# On graded Gorenstein injective dimension

Afsaneh Esmaeelnezhad and Parviz Sahandi  Department of Mathematics,
University of Kharazmi, Tehran, Iran<EMAIL_ADDRESS>Department of
Mathematics, University of Tabriz, Tabriz, Iran, and School of Mathematics,
Institute for Research in Fundamental Sciences (IPM), P.O. Box: 19395-5746,
Tehran Iran<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

There are nice relations between graded homological dimensions and ordinary
homological dimensions. We study the Gorenstein injective dimension of a
complex of graded modules denoted by ${}^{*}\operatorname{Gid}$, and derive
its properties. In particular we prove the Chouinard’s like formula for
${}^{*}\operatorname{Gid}$, and compare it with the usual Gorenstein injective
dimension.

###### Key words and phrases:

graded rings, graded modules, Chouinard’s formula, injective dimension,
Gorenstein injective dimension

###### 2010 Mathematics Subject Classification:

13D05,13D02,13A02

P. Sahandi was supported in part by a grant from IPM (No. 91130030).

##  1\. Introduction

Let $R$ be a Noetherian $\mathbb{Z}$-graded ring. In [9] and [10], Fossum and
Fossum-Foxby have studied the graded homological dimension of graded modules
and compare them with classical homological dimensions. They shown that for a
graded $R$-module $M$, one has
$\,{}^{*}\operatorname{id}_{R}M\leq\operatorname{id}_{R}M\leq\,^{*}\operatorname{id}_{R}M+1$
where $\operatorname{id}_{R}M$ (resp. $\,{}^{*}\operatorname{id}_{R}M$)
denotes for the injective dimension of $M$ in the category of $R$-modules
(resp. category of graded $R$-modules). It is natural to ask how these
inequalities hold for the Gorenstein injective dimension
$\operatorname{Gid}_{R}M$. In this paper we give an answer this question.
Section 2 of this paper is devoted to review some hyper-homological algebra
for the derived category of the graded ring $R$. In Section 3 we define the
∗injective dimension of complexes of graded modules and homogeneous
homomorphisms, and derived its properties. In the final section we define the
graded Gorenstein injective dimension of complexes of graded modules and
homogeneous homomorphisms denoted by $\,{}^{*}\operatorname{Gid}$. Among other
results we show that if the graded ring $R$ admits a $\\!{}^{*}$dualizing
complex, or is a non-negatively graded ring, then
$\,{}^{*}\operatorname{Gid}_{R}X\leq\operatorname{Gid}_{R}X\leq\,^{*}\operatorname{Gid}_{R}X+1$,
where $\operatorname{Gid}_{R}X$ is the Gorenstein injective dimension of $X$
over $R$ (see Corollary 4.19). Also in this case we prove a Chouinard’s like
formula for $\,{}^{*}\operatorname{Gid}_{R}X$ (see Theorem 4.18). Our source
of graded rings and modules are [3] and [10].

Throughout this paper $R$ is a commutative Noetherian $\mathbb{Z}$-graded
ring.

##  2\. Derived category of complexes of graded modules

We use the notation from the appendix of [5]. Let $X$ be a complex of
$R$-modules and $R$-homomorphisms. The _supremum_ and the _infimum_ of a
complex $X$, denoted by $\sup(X)$ and $\inf(X)$ are defined by the supremum
and infimum of $\\{i\in\mathbb{Z}|\mbox{H}_{i}(X)\neq 0\\}$. If $m$ is an
integer and $X$ is a complex, then $\Sigma^{m}X$ denotes the complex $X$
_shifted_ $m$ degrees to the left; it is given by

$(\Sigma^{m}X)_{\ell}=X_{\ell-m}\text{ and
}\partial^{\Sigma^{m}X}_{\ell}=(-1)^{m}\partial^{X}_{\ell-m}$

for $\ell\in\mathbb{Z}$.

The symbol ${\mathcal{D}}(R)$ denotes the _derived category_ of $R$-complexes.
The full subcategories ${\mathcal{D}}_{\sqsubset}(R)$,
${\mathcal{D}}_{\sqsupset}(R)$, ${\mathcal{D}}_{\square}(R)$ and
${\mathcal{D}}_{0}(R)$ of ${\mathcal{D}}(R)$ consist of $R$-complexes $X$
while $\mbox{H}_{\ell}(X)=0$, for respectively $\ell\gg 0$, $\ell\ll 0$,
$|\ell|\gg 0$ and $\ell\neq 0$. Homology isomorphisms are marked by the sign
$\simeq$. The right derived functor of the homomorphism functor of
$R$-complexes and the left derived functor of the tensor product of
$R$-complexes are denoted by ${\mathbf{R}}\operatorname{Hom}_{R}(-,-)$ and
$-\otimes^{\mathbf{L}}_{R}-$, respectively.

Let $M=\oplus_{n\in\mathbb{Z}}M_{n}$ and $N=\oplus_{n\in\mathbb{Z}}N_{n}$ be
two graded $R$-modules. The ${}^{*}\operatorname{Hom}$ functor is defined by
${}^{*}\operatorname{Hom}_{R}(M,N)=\bigoplus_{i\in\mathbb{Z}}\operatorname{Hom}_{i}(M,N)$,
such that $\operatorname{Hom}_{i}(M,N)$ is a $\mathbb{Z}$-submodule of
$\operatorname{Hom}_{R}(M,N)$ consisting of all $\varphi:M\to N$ such that
$\varphi(M_{n})\subseteq N_{n+i}$ for all $n\in\mathbb{Z}$. In general
${}^{*}\operatorname{Hom}_{R}(M,N)\neq\operatorname{Hom}_{R}(M,N)$ but
equality holds if M is finitely generated, see [3, Exercise 1.5.19]. Also the
tensor product $M\otimes_{R}N$ of $M$ and $N$ is a graded module with
$(M\otimes_{R}N)_{n}$ is generated (as a $\mathbb{Z}$-module) by elements
$m\otimes n$ with $m\in M_{i}$ and $n\in N_{j}$ where $i+j=n$.

Let $\\{M_{\alpha}\\}_{\alpha\in I}$ be a family of graded $R$-modules. Then
$\bigoplus_{\alpha}M_{\alpha}$ becomes a graded $R$-module with
$(\bigoplus_{\alpha}M_{\alpha})_{n}=\bigoplus_{\alpha}(M_{\alpha})_{n}$ for
all $n\in\mathbb{Z}$, see [10, Page 289]. Also recall that the direct product
exists in the category of graded modules. Then the direct product is denoted
by ${}^{*}\prod_{\alpha}M_{\alpha}$ and
$(^{*}\prod_{\alpha}M_{\alpha})_{n}=\prod_{\alpha}(M_{\alpha})_{n}$ for all
$n\in\mathbb{Z}$, see [10, Page 289]. In this case there are the following
bijections [10, Page 289]

${}^{*}\operatorname{Hom}_{R}(\bigoplus_{\alpha}M_{\alpha},-)\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\\!^{*}\prod_{\alpha}\\!^{*}\operatorname{Hom}_{R}(M_{\alpha},-),$
${}^{*}\operatorname{Hom}_{R}(-,\\!^{*}\prod_{\alpha}M_{\alpha})\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\\!^{*}\prod_{\alpha}\\!^{*}\operatorname{Hom}_{R}(-,M_{\alpha}).$

Likewise direct and inverse limits are exists in the category of graded
modules with

$(\\!^{*}\lim_{\longrightarrow}M_{\alpha})_{n}=\lim_{\longrightarrow}(M_{\alpha})_{n},$
$(\\!^{*}\lim_{\longleftarrow}M_{\alpha})_{n}=\lim_{\longleftarrow}(M_{\alpha})_{n},$

see [10, Page 289]. Let $(R,\mathfrak{m})$ be a ∗local (Noetherian) ring $R$,
that is, a graded ring with a unique homogeneous maximal ideal $\mathfrak{m}$.
The $\mathfrak{m}$-∗adic completion of $R$ is

$\\!{}^{*}\widehat{R}=\\!^{*}\lim_{\longleftarrow}R/\mathfrak{m}^{n},$

which is a Noetherian graded ring by [9, Corollary VIII.2]. It is known that
the $\mathfrak{m}$-∗adic completion $\,{}^{*}\widehat{R}$ is a flat $R$-module
[10, Corollary 3.3], and that if
$E:=\\!^{*}\operatorname{E}_{R}(R/\mathfrak{m})$ is the ∗ injective envelope
of $R/\mathfrak{m}$ over $R$, then
$\,{}^{*}\operatorname{Hom}_{R}(E,E)\cong\\!^{*}\widehat{R}$ [9, Theorem
VIII.3].

The symbol ${}^{*}{\mathcal{C}}(R)$ denotes the category of complexes of
graded $R$-modules and homogeneous differentials. Note that the category of
graded modules is an abelian category, hence ${}^{*}{\mathcal{C}}(R)$ has a
derived category, (see [15]), which will be denoted by
$\,{}^{*}{\mathcal{D}}(R)$. Analogously we have
$\,{}^{*}{\mathcal{C}}_{\sqsubset}(R)$,
$\,{}^{*}{\mathcal{C}}_{\sqsupset}(R)$, $\,{}^{*}{\mathcal{C}}_{\square}(R)$
and $\,{}^{*}{\mathcal{C}}_{0}(R)$ (resp.
$\,{}^{*}{\mathcal{D}}_{\sqsubset}(R)$,
$\,{}^{*}{\mathcal{D}}_{\sqsupset}(R)$, $\,{}^{*}{\mathcal{D}}_{\square}(R)$
and $\,{}^{*}{\mathcal{D}}_{0}(R)$) which are the full subcategories of
$\,{}^{*}{\mathcal{C}}(R)$ (resp. $\,{}^{*}{\mathcal{D}}(R)$). If we use the
notation $X\in^{*}\mathcal{C}_{(\sharp)}(R)$, we mean
$\mbox{H}(X)\in^{*}\mathcal{C}_{\sharp}(R)$.

For $R$-complexes $X$ and $Y$ of graded modules, with homogeneous
differentials $\partial^{X}$ and $\partial^{Y}$, we define the _homomorphism
complex_ ${}^{*}\operatorname{Hom}_{R}(X,Y)$ as follows:

${}^{*}\operatorname{Hom}_{R}(X,Y)_{\ell}=\\!^{*}\prod_{p\in\mathbb{Z}}\\!^{*}\operatorname{Hom}_{R}(X_{p},Y_{p+\ell})$

and when $\psi=(\psi_{p})_{p\in\mathbb{Z}}$ belongs to
${}^{*}\operatorname{Hom}_{R}(X,Y)_{\ell}$ the family
$\partial_{\ell}^{{}^{*}\operatorname{Hom}_{R}(X,Y)}(\psi)$ in
${}^{*}\operatorname{Hom}_{R}(X,Y)_{\ell-1}$ has $p$-th component

$\partial_{\ell}^{{}^{*}\operatorname{Hom}_{R}(X,Y)}(\psi)_{p}=\partial^{Y}_{p+\ell}\psi_{p}-(-1)^{\ell}\psi_{p-1}\partial^{X}_{p}.$

When $X\in\,^{*}{\mathcal{C}}_{\sqsupset}^{f}(R)$ and
$Y\in\,^{*}{\mathcal{C}}_{\sqsubset}(R)$ all the products
$\\!{}^{*}\prod_{p\in\mathbb{Z}}\\!^{*}\operatorname{Hom}_{R}(X_{p},Y_{p+\ell})$
are finite. Thus using [3, Exercise 1.5.19], we have

$\\!{}^{*}\prod_{p\in\mathbb{Z}}\\!^{*}\operatorname{Hom}_{R}(X_{p},Y_{p+\ell})=\bigoplus_{p\in\mathbb{Z}}\operatorname{Hom}_{R}(X_{p},Y_{p+\ell}),$

for every $\ell\in\mathbb{Z}$. Therefore
${}^{*}\operatorname{Hom}_{R}(X,Y)=\operatorname{Hom}_{R}(X,Y)$.

We also define the _tensor product complex_ $X\otimes_{R}Y$ as follows:

$(X\otimes_{R}Y)_{\ell}=\bigoplus_{p\in\mathbb{Z}}(X_{p}\otimes_{R}Y_{\ell-p})$

and the $\ell$-th differential $\partial_{\ell}^{X\otimes_{R}Y}$ is given on a
generator $x_{p}\otimes y_{\ell-p}$ in $(X\otimes_{R}Y)_{\ell}$, where $x_{p}$
and $y_{\ell-p}$ are homogeneous elements, by

$\partial_{\ell}^{X\otimes_{R}Y}(x_{p}\otimes
y_{\ell-p})=\partial^{X}_{p}(x_{p})\otimes
y_{\ell-p}+(-1)^{p}x_{p}\otimes\partial^{Y}_{\ell-p}(y_{\ell-p}).$

If $X$ and $Y$ are $R$-complexes of graded modules, then
$\,{}^{*}\operatorname{Hom}_{R}(X,-)$, $\,{}^{*}\operatorname{Hom}_{R}(-,Y)$,
and $X\otimes_{R}-$ are functors on $\,{}^{*}{\mathcal{C}}(R)$.

Note that any object of $\,{}^{*}{\mathcal{C}}_{\sqsubset}(R)$ has an
∗injective resolution by [15, Page 47], and any object of
$\,{}^{*}{\mathcal{C}}_{\sqsupset}(R)$ has an ∗projective resolution by [15,
Page 48]. The right derived functor of the $\\!{}^{*}\operatorname{Hom}$
functor in the category of graded complexes is denoted by
$\mathbf{R}\,^{*}\operatorname{Hom}_{R}(-,-)$ and set
$\,{}^{*}\operatorname{Ext}^{i}_{R}(-,-)=\mbox{H}_{-i}(\mathbf{R}\,^{*}\operatorname{Hom}_{R}(-,-))$.
It is easily seen that if $R$ is a Noetherian $\mathbb{Z}$-graded ring and $X$
a homologically finite complex of graded modules and
$Y\in\\!^{*}\mathcal{C}(R)$ then
$\mathbf{R}\,^{*}\operatorname{Hom}_{R}(X,Y)={\mathbf{R}}\operatorname{Hom}_{R}(X,Y)$.
Also the left derived functor of $-\otimes_{R}-$ in the category of graded
complexes is denoted by $-\otimes_{R}^{\mathbf{L}^{*}}-$. Since ∗projective
graded $R$-modules coincide with projective $R$-modules by [10, Proposition
3.1] we easily see that $-\otimes_{R}^{\mathbf{L}^{*}}-$ coincides with the
ordinary left derived functor of $-\otimes_{R}-$ in the category of complexes.
So we use $-\otimes^{\mathbf{L}}_{R}-$ instead of
$-\otimes_{R}^{\mathbf{L}^{*}}-$.

For the homomorphism and the tensor product functors we have the following
useful proposition, see [5, A.2.8, A.2.10, and A.2.11] for the ungraded case.
The proof is the same as the ungraded case so we omit it.

######  Proposition 2.1.

Let $S$ be a graded ring which is an $R$-algebra.

∗Adjointness. Let $Z,Y\in\,^{*}{\mathcal{C}}(S)$ and
$X\in\,^{*}{\mathcal{C}}(R)$. Then $Z\otimes_{S}Y\in\,^{*}{\mathcal{C}}(S)$
and $\\!{}^{*}\operatorname{Hom}_{R}(Y,X)\in\,^{*}{\mathcal{C}}(S)$, and there
is an isomorphism of $S$-complexes

$\rho_{ZYX}:^{*}\operatorname{Hom}_{R}(Z\otimes_{S}Y,X)\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\\!^{*}\operatorname{Hom}_{S}(Z,\\!^{*}\operatorname{Hom}_{R}(Y,X)),$

which is natural in $Z,Y$ and $X$.

∗Tensor-Evaluation. Let $Z,Y\in\,^{*}{\mathcal{C}}(S)$ and
$X\in\,^{*}{\mathcal{C}}(R)$. Then
${}^{*}\operatorname{Hom}_{S}(Z,Y)\in\,^{*}{\mathcal{C}}(R)$ and
$Y\otimes_{R}X\in\,^{*}{\mathcal{C}}(S)$, and there is a natural morphism of
$S$-complexes

$\omega_{ZYX}:^{*}\operatorname{Hom}_{S}(Z,Y)\otimes_{R}X\longrightarrow\\!^{*}\operatorname{Hom}_{S}(Z,Y\otimes_{R}X).$

The morphism is invertible under each of the next two extra conditions

  * (i)

$Z\in\\!^{*}\mathcal{C}_{\square}^{fp}(S)$,
$Y\in\\!^{*}\mathcal{C}_{\sqsupset}(S)$, and
$X\in\\!^{*}\mathcal{C}_{\sqsupset}(R)$; or

  * (ii)

$Z\in\\!^{*}\mathcal{C}_{\square}^{fp}(S)$,
$Y\in\\!^{*}\mathcal{C}_{\sqsubset}(S)$, and
$X\in\\!^{*}\mathcal{C}_{\square}(R)$.

∗Hom-Evaluation. Let $Z,Y\in\,^{*}{\mathcal{C}}(S)$ and
$X\in\,^{*}{\mathcal{C}}(R)$. Then
${}^{*}\operatorname{Hom}_{S}(Z,Y)\in\,^{*}{\mathcal{C}}(R)$ and
$\,{}^{*}\operatorname{Hom}_{R}(Y,X)\in\,^{*}{\mathcal{C}}(S)$, and there is a
natural morphism of $S$-complexes

$\theta_{ZYX}:Z\otimes_{S}\\!^{*}\operatorname{Hom}_{R}(Y,X)\longrightarrow\\!^{*}\operatorname{Hom}_{R}(\\!^{*}\operatorname{Hom}_{S}(Z,Y),X).$

The morphism is invertible under each of the next two extra conditions

  * (i)

$Z\in\\!^{*}\mathcal{C}_{\square}^{fp}(S)$,
$Y\in\\!^{*}\mathcal{C}_{\sqsupset}(S)$, and
$X\in\\!^{*}\mathcal{C}_{\sqsubset}(R)$; or

  * (ii)

$Z\in\\!^{*}\mathcal{C}_{\sqsupset}^{fp}(S)$,
$Y\in\\!^{*}\mathcal{C}_{\sqsubset}(S)$, and
$X\in\\!^{*}\mathcal{C}_{\square}(R)$.

By $Z\in\\!^{*}\mathcal{C}^{fp}(S)$ we mean that $Z$ consists of finitely
generated projective $S$-modules.

We recall the definition of the _depth_ and _width_ of complexes. Let
$\mathfrak{a}$ be an ideal in a ring $R$ and $X$ a complex of graded
$R$-modules. The $\mathfrak{a}$-$0pt$ and $\mathfrak{a}$-$0pt$ of $X$ over $R$
are defined respectively by

$\displaystyle 0pt(\mathfrak{a},X):=$
$\displaystyle-\sup{\mathbf{R}}\operatorname{Hom}_{R}(R/{\mathfrak{a}},X),$
$\displaystyle 0pt(\mathfrak{a},X):=$
$\displaystyle\inf(R/{\mathfrak{a}}\otimes^{\mathbf{L}}_{R}X).$

If $(R,\mathfrak{m})$ is a local ring then set $0pt_{R}X:=0pt(\mathfrak{m},X)$
and $0pt_{R}X:=0pt(\mathfrak{m},X)$.

Now let $(R,\mathfrak{m})$ be a $\,{}^{*}$local graded ring and $X$ be a
complex of graded $R$-modules. By [3, Proposition 1.5.15(c)],
$-\otimes_{R}R_{\mathfrak{m}}$ is a faithfully exact functor on the category
of graded $R$-modules. Then we have

$\displaystyle 0pt(\mathfrak{m},X)=$
$\displaystyle\inf\\{i|\mbox{H}_{i}(R/\mathfrak{m}\otimes^{\mathbf{L}}_{R}X)\neq
0\\}$ $\displaystyle=$
$\displaystyle\inf\\{i|\mbox{H}_{i}(R/\mathfrak{m}\otimes^{\mathbf{L}}_{R}X)\otimes_{R}R_{\mathfrak{m}}\neq
0\\}$ $\displaystyle=$
$\displaystyle\inf\\{i|\mbox{H}_{i}(R_{\mathfrak{m}}/\mathfrak{m}R_{\mathfrak{m}}\otimes^{\mathbf{L}}_{R_{\mathfrak{m}}}X_{\mathfrak{m}})\neq
0\\}$ $\displaystyle=$ $\displaystyle
0pt(\mathfrak{m}R_{\mathfrak{m}},X_{\mathfrak{m}})=0pt_{R_{\mathfrak{m}}}X_{\mathfrak{m}}.$

Likewise we have $0pt(\mathfrak{m},X)=0pt_{R_{\mathfrak{m}}}X_{\mathfrak{m}}$.

######  Proposition 2.2.

Let $(R,\mathfrak{m},k)$ be a ∗local ring and $X\in\mathcal{C}_{\sqsupset}(R)$
and set $E:=\\!^{*}\operatorname{E}_{R}(k)$. Then

$0pt(\mathfrak{m},X)=0pt(\mathfrak{m},\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(X,E)).$

###### Proof.

We have the following computations:

$\displaystyle 0pt(\mathfrak{m},X)=$
$\displaystyle\inf(k\otimes^{\mathbf{L}}_{R}X)=-\sup\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(k\otimes^{\mathbf{L}}_{R}X,E)$
$\displaystyle=$
$\displaystyle-\sup\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(k,\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(X,E))$
$\displaystyle=$ $\displaystyle
0pt(\mathfrak{m},\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(X,E)).$

The second equality hods since $E$ is faithfully ∗injective and the third one
uses the ∗adjointness isomorphism. ∎

The following lemma is the graded version of one of Foxby’s accounting
principles [5, Lemma A.7.9].

######  Lemma 2.3.

Let $(R,\mathfrak{m},k)$ be a ∗local ring. Then

  * (a)

If $X\in\,^{*}\mathcal{C}(k)$, then there is a quasiisomorphism
$\mbox{H}(X)\rightarrow X$.

  * (b)

If $X\,^{*}\mathcal{C}_{(\sqsubset)}(R)$ and $W\in\,^{*}\mathcal{C}(k)$ then

$\inf\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(W,X)=\inf\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(k,X)-\sup
W.$

###### Proof.

Part $(a)$ is easy since very graded $k$-module is free by [3, Exercise
1.5.20]. For part $(b)$ note that $W=k\otimes^{\mathbf{L}}_{k}W$. Then using
the ∗adjointness isomorphism we have

$\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(W,X)=\mathbf{R}\\!^{*}\operatorname{Hom}_{k}(W,\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(k,X))=\\!^{*}\operatorname{Hom}_{k}(W,\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(k,X)).$

Thus using part $(a)$ we have

$\displaystyle\inf\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(W,X)=$
$\displaystyle\inf\\{\ell|\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(W,X)_{\ell}\neq
0\\}$ $\displaystyle=$
$\displaystyle\inf\\{\ell|\,^{*}\prod_{i+j=\ell}\\!^{*}\operatorname{Hom}_{k}(W_{-j},\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(k,X)_{i})\neq
0\\}$ $\displaystyle=$
$\displaystyle\inf\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(k,X)-\sup W.$

This completes the proof. ∎

The following equality is the graded version of [20, Lemma 2.6].

######  Proposition 2.4.

Let $(R,\mathfrak{m},k)$ be a ∗local ring
$X,Y\in\,^{*}\mathcal{C}_{\square}(R)$, then

$0pt(\mathfrak{m},\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(X,Y))=0pt(\mathfrak{m},X)+0pt(\mathfrak{m},Y)-0pt(\mathfrak{m},R).$

###### Proof.

Use Lemma 2.3 and the same argument as [20, Lemma 2.6]. ∎

Let $a\in R$ be homogeneous and set $\alpha=\deg(a)$. Then the complex $0\to
R(-\alpha)\stackrel{{\scriptstyle a}}{{\to}}R\to 0$ concentrated in degrees 1
and 0 is called the _Koszul complex_ of $a$, and denoted by $K(a)$, where
$R(-\alpha)$ denotes the graded $R$-module with grading given by
$R(-\alpha)_{n}=R_{n-\alpha}$. Note that $K(a)\in^{*}\mathcal{C}(R)$. Now let
$\mathfrak{a}$ be a homogeneous ideal of $R$ and $a_{1},\cdots,a_{n}$ be a set
of generators of $\mathfrak{a}$ by homogeneous elements. The Koszul complex of
$\mathfrak{a}$, denoted by $K:=K(\mathfrak{a})$, and define as
$K(a_{1})\otimes_{R}\cdots\otimes_{R}K(a_{n})$. It is shown in [12] that
$0pt(\mathfrak{a},X)=\inf(K\otimes_{R}X)$.

Let $\alpha:X\to Y$ be a homogeneous morphism between $R$-complexes of graded
$R$-modules. The mapping cone of $\alpha$ is a complex of graded $R$-modules
given by

$\mathcal{M}(\alpha)_{\ell}:Y_{\ell}\oplus X_{\ell-1}$

and

$\delta_{\ell}^{\mathcal{M}(\alpha)}(y_{\ell},x_{\ell-1})=(\delta_{\ell}^{Y}(y_{\ell})+\alpha_{\ell-1}(x_{\ell-1}),-\delta_{\ell-1}^{X}(x_{\ell-1})).$

Note that $\delta_{\ell}^{\mathcal{M}(\alpha)}$ is a homogeneous
differentiation from $Y_{\ell}\oplus X_{\ell-1}$ to $Y_{\ell-1}\oplus
X_{\ell-2}$. It is easy to see that the morphism $\alpha:X\to Y$ between
complexes of graded $R$-modules is quasi-isomorphism if and only if the
mapping cone of $\alpha$ is homologically trivial (see [5, Lemma A.1.19]).
Also for the covariant and contravariant functors
${}^{*}\operatorname{Hom}_{R}(V,-)$ and ${}^{*}\operatorname{Hom}_{R}(-,W)$ we
have the following:

$\mathcal{M}(\,^{*}\operatorname{Hom}_{R}(V,\alpha))=\,^{*}\operatorname{Hom}_{R}(V,\mathcal{M}(\alpha)),$
$\mathcal{M}(\,^{*}\operatorname{Hom}_{R}(\alpha,W))=\Sigma^{1}\,{}^{*}\operatorname{Hom}_{R}(\mathcal{M}(\alpha),W).$

See [5, A2.1.2 and A.2.1.4].

The ungraded version of the following result contained in [7, Proposition
2.7].

######  Proposition 2.5.

Let $\mathfrak{B}$ be a class of graded $R$-modules, and $\alpha:X\to Y$ be a
quasiisomorphism between complexes of graded $R$-modules such that

${}^{*}\operatorname{Hom}_{R}(\alpha,V):\,^{*}\operatorname{Hom}_{R}(Y,V)\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}\,^{*}\operatorname{Hom}_{R}(X,V)$

is quasiisomorphism for every module $V\in\mathfrak{B}$. Let
$\widetilde{V}\in\,^{*}\mathcal{C}(R)$ be a complex consisting of modules from
$\mathfrak{B}$. Then the induced morphism,

${}^{*}\operatorname{Hom}_{R}(\alpha,\widetilde{V}):\,^{*}\operatorname{Hom}_{R}(Y,\widetilde{V})\longrightarrow\,^{*}\operatorname{Hom}_{R}(X,\widetilde{V}),$

is a quasiisomorphism, provided that either

  * (a)

$\widetilde{V}\in\,^{*}\mathcal{C}_{\sqsubset}(R)$, or

  * (b)

$X,Y\in\,^{*}\mathcal{C}_{\sqsubset}(R)$.

###### Proof.

In either cases it is enough to show that
${}^{*}\operatorname{Hom}_{R}(\mathcal{M}(\alpha),\widetilde{V})$ is
homologically trivial. On the other hand by graded version of [7, Lemma 2.5]
the result holds if
${}^{*}\operatorname{Hom}_{R}(\mathcal{M}(\alpha),\widetilde{V}_{\ell})$ is
homologically trivial for each $\ell\in\mathbb{Z}$ which this is our
assumption. ∎

##  3\. ∗injective dimension

The injective dimension of a complex $X$ is defined and studied in [1],
denoted by $\operatorname{id}_{R}X$. A graded module $J$ is called ∗injective
if it is an injective object in the category of graded modules. The injective
dimension of a graded module $M$ in the category of graded modules, is denoted
by ${}^{*}\operatorname{id}_{R}M$ (cf. [10, 15, 3]). The ∗injective dimension
of a complex of graded modules $X$ is studied in [15, Page 83]. Let
$n\in\mathbb{Z}$. A homologically left bounded complex of graded modules $X$,
is said to have ∗injective dimension at most $n$, denoted by
${}^{*}\operatorname{id}_{R}X\leq n$, if there exists an ∗injective resolution
$X\to I$, such that $I_{i}=0$ for $i<-n$. If ${}^{*}\operatorname{id}_{R}X\leq
n$ holds, but ${}^{*}\operatorname{id}_{R}X\leq n-1$ does not, we write
${}^{*}\operatorname{id}_{R}X=n$. If ${}^{*}\operatorname{id}_{R}X\leq n$ for
all $n\in\mathbb{Z}$ we write ${}^{*}\operatorname{id}_{R}X=-\infty$. If
${}^{*}\operatorname{id}_{R}X\leq n$ for no $n\in\mathbb{Z}$ we write
${}^{*}\operatorname{id}_{R}X=\infty$. The following theorem inspired by [1,
Theorem 2.4.I and Corollary 2.5.I].

######  Theorem 3.1.

For $X\in$ ${}^{*}\mathcal{D}_{\sqsubset}(R)$ and $n\in\mathbb{Z}$ the
following are equivalent:

  * (1)

${}^{*}\operatorname{id}_{R}X\leq n.$

  * (2)

$n\geq-\sup U-\inf(\mathbf{R}^{*}\operatorname{Hom}_{R}(U,X))$ for all $U\in$
${}^{*}\mathcal{D}_{\square}(R)$ and $\mbox{H}(U)\neq 0$.

  * (3)

$n\geq-\inf X$ and ${}^{*}\operatorname{Ext}^{n+1}_{R}(R/J,X)=0$ for every
homogeneous ideal $J$ of $R$.

  * (4)

$n\geq-\inf X$ and for any (resp. some) ∗injective resolution $I$ of $X$, the
graded $R$-module $\operatorname{Ker}(\partial_{-n}:I_{-n}\to I_{-n-1})$ is
∗injective.

Moreover the following hold:

${}^{*}\operatorname{id}_{R}X=$
$\displaystyle\sup\\{j\in\mathbb{Z}|^{*}\operatorname{Ext}^{j}_{R}(R/J,X)\neq
0\text{ for some homogeneous ideal }J\\}$ $\displaystyle=$
$\displaystyle\sup\\{-\sup(U)-\inf(\mathbf{R}^{*}\operatorname{Hom}_{R}(U,X))|U\ncong
0\text{ in }^{*}\mathcal{D}_{\square}(R)\\}.$

###### Proof.

$(1)\Rightarrow(2)$ Let $t:=\sup U$ and $I$ be an ∗injective resolution of
$X$, such that, for all $i<-n$, $I_{i}=0$. Then we have

${}^{*}\operatorname{Ext}^{i}_{R}(U,X)\cong\mbox{H}_{-i}(^{*}\operatorname{Hom}_{R}(U,I)).$

Since ${}^{*}\operatorname{Hom}_{R}(U,I)_{-i}=0$ for $-i<-n-t$, the assertion
follows.

$(2)\Rightarrow(3)$ It is trivial that
${}^{*}\operatorname{Ext}^{n+1}_{R}(R/J,X)=0$ for every homogeneous ideal $J$
of $R$. For the second assertion let $U=R$ in (2). So that
$\operatorname{Ext}^{i}_{R}(R,X)=\\!^{*}\operatorname{Ext}^{i}_{R}(R,X)=0$ for
$i>n$. Now by [1, Lemma 1.9(b)], we have $\mbox{H}_{-i}(X)=0$ for $-i<-n$.
This means that $n\geq-\inf X$.

$(3)\Rightarrow(4)$ By hypothesis of (4) $\mbox{H}_{i}(I)=0$ for $i<-n$. Thus
the complex

$\cdots\to 0\to 0\to I_{-n}\to I_{-n-1}\to\cdots\to I_{i}\to
I_{i-1}\to\cdots,$

gives an ∗injective resolution of $\operatorname{Ker}\partial_{-n}$. In
particular

$\\!{}^{*}\operatorname{Ext}^{1}_{R}(R/J,\operatorname{Ker}\partial_{-n})=\mbox{H}_{-n-1}\\!^{*}\operatorname{Hom}_{R}(R/J,I)=\\!^{*}\operatorname{Ext}^{n+1}_{R}(R/J,X)=0$

for every homogeneous ideal $J$ of $R$. Thus $\operatorname{Ker}\partial_{-n}$
is ∗injective by [10, Corollary 4.3].

$(4)\Rightarrow(1)$ Let $I$ be any ∗injective resolution of $X$. By (5) we
have $\operatorname{Ker}\partial_{-n}$ is ∗injective. Thus
$\\!{}^{*}\operatorname{id}_{R}X<-n$ by definition.

The next two equalities are trivial. ∎

For a local ring $(R,\mathfrak{m},k)$ and for an $R$-complex $X$ and
$i\in\mathbb{Z}$ the $i$th Bass number and Betti number of $X$ are defined
respectively by
$\mu^{i}_{R}(X):=\operatorname{dim}_{k}\mbox{H}_{-i}({\mathbf{R}}\operatorname{Hom}_{R}(k,X))$
and
$\beta_{i}^{R}(X):=\operatorname{dim}_{k}\mbox{H}_{i}(k\otimes_{R}^{\mathbf{L}}X).$
It is well-known that for $X\in{\mathcal{D}}_{\sqsubset}(R)$ one has (cf. [1,
Proposition 5.3.I])

$\operatorname{id}_{R}X=\sup\\{m\in\mathbb{Z}|\exists\mathfrak{p}\in\operatorname{Spec}(R);\mu^{m}_{R_{\mathfrak{p}}}(X_{\mathfrak{p}})\neq
0\\}.$

As a graded analogue we have:

######  Proposition 3.2.

For $X\in$ ${}^{*}\mathcal{D}_{\sqsubset}(R)$ we have the following equality

${}^{*}\operatorname{id}_{R}X=\sup\\{m\in\mathbb{Z}|\exists\mathfrak{p}\in^{*}\operatorname{Spec}(R);\mu^{m}_{R_{\mathfrak{p}}}(X_{\mathfrak{p}})\neq
0\\}.$

###### Proof.

The argument is the same as proof of [1, Proposition 5.3.I] with some changes.
Denote the supremum by $i$. By Theorem 3.1, we have
$\\!{}^{*}\operatorname{id}_{R}X\geq i$. Hence the equality holds if
$i=\infty$. Thus assume that $i$ is finite. By Theorem 3.1 we have to show
that if $\\!{}^{*}\operatorname{Ext}^{j}_{R}(M,X)\neq 0$ for some finitely
generated graded $R$-module $M$, then $j\leq i$; this implies that
$\\!{}^{*}\operatorname{id}_{R}X\leq i$. The elements of $Ass(M)$ are
homogeneous prime ideals. Thus we have a filtration $0=M_{0}\subset
M_{1}\subset\cdots\subset M_{t}=M$ of graded submodules of $M$ such that for
each $i$ we have $M_{i}/M_{i-1}\cong R/\mathfrak{p}_{i}$ with
$\mathfrak{p}_{i}\in\operatorname{Supp}M$ and is homogeneous. From the long
exact sequence of $\\!{}^{*}\operatorname{Ext}^{j}_{R}(-,X)\neq 0$ we have the
set

$\\{\mathfrak{q}\in\\!\operatorname{Spec}(R)|\text{ there is an }h\geq j\text{
such that }\\!^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{q},X)\neq 0\\},$

is not empty. Let $\mathfrak{p}$ maximal in this set, and for a homogeneous
$x\in R\backslash\mathfrak{p}$ consider the exact sequence

$0\to R/\mathfrak{p}\stackrel{{\scriptstyle x}}{{\to}}R/\mathfrak{p}\to
R/(\mathfrak{p}+Rx)\to 0.$

It induces an exact sequence

$\\!{}^{*}\operatorname{Ext}^{h}_{R}(R/(\mathfrak{p}+Rx),X)\to\\!^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{p},X)\stackrel{{\scriptstyle
x}}{{\to}}\\!^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{p},X)\to\\!^{*}\operatorname{Ext}^{h+1}_{R}(R/(\mathfrak{p}+Rx),X)$

in which the left-hand term is trivial because of the maximality of
$\mathfrak{p}$. Thus
$\\!{}^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{p},X)\stackrel{{\scriptstyle
x}}{{\to}}\\!^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{p},X)$ is injective
for all homogeneous elements $x\in R\backslash\mathfrak{p}$, hence so is the
homogeneous localization homomorphism
$\\!{}^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{p},X)\to\\!^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{p},X)_{(\mathfrak{p})}$.
Thus the free $R_{(\mathfrak{p})}/\mathfrak{p}R_{(\mathfrak{p})}$-module
$\\!{}^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{p},X)_{(\mathfrak{p})}$ is
nonzero. Consequently

$(\\!^{*}\operatorname{Ext}^{h}_{R}(R/\mathfrak{p},X)_{(\mathfrak{p})})_{\mathfrak{p}R_{(\mathfrak{p})}}\cong\\!^{*}\operatorname{Ext}^{h}_{R_{\mathfrak{p}}}(R_{\mathfrak{p}}/\mathfrak{p}R_{\mathfrak{p}},X_{\mathfrak{p}})$

is nonzero. This implies that $j\leq h\leq i$. ∎

######  Remark 3.3.

(1) A graded module is called ∗projective if it is a projective object in the
category of graded modules. By [10, Proposition 3.1] the ∗projective graded
$R$-modules coincide with projective $R$-modules. The projective dimension of
a graded module $M$ in the category of graded modules, is denoted by
${}^{*}\operatorname{pd}_{R}M$ (cf. [10]). Let $n\in\mathbb{Z}$. A
homologically right bounded complex of graded modules $X$, is said to have
∗projective dimension at most $n$, denoted by
${}^{*}\operatorname{pd}_{R}X\leq n$, if there exists a ∗projective resolution
$P\to X$, such that $P_{i}=0$ for $i>n$. If ${}^{*}\operatorname{pd}_{R}X\leq
n$ holds, but ${}^{*}\operatorname{pd}_{R}X\leq n-1$ does not, we write
${}^{*}\operatorname{pd}_{R}X=n$. If ${}^{*}\operatorname{pd}_{R}X\leq n$ for
all $n\in\mathbb{Z}$ we write ${}^{*}\operatorname{pd}_{R}X=-\infty$. If
${}^{*}\operatorname{pd}_{R}X\leq n$ for no $n\in\mathbb{Z}$ we write
${}^{*}\operatorname{pd}_{R}X=\infty$.

(2) For $X\in\\!^{*}\mathcal{D}_{\sqsupset}(R)$ by the same method as in [1,
Theorem 2.4.P and Corollary 2.5.P] we have

${}^{*}\operatorname{pd}_{R}X=$
$\displaystyle\sup\\{j\in\mathbb{Z}|^{*}\operatorname{Ext}^{j}_{R}(X,N)\neq
0\text{ for some graded }R\text{-module }N\\}$ $\displaystyle=$
$\displaystyle\sup\\{\inf(U)-\inf(\mathbf{R}^{*}\operatorname{Hom}_{R}(X,U))|U\ncong
0\text{ in }^{*}\mathcal{D}_{\square}(R)\\}.$

(3) It is easy to see that for $X\in\\!^{*}\mathcal{D}_{\sqsupset}(R)$, we
have $\\!{}^{*}\operatorname{pd}_{R}X\leq\operatorname{pd}_{R}X$.

The proof of the following proposition is easy so we omit it (see [3, Theorem
1.5.9]). Let $J$ be an ideal of the graded ring $R$. Then the graded ideal
$J^{*}$ is denoted to the ideal generated by all homogeneous elements of $J$.
It is well-known that if $\mathfrak{p}$ is a prime ideal of $R$, then
$\mathfrak{p}^{*}$ is a homogeneous prime ideal of $R$ by [3, Lemma 1.5.6].

######  Proposition 3.4.

Assume that $X\in\,^{*}{\mathcal{D}}_{\square}(R)$ and $\mathfrak{p}$ is a non
homogeneous prime ideal in $R$. Then
$\mu^{i+1}_{R_{\mathfrak{p}}}(X_{\mathfrak{p}})=\mu^{i}_{R_{\mathfrak{p}^{*}}}(X_{\mathfrak{p}^{*}})$
and
$\beta_{i}^{R_{\mathfrak{p}}}(X_{\mathfrak{p}})=\beta_{i}^{R_{\mathfrak{p}^{*}}}(X_{\mathfrak{p}^{*}})$for
any integer $i\geq 0$.

######  Corollary 3.5.

Let $X\in\,^{*}{\mathcal{D}}_{\square}(R)$ and $\mathfrak{p}$ be a non-
homogeneous prime ideal in $R$. Then

$0ptX_{\mathfrak{p}}=0ptX_{\mathfrak{p}^{*}}+1.$

###### Proof.

Using Proposition 3.4, we can assume that both $0ptX_{\mathfrak{p}}$ and
$0ptX_{\mathfrak{p}^{*}}$ are finite. So the equality follows from the fact
that over a local ring $(R,\mathfrak{m},k)$ we have
$0pt_{R}X=\inf\\{i\in\mathbb{Z}|\mu^{i}_{R}(X)\neq 0\\}$. ∎

Foxby in [11] defined the _small support_ of a homologically right bounded
complex $X$ over a Noetherian ring $R$, denoted by $\mbox{supp}_{R}X$, as

$\operatorname{supp}_{R}X=\\{\mathfrak{p}\in\operatorname{Spec}R|\exists
m\in\mathbb{Z}:\beta_{m}^{R_{\mathfrak{p}}}(X_{\mathfrak{p}})\neq 0\\}.$

Let $\,{}^{*}\operatorname{supp}_{R}X$ be a subset of
$\operatorname{supp}_{R}X$ consisting of homogeneous prime ideals of
$\operatorname{supp}_{R}X$. Then from Proposition 3.4 we see that
$\mathfrak{p}\in\operatorname{supp}_{R}X$ if and only if
$\mathfrak{p}^{*}\in\,^{*}\operatorname{supp}_{R}X$. Also using [11,
Proposition 2.8] and Corollary 3.5 (or directly from Proposition 3.4) we have

$0pt_{R_{\mathfrak{p}}}X_{\mathfrak{p}}<\infty\Leftrightarrow
0pt_{R_{\mathfrak{p}^{*}}}X_{\mathfrak{p}^{*}}<\infty.$

######  Proposition 3.6.

Assume that $X\in\,^{*}{\mathcal{D}}_{\square}(R)$, and $\mathfrak{p}$ is a
non homogeneous prime ideal in $R$. Then

$0pt_{R_{\mathfrak{p}}}X_{\mathfrak{p}}=0pt_{R_{\mathfrak{p}^{*}}}X_{\mathfrak{p}^{*}}$

###### Proof.

We can assume that both $0ptX_{\mathfrak{p}}$ and $0ptX_{\mathfrak{p}^{*}}$
are finite numbers. And the argument is dual to the proof of [3, Theorem
1.5.9]. ∎

The ungraded version of the following theorem was proved for modules by
Chouinard [4, Corollary 3.1] and extended to complexes by Yassemi [20, Theorem
2.10].

######  Theorem 3.7.

Let $X\in$ $\,{}^{*}\mathcal{D}_{\square}(R)$. If
$\,{}^{*}\operatorname{id}_{R}X<\infty$ then

${}^{*}\operatorname{id}_{R}X=\sup\\{0ptR_{\mathfrak{p}}-0ptX_{\mathfrak{p}}|\mathfrak{p}\in\,^{*}\\!\operatorname{Spec}(R)\\}.$

###### Proof.

We have the following computations

${}^{*}\operatorname{id}_{R}X=$
$\displaystyle\sup\\{m\in\mathbb{Z}|\exists\mathfrak{p}\in\,^{*}\operatorname{Spec}(R):\mu^{m}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\neq
0\\}$ $\displaystyle=$
$\displaystyle\sup\\{m\in\mathbb{Z}|\exists\mathfrak{p}\in\,^{*}\\!\operatorname{Spec}(R):\mbox{H}_{m}({\mathbf{R}}\operatorname{Hom}_{R_{\mathfrak{p}}}(\kappa(p),M_{\mathfrak{p}}))\neq
0\\}$ $\displaystyle=$
$\displaystyle\sup\\{-\inf{\mathbf{R}}\operatorname{Hom}_{R_{\mathfrak{p}}}(\kappa(\mathfrak{p}),M_{\mathfrak{p}})|\mathfrak{p}\in\,^{*}\\!\operatorname{Spec}(R)\\}$
$\displaystyle=$
$\displaystyle\sup\\{0ptR_{\mathfrak{p}}-0pt_{R_{\mathfrak{p}}}M_{\mathfrak{p}}|\mathfrak{p}\in\,^{*}\\!\operatorname{Spec}(R)\\}.$

The first equality is by Proposition 3.2 and the last one is by [20, Lemma
2.6(a)]. ∎

The following corollary was already known for graded modules in [10, Corollary
4.12].

######  Corollary 3.8.

For every $X\in\,^{*}\mathcal{D}_{\square}(R)$, we have

$\,{}^{*}\operatorname{id}_{R}X\leq\operatorname{id}_{R}X\leq\,^{*}\operatorname{id}_{R}X+1.$

###### Proof.

First of all note that by proposition 3.4, $\operatorname{id}_{R}X<\infty$ if
and only if $\,{}^{*}\operatorname{id}_{R}X<\infty$. The first inequality is
clear by Theorem 3.7 and [20, Theorem 2.10]. For the second one let
$\mathfrak{p}\in\operatorname{Spec}R$ be such that
$\operatorname{id}_{R}X=0ptR_{\mathfrak{p}}-0pt_{R_{\mathfrak{p}}}M_{\mathfrak{p}}$
by [20, Theorem 2.10]. By Corollary 3.5 and Proposition 3.6 we have

$0ptR_{\mathfrak{p}}-0pt_{R_{\mathfrak{p}}}M_{\mathfrak{p}}\leq
0ptR_{\mathfrak{p}^{*}}-0pt_{R_{\mathfrak{p}^{*}}}M_{\mathfrak{p}^{*}}+1\leq\,^{*}\operatorname{id}_{R}X+1,$

where the second inequality holds by Theorem 3.7. ∎

Here we define the $\\!{}^{*}$dualizing complex for a graded ring and prove
some related results that we need in the next section.

######  Definition 3.9.

A ∗dualizing complex for a graded ring $R$ is a homologically finite and
bounded complex $D$, of graded $R$-modules, such that
$\,{}^{*}\operatorname{id}_{R}D<\infty$ and the homothety morphism
$\psi:R\to\textbf{R}\,^{*}\operatorname{Hom}_{R}(D,D)$ is invertible in
$\,{}^{*}\mathcal{D}(R)$.

######  Corollary 3.10.

Any ∗dualizing complex for $R$ is a dualizing complex for $R$.

The proof of the following lemma is the same as [15, Chapter V, Proposition
3.4].

######  Lemma 3.11.

Let $(R,\mathfrak{m},k)$ be a ∗local ring and that $D$ is a ∗dualizing complex
for $R$. Then there exists an integer $t$ such that
$\mbox{H}^{t}(\textbf{R}\,^{*}\operatorname{Hom}_{R}(k,D))\cong k$ and
$\mbox{H}^{i}(\textbf{R}\,^{*}\operatorname{Hom}_{R}(k,D))=0$ for $i\neq t$.

Assume that $(R,\mathfrak{m})$ is a ∗local ring. A ∗dualizing complex $D$ is
said to be _normalized ∗dualizing complex_, if $t=0$ in the lemma. It is easy
to see that a suitable shift of any ∗dualizing complex is a normalized one.
Also using [15, Chapter V, Proposition 3.4] we see that if $D$ is a normalized
∗dualizing complex for $(R,\mathfrak{m})$, then $D_{\mathfrak{m}}$ is a
normalized dualizing complex for $R_{\mathfrak{m}}$.

######  Lemma 3.12.

Let $(R,\mathfrak{m},k)$ be a ∗local ring and that $D$ is a normalized
∗dualizing complex for $R$. Then there exists a natural functorial isomorphism
on the category of graded modules of finite length to itself

$\phi:\mbox{H}^{0}(\textbf{R}\,^{*}\operatorname{Hom}_{R}(-,D))\to\,^{*}\operatorname{Hom}_{R}(-,\\!^{*}\operatorname{E}_{R}(k)),$

where $\\!{}^{*}E_{R}(k)$ is the ∗injective envelope of $k$ over $R$.

###### Proof.

Since $D$ is a normalized ∗dualizing complex for $R$,
$T:=\mbox{H}^{0}(\textbf{R}\,^{*}\operatorname{Hom}_{R}(-,D))$ is an additive
contravariant exact functor from the category of graded modules of finite
length to itself. Let $M$ be a graded $R$-module and $m\in M$ is homogeneous
element of degree $\alpha$. Then $\epsilon_{m}:R(-\alpha)\to M$ is a
homogeneous morphism which sends $1$ into $m$. Thus we have a homogeneous
morphism $\phi(M):T(M)\to\\!^{*}\operatorname{Hom}_{R}(M,T(R))$ which sends a
homogeneous element $x\in T(M)$ to a morphism
$f_{x}\in^{*}\operatorname{Hom}_{R}(M,T(R))$ such that
$f_{x}(m)=T(\epsilon_{m})(x)$ for every homogeneous element $m\in M$. It is
easy to see that it is functorial on $M$. Thus we showed that there is a
natural functorial morphism $\phi:T\to\\!^{*}\operatorname{Hom}_{R}(-,T(R))$.
Therefore by the same method of [14, Lemma 4.4 and Propositions 4.5], there is
a functorial isomorphism

$\phi:\mbox{H}^{0}(\textbf{R}\,^{*}\operatorname{Hom}_{R}(-,D))\to\,^{*}\operatorname{Hom}_{R}(-,\,^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n})),$

from the category of graded modules of finite length to itself. Using the
technique of proof of [14, Proposition 4.7] in conjunction with [10, Corollary
4.3], we see that
$\,{}^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n})$ is an
∗injective $R$-module. Since $D$ is a normalized ∗dualizing complex for $R$ we
have

$\,{}^{*}\operatorname{Hom}_{R}(k,\,^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n}))\cong\mbox{H}^{0}(\textbf{R}\,^{*}\operatorname{Hom}_{R}(k,D))\cong
k.$

Thus in particular we can embed $k$ to
$\,{}^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n})$. To show
that $\,{}^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n})$ is an
∗essential extension of $k$, let $Q$ be a graded submodule of
$\,{}^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n})$ such that
$k\cap Q=0$. Then ${}^{*}\operatorname{Hom}_{R}(k,Q)$ can be embed in

${}^{*}\operatorname{Hom}_{R}(k,\,^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n}))\cong
k.$

Therefore ${}^{*}\operatorname{Hom}_{R}(k,Q)=0$. On the other hand
$\operatorname{Ass}(T(R/\mathfrak{m}^{n}))$ is in $V(\mathfrak{m})$ for each
$n\in\mathbb{N}$. Now by [19, Proposition 2.1], the fact that each prime ideal
of
$\operatorname{Ass}(\,^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n}))$
is the annihilator of a homogeneous element [3, Lemma 1.5.6], and the
definition of $\,{}^{*}\displaystyle\lim_{\longrightarrow}$, we have

$\operatorname{Ass}(\,^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n}))\subseteq\bigcup_{n\in\mathbb{N}}\operatorname{Ass}(T(R/\mathfrak{m}^{n}))\subseteq
V(\mathfrak{m}).$

Consequently $Q$ has support in $V(\mathfrak{m})$, so that $Q=0$. Therefore
$\,{}^{*}\displaystyle\lim_{\longrightarrow}T(R/\mathfrak{m}^{n})\cong\\!^{*}\operatorname{E}_{R}(k)$.
∎

Let $\mathfrak{a}$ be an ideal of $R$. The right derived _local cohomology
functor_ with support in $\mathfrak{a}$ is denoted by
$\mathbf{R}\Gamma_{\mathfrak{a}}(-)$. Its right adjoint,
$\mathbf{L}\Lambda^{\mathfrak{a}}(-)$, is the left derived _local homology
functor_ with support in $\mathfrak{a}$ (see [13] for detail).

Now we have the following proposition, which its proof use Lemma 3.12, and the
argument is the same as [15, Chapter V, Proposition 6.1].

######  Proposition 3.13.

Let $(R,\mathfrak{m},k)$ be a ∗local ring and that $D$ be a normalized
∗dualizing complex for $R$. Then
${\mathbf{R}}\Gamma_{\mathfrak{m}}(D)\simeq\\!^{*}\operatorname{E}_{R}(k)$.

##  4\. ∗Gorenstein injective dimension

In this section we introduce the concept of $\\!{}^{*}$Gorenstein injective
dimension of complexes and we derive its main properties. In particular we
prove a Chouinard’s like formula for this dimension, and compare it with the
usual Gorenstein injective dimension.

######  Definition 4.1.

A graded $R$-module $N$ is called $\,{}^{*}$Gorenstein injective, if there
exists an acyclic complex I of $\,{}^{*}$injective $R$-modules and homogeneous
homomorphisms such that $M\cong\operatorname{Ker}(I^{0}\to I^{1})$ and for
every $\,{}^{*}$injective module $E$, the complex
$\,{}^{*}\operatorname{Hom}_{R}(E,\textbf{I})$ is exact.

It is clear that every ∗injective $R$-module is $\,{}^{*}$Gorenstein
injective. So that every $Y\in\,^{*}\mathcal{D}_{\sqsubset}(R)$ has a
$\,{}^{*}$Gorenstein injective resolution. The ∗Gorenstein injective dimension
of $Y\in\,^{*}\mathcal{D}_{\sqsubset}(R)$ denoted by
$\,{}^{*}\operatorname{Gid}_{R}Y$, is define as:

$\,{}^{*}\operatorname{Gid}_{R}Y:=\inf\left\\{\sup\\{\ell\in Z|B_{-\ell}\neq
0\\}\bigg{|}\begin{array}[]{l}B_{\ell}\text{ is }\,^{*}\text{Gorenstein
injective and}\\\ B\in\,^{*}\mathcal{D}_{\sqsubset}(R)\text{ is isomorphic to
}Y\end{array}\right\\}.$

By a careful revision of the proof of dual version of [16, Theorems 2.5 and
2.20] we have the following two results.

######  Theorem 4.2.

The class of ∗Gorenstein injective $R$-modules is ∗injectively resolving, that
is for any short exact sequence $0\to X^{\prime}\to X\to X^{\prime\prime}\to
0$ with $X^{\prime}$ ∗Gorenstein injective $R$-module, $X^{\prime\prime}$ is
∗Gorenstein injective if and only if $X$ is ∗Gorenstein injective.

######  Proposition 4.3.

Let $N$ be a graded $R$-module with finite ∗Gorenstein injective dimension,
and let $n$ be an integer. Then the following conditions are equivalent:

  * (1)

${}^{*}\\!\operatorname{Gid}_{R}N\leq n$.

  * (2)

${}^{*}\\!\operatorname{Ext}^{i}_{R}(L,N)=0$ for all $i>n$, and all
$R$-modules $L$ with finite ${}^{*}\operatorname{id}_{R}L$.

  * (3)

${}^{*}\\!\operatorname{Ext}^{i}_{R}(I,N)=0$ for all $i>n$, and all ∗injective
$R$-modules $I$.

  * (4)

For every exact sequence $0\rightarrow N\rightarrow
H^{0}\rightarrow\cdots\rightarrow H^{n-1}\rightarrow C^{n}\rightarrow 0$ where
$H^{0},\cdots,H^{n-1}$ are ∗Gorenstein injectives, then $C^{n}$ is also
∗Gorenstein injective.

Consequently, the ∗Gorenstein injective dimension of $M$ is determined by the
formulas:

${}^{*}\operatorname{Gid}_{R}N=$
$\displaystyle\sup\\{i\in\mathbb{N}_{0}|\,^{*}\operatorname{Ext}^{i}_{R}(L,N)\neq
0\text{ for some graded }R\text{-module }L\text{ with
finite}\,^{*}\operatorname{id}_{R}L\\}$ $\displaystyle=$
$\displaystyle\sup\\{i\in\mathbb{N}_{0}|\,^{*}\operatorname{Ext}^{i}_{R}(I,N)\neq
0\text{ for some graded }^{*}\text{injective module }I\\}.$

The ungraded version of the following theorem is in [7, Theorem 2.8]. The
proof uses Proposition 2.5 and the same technique of proof of [7, Theorem
2.8].

######  Theorem 4.4.

Let $V\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}W$ be a
quasiisomorphism between complexes of graded $R$-modules, where each module in
$V$ and $W$ has finite ∗projective dimension or finite ∗injective dimension.
If $B\in\,^{*}\mathcal{C}_{\sqsubset}(R)$ is a complex of ∗Gorenstein
injective modules, then the induced morphism

${}^{*}\operatorname{Hom}_{R}(W,B)\to\,^{*}\operatorname{Hom}_{R}(V,B)$

is a quasiisomorphism under each of the next two condition:

  * (a)

$V,W\in\,^{*}\mathcal{C}_{\sqsupset}(R)$, or

  * (b)

$V,W\in\,^{*}\mathcal{C}_{\sqsubset}(R)$.

######  Corollary 4.5.

Assume that $Y\simeq B$ where $B\in$ ${}^{*}\mathcal{C}_{\sqsubset}(R)$ is a
complex of ∗Gorenstein injective modules. If $U\simeq V$, where $V\in$
${}^{*}\mathcal{C}_{\sqsupset}(R)$ is a complex in which each module has
finite ∗projective dimension or finite ∗injective dimension, then

$\mathbf{R}^{*}\\!\operatorname{Hom}_{R}(U,Y)\simeq\,^{*}\operatorname{Hom}_{R}(V,B).$

###### Proof.

Is the same as [7, Corollary 2.10] using Theorem 4.4(a). ∎

The ungraded version of the following theorem is contained in [7, Theorem
3.3], and its proof is dual of [7, Theorem 3.1]. We present the proof of
(3)$\Rightarrow$(4) for later use.

######  Theorem 4.6.

Let $Y\in$ ${}^{*}\mathcal{D}_{\sqsubset}(R)$ be a complex of finite
∗Gorenstein injective dimension. For $n\in\mathbb{Z}$ the following are
equivalent:

  * (1)

${}^{*}\operatorname{Gid}_{R}Y\leq n.$

  * (2)

$n\geq-\sup U-\inf\mathbf{R}^{*}\\!\operatorname{Hom}_{R}(U,Y)$ for all $U\in$
${}^{*}\mathcal{D}_{\square}(R)$ of finite ∗projective or finite ∗injective
dimension with $H(U)\neq 0$.

  * (3)

$n\geq-\inf\mathbf{R}^{*}\\!\operatorname{Hom}_{R}(J,Y)$ for all ∗injective
$R$-modules $J$.

  * (4)

$n\geq-\inf Y$ and for any left-bounded complex $B\simeq Y$ of ∗Gorenstein
injective modules, the $\operatorname{Ker}(B_{-n}\rightarrow B_{-(n+1)})$ is a
∗Gorenstein injective module.

Moreover the following hold:

${}^{*}\operatorname{Gid}_{R}Y=$
$\displaystyle\sup\\{-\sup(U)-\inf\mathbf{R}^{*}\operatorname{Hom}_{R}(U,Y)|^{*}\operatorname{id}_{R}(U)<\infty\text{
and }H(U)\neq 0\\}$ $\displaystyle=$
$\displaystyle\sup\\{-\inf\mathbf{R}^{*}\\!\operatorname{Hom}_{R}(J,Y)|J\text{
is }^{*}\text{injective}\\}.$

###### Proof.

(2)$\Rightarrow$(3) and (4)$\Rightarrow$(1) are clear. $(1)\Rightarrow(2)$ is
dual of $(1)\Rightarrow(2)$ in [7, Theorem 3.1].

(3)$\Rightarrow$(4): To establish $n\geq-\inf Y$, it is sufficient to show
that

$\sup\\{-\inf\mathbf{R}^{*}\\!\operatorname{Hom}_{R}(J,Y)|J\text{ is
}^{*}\text{injective}\\}\geq-\inf Y.\quad(*)$

By assumption $g:=\\!^{*}\operatorname{Gid}_{R}(Y)$ is finite, so $Y\simeq B$
for some complex of ∗Gorenstein injective modules:

$B=0\to B_{s}\to B_{s-1}\to\cdots\to B_{-g+1}\to B_{-g}\to 0.$

Now it is clear that $-g\leq\inf Y$. By Lemma 4.5, for any ∗injective module
$J$, the complex $\\!{}^{*}\operatorname{Hom}_{R}(J,B)$ is isomorphic to
$\mathbf{R}^{*}\operatorname{Hom}_{R}(J,Y)$ in ${}^{*}\mathcal{D}(R)$. If
$g=-\inf Y$ then the differential $\partial_{-g+1}:B_{-g+1}\to B_{-g}$ is not
surjective. Now by the definition of ∗Gorenstein injective modules there
exists an ∗injective module $J$ such that $J\to B_{-g}$ is surjective. Notice
that the differential ${}^{*}\operatorname{Hom}_{R}(J,\partial_{-g+1})$ in the
complex ${}^{*}\operatorname{Hom}_{R}(J,B)$ is not surjective, for otherwise,
for any $\phi\in^{*}\operatorname{Hom}_{R}(J,B_{-g})$ there exists a
$\psi\in^{*}\operatorname{Hom}_{R}(J,B_{-g+1})$ such that
$\phi=\partial_{-g+1}\psi$. This implies that $\partial_{-g+1}$ is surjective
which is a contradiction. Therefore ${}^{*}\operatorname{Hom}_{R}(J,B)$ has
nonzero homology in degree $-g=\inf Y$. This gives $(*)$. Next assume that
$g>-\inf Y=-t$ and consider the exact sequence

$B:0\to Z_{t}^{B}\to B_{t}\to B_{t-1}\to\cdots\to B_{-g+1}\to B_{-g}\to
0.\quad(**)$

It shows that ${}^{*}\operatorname{Gid}_{R}Z^{B}_{t}\leq t+g$ and it is not
difficult to see that the equality must hold. For otherwise
${}^{*}\operatorname{Gid}_{R}Y<g$ and by Proposition 4.3,
$\mbox{H}_{-g}(\mathbf{R}^{*}\operatorname{Hom}_{R}(J,Y))\cong\\!^{*}\operatorname{Ext}^{t+g}(J,Z^{B}_{t})\neq
0$ for some ∗injective module $J$. Therefore $(*)$ follows, which gives the
inequality $n\geq-\inf Y$.

To prove the second part of (4) let $B$ be a left-bounded complex of
∗Gorenstein injective modules such that $B\simeq Y$. By assumption
${}^{*}\operatorname{Gid}_{R}Y$ is finite, so there exists a bounded complex
$\tilde{B}$ of ∗Gorenstein injective modules such that $\tilde{B}\simeq Y$.
Since $n\geq-\inf Y=-\inf\tilde{B}$, the kernel $Z^{\tilde{B}}_{n}$ fits in an
exact sequence

$0\to
Z^{\tilde{B}}_{-n}\to\tilde{B}_{-n}\to\tilde{B}_{-n-1}\to\cdots\to\tilde{B}_{t}\to
0.$

By Proposition 4.3 and the isomorphism
${}^{*}\operatorname{Ext}^{i}(J,Z^{\tilde{B}}_{-n})\simeq\mbox{H}_{-(n+i)}(\mathbf{R}^{*}\operatorname{Hom}_{R}(J,Y))=0$
for $i>0$ we get that $Z^{\tilde{B}}_{-n}$ is ∗Gorenstein injective.

Now it is enough to prove that if $I$ and $B$ are left bounded complexes of
respectively ∗injective and ∗Gorenstein injective modules and $B\simeq Y\simeq
I$ then the kernel $Z^{I}_{-n}$ is ∗Gorenstein injective if and only if
$Z^{B}_{-n}$ is so. Let $B$ and $I$ be such complexes. As $I$ consists of
∗injectives by [15, Chapter I, Lemma 4.5] there is a quasi isomorphism
$\pi:B\rightarrow I$ which induces a quasi isomorphism between the complexes
$\pi\supset_{n}:B\supset_{n}\rightarrow I\supset_{n}$. The mapping cone

$\mathcal{M}(\pi\supset_{n})=\cdots\to B_{n-1}\oplus I_{n}\to B_{n}\oplus
I_{n+1}\to B_{n+1}\oplus Z^{B}_{n}\to Z^{I}_{n}\to 0$

is bounded exact such that all modules but the two right-most ones are
∗Gorenstein injective modules. It follows by Theorem 4.2 that $Z^{I}_{n}$ is
∗Gorenstein injective if and only if $B_{n+1}\oplus Z^{B}_{n}$ is so, which is
tantamount to $Z^{B}_{n}$ being ∗Gorenstein injective.

The two equalities are immediate consequences of the equivalence of (1)-(4). ∎

######  Corollary 4.7.

Let $Y\in$ ${}^{*}\mathcal{D}_{\sqsubset}(R)$. Then

${}^{*}\operatorname{Gid}_{R}Y\leq^{*}\operatorname{id}_{R}Y,$

with equality if ${}^{*}\operatorname{id}_{R}Y$ is finite.

###### Proof.

Is the same as [5, Proposition 6.2.6] using Theorem 4.6 and Corollary 3.8. ∎

Recall the _finitistic injective dimension_ of $R$ which defined as

$\operatorname{FID}(R):=\sup\\{\operatorname{id}_{R}M|M\text{ is an
}R\text{-module with }\operatorname{id}_{R}M<\infty\\}.$

The _finitistic Gorenstein injective dimension_ $\mbox{FGID}(R)$, _finitistic
∗injective dimension_ ${}^{*}\operatorname{FID}(R)$ and _finitistic
∗Gorenstein injective dimension_ ${}^{*}\mbox{FGID}(R)$ are define similarly.
In [16, Theorem 2.29], Holm proved that
$\mbox{FGID}(R)=\operatorname{FID}(R)$. Also it is known that over a
commutative Noetherian ring $R$ we have
$\operatorname{FID}(R)\leq\operatorname{dim}R$ by [2, Corollary 5.5] and [17,
II. Theorem 3.2.6].

The following theorem is the graded version of Holm’s result [16, Theorem
2.29]. Its proof is dual of [16, Theorem 2.28]. We give the sketch of proof.

######  Theorem 4.8.

If $R$ is any graded ring
${}^{*}\operatorname{FGID}(R)=^{*}\operatorname{FID}(R)$.

###### Proof.

Using Corollary 4.7 we have
${}^{*}\operatorname{FID}(R)\leq\\!^{*}\operatorname{FGID}(R)$. If $N$ is a
graded $R$-module with $0<\,^{*}\operatorname{Gid}_{R}N<\infty$, then the
graded version of [16, Theorem 2.15] gives a graded $R$-module $C$ such that
${}^{*}\operatorname{id}_{R}C=\,^{*}\operatorname{Gid}_{R}N-1$. Therefore
${}^{*}\operatorname{FGID}(R)\leq\,^{*}\operatorname{FID}(R)+1$. Now for the
reverse inequality
$\\!{}^{*}\operatorname{FGID}(R)\leq\,^{*}\operatorname{FID}(R)$, assume that
$0<\,^{*}\operatorname{FGID}(R)=m<\infty$. Pick a module $N$ with
${}^{*}\operatorname{Gid}_{R}N=m$. Hence the same technique of [18, Lemma
2.2], gives a graded module $T$ such that ${}^{*}\operatorname{id}_{R}T=m$. ∎

######  Corollary 4.9.

Let $Y\in$ ${}^{*}\mathcal{D}_{\sqsubset}(R)$ be a complex of finite
∗Gorenstein injective dimension. Then

${}^{*}\operatorname{Gid}_{R}Y\leq\operatorname{FID}(R)-\inf Y.$

###### Proof.

Note that by $(**)$ in the proof of Theorem 4.6 above we have
$\\!{}^{*}\operatorname{Gid}_{R}Y=\\!^{*}\operatorname{Gid}_{R}M-\inf Y$ for
some graded $R$-module $M$. Thus
$\\!{}^{*}\operatorname{Gid}_{R}Y\leq\\!^{*}\mbox{FGID}(R)-\inf
Y=\\!^{*}\operatorname{FID}(R)-\inf Y\leq\operatorname{FID}(R)-\inf Y$. ∎

Let $D$ be a $\\!{}^{*}$dualizing complex of $R$. Then $D$ is a dualizing
complex of $R$, so we have the Bass category $B(R)$ with respect to $D$ (cf.
[7, Page 237]). It is known that for $Y\in\mathcal{D}_{\sqsubset}(R)$, we have
$Y\in$ $B(R)$ if and only if $\operatorname{Gid}_{R}Y<\infty$, [7, Theorem
4.4].

######  Definition 4.10.

Let $D$ be a $\\!{}^{*}$dualizing complex of $R$. The ∗Bass category
${}^{*}B(R)$ with respect to $D$ is define as:

${}^{*}B(R):=\left\\{Y\in^{*}\mathcal{D}_{\square}(R)\bigg{|}\begin{array}[]{l}\epsilon_{Y}:D\otimes_{R}^{\mathbf{L}}\mathbf{R}^{*}\\!\operatorname{Hom}_{R}(D,Y)\to
Y\text{ is an iso-}\\\ \text{morphism and
}\mathbf{R}^{*}\operatorname{Hom}_{R}(D,Y)\in\\!^{*}\mathcal{D}_{\square}(R)\end{array}\right\\}.$

It is easily seen that a complex $Y\in^{*}\mathcal{D}_{\square}(R)$ is in
$B(R)$ if and only if is in $\\!{}^{*}B(R)$.

######  Theorem 4.11.

Assume that $R$ admits a $\\!{}^{*}$dualizing complex $D$. Then For $Y\in$
${}^{*}\mathcal{D}_{\sqsubset}(R)$ the following are equivalent:

  * (1)

$Y\in\\!^{*}B(R)$.

  * (2)

${}^{*}\operatorname{Gid}_{R}Y<\infty$.

###### Proof.

It is dual to the proof of [7, Theorem 4.1]. In fact the proof uses that if
$N$ is a graded $R$-module satisfying both $N\in$ ${}^{*}B(R)$ and
${}^{*}\operatorname{Ext}^{m}_{R}(J,N)=0$ for all integer $m>0$ and all
∗injective $R$-module $J$, then $N$ is ∗Gorenstein injective, which its proof
is dual to [7, Lemma 4.6]. ∎

######  Theorem 4.12.

Let $(R,\mathfrak{m},k)$ be a ∗local ring that admits a ∗dualizing complex
$D$. For a complex $Y\in$ ${}^{*}\mathcal{D}_{\square}(R)$ of finite
∗Gorenstein injective dimension, we have

$0pt(\mathfrak{m},Y)=0pt(\mathfrak{m},R)+\inf\textbf{R}^{*}\operatorname{Hom}_{R}(\\!^{*}\operatorname{E}_{R}(k),Y).$

###### Proof.

We follow the method of [7, Theorem 6.5]. By Theorem 4.11, $Y\in\\!^{*}B(R)$;
in particular $Y\simeq
D\otimes^{\mathbf{L}}_{R}\textbf{R}^{*}\operatorname{Hom}_{R}(D,Y)$.
Furthermore, we can assume that $D$ is a normalized ∗dualizing complex, so
that by Proposition 3.13 we have
${\mathbf{R}}\Gamma_{\mathfrak{m}}(D)\cong\\!^{*}\operatorname{E}_{R}(k)$ and
that $D_{\mathfrak{m}}$ is a normalized dualizing complex for
$R_{\mathfrak{m}}$. We compute as follows:

$\displaystyle 0pt(\mathfrak{m},Y)=$ $\displaystyle
0pt(\mathfrak{m},D\otimes^{\mathbf{L}}_{R}\textbf{R}^{*}\operatorname{Hom}_{R}(D,Y))$
$\displaystyle=$ $\displaystyle
0pt(\mathfrak{m},D)+0pt(\mathfrak{m},\textbf{R}^{*}\operatorname{Hom}_{R}(D,Y))$
$\displaystyle=$ $\displaystyle\inf
D_{\mathfrak{m}}+\inf\textbf{L}\Lambda^{\mathfrak{m}}\mathbf{R}\\!^{*}\operatorname{Hom}_{R}(D,Y)$
$\displaystyle=$ $\displaystyle
0ptR_{\mathfrak{m}}+\inf\textbf{R}^{*}\operatorname{Hom}_{R}({\mathbf{R}}\Gamma_{\mathfrak{m}}(D),Y)$
$\displaystyle=$ $\displaystyle
0pt(\mathfrak{m},R)+\inf\textbf{R}^{*}\operatorname{Hom}_{R}(\\!^{*}\operatorname{E}_{R}(k),Y).$

The second equality is by [20, Theorem 2.4(b)], the third one by the fact that
$D_{\mathfrak{m}}$ is homologically finite and by [13, Theorem 2.11], and the
forth one by [13, 2.6], and the fact that
$\textbf{R}^{*}\operatorname{Hom}_{R}(D,Y)={\mathbf{R}}\operatorname{Hom}_{R}(D,Y)$.
∎

The ungraded version of the following result is in [7, Proposition 5.5].

######  Proposition 4.13.

Assume that $R$ admits a $\\!{}^{*}$dualizing complex and let $Y\in$
${}^{*}\mathcal{D}_{\square}(R)$. Then for any homogeneous prime ideal
$\mathfrak{p}\in R$ there is an inequality

${}^{*}\operatorname{Gid}_{R_{(\mathfrak{p})}}Y_{(\mathfrak{p})}\leq^{*}\operatorname{Gid}_{R}Y.$

###### Proof.

It is enough to show that if $N$ is a $\\!{}^{*}$Gorenstein injective, then
$N_{(\mathfrak{p})}$ is $\\!{}^{*}$Gorenstein injective over
$R_{(\mathfrak{p})}$. This is similar to the proof of [5, Theorem 6.2.13]
using Corollary 4.9. ∎

The following proposition is the graded version of [8, Lemma 2.1]. For part
$(b)$ we follow the technique of [8, Lemma 2.1]. We present the proof to give
some hints for the graded analogues. Before doing that we need a lemma.

######  Lemma 4.14.

Let $(R,\mathfrak{m})$ be a ∗local non-negatively graded ring. Then the
$\mathfrak{m}$-∗adic completion ${}^{*}\widehat{R}$ of $R$, is a ∗faithfully
flat $R$-module, that is ${}^{*}\widehat{R}$ is $R$-flat and for any graded
$R$-module $M$, $M=0$ if and only if $M\otimes_{R}\,^{*}\widehat{R}=0$.

###### Proof.

It is well known that ${}^{*}\widehat{R}$ is a flat $R$-module by [10,
Corollary 3.3]. Now let $M$ be a graded $R$-module such that
$M\otimes_{R}\,^{*}\widehat{R}=0$. Note that
$\mathfrak{m}=\mathfrak{m}_{0}\oplus R_{1}\oplus R_{2}\oplus\cdots$, where
$\mathfrak{m}_{0}$ is the unique maximal ideal of $R_{0}$. So that
$(^{*}\widehat{R})_{0}=\displaystyle\lim_{\longleftarrow}(R/\mathfrak{m}^{n})_{0}=\displaystyle\lim_{\longleftarrow}R_{0}/\mathfrak{m}_{0}^{n}=\widehat{R_{0}}$,
where $\widehat{R_{0}}$ is the $\mathfrak{m}_{0}$-adic completion of $R_{0}$.
Let $t$ be an integer and set $M_{(t)}:=M_{t}\oplus M_{t+1}\oplus
M_{t+2}\oplus\cdots$, which is a graded submodule of $M$. Hence
$M_{(t)}\otimes_{R}\,^{*}\widehat{R}=0$. Therefore
$(M_{(t)}\otimes_{R}\,^{*}\widehat{R})_{t}=0$. Since
$(M_{(t)}\otimes_{R}\,^{*}\widehat{R})_{t}$ is generated as a
$\mathbb{Z}$-module by $x\otimes r$ for $x\in M_{t}$ and
$r\in(^{*}\widehat{R})_{0}=\widehat{R_{0}}$ and $t$ is arbitrary, we see that
$M\otimes_{R}\widehat{R_{0}}=0$. On the other hand we have
$R_{0}\otimes_{R}\widehat{R_{0}}=R_{0}\otimes_{R}(R_{0}\otimes_{R_{0}}\widehat{R_{0}})=(R_{0}\otimes_{R}R_{0})\otimes_{R_{0}}\widehat{R_{0}}=(R_{0}\otimes_{R_{0}}R_{0})\otimes_{R_{0}}\widehat{R_{0}}=R_{0}\otimes_{R_{0}}\widehat{R_{0}}=\widehat{R_{0}}$.
Hence
$0=M\otimes_{R}\widehat{R_{0}}=(M\otimes_{R_{0}}R_{0})\otimes_{R}\widehat{R_{0}}=M\otimes_{R_{0}}(R_{0}\otimes_{R}\widehat{R_{0}})=M\otimes_{R_{0}}\widehat{R_{0}}$.
Since $\widehat{R_{0}}$ is a faithfully flat $R_{0}$-module, we get that
$M=0$. This completes the proof. ∎

######  Proposition 4.15.

Let $N$ be a ∗Gorenstein injective $R$-module. Then under each of the
following conditions

  * (a)

$R$ admits a ∗dualizing complex; or

  * (b)

$R$ is a non-negatively graded ring,

one has

$0ptR_{\mathfrak{p}}\leqslant 0pt_{R_{\mathfrak{p}}}N_{\mathfrak{p}}$

for every $\mathfrak{p}$ in ${}^{*}\operatorname{Spec}R$, and equality holds
if $\mathfrak{p}$ is a maximal element in ${}^{*}\operatorname{supp}_{R}N$.

###### Proof.

For part $(a)$ let $\mathfrak{p}$ be a homogenous prime ideal. Consider the
∗local ring
$(S,\mathfrak{n}):=(R_{(\mathfrak{p})},\mathfrak{p}R_{(\mathfrak{p})})$. By
Proposition 4.13, $B:=N_{(\mathfrak{p})}$ is a ∗Gorenstein injective
$S$-module. Thus

${}^{*}\operatorname{Ext}^{i}_{S}(\\!^{*}\operatorname{E}_{S}(S/\mathfrak{n}),B)=0$

for all $i>0$, that is
$\inf\textbf{R}^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{E}_{S}(S/\mathfrak{n}),B)\geq
0$. On the other hand since $B_{\mathfrak{n}}\cong N_{\mathfrak{p}}$ and
$S_{\mathfrak{n}}\cong R_{\mathfrak{p}}$, by Theorem 4.12 we have

$0ptN_{\mathfrak{p}}=0ptR_{\mathfrak{p}}+\inf\textbf{R}^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{E}_{S}(S/\mathfrak{n}),B).$

Thus $0ptR_{\mathfrak{p}}\leqslant 0pt_{R_{\mathfrak{p}}}N_{\mathfrak{p}}$.

Now if $0pt_{R_{\mathfrak{p}}}N_{\mathfrak{p}}$ is finite, we have
$\inf\textbf{R}^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{E}_{S}(S/\mathfrak{n}),B)<\infty$.
Using Lemma 4.5 we have

$\textbf{R}^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{E}_{S}(S/\mathfrak{n}),B)\simeq^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{E}_{S}(S/\mathfrak{n}),B).$

Therefore the infimum must be zero. This proves the second statement.

For $(b)$ assume that $\mathfrak{p}$ is a homogeneous prime ideal and $T$ is a
graded $R_{(\mathfrak{p})}$-module with
$\,{}^{*}\operatorname{pd}_{R_{(\mathfrak{p})}}T<\infty$. A standard dimension
shifting argument shows that
$\,{}^{*}\operatorname{Ext}^{i}_{R_{(\mathfrak{p})}}(T,N_{(\mathfrak{p})})=0$
for all $i>0$. Set $d=0pt(\mathfrak{p}R_{(\mathfrak{p})},R_{(\mathfrak{p})})$
and choose a homogeneous maximal $R_{(\mathfrak{p})}$-regular sequence ${\bf
x}$ in $\mathfrak{p}R_{(\mathfrak{p})}$ by [3, Proposition 1.5.11]. Since the
∗projective dimension of $R_{(\mathfrak{p})}/({\bf x})$ is finite we have

$\displaystyle 0\leq$
$\displaystyle\inf\mathbf{R}\\!^{*}\operatorname{Hom}_{R_{(\mathfrak{p})}}(R_{(\mathfrak{p})}/({\bf
x}),N_{(\mathfrak{p})})$ $\displaystyle\leq$ $\displaystyle
0pt(\mathfrak{p}R_{(\mathfrak{p})},\mathbf{R}\\!^{*}\operatorname{Hom}_{R_{(\mathfrak{p})}}(R_{(\mathfrak{p})}/({\bf
x}),N_{(\mathfrak{p})}))$ $\displaystyle=$ $\displaystyle
0pt(\mathfrak{p}R_{(\mathfrak{p})},N_{(\mathfrak{p})})-d=0pt_{R_{\mathfrak{p}}}N_{\mathfrak{p}}-d,$

where the second inequality holds by [6, 4.3], and the first equality follows
from Proposition 2.4.

Now let $\mathfrak{p}$ be a maximal element in
${}^{*}\operatorname{supp}_{R}N$. Set $S=R_{(\mathfrak{p})}$ which is a ∗local
ring with depth $d$, homogeneous maximal ideal
$\mathfrak{n}=\mathfrak{p}R_{(\mathfrak{p})}$, $B=N_{(\mathfrak{p})}$, and
$l=R_{(\mathfrak{p})}/\mathfrak{p}R_{(\mathfrak{p})}$. One has
$\mathfrak{n}\in\operatorname{supp}_{S}B$ and by exactly the same method of
proof of [8, Lemma 2.1] we have
$\,{}^{*}\operatorname{Ext}^{i}_{S}(T,B)=0=\,^{*}\operatorname{Ext}^{i}_{S}(E,B)$
for all $i>0$ and every graded $S$-module $T$ with
$\,{}^{*}\operatorname{pd}_{S}T$ finite and
$E:=\\!^{*}\operatorname{E}_{S}(l)$.

Let $K$ denote the Koszul complex on a homogeneous system of generators for
$\mathfrak{n}$. Since $\mbox{H}_{i}(K\otimes_{S}E)$ are Artinian, and by [3,
Corollary 1.6.13], we have $\mathfrak{n}\mbox{H}_{i}(K\otimes_{S}E)=0$, we see
that $\mbox{H}_{i}(K\otimes_{S}E)$ are finitely generated. So there is a
resolution $L\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}K\otimes_{S}E$
by finitely generated free S-modules. Using the ∗Hom-evaluation we have

$K\otimes_{S}(E\otimes^{\mathbf{L}}_{S}\\!{}^{*}\operatorname{Hom}_{S}(E,B))\simeq
L\otimes_{S}\\!^{*}\operatorname{Hom}_{S}(E,B)\cong\\!^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{Hom}_{S}(L,E),B).$

The free resolution above induces a quasiisomorphism $\alpha$ from
$\,{}^{*}\operatorname{Hom}_{S}(K\otimes_{S}E,E)\cong\\!^{*}\operatorname{Hom}_{S}(K,\\!^{*}\widehat{S})$
to $\,{}^{*}\operatorname{Hom}_{S}(L,E)$. The mapping cone
$C=\mathcal{M}(\alpha)$ is a bounded complex of direct sums of
${}^{*}\widehat{S}$ and $E$. Thus
$\,{}^{*}\operatorname{Ext}^{i}_{S}(C_{j},B)=0$ for all $i>0$ and all $j$.
Hence, an application of $\,{}^{*}\operatorname{Hom}_{S}(-,B)$ yields a
quasiisomorphism

$\,{}^{*}\operatorname{Hom}_{S}(\alpha,B):\\!^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{Hom}_{S}(L,E),B)\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}\\!^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{Hom}_{S}(K,\\!^{*}\widehat{S}),B).$

The modules in the complex $\,{}^{*}\operatorname{Hom}_{S}(K,^{*}\widehat{S})$
are Ext-orthogonal to the modules in the mapping cone of an injective
resolution $B\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}H$. Therefore,
one has

$\\!{}^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{Hom}_{S}(K,\\!^{*}\widehat{S}),B)\simeq\\!^{*}\operatorname{Hom}_{S}(\\!^{*}\operatorname{Hom}_{S}(K,\\!^{*}\widehat{S}),H)$

by the graded analogue of [7, Lemma 2.4]. Now piece together the last three
quasiisomorphisms, and use ∗Hom-evaluation to obtain

$K\otimes_{S}(E\otimes^{\mathbf{L}}_{S}\\!{}^{*}\operatorname{Hom}_{S}(E,B))\simeq
K\otimes_{S}\mathbf{R}^{*}\operatorname{Hom}_{S}(\\!^{*}\widehat{S},B).$

Therefore by [6, (4.2) and (4.11)], the complexes
$E\otimes^{\mathbf{L}}_{S}\\!{}^{*}\operatorname{Hom}_{S}(E,B)$ and
$\mathbf{R}^{*}\operatorname{Hom}_{S}(\\!^{*}\widehat{S},B)$ have the same
width. From [20, Theorem 2.4(b)] and Proposition 2.4 we have

$0pt(\mathfrak{n},E)+0pt(\mathfrak{n},\,^{*}\operatorname{Hom}_{S}(E,B))=0pt(\mathfrak{n},B).$

Now we can see that $0pt(\mathfrak{n},\,^{*}\operatorname{Hom}_{S}(E,B))=0$
(see proof of [8, Lemma 2.1]). Consequently
$0pt_{R_{\mathfrak{p}}}N_{\mathfrak{p}}=0pt(\mathfrak{n},B)=0pt(\mathfrak{n},E)=0pt(\mathfrak{n},\\!^{*}\widehat{S})$
using Proposition 2.2. Now since $\\!{}^{*}\widehat{S}$ is a flat $S$-module
we have

$\operatorname{Ext}^{i}_{S}(S/\mathfrak{n},\\!^{*}\widehat{S})\cong\operatorname{Ext}^{i}_{S}(S/\mathfrak{n},S)\otimes_{S}\\!^{*}\widehat{S}.$

Keep in mind that
$\operatorname{Ext}^{i}_{S}(S/\mathfrak{n},S)=\,^{*}\operatorname{Ext}^{i}_{S}(S/\mathfrak{n},S)$
is a graded $S$-module. Hence using Lemma 4.14 we have
$\operatorname{Ext}^{i}_{S}(S/\mathfrak{n},\\!^{*}\widehat{S})=0$ if and only
if $\operatorname{Ext}^{i}_{S}(S/\mathfrak{n},S)=0$. Therefore
$0pt(\mathfrak{n},\\!^{*}\widehat{S})=0pt(\mathfrak{n},S)=d$. ∎

######  Lemma 4.16.

Let $X$ be a complex of graded $R$-modules. For any homogeneous surjective
homomorphism $i:Y_{n}\longrightarrow X_{n}$ of graded $R$-modules there is a
commutative diagram

$\textstyle{Y=\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n+2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\scriptstyle{=}$$\textstyle{Y_{n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\scriptstyle{i^{\prime}}$$\textstyle{Y_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{i}$$\textstyle{X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{=}$$\textstyle{\cdots}$$\textstyle{X=\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n+2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma^{\prime}}$$\textstyle{X_{n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta^{\prime}}$$\textstyle{X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha^{\prime}}$$\textstyle{X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots}$

such that $Y$ is a complex of graded $R$-modules, that $i^{\prime}$ is
surjective and $\operatorname{Ker}i\cong\operatorname{Ker}i^{\prime}$ and the
induced map $\mbox{H}(Y)\longrightarrow\mbox{H}(X)$ is an isomorphism.

###### Proof.

Let $\alpha=\alpha^{\prime}i$ and $\beta:Y_{n+1}\longrightarrow Y_{n}$ be the
pullback of $\beta^{\prime}$ along $i$ thus

$Y_{n+1}=\\{(x,y)|x\in X_{n+1},y\in Y_{n}\text{ and
}\beta^{\prime}(x)=i(y)\\}.$

It is clear that $Y_{n+1}$ is a graded $R$-module. Let $i^{\prime}:Y_{n+1}\to
X_{n+1}$ be a map defined by $i^{\prime}(x,y)=x$ which is seen to be
surjective. Define $\gamma:X_{n+2}\to Y_{n+1}$ by
$\gamma(x)=(\gamma^{\prime}(x),0)$ for every $x\in X_{n+2}$. It is clear that
$Y$ is a complex and $\operatorname{Ker}i\cong\operatorname{Ker}i^{\prime}$.
Therefore the induced map $\mbox{H}(Y)\longrightarrow\mbox{H}(X)$ is an
isomorphism. ∎

######  Corollary 4.17.

Let $Y\in$ ${}^{*}\mathcal{D}_{\square}(R)$ such that
${}^{*}\operatorname{Gid}_{R}Y=g>0$. Then there is an exact triangle $G\to
I\to Y\to\Sigma G$ such that $G$ is a ∗Gorenstein injective module and $I$ is
a complex of graded $R$-modules such that ${}^{*}\operatorname{id}_{R}I=g$.

###### Proof.

Without loss of generality we can assume that $Y$ has the form

$0\to I_{0}\to I_{-1}\to\cdots\to I_{-g+1}\to B_{-g}\to 0,$

where the $I_{j}$s are ∗injective and $B_{-g}$ is ∗Gorenstein injective. By
definition $B_{-g}$ is homomorphic image of some $\\!{}^{*}$injective module
$J_{0}$. Thus by Lemma 4.16 we have a commutative diagram

---
$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{-g+2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G_{-g+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i^{\prime}}$$\textstyle{J_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{-g+2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{-g+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{-g}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$$\textstyle{0}$$\textstyle{0}$

such that $\operatorname{Ker}i\cong\operatorname{Ker}i^{\prime}$. Since
$\operatorname{Ker}i$ is a ∗Gorenstein injective module, using Theorem 4.2,
$G_{-g+1}$ is also ∗Gorenstein injective. By repeating this argument $g$ times
we get the desired exact triangle $G\to I\to Y\to\Sigma G$. ∎

The following equality extends Chouinard’s formula [4] and [8, Theorem 2.2] to
the ∗Gorenstein injective dimension.

######  Theorem 4.18.

Assume that $R$ admits a $\\!{}^{*}$dualizing complex or $R$ is a non-
negatively graded ring. Let $Y\in$ ${}^{*}\mathcal{D}_{\square}(R)$ of finite
∗Gorenstein injective dimension. Then there is an equality

${}^{*}\operatorname{Gid}_{R}Y=\sup\\{0ptR_{\mathfrak{p}}-0pt_{R_{\mathfrak{p}}}Y_{\mathfrak{p}}|\mathfrak{p}\in^{*}\operatorname{Spec}R\\}.$

###### Proof.

The argument is similar to [8, Theorem 2.2], just use Proposition 4.15 and
Corollary 4.17. ∎

In the following corollary we compare the ∗Gorenstein injective dimension with
the usual Gorenstein injective dimension.

######  Corollary 4.19.

Let $Y\in$ ${}^{*}\mathcal{D}_{\square}(R)$.Then under each of the following
conditions

  * (a)

$R$ admits a ∗dualizing complex; or

  * (b)

$R$ is a non-negatively graded ring, and ${}^{*}\operatorname{Gid}_{R}Y$ and
$\operatorname{Gid}_{R}Y$ are finite,

one has

${}^{*}\operatorname{Gid}_{R}Y\leq\operatorname{Gid}_{R}Y\leq^{*}\operatorname{Gid}_{R}Y+1.$

###### Proof.

If $R$ admits a ∗dualizing complex, then Theorem 4.11, implies that
${}^{*}\operatorname{Gid}_{R}Y<\infty$, if and only if
$\operatorname{Gid}_{R}Y<\infty$. In each both cases the inequalities follow
from Theorem 4.18 and [8, Theorem 2.2]. ∎

## References

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