\mathfrak{m},M)}(k,M)\right)$.
Then $M$ is an injective $R$-module if and only if $M\cong
E_{R}(k)^{\operatorname{type}_{R}(M)}$.

###### Proof.

Suppose that $M$ is injective. If $M$ is finitely generated, then $R$ is
artinian; see [BH, Exercise 3.1.23]. Thus in any case, $M$ is artinian, so
$\operatorname{Ass}_{R}(M)=\\{\mathfrak{m}\\}$, whence
$0pt_{R}(\mathfrak{m},M)=0$. But then the Bass number
$\mu_{0}^{R}(\mathfrak{m},M)=\operatorname{rank}_{k}\left(\operatorname{Ext}_{R}^{0}(k,M)\right)=\operatorname{type}_{R}(M)$,
so $M\cong E_{R}(M)\cong E_{R}(k)^{\operatorname{type}_{R}(M)}$. The converse
is clear. ∎

######  Lemma 3.2.

Let $f:R\rightarrow S$ be a homomorphism of noetherian rings, $X$ an
$S$-complex that is considered an $R$-complex via $f$, and
$X^{\vee}=\operatorname{Hom}_{S}\left(X,E_{S}(l)\right)$ where $l$ is the
residue field of $S$. Then the following assertions hold:

  1. (i)

$\operatorname{id}_{R}(X)=\operatorname{fd}_{R}(X^{\vee})$ provided
$H_{i}(X)=0$ for $|i|\gg 0$.

  2. (ii)

$\operatorname{fd}_{R}(X)=\operatorname{id}_{R}(X^{\vee})$.

###### Proof.

(i): Let $\mathfrak{a}$ be an ideal of $R$, and $L$ a free resolution of
$R/\mathfrak{a}$ in which $L_{i}$ is finitely generated for every $i\geq 0$.
Then using the Hom Evaluation (see [AF, Lemma 4.4 (I)]), we get

$\begin{split}\operatorname{Ext}_{R}^{i}(R/\mathfrak{a},X)^{\vee}&=\operatorname{Hom}_{S}\left(\operatorname{Ext}_{R}^{i}(R/\mathfrak{a},X),E_{S}(l)\right)\\\
&=\operatorname{Hom}_{S}\left(H_{-i}\left(\operatorname{Hom}_{R}(L,X)\right),E_{S}(l)\right)\\\
&\cong
H_{i}\left(\operatorname{Hom}_{S}\left(\operatorname{Hom}_{R}(L,X),E_{S}(l)\right)\right)\\\
&\cong
H_{i}\left(L\otimes_{R}\operatorname{Hom}_{S}\left(X,E_{S}(l)\right)\right)\\\
&=\operatorname{Tor}_{i}^{R}\left(R/\mathfrak{a},\operatorname{Hom}_{S}\left(X,E_{S}(l)\right)\right)\\\
&=\operatorname{Tor}_{i}^{R}\left(R/\mathfrak{a},X^{\vee}\right)\end{split}$

for every $i\in\mathbb{Z}$. Since $E_{S}(l)$ is faithfully injective, the
isomorphism implies that for any $i\in\mathbb{Z}$, we have
$\operatorname{Ext}_{R}^{i}(R/\mathfrak{a},X)=0$ if and only if
$\operatorname{Tor}_{i}^{R}\left(R/\mathfrak{a},X^{\vee}\right)=0$. Now the
result follows from [AF, Corollaries 2.5.I and 2.5.F].

(ii): Let $\mathfrak{a}$ be an ideal of $R$, and $L$ a free resolution of
$R/\mathfrak{a}$. Then using the Hom-Tensor Adjunction, see [Ch, A.2.8], we
get

$\begin{split}\operatorname{Tor}_{i}^{R}(R/\mathfrak{a},X)^{\vee}&=\operatorname{Hom}_{S}\left(\operatorname{Tor}_{i}^{R}(R/\mathfrak{a},X),E_{S}(l)\right)\\\
&=\operatorname{Hom}_{S}\left(H_{i}\left(L\otimes_{R}X\right),E_{S}(l)\right)\\\
&\cong
H_{-i}\left(\operatorname{Hom}_{S}\left(L\otimes_{R}X,E_{S}(l)\right)\right)\\\
&\cong
H_{-i}\left(\operatorname{Hom}_{R}\left(L,\operatorname{Hom}_{S}\left(X,E_{S}(l)\right)\right)\right)\\\
&=\operatorname{Ext}_{R}^{i}\left(R/\mathfrak{a},\operatorname{Hom}_{S}\left(X,E_{S}(l)\right)\right)\\\
&=\operatorname{Ext}_{R}^{i}\left(R/\mathfrak{a},X^{\vee}\right)\end{split}$

for every $i\in\mathbb{Z}$. As in (i), the isomorphism implies that for any
$i\in\mathbb{Z}$, we have $\operatorname{Tor}_{i}^{R}(R/\mathfrak{a},X)=0$ if
and only if
$\operatorname{Ext}_{R}^{i}\left(R/\mathfrak{a},X^{\vee}\right)=0$. Now the
result follows from [AF, Corollaries 2.5.I and 2.5.F]. ∎

######  Theorem 3.3.

Let $f:(R,\mathfrak{m},k)\rightarrow(S,\mathfrak{n},l)$ be a local
homomorphism of noetherian local rings, and $M$ a non-zero artinian $S$-module
with
$\operatorname{id}_{R}(M)\leq\operatorname{edim}(R)-\operatorname{edim}(S)$.
Then the following assertions hold:

  1. (i)

$M$ is an injective $S$-module, i.e. $M\cong
E_{S}(l)^{\operatorname{type}_{S}(M)}$.

  2. (ii)

$f$ is an exceptional complete intersection map.

  3. (iii)

$\operatorname{id}_{R}(M)=\operatorname{edim}(R)-\operatorname{edim}(S)=\operatorname{fd}_{R}(S)$.

In particular, if $M$ is injective as an $R$-module, then
$\operatorname{edim}(R)=\operatorname{edim}(S)$ and $f$ is flat.

###### Proof.

If $M$ is an injective $S$-module, then by Lemma 3.1, $M\cong
E_{S}(l)^{\operatorname{type}_{S}(M)}$, so Lemma 3.2 (ii) implies that

$\operatorname{id}_{R}(M)=\operatorname{id}_{R}\left(E_{S}(l)\right)=\operatorname{id}_{R}(S^{\vee})=\operatorname{fd}_{R}(S).$

That is to say, the second equality in (iii) follows from (i).

We next reduce to the case where $S$ is $\mathfrak{n}$-adically complete. As
$M$ is an artinian $S$-module, hence $\mathfrak{n}$-torsion, it has an
$\widehat{S}$-module structure that restricts to its original $S$-module
structure via the completion map $S\rightarrow\widehat{S}$, and $M\cong
M\otimes_{S}\widehat{S}$ both as $S$-modules and $\widehat{S}$-modules; see
[SS, Proposition 2.1.15 and Corollary 2.2.6]. Also, $M$ is an artinian
$\widehat{S}$-module. We further note that
$\operatorname{edim}\left(\widehat{S}\right)=\operatorname{edim}(S)$,

$E_{\widehat{S}}\left(\widehat{S}/\mathfrak{n}\widehat{S}\right)\cong
E_{S}(S/\mathfrak{n})\cong E_{S}(l),$

and

$\operatorname{type}_{\widehat{S}}(M)=\operatorname{type}_{\widehat{S}}\left(M\otimes_{S}\widehat{S}\right)=\operatorname{type}_{S}(M).$

By Lemma 3.1, $M$ is an injective $S$-module if and only if $M\cong
E_{S}(l)^{\operatorname{type}_{S}(M)}$. Similarly, $M$ is an injective
$\widehat{S}$-module if and only if $M\cong
E_{\widehat{S}}\left(\widehat{S}/\mathfrak{n}\widehat{S}\right)^{\operatorname{type}_{\widehat{S}}(M)}$.
It follows that $M$ is an injective $S$-module if and only if $M$ is an
injective $\widehat{S}$-module. Therefore, we can replace $S$ with
$\widehat{S}$ and assume that $S$ is $\mathfrak{n}$-adically complete.
Consequently, the Matlis duality theory implies that
$M^{\vee}=\operatorname{Hom}_{S}\left(M,E_{S}(l)\right)$ is a finitely
generated $S$-module.

Now by Lemma 3.2 (i), we have

$\operatorname{fd}_{R}(M^{\vee})=\operatorname{id}_{R}(M)\leq\operatorname{edim}(R)-\operatorname{edim}(S).$

Therefore, [BIK, Theorem 3.1] implies that $M^{\vee}$ is a free $S$-module,
$f$ is an exceptional complete intersection map, and

$\operatorname{id}_{R}(M)=\operatorname{fd}_{R}(M^{\vee})=\operatorname{edim}(R)-\operatorname{edim}(S).$

Furthermore, $\operatorname{id}_{S}(M)=\operatorname{fd}_{S}(M^{\vee})=0$, so
$M$ is an injective $S$-module. ∎

We next need a series of lemmas before we can establish the finitely generated
case of the theorem outlined in the introduction.

######  Lemma 3.4.

Let $f:(R,\mathfrak{m})\rightarrow S$ be a flat local homomorphism of
noetherian local rings, and $M$ a non-zero finitely generated $R$-module. Then
we have:

$\operatorname{id}_{S}(M\otimes_{R}S)=\operatorname{id}_{R}(M)+\operatorname{id}_{S/\mathfrak{m}S}(S/\mathfrak{m}S)$

###### Proof.

See [FT, Corollary 1]. ∎

######  Corollary 3.5.

Let $(R,\mathfrak{m},k)$ be a noetherian local ring, and $M$ a finitely
generated $R$-module. Then
$\operatorname{id}_{\widehat{R}}\left(M\otimes_{R}\widehat{R}\right)=\operatorname{id}_{R}(M)$.
In particular, $M$ is an injective $R$-module if and only if
$M\otimes_{R}\widehat{R}$ is an injective $\widehat{R}$-module.

###### Proof.

Since the completion map $R\rightarrow\widehat{R}$ is flat and local, Lemma
3.4 yields

$\operatorname{id}_{\widehat{R}}\left(M\otimes_{R}\widehat{R}\right)=\operatorname{id}_{R}(M)+\operatorname{id}_{\widehat{R}/\mathfrak{m}\widehat{R}}\left(\widehat{R}/\mathfrak{m}\widehat{R}\right).$

However, $\widehat{R}/\mathfrak{m}\widehat{R}\cong R/\mathfrak{m}\cong k$, so
$\operatorname{id}_{\widehat{R}/\mathfrak{m}\widehat{R}}\left(\widehat{R}/\mathfrak{m}\widehat{R}\right)=\operatorname{id}_{k}(k)=0$
as any vector space is injective. Thus the result follows. ∎

######  Lemma 3.6.

Let $R$ be a noetherian local ring, and $M$ a finitely generated $R$-module
with $\operatorname{id}_{R}(M)<\infty$. Then $\operatorname{ann}_{R}(M)$ is
either zero or contains a non-zerodivisor of $R$.

###### Proof.

See [LV, Theorem 4.1]. ∎

######  Lemma 3.7.

Let $(R,\mathfrak{m})$ be a noetherian local ring,
$a\in\mathfrak{m}\backslash\mathfrak{m}^{2}$ a non-zerodivisor of $R$, and $M$
a finitely generated $R/\langle a\rangle$-module. Then the following
assertions hold:

  1. (i)

$\operatorname{pd}_{R/\langle a\rangle}(M)=\operatorname{pd}_{R}(M)-1$.

  2. (ii)

$\operatorname{id}_{R/\langle a\rangle}(M)=\operatorname{id}_{R}(M)-1$.

###### Proof.

See [Na, Corollary 27.4] and [LV, Theorem 3.1]. ∎

######  Lemma 3.8.

Let $R$ be a noetherian ring, $M$ a finitely generated $R$-module, and
$\underline{a}=a_{1},...,a_{n}\in R$ a regular sequence on $M$. If $L$ is a
free resolution of $M$, then $K^{R}(\underline{a})\otimes_{R}L$ is a free
resolution of $M/\langle\underline{a}\rangle M$.

###### Proof.

Since $L$ is a free resolution of $M$, there is a quasi-isomorphism
$L\rightarrow M$. As the Koszul complex $K^{R}(\underline{a})$ is a bounded
complex of free $R$-modules, we get a quasi-isomorphism
$K^{R}(\underline{a})\otimes_{R}L\rightarrow
K^{R}(\underline{a})\otimes_{R}M$. Since $\underline{a}=a_{1},...,a_{n}$ is a
regular sequence on $M$, [BH, Corollary 1.6.14 (a)] implies that

$H_{i}\left(K^{R}(\underline{a})\otimes_{R}M\right)=H_{i}(\underline{a};M)\cong\begin{cases}M/\langle\underline{a}\rangle
M&\text{if }i=0\\\ 0&\text{if }i>0,\end{cases}$

so there is a quasi-isomorphism $K^{R}(\underline{a})\otimes_{R}M\rightarrow
M/\langle\underline{a}\rangle M$. As a result, we get a quasi-isomorphism
$K^{R}(\underline{a})\otimes_{R}L\rightarrow M/\langle\underline{a}\rangle M$,
so $K^{R}(\underline{a})\otimes_{R}L$ is a free resolution of
$M/\langle\underline{a}\rangle M$. ∎

######  Remark 3.9.

Let $R$ and $S$ be two rings, $M$ a symmetric $R$-$R$-bimodule, i.e. $ax=xa$
for every $a\in R$ and $x\in M$, and $N$ and $K$ two $R$-$S$-bimodules.
Considering $M$ as a left $R$-module, $N$ as a right $S$-module, and $K$ as an
$R$-$S$-bimodule, we have the Hom Swap Isomorphism

$\operatorname{Hom}_{R}\left(M,\operatorname{Hom}_{S}(N,K)\right)\cong\operatorname{Hom}_{S}\left(N,\operatorname{Hom}_{R}(M,K)\right);$

see [Ch, A.2.9]. On the other hand, considering $M$ as a right $R$-module, $N$
as an $R$-$S$-bimodule, and $K$ as a right $S$-module, we have the Hom-Tensor
Adjoint Isomorphism

$\operatorname{Hom}_{R}\left(M,\operatorname{Hom}_{S}(N,K)\right)\cong\operatorname{Hom}_{S}(M\otimes_{R}N,K);$

see [Ch, A.2.8]. One might be tempted to merge the two isomorphisms as their
left hand sides seem to coincide. However, in the first isomorphism, $M$ is a
left $R$-module and $\operatorname{Hom}_{S}(N,K)$ is a left $R$-module via the
left $R$-module structure of $K$. Similarly, in the second isomorphism, $M$ is
a right $R$-module and $\operatorname{Hom}_{S}(N,K)$ is a right $R$-module via
the left $R$-module structure of $N$. As a result, if this $R$-$R$-bimodule
structure of $\operatorname{Hom}_{S}(N,K)$ is symmetric, then we conclude that
the left hand sides of the above isomorphisms coincide, so we get:

$\operatorname{Hom}_{S}(M\otimes_{R}N,K)\cong\operatorname{Hom}_{S}\left(N,\operatorname{Hom}_{R}(M,K)\right)$

But if the $R$-$R$-bimodule structure of $\operatorname{Hom}_{S}(N,K)$ is not
symmetric, then the above isomorphism need not hold.

For example, let $K$ be a field. Considering the polynomial ring $K[X]$ as a
$K[X]$-$K$-bimodule in a natural way, we notice that the
$K[X]$-$K[X]$-bimodule $\operatorname{Hom}_{K}\left(K[X],K[X]\right)$ is not
symmetric. Indeed, the left $K[X]$-module structure of
$\operatorname{Hom}_{K}\left(K[X],K[X]\right)$ comes from the left
$K[X]$-module structure of $K[X]$ in the second argument, and the right
$K[X]$-module structure of $\operatorname{Hom}_{K}\left(K[X],K[X]\right)$
comes from the left $K[X]$-module structure of $K[X]$ in the first argument.
Consider the elements $X\in K[X]$ and
$f\in\operatorname{Hom}_{K}\left(K[X],K[X]\right)$ defined by $f(1)=1$ and
$f(X^{n})=X^{n+1}$ for every $n\geq 1$. Then we have

$(Xf)(1)=Xf(1)=X\neq X^{2}=f(X)=f(X1)=(fX)(1),$

so $Xf\neq fX$. We further observe that:

$\begin{split}\operatorname{Hom}_{K}\left(K[X]\otimes_{K[X]}K[X]/\langle
X\rangle,K[X]\right)&\cong\operatorname{Hom}_{K}\left(K[X]/\langle
X\rangle,K[X]\right)\\\ &\cong\operatorname{Hom}_{K}\left(K,K[X]\right)\\\
&\cong K[X]\end{split}$

and

$\begin{split}\operatorname{Hom}_{K}\left(K[X],\operatorname{Hom}_{K[X]}\left(K[X]/\langle
X\rangle,K[X]\right)\right)&\cong\operatorname{Hom}_{K}\left(K[X],\left(0:_{K[X]}\langle
X\rangle\right)\right)\\\ &=\operatorname{Hom}_{K}(K[X],0)\\\ &=0\end{split}$

Thus the left hand sides cannot be isomorphic.

######  Lemma 3.10.

Let $f:(R,\mathfrak{m},k)\rightarrow S$ be a local homomorphism of noetherian
local rings, and $M$ a finitely generated $S$-module that is considered an
$R$-module via $f$. Then the following assertions are equivalent for any given
$n\geq 0$:

  1. (i)

$\operatorname{id}_{R}(M)\leq n$.

  2. (ii)

$\operatorname{Ext}_{R}^{i}(k,M)=0$ for every $i\geq n+1$.

###### Proof.

See [AF, Proposition 5.5 (I)]. ∎

######  Lemma 3.11.

Let $f:(R,\mathfrak{m},k)\rightarrow(S,\mathfrak{n},l)$ be a flat local
homomorphism of noetherian local rings with $S/\mathfrak{m}S$ regular,
$g:S\rightarrow T$ a local homomorphism of noetherian local rings, and $M$ a
finitely generated $T$-module. Then we have:

$\operatorname{id}_{S}(M)\leq\operatorname{id}_{R}(M)+\operatorname{edim}(S/\mathfrak{m}S)$

###### Proof.

Since $f$ is local, we have $\mathfrak{m}S\subseteq\mathfrak{n}$. Considering
the noetherian local ring $(S/\mathfrak{m}S,\mathfrak{n}/\mathfrak{m}S)$, we
set $\mu(\mathfrak{n}/\mathfrak{m}S)=\operatorname{edim}(S/\mathfrak{m}S)=n$.
Let $\underline{a}=a_{1},...,a_{n}\in\mathfrak{n}$ be such that
$\\{a_{1}+\mathfrak{m}S,...,a_{n}+\mathfrak{m}S\\}$ is a minimal generating
set for $\mathfrak{n}/\mathfrak{m}S$. In particular,
$\mathfrak{n}=\langle\underline{a}\rangle+\mathfrak{m}S$. As $S/\mathfrak{m}S$
is regular, we conclude that $a_{1}+\mathfrak{m}S,...,a_{n}+\mathfrak{m}S$ is
a regular sequence on the ring $S/\mathfrak{m}S$, so
$\underline{a}=a_{1},...,a_{n}$ is a regular sequence on the $S$-module
$S/\mathfrak{m}S$. Set the Koszul complex $K=K^{S}(\underline{a})$. Moreover,
let $L$ be a free resolution for the $R$-module $k$. Since $f$ is flat, we
infer that $S\otimes_{R}L$ is a free resolution for the $S$-module
$S\otimes_{R}k\cong S/\mathfrak{m}S$. It follows from Lemma 3.8 that
$K\otimes_{R}L\cong K\otimes_{S}(S\otimes_{R}L)$ is a free resolution for the
$S$-module

$(S/\mathfrak{m}S)/\langle\underline{a}\rangle(S/\mathfrak{m}S)\cong
S/(\langle\underline{a}\rangle+\mathfrak{m}S)=S/\mathfrak{n}\cong l.$

Consider the third quadrant $S$-bicomplex $W$ with
$W_{p,q}=\operatorname{Hom}_{S}\left(K_{-p},\operatorname{Hom}_{R}(L_{-q},M)\right)$
for every $p,q\in\mathbb{Z}$ and the differentials induced by those of $K$ and
$L$. It can be seen by inspection that
$\operatorname{Tot}(W)=\operatorname{Hom}_{S}\left(K,\operatorname{Hom}_{R}(L,M)\right)$.
We compute the first spectral sequence arising from $W$. For any $p\leq 0$,
$K_{-p}$ is a free $S$-module, so the functor
$\operatorname{Hom}_{S}\left(K_{-p},-\right)$ is exact. Thus we get

$\begin{split}E^{1}_{p,q}&=H_{p,q}^{v}(W)\\\
&=H_{q}\left(\operatorname{Hom}_{S}\left(K_{-p},\operatorname{Hom}_{R}(L,M)\right)\right)\\\
&\cong\operatorname{Hom}_{S}\left(K_{-p},H_{q}\left(\operatorname{Hom}_{R}(L,M)\right)\right)\\\
&=\operatorname{Hom}_{S}\left(K_{-p},\operatorname{Ext}_{R}^{-q}(k,M)\right)\end{split}$

for every $p,q\leq 0$. Next, we obtain

$\begin{split}E^{2}_{p,q}&=H_{p,q}^{h}\left(H^{v}(W)\right)\\\
&=H_{p}\left(\operatorname{Hom}_{S}\left(K,\operatorname{Ext}_{R}^{-q}(k,M)\right)\right)\\\
&\cong
H_{p+n}\left(\underline{a};\operatorname{Ext}_{R}^{-q}(k,M)\right)\end{split}$

for every $p,q\leq 0$. We note that $L$ is a complex of symmetric
$R$-$R$-bimodules. Moreover, by restriction of scalars via $f$, we notice that
$K$ is a complex of $R$-$S$-bimodules and $M$ is an $R$-$S$-bimodule.
Accordingly, for any $i\geq 0$, we deduce that
$\operatorname{Hom}_{S}(K_{i},M)$ is a left $R$-module via the left $R$-module
structure of $M$, and it is a right $R$-module via the left $R$-module
structure of $K_{i}$. If $\varphi_{i}\in\operatorname{Hom}_{S}(K_{i},M)$,
$x\in K_{i}$, and $r\in R$, then we have:

$(r\varphi_{i})(x)=r\varphi_{i}(x)=\varphi_{i}(x)f(r)=\varphi_{i}\left(xf(r)\right)=\varphi_{i}(rx)=(\varphi_{i}r)(x)$

Hence $r\varphi_{i}=\varphi_{i}r$. This shows that
$\operatorname{Hom}_{S}(K_{i},M)$ is a symmetric $R$-$R$-bimodule for every
$i\geq 0$. Therefore, by Remark 3.9, the following isomorphism of
$S$-complexes holds:

$\operatorname{Tot}(W)=\operatorname{Hom}_{S}\left(K,\operatorname{Hom}_{R}(L,M)\right)\cong\operatorname{Hom}_{S}(L\otimes_{R}K,M)$

Bearing in mind that $L\otimes_{R}K\cong K\otimes_{R}L$ is a free resolution
for the $S$-module $l$, we get

$H_{i}\left(\operatorname{Tot}(W)\right)\cong
H_{i}\left(\operatorname{Hom}_{S}(L\otimes_{R}K,M)\right)=\operatorname{Ext}_{S}^{-i}(l,M)$

for every $i\in\mathbb{Z}$. Therefore, we wind up with the following third
quadrant spectral sequence:

$E^{2}_{p,q}=H_{p+n}\left(\underline{a};\operatorname{Ext}_{R}^{-q}(k,M)\right)\Rightarrow\operatorname{Ext}_{S}^{-p-q}(l,M)$

Now if $p+q<-\operatorname{id}_{R}(M)-n$, then either $p<-n$ or
$q<-\operatorname{id}_{R}(M)$. If $p<-n$, then $E^{2}_{p,q}=0$, and if
$q<-\operatorname{id}_{R}(M)$, then $\operatorname{Ext}^{-q}_{R}(k,M)=0$, so
$E^{2}_{p,q}=0$. It follows that $\operatorname{Ext}^{i}_{S}(l,M)=0$ for every
$i>\operatorname{id}_{R}(M)+n$, so by Lemma 3.10,
$\operatorname{id}_{S}(M)\leq\operatorname{id}_{R}(M)+n$. ∎

We are now ready to deal with the finitely generated case.

######  Theorem 3.12.

Let $f:(R,\mathfrak{m},k)\rightarrow(S,\mathfrak{n},l)$ be a local
homomorphism of noetherian local rings, and $M$ a non-zero finitely generated
$S$-module with
$\operatorname{id}_{R}(M)\leq\operatorname{edim}(R)-\operatorname{edim}(S)$.
Then the following assertions hold:

  1. (i)

$M$ is an injective $S$-module, i.e. $M\cong
E_{S}(l)^{\operatorname{type}_{S}(M)}$.

  2. (ii)

$f$ is an exceptional complete intersection map.

  3. (iii)

$\operatorname{id}_{R}(M)=\operatorname{edim}(R)-\operatorname{edim}(S)=\operatorname{fd}_{R}(S)$.

In particular, if $M$ is injective as an $R$-module, then
$\operatorname{edim}(R)=\operatorname{edim}(S)$ and $f$ is flat.

###### Proof.

If $M$ is an injective $S$-module, then by Lemma 3.1, $M\cong
E_{S}(l)^{\operatorname{type}_{S}(M)}$, so Lemma 3.2 (ii) implies that

$\operatorname{id}_{R}(M)=\operatorname{id}_{R}\left(E_{S}(l)\right)=\operatorname{id}_{R}(S^{\vee})=\operatorname{fd}_{R}(S).$

That is to say, the second equality in (iii) follows from (i).

We first reduce to the case where $S$ is $\mathfrak{n}$-adically complete. We
note that $M\otimes_{S}\widehat{S}$ is a non-zero finitely generated
$\widehat{S}$-module. Let $\mathfrak{a}$ be an ideal of $R$, and $L$ a free
resolution of $R/\mathfrak{a}$ in which $L_{i}$ is finitely generated for
every $i\geq 0$. Then using the Tensor Evaluation (see [Ch, Page 12]), we get

$\begin{split}\operatorname{Ext}_{R}^{i}(R/\mathfrak{a},M)\otimes_{S}\widehat{S}&=H_{-i}\left(\operatorname{Hom}_{R}(L,M)\right)\otimes_{S}\widehat{S}\\\
&\cong H_{-i}\left(\operatorname{Hom}_{R}(L,M)\otimes_{S}\widehat{S}\right)\\\
&\cong
H_{-i}\left(\operatorname{Hom}_{R}\left(L,M\otimes_{S}\widehat{S}\right)\right)\\\
&=\operatorname{Ext}_{R}^{i}\left(R/\mathfrak{a},M\otimes_{S}\widehat{S}\right)\end{split}$

for every $i\geq 0$. Since the completion map $S\rightarrow\widehat{S}$ is
faithfully flat, the above isomorphism implies that for any $i\geq 0$, we have
$\operatorname{Ext}_{R}^{i}(R/\mathfrak{a},M)=0$ if and only if
$\operatorname{Ext}_{R}^{i}\left(R/\mathfrak{a},M\otimes_{S}\widehat{S}\right)=0$.
Therefore, we infer from [BH, Proposition 3.1.10] that
$\operatorname{id}_{R}(M)=\operatorname{id}_{R}\left(M\otimes_{S}\widehat{S}\right)$.
On the other hand, Corollary 3.5 implies that $M$ is an injective $S$-module
if and only if $M\otimes_{S}\widehat{S}$ is an injective $\widehat{S}$-module.
All in all, we can replace $S$ with $\widehat{S}$ and $M$ with
$M\otimes_{S}\widehat{S}$, and assume that $S$ is $\mathfrak{n}$-adically
complete.

We next reduce to the case where $f$ is surjective. By Remark 2.3, $f$ has a
Cohen factorization

$R\xrightarrow{\dot{f}}R^{\prime}\xrightarrow{f^{\prime}}S$

in which $\dot{f}$ is flat with regular closed fiber
$R^{\prime}/\mathfrak{m}R^{\prime}$ and $f^{\prime}$ is surjective. Moreover,
$f$ is exceptional complete intersection if and only if $f^{\prime}$ is
exceptional complete intersection. In addition, in light of Lemma 3.11, Lemma
2.2, and our hypothesis, we see that:

$\begin{split}\operatorname{id}_{R^{\prime}}(M)&\leq\operatorname{id}_{R}(M)+\operatorname{edim}\left(R^{\prime}/\mathfrak{m}R^{\prime}\right)\\\
&=\operatorname{id}_{R}(M)+\operatorname{edim}(R^{\prime})-\operatorname{edim}(R)\\\
&\leq\operatorname{edim}(R)-\operatorname{edim}(S)+\operatorname{edim}(R^{\prime})-\operatorname{edim}(R)\\\
&=\operatorname{edim}(R^{\prime})-\operatorname{edim}(S)\end{split}$

If
$\operatorname{id}_{R^{\prime}}(M)=\operatorname{edim}(R^{\prime})-\operatorname{edim}(S)$,
then the above display shows that
$\operatorname{id}_{R}(M)=\operatorname{edim}(R)-\operatorname{edim}(S)$.
Hence it suffices to prove the result for $f^{\prime}$. Therefore, we can
assume that $f$ is surjective, so that $M$ is a finitely generated $R$-module
as well.

We now consider the special case in which $M$ is a faithful $S$-module, i.e.
$\operatorname{ann}_{S}(M)=0$. Then
$\operatorname{ann}_{R}(M)=f^{-1}\left(\operatorname{ann}_{S}(M)\right)=\operatorname{Ker}(f)$.
We argue by induction on $\operatorname{edim}(R)-\operatorname{edim}(S)$.

If $\operatorname{edim}(R)-\operatorname{edim}(S)=0$, then our assumption
implies that $\operatorname{id}_{R}(M)=0$, i.e. $M$ is an injective
$R$-module. As $M$ is a finitely generated $R$-module, we deduce from Lemma
3.1 that $M\cong E_{R}(k)^{\operatorname{type}_{R}(M)}$. But $E_{R}(k)\cong
R^{\vee}=\operatorname{Hom}_{R}\left(R,E_{R}(k)\right)$, so
$\operatorname{ann}_{R}\left(E_{R}(k)\right)=\operatorname{ann}_{R}(R)=0$. It
follows that
$\operatorname{Ker}(f)=\operatorname{ann}_{R}(M)=\operatorname{ann}_{R}\left(E_{R}(k)\right)=0$,
so $f$ is an isomorphism. As a result, $M$ is an injective $S$-module and $f$
is an exceptional complete intersection map.

Next suppose that $\operatorname{edim}(R)-\operatorname{edim}(S)\geq 1$. Then
$f$ is not an isomorphism, so
$\operatorname{ann}_{R}(M)=\operatorname{Ker}(f)\neq 0$. As
$\operatorname{id}_{R}(M)<\infty$, Lemma 3.6 implies that there exists a non-
zerodivisor in $\operatorname{Ker}(f)$, so
$\operatorname{Ker}(f)\nsubseteq\mathcal{Z}(R)=\bigcup_{\mathfrak{p}\in\operatorname{Ass}(R)}\mathfrak{p}$.
On the other hand,
$\operatorname{edim}(S)=\operatorname{edim}\left(R/\operatorname{Ker}(f)\right)\leq\operatorname{edim}(R)$
with equality if and only if $\operatorname{Ker}(f)\subseteq\mathfrak{m}^{2}$.
But $\operatorname{edim}(R)-\operatorname{edim}(S)\geq 1$, so
$\operatorname{Ker}(f)\nsubseteq\mathfrak{m}^{2}$. By the Prime Avoidance
Lemma (see [Sh, Theorem 3.61]), we infer that
$\operatorname{Ker}(f)\nsubseteq\mathfrak{m}^{2}\cup\bigcup_{\mathfrak{p}\in\operatorname{Ass}(R)}\mathfrak{p}$,
so there exists a non-zerodivisor
$a\in\operatorname{Ker}(f)\backslash\mathfrak{m}^{2}$. Consider the following
factorization of $f$:

$R\rightarrow R/\langle a\rangle\xrightarrow{\bar{f}}S$

As $a\in\mathfrak{m}\backslash\mathfrak{m}^{2}$, Lemma 3.7 (ii) implies that
$\operatorname{id}_{R/\langle a\rangle}(M)=\operatorname{id}_{R}(M)-1$. Also,
$\operatorname{edim}\left(R/\langle a\rangle\right)=\operatorname{edim}(R)-1$
and $\dim\left(R/\langle a\rangle\right)=\dim(R)-1$. Apply the induction
hypothesis to $\bar{f}$ to conclude that $M$ is an injective $S$-module and
$\bar{f}$ is an exceptional complete intersection map. Then by the choice of
$a$, it follows that $f$ is an exceptional complete intersection map as well.

We finally consider the general case. Set
$\mathfrak{a}=\operatorname{ann}_{S}(M)$, $T=S/\mathfrak{a}$, and consider the
ring homomorphisms

$R\xrightarrow{f}S\xrightarrow{\pi}T$

where $\pi$ is the quotient map. Clearly, $M$ is a finitely generated faithful
$T$-module with

$\operatorname{id}_{R}(M)\leq\operatorname{edim}(R)-\operatorname{edim}(S)\leq\operatorname{edim}(R)-\operatorname{edim}(T).$

By the special case in the previous paragraph, $M$ is an injective $T$-module,
$\pi f$ is an exceptional complete intersection map, and
$\operatorname{id}_{R}(M)=\operatorname{edim}(R)-\operatorname{edim}(T)$. Thus
the above inequalities imply that
$\operatorname{edim}(S)=\operatorname{edim}(T)$.

Our goal is to show that $\mathfrak{a}=0$, and then we are done. Recall that
$f$ is assumed to be surjective, so $k$ can be considered as the residue field
of $S$ as well as $T$. Consider the commutative diagram

${R}$${S}$${T}$${R}$${S}$${k}$${R}$${T}$${k}$$\scriptstyle{f}$$\scriptstyle{\pi}$$\scriptstyle{\rho}$$\scriptstyle{f}$$\scriptstyle{\varpi}$$\scriptstyle{\pi}$$\scriptstyle{\pi
f}$$\scriptstyle{\rho}$

in which $\varpi$ and $\rho$ are quotient maps. In view of Remark 2.1, this
diagram provides us with the commutative diagram

${D_{1}(S|R;k)}$${D_{1}(T|R;k)}$${D_{1}(T|S;k)}$${D_{0}(S|R;k)=0}$${D_{1}(S|R;k)}$${D_{1}(k|R;k)}$${D_{1}(k|S;k)}$${D_{0}(S|R;k)=0}$${D_{1}(T|R;k)}$${D_{1}(k|R;k)}$${D_{1}(k|T;k)}$${D_{0}(T|R;k)=0}$$\scriptstyle{\pi_{\ast}=D_{1}(\pi|R;k)}$$\scriptstyle{\rho_{\ast}=D_{1}(\rho|R;k)}$$\scriptstyle{\varpi_{\ast}=D_{1}(\varpi|R;k)}$$\scriptstyle{\pi_{\ast}=D_{1}(\pi|R;k)}$$\scriptstyle{\rho_{\ast}=D_{1}(\rho|R;k)}$

of vector spaces over $k$ with exact rows in which the vanishings follow from
the surjectivity of $f$ and $\pi$. From the exactness of the second row, we
get:

$\begin{split}\operatorname{edim}(S)&=\operatorname{rank}_{k}\left(D_{1}(k|S;k)\right)\\\
&=\operatorname{rank}_{k}\left(\operatorname{Coker}(\varpi_{\ast})\right)\\\
&=\operatorname{rank}_{k}\left(D_{1}(k|R;k)\right)-\operatorname{rank}_{k}\left(\operatorname{Im}(\varpi_{\ast})\right)\\\
&=\operatorname{edim}(R)-\operatorname{rank}_{k}\left(\operatorname{Im}(\varpi_{\ast})\right)\end{split}$

Similarly, from the exactness of the third row, we obtain:

$\begin{split}\operatorname{edim}(T)&=\operatorname{rank}_{k}\left(D_{1}(k|T;k)\right)\\\
&=\operatorname{rank}_{k}\left(\operatorname{Coker}(\rho_{\ast})\right)\\\
&=\operatorname{rank}_{k}\left(D_{1}(k|R;k)\right)-\operatorname{rank}_{k}\left(\operatorname{Im}(\rho_{\ast})\right)\\\
&=\operatorname{edim}(R)-\operatorname{rank}_{k}\left(\operatorname{Im}(\rho_{\ast})\right)\end{split}$

These relations conspire to yield:

$\operatorname{rank}_{k}\left(\operatorname{Im}(\rho_{\ast}\pi_{\ast})\right)=\operatorname{rank}_{k}\left(\operatorname{Im}(\varpi_{\ast})\right)=\operatorname{rank}_{k}\left(\operatorname{Im}(\rho_{\ast})\right)$

But
$\operatorname{Im}(\rho_{\ast}\pi_{\ast})\subseteq\operatorname{Im}(\rho_{\ast})$,
so we conclude that
$\operatorname{Im}(\rho_{\ast}\pi_{\ast})=\operatorname{Im}(\rho_{\ast})$. As
$\pi f$ is a surjective exceptional complete intersection map, Lemma 2.4
implies that $\rho_{\ast}$ is injective, so we deduce that $\pi_{\ast}$ is
surjective. Thus from the exactness of the first row, we achieve
$D_{1}(T|S;k)=0$, so
$\mu(\mathfrak{a})=\operatorname{rank}_{k}\left(D_{1}(T|S;k)\right)=0$, whence
$\mathfrak{a}=0$. This completes the proof. ∎

The following corollary is an analogue of [BIK, Theorem 3.1] for artinian
modules.

######  Corollary 3.13.

Let $f:(R,\mathfrak{m},k)\rightarrow(S,\mathfrak{n},l)$ be a local
homomorphism of noetherian local rings, and $M$ a non-zero artinian $S$-module
with
$\operatorname{fd}_{R}(M)\leq\operatorname{edim}(R)-\operatorname{edim}(S)$.
Then the following assertions hold:

  1. (i)

$M$ is a free $S$-module, i.e. $M\cong S^{\mu_{S}(M)}$.

  2. (ii)

$f$ is an exceptional complete intersection map.

  3. (iii)

$\operatorname{fd}_{R}(M)=\operatorname{edim}(R)-\operatorname{edim}(S)=\operatorname{fd}_{R}(S)$.

In particular, if $M$ is flat as an $R$-module, then
$\operatorname{edim}(R)=\operatorname{edim}(S)$ and $f$ is flat.

###### Proof.

As we have observed before, since $M$ is artinian, it has an
$\widehat{S}$-module structure that restricts to its $S$-module structure via
the completion map $S\rightarrow\widehat{S}$. Since the completion map is
faithfully flat, we have
$\operatorname{fd}_{\widehat{S}}(M)=\operatorname{fd}_{\widehat{S}}\left(M\otimes_{S}\widehat{S}\right)=\operatorname{fd}_{S}(M)$,
so $M$ is a flat $S$-module if and only $M$ is a flat $\widehat{S}$-module. By
[Ma, Theorem 18.6 (iii)], $E_{S}(l)$ is an $\widehat{S}$-module, so we have:

$\begin{split}M^{\vee}&=\operatorname{Hom}_{S}\left(M,E_{S}(l)\right)\\\
&\cong\operatorname{Hom}_{S}\left(M,\operatorname{Hom}_{\widehat{S}}\left(\widehat{S},E_{S}(l)\right)\right)\\\
&\cong\operatorname{Hom}_{\widehat{S}}\left(M\otimes_{S}\widehat{S},E_{S}(l)\right)\\\
&\cong\operatorname{Hom}_{\widehat{S}}\left(M,E_{\widehat{S}}\left(\widehat{S}/\mathfrak{n}\widehat{S}\right)\right)\end{split}$

Therefore, if $M$ happens to be a flat $S$-module, then $M$ is a flat
$\widehat{S}$-module, so by Lemma 3.2 (ii) and the above isomorphism,
$M^{\vee}$ is an injective $\widehat{S}$-module. On the other hand, by the
Matlis duality theory, $M^{\vee}$ is a finitely generated
$\widehat{S}$-module, so by Lemma 3.1, we get:

$M^{\vee}\cong
E_{\widehat{S}}\left(\widehat{S}/\mathfrak{n}\widehat{S}\right)^{\operatorname{type}_{\widehat{S}}(M^{\vee})}\cong
E_{S}(l)^{\operatorname{type}_{\widehat{S}}(M^{\vee})}$

Moreover, by [BH, Exercise 3.1.23], $\widehat{S}$ is artinian, so $S$ is
artinian and we have

$\ell_{S}(M)=\ell_{\widehat{S}}\left(M\otimes_{S}\widehat{S}\right)=\ell_{\widehat{S}}(M)=\ell_{\widehat{S}}(M^{\vee})<\infty,$

which shows that $\mu_{S}(M)$ is well-defined. Now [BH, Proposition 3.2.12
(d)] implies that

$\operatorname{type}_{\widehat{S}}(M^{\vee})=\mu_{\widehat{S}}(M)=\mu_{\widehat{S}}\left(M\otimes_{S}\widehat{S}\right)=\mu_{S}(M),$

so we get $M^{\vee}\cong E_{S}(l)^{\mu_{S}(M)}$. As $M$ is Matlis reflexive,
we obtain:

$M\cong
M^{\vee\vee}\cong\left(E_{S}(l)^{\mu_{S}(M)}\right)^{\vee}\cong\left(E_{S}(l)^{\vee}\right)^{\mu_{S}(M)}\cong\widehat{S}^{\mu_{S}(M)}\cong
S^{\mu_{S}(M)}$

In particular, $\operatorname{fd}_{R}(M)=\operatorname{fd}_{R}(S)$. That is,
the second equality in (iii) follows from (i).

As in the proof of Theorem 3.3, we can reduce to the case where $S$ is
$\mathfrak{n}$-adically complete, so the Matlis duality theory implies that
$M^{\vee}=\operatorname{Hom}_{S}\left(M,E_{S}(l)\right)$ is a finitely
generated $S$-module. Now by Lemma 3.2 (ii), we have

$\operatorname{id}_{R}(M^{\vee})=\operatorname{fd}_{R}(M)\leq\operatorname{edim}(R)-\operatorname{edim}(S).$

Therefore, Theorem 3.12 implies that $M^{\vee}$ is an injective $S$-module,
$f$ is an exceptional complete intersection map, and

$\operatorname{fd}_{R}(M)=\operatorname{id}_{R}(M^{\vee})=\operatorname{edim}(R)-\operatorname{edim}(S).$

Furthermore, $\operatorname{fd}_{S}(M)=\operatorname{id}_{S}(M^{\vee})=0$, so
$M$ is a flat $S$-module. ∎

We need the following lemma in order to derive another corollary from our
theorem.

######  Lemma 3.14.

Let $R\rightarrow S\rightarrow T$ be local homomorphisms of noetherian local
rings, and $X$ a non-exact $T$-complex such that $H_{i}(X)=0$ for $|i|\gg 0$
and $H_{i}(X)$ is a finitely generated $T$-module for every $i\in\mathbb{Z}$.
If $\operatorname{id}_{R}(X)<\infty$ and $\operatorname{id}_{S}(X)<\infty$,
then $\operatorname{fd}_{R}(S)<\infty$.

###### Proof.

Let $k$ denote the residue field of $T$. For any $i\in\mathbb{Z}$, $H_{-i}(X)$
is a finitely generated $T$-module, so the Matlis duality theory implies that

$H_{i}(X^{\vee})=H_{i}\left(\operatorname{Hom}_{T}\left(X,E_{T}(k)\right)\right)\cong\operatorname{Hom}_{T}\left(H_{-i}(X),E_{T}(k)\right)=H_{-i}(X)^{\vee}$

is an artinian $T$-module. By Lemma 3.2 (i),
$\operatorname{fd}_{S}(X^{\vee})=\operatorname{id}_{S}(X)<\infty$, so by [DGI,
Theorem 4.9], $X^{\vee}$ is virtually small over $S$ in the sense of [DGI,
4.1]. On the other hand,
$\operatorname{fd}_{R}(X^{\vee})=\operatorname{id}_{R}(X)<\infty$, so [DGI,
Theorem 5.5] implies that $\operatorname{fd}_{R}(S)<\infty$. ∎

######  Corollary 3.15.

Let $f:(R,\mathfrak{m})\rightarrow(S,\mathfrak{n})$ be a local homomorphism of
noetherian local rings, and $M$ a non-zero finitely generated $S$-module. Then
the following assertions hold:

  1. (i)

If $\operatorname{id}_{S}(M)<\infty$, then we have:

$\operatorname{edim}(S)+\operatorname{id}_{R}(M)\geq\operatorname{edim}(R)+\operatorname{id}_{S}(M)$

Moreover, if equality holds, then $f$ is an exceptional complete intersection
map.

  2. (ii)

If $f$ is an exceptional complete intersection map, then we have:

$\operatorname{edim}(S)+\operatorname{id}_{R}(M)\geq\operatorname{edim}(R)+\operatorname{id}_{S}(M)$

###### Proof.

(i): If $\operatorname{id}_{R}(M)=\infty$, then there is nothing to prove, so
we may assume that $\operatorname{id}_{R}(M)<\infty$. As a result, Lemma 3.14
yields $\operatorname{fd}_{R}(S)<\infty$.

We first handle the special case where $f$ is surjective. Then $M$ is a
finitely generated $R$-module as well as $S$-module, $S$ is a finitely
generated $R$-module, $\operatorname{id}_{R}(M)<\infty$,
$\operatorname{id}_{S}(M)<\infty$, and
$\operatorname{pd}_{R}(S)=\operatorname{fd}_{R}(S)<\infty$. These assumptions
allow us to take advantage of Auslander-Buchsbaum and Bass formulas (see [BH,
Theorems 1.3.3 and 3.1.17]) to write:

$\begin{split}\operatorname{id}_{R}(M)-\operatorname{id}_{S}(M)&=0pt(\mathfrak{m},R)-0pt(\mathfrak{n},S)\\\
&=0pt(\mathfrak{m},R)-0pt_{R}(\mathfrak{m},S)\\\
&=\operatorname{pd}_{R}(S)\end{split}$

Therefore, it suffices to show that
$\operatorname{pd}_{R}(S)\geq\operatorname{edim}(R)-\operatorname{edim}(S)$.
To this end, we argue by induction on
$\operatorname{edim}(R)-\operatorname{edim}(S)$.

If $\operatorname{edim}(R)-\operatorname{edim}(S)=0$, then there is nothing to
prove. Now suppose that $\operatorname{edim}(R)-\operatorname{edim}(S)\geq 1$.
As in the proof of Theorem 3.12, we can obtain a non-zerodivisor
$a\in\operatorname{Ker}(f)\backslash\mathfrak{m}^{2}$. In light of Lemma 3.7
(i) and the induction hypothesis applied to $\bar{f}:R/\langle
a\rangle\rightarrow S$, we see that:

$\begin{split}\operatorname{pd}_{R}(S)&=\operatorname{pd}_{R/\langle
a\rangle}(S)+1\\\ &\geq\operatorname{edim}\left(R/\langle
a\rangle\right)-\operatorname{edim}(S)+1\\\
&=\operatorname{edim}(R)-\operatorname{edim}(S)\end{split}$

Now if equality holds above, i.e.
$\operatorname{fd}_{R}(S)=\operatorname{pd}_{R}(S)=\operatorname{edim}(R)-\operatorname{edim}(S)$,
then [BIK, Theorem 3.1] implies that $f$ is an exceptional complete
intersection map.

Next consider the general case. It is clear that $M\otimes_{S}\widehat{S}$ is
a non-zero finitely generated $\widehat{S}$-module. By Corollary 3.5, we have
$\operatorname{id}_{\widehat{S}}\left(M\otimes_{S}\widehat{S}\right)=\operatorname{id}_{S}(M)$.
In addition, we observed in the proof of Theorem 3.12 that
$\operatorname{id}_{R}\left(M\otimes_{S}\widehat{S}\right)=\operatorname{id}_{R}(M)$
and $\operatorname{edim}\left(\widehat{S}\right)=\operatorname{edim}(S)$.
Hence we can replace $S$ with $\widehat{S}$ and $M$ with
$M\otimes_{S}\widehat{S}$, and assume accordingly that $S$ is
$\mathfrak{n}$-adically complete. Then by Remark 2.3, $f$ has a Cohen
factorization

$R\xrightarrow{\dot{f}}R^{\prime}\xrightarrow{f^{\prime}}S$

in which $\dot{f}$ is flat with regular closed fiber
$R^{\prime}/\mathfrak{m}R^{\prime}$ and $f^{\prime}$ is surjective. By Lemmas
3.11 and 2.2, we have:

$\begin{split}\operatorname{id}_{R^{\prime}}(M)&\leq\operatorname{id}_{R}(M)+\operatorname{edim}\left(R^{\prime}/\mathfrak{m}R^{\prime}\right)\\\
&=\operatorname{id}_{R}(M)+\operatorname{edim}(R^{\prime})-\operatorname{edim}(R)\end{split}$

Applying the special case to $f^{\prime}$, we get:

$\begin{split}\operatorname{id}_{R}(M)-\operatorname{id}_{S}(M)&\geq\operatorname{id}_{R^{\prime}}(M)-\operatorname{id}_{S}(M)+\operatorname{edim}(R)-\operatorname{edim}(R^{\prime})\\\
&\geq\operatorname{edim}(R^{\prime})-\operatorname{edim}(S)+\operatorname{edim}(R)-\operatorname{edim}(R^{\prime})\\\
&=\operatorname{edim}(R)-\operatorname{edim}(S)\end{split}$

Now if

$\operatorname{id}_{R}(M)-\operatorname{id}_{S}(M)=\operatorname{edim}(R)-\operatorname{edim}(S),$

then the above display shows that

$\operatorname{id}_{R^{\prime}}(M)-\operatorname{id}_{S}(M)=\operatorname{edim}(R^{\prime})-\operatorname{edim}(S),$

so by the special case, $f^{\prime}$ is exceptional complete intersection,
thereby Remark 2.3 implies that $f$ is exceptional complete intersection.

(ii): If $\operatorname{id}_{R}(M)=\infty$, then there is nothing to prove, so
we may assume that $\operatorname{id}_{R}(M)<\infty$. Since $f$ is exceptional
complete intersection, [ILP, Corollary 4.2] yields

$\operatorname{id}_{S}(M)=\operatorname{id}_{R}(M)-\dim(R)+\dim(S)<\infty.$

Now the result follows from (i). ∎

## Acknowledgement

It is a pleasure to express my sincerest gratitude to professor Srikanth
Iyengar for his invaluable comments and suggestions on this manuscript which
is part of my Ph.D. thesis at Clemson University.

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