# The Grothendieck Ring of Certain Non-Noetherian Group-Graded Algebras via
$K$-Series

Nathaniel Gallup

###### Abstract.

With the goal of computing the Grothendieck group of certain multigraded
infinite polynomial rings and the $K$-series of infinite matrix Schubert
spaces, we introduce a new type of $\Gamma$-graded $k$-algebra (which we call
a PDCF algebra) and a new type of graded module (a BDF module) over said
algebra. Since infinite polynomial rings are not Noetherian and the modules of
interest are not finitely generated, we compensate by requiring certain
finiteness properties of the grading group $\Gamma$ and of the grading itself.
If $R$ is a PDCF $\Gamma$-graded $k$-algebra, we prove that every projective
BDF $\Gamma$-graded $R$-module is free, and that when the graded maximal ideal
of $R$ is generated by a regular sequence, a BDF analog of the Hilbert Syzygy
Theorem holds: every BDF $R$-module has a free resolution of BDF $R$-modules
which, though not finite, has the property that each graded piece is
eventually zero. We use this to show that the Grothendieck group of projective
BDF $R$-modules (defined similarly to $K_{0}(R)$) is isomorphic to the
Grothendieck group of all BDF $R$-modules (defined similarly to $G_{0}(R)$).
We describe this Grothendieck group explicitly by using $K$-series to give an
isomorphism with a certain space of formal Laurent series. Finally we give a
BDF version of Serre’s formula for the product in the Grothendieck group,
making it into a ring.

##  1\. Introduction

In [Gal21] we introduced a type of infinite-dimensional flag space
$\text{Fl}_{\mathbb{N}}$ which is a $k$-functor (i.e. a functor from the
category of $k$-algebras to the category of sets, for $k$ a field). This
functor is covered disjointly by Schubert subfunctors $X^{\circ}_{\sigma}$
which are indexed by permutations $\sigma$ of $\mathbb{N}$, and which are
quotients $\text{sh}(B\sigma B)/B$ of (the fppf-sheafification of) the coset
spaces $B\sigma B$. The affine analogs of these spaces, called matrix Schubert
spaces, are quotients of a Cox ring with infinitely many columns (i.e. a
polynomial ring $k[x_{i,j}]_{n}$ where $n_{1},n_{2},\ldots$ is an increasing
sequence of natural numbers, and $i,j\in\mathbb{N}$ are such that $i\leq
n_{j}$) by an ideal $\mathfrak{d}_{\sigma}$ generated by certain minors. There
is a torus action of $\mathbb{Z}^{(\mathbb{N})}$ (the abelian group of
$\mathbb{N}$-indexed integer sequences with finitely many nonzero entries) on
the $k$-functor represented by $k[x_{i,j}]_{n}$ which corresponds to a
multigrading of $k[x_{i,j}]_{n}$ by $\mathbb{Z}^{(\mathbb{N})}$ and since the
ideal $\mathfrak{d}_{\sigma}$ is also multigraded, it corresponds to an
equivariant sheaf on this $k$-functor. This makes
$k[x_{i,j}]_{n}/\mathfrak{d}_{\sigma}$ into a
$\mathbb{Z}^{(\mathbb{N})}$-graded $k[x_{i,j}]_{n}$-module. The goal of this
paper is to compute the $K$-theory (i.e. Grothendieck group) of this
multigraded polynomial ring. A forthcoming paper will then compute the
$K$-series of $k[x_{i,j}]_{n}/\mathfrak{d}_{\sigma}$.

Much of the classical theory in the case of finite-dimensional flag varieties
(see [MS05] for a thorough treatment) uses the fact that a polynomial ring in
finitely many variables over a field is Noetherian and hence the matrix
Schubert variety $k[x_{i,j}\mid 1\leq i,j\leq n]/\mathfrak{d}_{\sigma}$ (with
$\sigma\in S_{n}$) is finitely generated. But of course polynomial rings in
infinitely many variables are not Noetherian and the ideal
$\mathfrak{d}_{\sigma}$ in this case is not finitely generated. Miraculously,
however, the $\mathbb{Z}^{(\mathbb{N})}$-grading we care about has some other
finiteness properties which still allow us to define a well-behaved graded
Grothendieck group containing these modules. In particular, every graded piece
of $k[x_{i,j}]_{n}$ is finite-dimensional (such rings and modules are called
_finite-dimensionally graded_), the $0$th graded piece of $k[x_{i,j}]_{n}$ is
just $k$ (such graded rings are called _connected_), the graded support of
$k[x_{i,j}]_{n}$ is $\mathbb{Z}_{\geq 0}^{(\mathbb{N})}$ which is a pointed
submonoid of $\mathbb{Z}^{(\mathbb{N})}$, and for every
$\alpha\in\mathbb{Z}^{(\mathbb{N})}$, there are only finitely many
$\beta\in\mathbb{Z}^{(\mathbb{N})}$ which are in the support of
$k[x_{i,j}]_{n}$ and satisfy $\beta_{i}\leq\alpha_{i}$ for every
$i\in\mathbb{N}$ (we call this property _downward finite_). Finally, the
graded maximal ideal $\bigoplus_{\alpha>0}(k[x_{i,j}]_{n})_{\alpha}$ is
generated by the regular sequence consisting of all of the variables
$x_{i,j}$.

It turns out that these properties are all that is needed define a
Grothendieck group of a (not necessarily Noetherian) graded $k$-algebra which
is an integral domain. Specifically, if $\Gamma$ is an abelian group and $R$
is such a $\Gamma$-graded $k$-algebra with support $Q\subseteq\Gamma$, then we
say that $R$ is _pointed_ if $q+q^{\prime}=0$ implies $q=q^{\prime}=0$ for
$q,q^{\prime}\in Q$, _downward finite_ if for all $g\in\Gamma$ the set
$\\{h\in Q\mid h\leq_{Q}g\\}$ is finite (the partial order on $\Gamma$ is
defined by setting $Q$ to be the nonnegative elements), _connected_ if
$R_{0}=k$, and _finite-dimensionally graded_ if $\dim_{k}(R_{g})$ is finite
for all $g\in\Gamma$ (this condition is needed to define Hilbert series and
$K$-series and is automatic if $R$ is finitely generated as a $k$-algebra). We
call algebras satisfying these conditions _PDCF $\Gamma$-graded $k$-algebras_.
Furthermore if $M$ is a (not necessarily finitely generated) $\Gamma$-graded
$R$-module, we say that it is _finitely bounded below_ if its $\Gamma$-support
is bounded below by finitely many elements (the purpose of this condition is
to guarantee closure of our category under tensor product), and we call such a
module $M$ a _BDF $\Gamma$-graded $R$-module_ if additionally its support is
downward finite and it is also finite-dimensionally graded (these conditions
are automatic for finitely generated modules).

BDF $\Gamma$-graded $R$-modules over a PDCF $\Gamma$-graded $k$-algebra share
a shocking number of properties with finitely generated $\mathbb{Z}$-graded
modules over a Noetherian $\mathbb{Z}_{\geq 0}$-graded $k$-algebra.
Preliminarily, the category $R\text{-}\textbf{Mod}^{\text{BDF}}_{\Gamma}$ of
such modules is abelian and closed under tensor product in the category of all
$R$-modules. Furthermore, recall that the Quillen–Suslin theorem ([Qui76],
[Sus76]) says that every finitely-generated projective module over a
polynomial ring is free. If one is satisfied with just finitely generated
graded modules over a positively graded ring, then the proof that all such
projective modules are free is substantially easier (see [MM11], Section 2).
We use a similar technique to that found in the latter reference to prove the
following analogous result for BDF modules.

######  Theorem.

Every projective BDF $\Gamma$-graded $R$-module is free.

Recall the Hilbert Syzygy Theorem states that every finitely generated module
over a Noetherian regular local ring or over a polynomial ring in finitely
many variables has a finite free resolution. This is of course not true for
BDF modules, but using the existence of minimal free resolutions of BDF
modules, we adapt the proof given in [Eis13] (which uses $\operatorname{Tor}$)
to obtain the following analog.

######  Theorem (BDF Hilbert Syzygy Theorem).

If $R$ is a PDCF $\Gamma$-graded $k$-algebra with graded maximal ideal
generated by a regular sequence, then every BDF $\Gamma$-graded $R$-module has
a resolution of BDF $\Gamma$-graded free modules with the property that each
graded piece is eventually zero.

Classically, there are two common constructions of the Grothendieck group of a
Noetherian ring. In the first, $K_{0}(R)$ is defined to be the Grothendieck
group completion of the set of isomorphism classes of finitely generated
projective $R$-modules which is a commutative monoid under direct sum. In the
second, $G_{0}(R)$ is defined to be the free abelian group on the set of
isomorphism classes of all finitely generated $R$-modules, modulo the relation
$[M]-[N]+[L]=0$ whenever there is a short exact sequence $0\to M\to N\to L\to
0$. Then there is an obvious map $K_{0}(R)\to G_{0}(R)$, and when $R$ is
either Noetherian regular local or a polynomial ring in finitely many
variables, then the existence of a finite projective resolution gives an
inverse to this map (see [Wei13]).

We make analogous definitions for BDF modules and prove the analogous result.
Specifically we define $K^{\text{BDF}}_{\Gamma,0}(R)$ to be the Grothendieck
group completion of the set of isomorphism classes of BDF $\Gamma$-graded
projective $R$-modules, and $G^{\text{BDF}}_{\Gamma,0}(R)$ to be the free
abelian group on the set of isomorphism classes of all BDF $\Gamma$-graded
$R$-modules, modulo the same relation as before. We then prove:

######  Theorem.

If $R$ is a PDCF $\Gamma$-graded $k$-algebra with graded maximal ideal
generated by a regular sequence, the map $K^{\text{BDF}}_{\Gamma,0}(R)\to
G^{\text{BDF}}_{\Gamma,0}(R)$ is an isomorphism.

Recall that homotopy invariance tells us that the Grothendieck ring $G_{0}(S)$
of a polynomial ring $S=k[x_{1},\ldots,x_{n}]$ in finitely many variables is
isomorphic to $\mathbb{Z}$. If one is interested instead in graded modules
over $S$ with the standard $\mathbb{Z}$-grading, then the map
$G_{\mathbb{Z},0}(S)\to\mathbb{Z}[t,t^{-1}]$ given by sending $[M]$ to its
$K$-polynomial is an isomorphism, where the latter denotes all Laurent
polynomials. More generally, if $S$ is $\mathbb{Z}^{d}$-multigraded with
graded support $Q$, the same $K$-polynomial map
$G_{\mathbb{Z}^{d},0}(S)\to\mathbb{Z}[Q][\mathbb{Z}^{d}]$ is again an
isomorphism, where the latter denotes all Laurent polynomials in variables
indexed by $\mathbb{Z}^{d}$ (see [MS05] Chapter 8). We give the following
analogous result for BDF modules, with $\mathbb{Z}[Q][\mathbb{Z}^{d}]$
replaced by $\mathbb{Z}[[Q]]\\{\Gamma\\}$, the formal Laruent series with
variables indexed by $\Gamma$ still with finitely bounded below, but now also
with downward finite, support. Note that in the Noetherian case,
$K$-polynomials are honest Laurent polynomials, but because our modules are
not finitely generated, we must pass to $K$-series.

######  Theorem.

The map $K^{\text{BDF}}_{\Gamma,0}(R)\to\mathbb{Z}[[Q]]\\{\Gamma\\}$, which
sends $[F]$ to its $K$-series, is an isomorphism.

Classically, the tensor product makes $K_{0}(R)$ into a ring (called the
Grothendieck ring) and the corresponding product in $G_{0}(R)$ is given by
Serre’s formula
$[M]\cdot[N]=\sum_{i=0}^{\infty}(-1)^{i}[\operatorname{Tor}_{i}(M,N)]$ (see
[Ser00]), the latter sum begin finite by the Hilbert Syzygy Theorem. For BDF
modules, the tensor product does still make $K^{\text{BDF}}_{\Gamma,0}(R)$
into a ring, but in order to interpret Serre’s formula, we must make sense of
an in infinite sum of elements of $G^{\text{BDF}}_{\Gamma,0}(R)$. To that end
we define a certain directed filtration of subgroups on
$G^{\text{BDF}}_{\Gamma,0}(R)$ making it into a linear topological group, and
show that the sum $\sum_{i=0}^{\infty}(-1)^{i}[\operatorname{Tor}_{i}(M,N)]$
converges. Finally we prove the following theorem:

######  Theorem.

The isomorphisms $\mathbb{Z}[[Q]]\\{\Gamma\\}\cong
K^{\text{BDF}}_{\Gamma,0}(R)\cong G^{\text{BDF}}_{\Gamma,0}(R)$ are ring
isomorphisms with the product in the first given by the usual product of
formal Laurent series, in the second given by the tensor product, and in the
third given by Serre’s formula.

##  2\. Background

In this section we fix notation and give some necessary background on graded
rings and graded modules. See also [Haz16] and [NVO04] for excellent
treatments of this material (and much more).

###  2.1. Families

In this paper, a _family_ of elements of a set $X$ is a function $a:I\to X$
where $I$ is an index set. We denote the image of $i\in I$ under $a$ by
$a_{i}$ and, to distinguish families from sets, we denote this family by
$(a_{i}\mid i\in I)$.

###  2.2. Graded Rings

Throughout this paper $k$ will denote a field and $\Gamma$ will denote an
abelian group. A _$\Gamma$ -grading_ of a commutative $k$-algebra $R$ is a
function which assigns to every element $g\in\Gamma$ a $k$-submodule
$R_{g}\subseteq R$ such that $R=\bigoplus_{g\in\Gamma}R_{g}$ and for all
$g,h\in\Gamma$ we have $R_{g}R_{h}\subseteq R_{g+h}$. A _$\Gamma$ -graded
$k$-algebra_ is a $k$-algebra $R$ together with a $\Gamma$-grading of $R$.
Hence we can write any element $r\in R$ as a sum $r=\sum_{g\in\Gamma}r_{g}$
where $r_{g}\in R_{g}$ and only finitely many of the $r_{g}$’s are nonzero.
Sine we will never put a topology on $R$ and therefore cannot take sums with
more than finitely many nonzero terms, we will often fail to mention the
nonzero condition explicitly. Throughout this paper $R$ will denote a
$\Gamma$-graded $k$-algebra which, for reasons explained in Section 3, we will
always assume is an integral domain (unless otherwise stated).

######  Example 1.

When $\Gamma=\mathbb{Z}$ we obtain the usual notion of a graded ring.

######  Proposition 1.

$R_{0}$ is a subalgebra of $R$.

###### Proof.

By definition $R_{0}$ is a $k$-subspace and we also have $R_{0}\cdot
R_{0}\subseteq R_{0}$ hence $R_{0}$ is closed under multiplication.
Furthermore we can write $1=\sum_{g\in\Gamma}r_{g}$ where $r_{g}\in R_{g}$ and
so for any $h\in\Gamma$ we have $r_{h}=r_{h}\cdot
1=\sum_{g\in\Gamma}r_{h}r_{g}$. The LHS is in $R_{h}$, so the RHS must also be
by the uniqueness of the direct sum. But $r_{h}r_{g}\in R_{h+g}$ and the
latter is equal to $R_{h}$ if and only if $g=0$. Therefore $r_{g}r_{h}=0$ for
all $g\neq 0$ and $r_{0}r_{h}=r_{h}$. In particular we have

$\displaystyle r_{0}=r_{0}\cdot
1=\sum_{g\in\Gamma}r_{0}r_{g}=\sum_{g\in\Gamma}r_{g}=1.$

Hence $1\in R_{0}$ and so $R_{0}$ is a subalgebra as desired. ∎

###  2.3. Graded Modules

A _$\Gamma$ -grading_ of an $R$-module $M$ is a function which assigns to
every element $g\in\Gamma$ a $k$-submodule $M_{g}\subseteq M$ such that
$M=\bigoplus_{g\in\Gamma}M_{g}$ and for all $g,h\in\Gamma$ we have
$R_{g}M_{h}\subseteq M_{g+h}$. In particular $R$ is itself a $\Gamma$-graded
$R$-module. Elements of $M$ which are in $M_{g}$ for some $g\in\Gamma$ are
called _homogeneous of degree $g$_ and if $m\in M_{g}$ we write $\deg(m)=g$.

One can shift the grading of any graded module: if $M$ is a $\Gamma$-graded
$R$-module and $g\in\Gamma$, then we denote by $M(g)$ the $\Gamma$-graded
$R$-module which is equal to $M$ as an $R$-module but whose grading is given
by $M(g)_{h}=M_{g+h}$.

######  Proposition 2.

Let $M$ be a $\Gamma$-graded $R$-module. The following are equivalent for a
submodule $N\subseteq M$.

  1. (1)

$N$ is generated by homogeneous elements.

  2. (2)

$N=\bigoplus_{g\in\Gamma}N\cap M_{g}$.

  3. (3)

For any $n\in N$ if we write $n=\sum_{g\in\Gamma}m_{g}$ where $m_{g}\in M_{g}$
then $m_{g}\in N$ for all $g\in\Gamma$.

###### Proof.

((1) $\implies$ (2)). By hypothesis the family $(N\cap M_{g}\mid g\in\Gamma)$
is independent, and if $N$ is generated by homogeneous elements, then every
element of $N$ can be written as a sum of homogeneous elements of $N$, so this
family spans as well.

((2) $\implies$ (3)). If we have $n=\sum_{g\in\Gamma}m_{g}$ where $m_{g}\in
M_{g}$, then because the family $(N\cap M_{g}\mid g\in\Gamma)$ spans $M$, we
can also write $n=\sum_{g\in\Gamma}n_{g}$ where $n_{g}\in N\cap M_{g}$.
Subtraction yields $0=\sum_{g\in\Gamma}m_{g}-n_{g}$ with $m_{g}-n_{g}\in
M_{g}$. But because the family $(M_{g}\mid g\in\Gamma)$ is independent, it
follows that $m_{g}=n_{g}\in N\cap M_{g}$ as desired.

((3) $\implies$ (1)). Given $n\in N$, since $M=\bigoplus_{g\in\Gamma}M_{g}$,
we can write $n=\sum_{g\in\Gamma}m_{g}$ where $m_{g}\in M_{g}$. Then by
hypothesis we have $m_{g}\in N\cap M_{g}$, so $N$ is generated by the
homogeneous elements $N\cap M_{g}$ for $g\in\Gamma$. ∎

A submodule which satisfies the conditions of Proposition 2 is called
_$\Gamma$ -graded_ or just _graded_ for short. An ideal of $R$ is _$\Gamma$
-graded_ if it is as a submodule of $R$. We now collect a technical results
that we will need later.

######  Proposition 3.

Let $M$ be a $\Gamma$-graded $R$-module, and $N\subseteq M$ a graded
submodule. Then there is a $\Gamma$-grading of $M/N$ given by
$M/N=\bigoplus_{g\in\Gamma}(M_{g}+N)/N$, where we identify $(M_{g}+N)/N$ with
the submodule of $M/N$ spanned by elements of the form $m_{g}+N$ for $m_{g}\in
M_{g}$.

###### Proof.

Given $g,h\in\Gamma$, it is clear that $(M_{h}+N)/N$ is a $k$-subspace of
$M/N$ and that $R_{g}\cdot(M_{g}+N)/N\subseteq(M_{g+h}+N)/N$. Furthermore it
is also obvious that $M/N=\sum_{g\in\Gamma}(M_{g}+N)/N$. To show that the
family $((M_{g}+N)/N\mid g\in\Gamma)$ of $k$-subspaces of $M/N$ is
independent, we compute:

$0=\sum_{g\in\Gamma}m_{g}+N=\left[\sum_{g\in\Gamma}m_{g}\right]+N\implies\sum_{g\in\Gamma}m_{g}\in
N$

Because $N$ is graded, for all $g\in\Gamma$ it follows that $m_{g}\in N$ and
hence $m_{g}+N=0$ as desired. ∎

######  Proposition 4.

Let $M$ be a $\Gamma$-graded $R$-module, and $I\subseteq R$ a graded ideal.
Then $IM$ is a graded submodule of $M$.

###### Proof.

Given $x=\sum_{j=1}^{\ell}i_{j}m_{j}\in IM$, because $I$ is graded, we can
write $i_{j}=\sum_{g\in\Gamma}(i_{j})_{g}$ and because $M$ is graded we can
write $m_{j}=\sum_{g\in\Gamma}(m_{j})_{g}$. Then we can write
$x=\sum_{j=1}^{\ell}\sum_{g,h\in\Gamma}(i_{j})_{g}(m_{j})_{h}$ and
$(i_{j})_{g}(m_{j})_{g}\in IM\cap M_{g+h}$, hence $IM$ is generated by
homogeneous elements of the form $im$ where $i\in I$ is homogeneous and $m\in
M$ is homogeneous, so it is a graded submodule. ∎

###  2.4. Homomorphisms of Graded Modules

If $M$ and $N$ are two $\Gamma$-graded $R$-modules we say that an $R$-module
homomorphism $\varphi:M\to N$ is _$\Gamma$ -graded_ (or just _graded_) if for
all $g\in\Gamma$ we have $\varphi(M_{g})\subseteq N_{g}$.

######  Proposition 5.

Let $M,N$ be $\Gamma$-graded $R$-modules and $\varphi:M\to N$ be a graded
$R$-module homomorphism. Then $\ker\varphi$ is a graded submodule of $M$ and
$\operatorname{im}\varphi$ is a graded submodule of $N$. Furthermore if
$y_{g}\in(\operatorname{im}\varphi)_{g}$ for some $g\in\Gamma$ then there
exists $x_{g}\in M_{g}$ such that $\varphi(x_{g})=y_{g}$.

###### Proof.

If $x\in\ker\varphi$ and we write $x=\sum_{g\in\Gamma}x_{g}$ then we have
$0=\varphi(x)=\sum_{g\in\Gamma}\varphi(x_{g})$ and since $\varphi$ is graded,
we have $\varphi(x_{g})\in N_{g}$. Because the family $(N_{g}\mid g\in\Gamma)$
is independent, it follows that $\varphi(x_{g})=0$, and hence $x_{g}\in\ker f$
as desired.

If $y\in\operatorname{im}\varphi$ and we write $y=\sum_{g\in\Gamma}y_{g}$,
there exists some $x\in M$ such that $\varphi(x)=y$. We can also write
$x=\sum_{g\in\Gamma}x_{g}$ and then we have
$\sum_{g\in\Gamma}y_{a}=y=\varphi(x)=\sum_{g\in\Gamma}\varphi(x_{g})$. Since
$\varphi$ is graded, we have $\varphi(x_{g})\in N_{g}$, so because the family
of subspaces $(N_{g}\mid g\in\Gamma)$ is independent, it follows that
$y_{g}=\varphi(x_{g})$ for all $g\in\Gamma$ so
$y_{g}\in\operatorname{im}\varphi$ as desired.

Taking $y=y_{g}$ in the previous paragraph, we obtain that
$y_{g}=\varphi(x_{g})$ and $\varphi(x_{h})=0$ for $h\neq g$ proving the final
statement. ∎

###  2.5. Multigraded Polynomial Rings

In this paper we will mainly be concerned with $\Gamma$-graded polynomial
rings $S:=k[x_{i}\mid i\in I]$ such that the monomials $x^{\alpha}$ (where
$\alpha\in\mathbb{Z}_{\geq 0}^{(I)}$) are all homogeneous (a _$\Gamma$
-multigrading_). Note that the degree of a monomial in the multigraded setting
is completely determined by the degrees of the variables themselves.

######  Example 2.

The main example we will be interested in throughout the paper involves the
following polynomial ring which can be thought of as a Cox ring (see [HK00])
with infinitely many columns in the variable matrix. Suppose that
$n_{1},n_{2},\ldots$ is any sequence of natural numbers. We denote by
$k[x_{i,j}]_{n}$ the ring of polynomials in the variables $x_{i,j}$ where
$i,j\in\mathbb{N}$ and $i\leq n_{j}$. We picture this as a matrix $X$ with
$(i,j)$ entry equal to $x_{i,j}$ whose $j$th column has zeros after the
$(n_{j},j)$ entry.

Define a $\mathbb{Z}^{(\mathbb{N})}$-multigrading on $k[x_{i,j}\mid
i,j\in\mathbb{N}]_{n}$ by $\deg(x_{i,j})=e_{j}$, where we denote the standard
basis on $\mathbb{Z}^{(\mathbb{N})}$ by $(e_{i}\mid i\in\mathbb{N})$. We
denote the elements of $\mathbb{Z}^{(\mathbb{N})}$ by
$\vec{a}=(a_{1},a_{2},\ldots)$ and recall that the parentheses in the notation
“$\mathbb{N}$” require that $\vec{a}$ has only finitely many nonzero entries.

##  3\. Finite Dimensionally Graded Rings and Modules

In this paper we will study polynomial rings in infinitely many variables
which, we note, are not finitely generated as algebras over $k$. Because of
this we will need to place other finiteness conditions on our rings and
modules. In particular we will study the Grothendieck group of $R$ using the
Hilbert series of a $\Gamma$-graded $R$-module $M$ and will therefore need
$M_{g}$ to be a finite-dimensional $k$-vector space for all $g\in\Gamma$. We
call such a $\Gamma$-graded $R$-module _finite-dimensionally graded_ (some
authors call this _modest_). We certainly want $R$ to be a finite-
dimensionally graded module over itself and in this section we give a
combination of (typically easy to check) conditions on $R$ that will guarantee
this.

If $M$ is a $\Gamma$-graded $R$-module, we define the _$\Gamma$ -support_ (or
just support) of $M$ to be the set $\text{Supp}_{\Gamma}(M)=\\{g\in\Gamma\mid
M_{g}\neq 0\\}$. In the case when $M=R$, we define the _image monoid_ $Q$ of
the grading to be the submonoid of $\Gamma$ generated by the set
$\text{Supp}_{\Gamma}(R)$. Note that by Proposition 1, $1\in R_{0}$, hence
$0\in\text{Supp}_{\Gamma}(R)$. However if $R$ is not a domain, it is not
always true that $\text{Supp}_{\Gamma}(R)$ is closed under addition, as the
following example shows.

######  Example 3.

Give $k[x,y]$ the coarse $\mathbb{Z}^{2}$-grading with $\deg(x)=(1,0)$ and
$\deg(y)=(0,1)$. Then the ideal $(xy)$ is graded (it is generated by $xy$
which has degree $(1,1)$) and hence the quotient $k[x,y]/(xy)$ is also
$\mathbb{Z}^{2}$-graded. However we have that
$\left(k[x,y]/(xy)\right)_{(1,0)}=\text{span}_{k}(x+(xy))$ and
$\left(k[x,y]/(xy)\right)_{(0,1)}=\text{span}_{k}(y+(xy))$ while
$\left(k[x,y]/(xy)\right)_{(1,1)}=0$ and hence $\text{Supp}_{\Gamma}(R)$ is
not closed under addition.

######  Proposition 6.

If $R$ is a $\Gamma$-graded $k$-algebra which is an integral domain then the
image monoid $Q$ of the grading is just equal to $\text{Supp}_{\Gamma}(R)$.

###### Proof.

We show that when $R$ is a domain, $\text{Supp}_{\Gamma}(R)$ is a submonoid of
$\Gamma$. It follows from Proposition 1 that $0\in\text{Supp}_{\Gamma}(R)$.
Given $g,h\in\text{Supp}_{\Gamma}(R)$, by definition there exist nonzero $r\in
R_{g}$ and $s\in R_{h}$, and hence $rs\in R_{g+h}$ but because $R$ is a domain
$rs\neq 0$ so $g+h\in\text{Supp}_{\Gamma}(R)$. ∎

Proposition 6 is the reason we requite $R$ to be a domain throughout the
paper.

###  3.1. Conditions on the Image Monoid

We now discuss the finiteness conditions we require on the image monoid of
$R$, and begin with a brief summary of some necessary monoid terminology.
Recall that a _monoid_ consists of a set $Q$ together with a binary operation
on $Q$ which is associative and has a (necessarily unique) identity. The
monoid is called _commutative_ if the binary operation is commutative. In this
paper all monoids will be commutative, hence we use “$+$” to denote the binary
operation and “$0$” to denote the identity.

A monoid $Q$ is called _cancelative_ if $p+u=q+u$ implies $p=q$ for all
$p,q,u\in Q$. Any abelian group has an underlying commutative monoid and if
$Q$ is a submonoid of an abelian group $\Gamma$, it is automatically
cancelative since every element of $Q$ has an inverse in $\Gamma$. Conversely
any cancelative commutative monoid can be embedded as a submonoid of an
abelian group (see [Wei13] Chapter II Section 1).

######  Definition 1.

A monoid $Q$ is called _pointed_ if $p+q=0$ implies $p=q=0$ for all $p,q\in
Q$.

######  Proposition 7.

Let $Q$ be a pointed submonoid of an abelian group $\Gamma$. Define a relation
$\leq_{Q}$ on $\Gamma$ by $g\leq_{Q}h$ if and only if there exists $q\in Q$
such that $g+q=h$. Then $\leq_{Q}$ is a partial order.

###### Proof.

To check reflexivity, note that $g=g+0$ and $0\in Q$, so $g\leq_{Q}g$. For
antisymmetry note that if $g\leq_{Q}h$ and $h\leq_{Q}g$ then there exists
$p,q\in Q$ such that $b=a+p$ and $a=b+q$. Then we have $a=a+p+q$ so by
cancelativity we have $0=p+q$ and by pointedness of $Q$ we have $p=q=0$, hence
$g=h$. Finally for transitivity, if $g\leq_{Q}h$ and $h\leq_{Q}u$ then there
exist $p,q\in Q$ such that $h=g+p$ and $u=b+q$, hence we have $u=g+(p+q)$ and
since $Q$ is closed under addition, $p+q\in Q$, hence $g\leq_{Q}u$ as desired.
∎

Note that for any $q\in Q$, we have $0\leq_{Q}q$. Furthermore, if $Q$ is a
pointed submonoid of an abelian group $\Gamma$ and $g\leq_{Q}h$ then $g+q=h$
for some $q\in Q$ and hence for any $u\in\Gamma$ we have $g+u+q=h+u$, so
$g+u\leq_{Q}h+u$. Thus $(\Gamma,\leq_{Q})$ is a partially ordered group.

######  Definition 2.

We say that a subset $U$ of a poset $(P,\leq)$ is _downward finite_ if for all
$p\in P$ the set $\\{u\in U\mid u\leq p\\}$ is finite. If $\Gamma$ is an
abelian group and $Q\subseteq\Gamma$ is a pointed submonoid, we say that a
subset $U\subseteq\Gamma$ is _downward finite_ if it is with respect to the
partial order $\leq_{Q}$.

######  Proposition 8.

Let $\Gamma$ be an abelian group and $Q\subseteq\Gamma$ a pointed submonoid.
Then $Q$ itself is downward finite if and only if for all $p\in Q$ the set
$\\{q\in Q\mid q\leq_{Q}p\\}$ is finite.

###### Proof.

The forward direction is clear. For the reverse, given $g\in\Gamma$ if
$\\{q\in Q\mid q\leq_{Q}g\\}$ is empty it is certainly finite and if it is
nonempty then there exists $q\in Q$ such that $q\leq_{Q}g$ so by definition
there is some $q^{\prime}\in Q$ with $q+q^{\prime}=g$ which implies $g\in Q$,
hence by hypothesis the set $\\{q\in Q\mid q\leq_{Q}g\\}$ is finite. ∎

######  Definition 3.

We say that a $\Gamma$-graded $k$-algebra $R$ is _connected_ if $R_{0}=k$.

######  Theorem 1.

Let $R$ be a connected $\Gamma$-graded $k$-algebra with image monoid $Q$ which
is pointed and downward finite. Let $S\subseteq R$ be a set of homogeneous
elements which generate $R$ as a $k$-algebra. If $S_{g}:=R_{g}\cap S$ is
finite for all $g\in\Gamma$, then $R$ is finite-dimensionally graded.

###### Proof.

Given $q\in Q$, every $r\in R_{q}$ can be written as a polynomial in finitely
many of the elements of $S$, say $s_{1},\ldots,s_{n}$, with coefficients in
$k$. Because each element of $S$ is homogeneous, each monomial in this
polynomial is also homogeneous, hence we can write
$r=f_{1}(s_{1},\ldots,s_{n})+\ldots+f_{m}(s_{1},\ldots,s_{n})$ where
$f_{i}(s_{1},\ldots,s_{n})$ is a $k$-linear combination of monomials each of
which lies in $R_{q_{i}}$ for $q_{1},\ldots,q_{n}\in Q$ distinct. We may
assume without loss of generality that $q_{1}=q$. By independence of the
family $(R_{g}:g\in\Gamma)$ of $k$-subspaces of $R$, we must have
$r=f_{1}(s_{1},\ldots,s_{n})$ and $f_{i}(s_{1},\ldots,s_{n})=0$ for all $i>1$.
Suppose that $s_{i}\in R_{q_{i}}$. Given any $1\leq i\leq n$, we may assume
that $s_{i}$ appears with nonzero exponent in at least one of the monomials in
the expansion $r=f_{1}(s_{1},\ldots,s_{n})$ (otherwise we can simply leave it
out of the list $s_{1},\ldots,s_{n}$ at the beginning). Then if $\gamma
s_{1}^{\ell_{1}}\ldots s_{n}^{\ell_{n}}$ for $\gamma\in k$ is such a monomial
(i.e. with $\ell_{i}>0$), we have $q=\ell_{1}q_{1}+\ldots+\ell_{n}q_{n}$, and
hence $\ell_{i}q_{i}\leq_{Q}q$, so since $\ell_{i}>0$, we have
$q_{i}\leq_{Q}q$.

This implies that the monomials $s_{1}^{\ell_{1}}\ldots s_{n}^{\ell_{n}}$ with
$q=\ell_{1}q_{1}+\ldots+\ell_{n}q_{n}$ and $s_{i}\in\bigcup_{p<_{Q}q}S_{p}$
form a $k$-spanning set for $R_{q}$. Since $Q$ is downward finite, the set
$\\{p\in Q\mid p<_{Q}q\\}$ is finite, and by hypothesis $S_{p}$ is finite for
all $p\in Q$, so $\dim_{k}(R_{q})$ is finite. ∎

######  Definition 4.

We call a $\Gamma$-graded $k$-algebra $R$ that is connected and finite-
dimensionally graded and whose image monoid is pointed and downward finite a
_PDCF $\Gamma$-graded $k$-algebra_.

######  Example 4.

Let $R$ be a PDCF $\Gamma$-graded $k$-algebra with support monoid $Q$. We make
$k$ into a $\Gamma$-graded $k$-algebra by defining $k_{0}=k$ and $k_{g}=0$ for
all $g\neq 0$. Then $k$ is PDCF because $\text{Supp}_{\Gamma}(k)=\\{0\\}$
which is obviously a pointed and downward finite submonoid of $\Gamma$,
$k_{0}=k$ so $k$ is connected, and $\dim_{k}(k_{g})$ is either $0$ or $1$ so
$k$ is finite-dimensionally graded.

###  3.2. Multigradings on the Main Example

######  Example 5.

We now check that the polynomial ring $k[x_{i,j}\mid i,j\in\mathbb{N}]_{n}$
together with the $\mathbb{Z}^{(\mathbb{N})}$-grading given by
$\deg(x_{i,j})=e_{i}$ is a PDCF $\mathbb{Z}^{(\mathbb{N})}$-graded
$k$-algebra. The grading is clearly connected. The image monoid of this
grading is $\mathbb{Z}_{\geq 0}^{(\mathbb{N})}$ which is obviously pointed and
also downward finite because given $\vec{u}\in\mathbb{Z}_{\geq
0}^{(\mathbb{N})}$, we have that $\vec{g}\leq_{\mathbb{Z}_{\geq
0}^{(\mathbb{N})}}\vec{u}$ if and only if $g_{i}\leq u_{i}$ for all
$i\in\mathbb{N}$, and since $\vec{u}$ has only finitely many nonzero entries,
there are only finitely many such $\vec{g}$ in $\mathbb{Z}_{\geq
0}^{(\mathbb{N})}$. Therefore the desired result follows from Theorem 1.

##  4\. Finiteness Conditions on Graded Modules

Having defined, in the previous section, the finiteness conditions on
$k$-algebra $R$ that we will need, we now turn to finiteness conditions on
modules. These will be important to guarantee that the resulting category of
modules is closed under tensor product. We begin with the conditions on the
support.

######  Definition 5.

We say that a subset $U$ of a poset $(P,\leq)$ is _finitely bounded below_ if
there exists $p_{1},\ldots,p_{n}\in P$ such that for all $u\in U$ there exists
$1\leq i\leq n$ such that $p_{i}\leq u$. In this case we say that $U$ is
_bounded below by $\\{p_{1},\ldots,p_{n}\\}$_. If $Q$ is a pointed submonoid
of an abelian group $\Gamma$, then we say that $U\subseteq\Gamma$ is _finitely
bounded below_ if it is with respect to the partial order $\leq_{Q}$.

######  Note 1.

In Section 8.2 of [MS05], for $Q$ a submonoid of an abelian group $\Gamma$, a
Laurent series $a=\sum_{g\in\Gamma}a_{g}\textbf{t}^{g}$ is said to be
_supported on finitely many translates of $Q$_ if the support of $a$ is
contained in $(h_{1}+Q)\cup\ldots\cup(h_{n}+Q)$ for some
$h_{1},\ldots,h_{n}\in\Gamma$. As long as $Q$ is pointed, this is equivalent
to the support of $a$ being a finitely bounded below subset of $\Gamma$ with
respect to $\leq_{Q}$. We use the latter terminology because we often refer to
finitely bounded below subsets in a context broader than the support of a
series or module.

######  Example 6.

Consider the abelian group $\Gamma=\mathbb{Z}^{\mathbb{N}}$ with pointed
submonoid $Q=\mathbb{Z}_{\geq 0}^{\mathbb{N}}$. The subset $\mathbb{Z}_{\geq
0}^{\mathbb{N}}$ is finitely bounded below, since $(0,0,\ldots)$ is a lower
bound, but not downward finite since the set of elements less than
$(1,1,\ldots)$ contains all of the standard basis vectors
$e_{i}=(0,\ldots,0,1,0,\ldots)$.

######  Example 7.

Consider the abelian group $\mathbb{Z}^{(\mathbb{N})}$ with pointed submonoid
$Q=\mathbb{Z}_{\geq 0}^{(\mathbb{N})}$. The subset $U=\\{e_{i}-e_{i+1}\mid
i\in\mathbb{N}\\}$ is downward finite because given any
$(n_{1},n_{2},\ldots)\in\mathbb{Z}^{(\mathbb{N})}$ there exists
$\ell\in\mathbb{N}$ such that $n_{i}=0$ for all $i\geq\ell$ and then it cannot
be that $e_{i}-e_{i+1}\leq_{Q}(n_{1},n_{2},\ldots)$ for $i\geq\ell$. But $U$
is not finitely bounded below since if
$(m_{1},m_{2},\ldots)\in\mathbb{Z}^{(\mathbb{N})}$ were smaller than
infinitely many elements of $U$, infinitely many of the $m_{i}$’s would have
to be negative.

If $R$ is a PDCF $\Gamma$-graded $k$-algebra and $M$ a $\Gamma$-graded
$R$-module, we say that $M$ is _finitely bounded below_ or that it is
_downward finite_ if $\text{Supp}_{\Gamma}(M)\subseteq\Gamma$ satisfies the
corresponding property.

######  Definition 6.

If $M$ is finitely bounded below, downward finite, and finite-dimensionally
graded, we call it a _BDF_ $\Gamma$-graded $R$-module. We denote by
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ the full subcategory of the
category of $\Gamma$-graded $R$-modules consisting of the BDF modules.

######  Note 2.

In Theorem 8.20 of [MS05] it is proved that in the multigraded case, finitely
generated modules are finitely bounded below. It is easy to see in the more
general case of modules over a PDCF $\Gamma$-graded $k$-algebra that finitely
generated modules are in fact BDF.

In the rest of this section, we will prove various results about the category
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$, and we begin by showing that it
contains the ring $R$ itself.

######  Proposition 9.

If $R$ is a PDCF $\Gamma$-graded $k$-algebra then $R$ is a BDF $\Gamma$-graded
$R$-module.

###### Proof.

By definition of PDCF, $Q=\text{Supp}_{\Gamma}(R)$ is downward finite.
Furthermore for any $q\in Q$, $0\leq_{Q}q$ and therefore $Q$ is finitely
bounded below. Again by definition of PDCF, $R$ is a finite-dimensionally
graded $k$-algebra, hence it is a finite-dimensionally graded $R$-module. ∎

Next we show that $R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is closed
under shifting of the grading.

######  Proposition 10.

If $R$ is a PDCF $\Gamma$-graded $k$-algebra and $M$ is a BDF $\Gamma$-graded
$R$-module and $g\in\Gamma$ then $M(g)$ is also a BDF $\Gamma$-graded
$R$-module.

###### Proof.

Note that $\text{Supp}_{\Gamma}(M(g))=\text{Supp}_{\Gamma}(M)-g$. Therefore if
$\text{Supp}_{\Gamma}(M)$ is bounded below by $\\{g_{1},\ldots,g_{n}\\}$,
$\text{Supp}_{\Gamma}(M(g))$ is bounded below by
$\\{g_{1}-g,\ldots,g_{n}-g\\}$. Similarly if $h\in\Gamma$, then because
$\text{Supp}_{\Gamma}(M)$ is downward finite, the set
$\\{u\in\text{Supp}_{\Gamma}(M)\mid u\leq_{Q}h+g\\}$ is finite, hence the
translation $\\{u\in\text{Supp}_{\Gamma}(M)\mid u\leq_{Q}h+g\\}-g$ is also
finite, but this set is equal to $\\{v\in\text{Supp}_{\Gamma}(M(g))\mid
v\leq_{Q}h\\}$, which is therefore also finite. Finally given any
$h\in\Gamma$, $M(g)_{h}=M_{g+h}$ which is a finite dimensional $k$-vector
space by hypothesis. ∎

Finally we show that $R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is an
abelian category, preceded by several necessary lemmas.

######  Lemma 1.

Let $(P,\leq)$ be a poset.

  1. (1)

If $U\subseteq P$ is finitely bounded below and $U^{\prime}\subseteq U$ then
$U^{\prime}$ is also finitely bounded below.

  2. (2)

If $U\subseteq P$ is downward finite and $U^{\prime}\subseteq U$ then
$U^{\prime}$ is also downward finite.

###### Proof.

  1. (1)

If $U\subseteq P$ is finitely bounded below then there exists
$p_{1},\ldots,p_{n}$ such that for all $u\in U$ there exists $1\leq i\leq n$
with $p_{i}\leq u$. As $U^{\prime}\subseteq U$, it is true the for all
$u^{\prime}\in U^{\prime}$ there exists $1\leq i\leq n$ such that $p_{i}\leq
u^{\prime}$.

  2. (2)

Given $p\in P$, we have that the set $\\{u\in U\mid u\leq p\\}$ is finite, but
because $U^{\prime}\subseteq U$ it follows that $\\{u^{\prime}\in
U^{\prime}\mid u^{\prime}\leq p\\}$ is a subset of the aforementioned set and
is therefore also finite.

∎

######  Lemma 2.

If $R$ is a PDCF $\Gamma$-graded $k$-algebra, $M$ is a BDF $R$-module, and
$N\subseteq M$ is a graded submodule, then $N$ and $M/N$ are also BDF
$R$-modules.

###### Proof.

For any $g\in\Gamma$, $N_{g}=N\cap M_{g}$ is a subspace of $M_{g}$. Therefore
$\dim_{k}(N_{g})\leq\dim_{k}(M_{g})$, so since $M$ is finite-dimensionally
graded, so is $N$. Furthermore
$\text{Supp}_{\Gamma}(N)\subseteq\text{Supp}_{\Gamma}(M)$, so by Lemma 1, $N$
is finitely bounded below and downward finite.

Next, for any $g\in\Gamma$, there is a surjective $k$-linear map from $M_{g}$
to $(M/N)_{g}=(M_{g}+N)/N$ which sends $m_{g}\mapsto m_{g}+N$, the kernel of
which is clearly $N_{g}$. Therefore $(M/N)_{g}$ is isomorphic as a $k$-vector
space to $M_{g}/N_{g}$, which is a quotient of a finite-dimensional vector
space. Hence $M/N$ is finite-dimensionally graded. Furthermore, this implies
that $\text{Supp}_{\Gamma}(M/N)\subseteq\text{Supp}_{\Gamma}(M)$, so again by
Lemma 1, $M/N$ is finitely bounded below and downward finite. ∎

######  Theorem 2.

Let $R$ be a PDCF $\Gamma$-graded $k$-algebra. Then
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is an abelian category (see
[Sta21] for the definition of an abelian category).

###### Proof.

The category of $\Gamma$-graded $R$-modules is an abelian category (see
[NVO04] Section 2.2). Since $R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is a
full subcategory, it is a pre-additive, meaning its sets of morphisms are
abelian groups and composition is bilinear. The zero module with the unique
grading is clearly BDF. Finally finite direct sums of BDF modules are BDF
since the support of a direct sum of graded modules is a union of the supports
and finite union preserves being finitely bounded below and downward finite.
Therefore $R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is an additive
category.

Now if $M,N$ are BDF $\Gamma$-graded $R$-modules and $\varphi:M\to N$ is a
$\Gamma$-graded $R$-module homomorphism, by Proposition 5,
$\ker\varphi\subseteq M$ and $\operatorname{im}\varphi\subseteq N$ are graded
submodules, therefore by Lemma 2, $\ker\varphi$ and
$\operatorname{coker}\varphi$ are BDF $\Gamma$-graded $R$-modules. Thus
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is closed under kernels and
cokernels and hence is a pre-abelian category. Finally, the natural map
$M/\ker\varphi\to\operatorname{im}\varphi$ is an isomorphism in the category
of $\Gamma$-graded $R$-modules (because this category is abelian) and because
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is a full subcategory, the
inverse of this isomorphism is also in
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$, so abelianity follows. ∎

######  Note 3.

We will henceforth assume that $R$ denotes a PDCF $\Gamma$-graded $k$-algebra
that is a domain unless otherwise stated.

######  Example 8.

We now show that if $\mathfrak{d}$ is any ideal of $k[x_{i,j}]_{n}$ generated
by minors of $X$ then the quotient $k[x_{i,j}]_{n}/\mathfrak{d}$ is a BDF
$\mathbb{Z}^{(\mathbb{N})}$-graded $R$-module. First of all, any generating
minor of $\mathfrak{d}$ is an alternating sum of monomials which each have a
degree of the form $\vec{a}$ with $a_{i}\in\\{0,1\\}$ in the
$\mathbb{Z}^{(\mathbb{N})}$ grading. Therefore each minor is homogeneous, so
$\mathfrak{d}$ is a $\mathbb{Z}^{(\mathbb{N})}$-graded ideal. Hence
$\mathfrak{d}$ is a graded submodule of the BDF module $k[x_{i,j}]_{n}$, so by
Lemma 2, $\mathfrak{d}$ and $k[x_{i,j}]_{n}/\mathfrak{d}$ are also BDF
modules.

##  5\. Tensor Product of BDF Modules

The goal of this section is to show that
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is not just an abelian category
but is closed in the category of $R$-modules under tensor product. This will
be important for two reasons in future sections: the first is that it will be
needed to define minimal free resolutions of BDF modules and the second is
that it will give us a product on the Grothendieck group.

###  5.1. Hilbert Series

In order to show that the support of the tensor product of two BDF modules is
again finitely bounded below and downward finite, we need to give the
definition of the Hilbert series of a graded module, and prove that the
product of two such series is well-defined.

Given an abelian group $\Gamma$, we define $\mathbb{Z}^{\Gamma}$ to be the set
of functions $a:\Gamma\to\mathbb{Z}$ and we identify $a$ with its generating
function $a(\textbf{t})=\sum_{g\in\Gamma}a_{g}\textbf{t}^{g}$. We would like
to define the product of two such series by the following formula.

(1)
$\left(\sum_{g\in\Gamma}a_{g}\textbf{t}^{g}\right)\left(\sum_{h\in\Gamma}b_{h}\textbf{t}^{h}\right)=\left(\sum_{u\in\Gamma}c_{u}\textbf{t}^{u}\right)$

where $c_{u}=\sum_{g+h=u}a_{g}b_{h}$, but this is not well-defined in general
since the sum defining $c_{u}$ could be infinite. We restrict to a subgroup of
$\mathbb{Z}^{\Gamma}$ on which this multiplication is actually well-defined.

######  Definition 7.

Define $\mathbb{Z}[[Q]]\\{\Gamma\\}$ to be the subset of elements
$a(\textbf{t})=\sum_{g\in\Gamma}a_{g}\textbf{t}^{g}\in\mathbb{Z}^{\Gamma}$
such that $\text{Supp}_{\Gamma}(a):=\\{g\in\Gamma\mid a_{g}\neq 0\\}$ is
downward finite and finitely bounded below.

We think of elements of $\mathbb{Z}[[Q]]\\{\Gamma\\}$ as analogs of formal
Laurent series. We now show that multiplication of two such series is well-
defined and that $\mathbb{Z}[[Q]]\\{\Gamma\\}$ is closed under this product,
but first we need several lemmas.

######  Lemma 3.

Let $V,W\subseteq\Gamma$ be finitely bounded below and downward finite. Given
any $u\in\Gamma$, the sets $U=\\{(g,h)\mid g\in V,h\in
W,g+h\leq_{Q}u\\}\subseteq\Gamma\times\Gamma$, $U_{1}=\\{g\in V\mid\exists
h\in W,g+h\leq_{Q}u\\}$, and $U_{2}=\\{h\in W\mid\exists g\in
V,g+h\leq_{Q}u\\}$ are all finite.

###### Proof.

Consider the projection functions $\pi_{1},\pi_{2}:U\to\Gamma$ which send
$(g,h)$ to $g$ and $h$ respectively. Given any $h\in\pi_{2}(U)$, there exists
$g\in V$ such that $g+h\leq_{Q}u$ and since $V$ is finitely bounded below,
there exists $1\leq i\leq n$ such that $g_{i}\leq_{Q}g$, which gives
$g_{i}+h\leq_{Q}g+h\leq_{Q}u$ and hence that $h\leq_{Q}u-g_{i}$. Since $W$ is
downward finite, there are only finitely many elements of $\\{h\in W\mid
h\leq_{Q}u-g_{i}\\}$ and since there are only finitely many $i$’s, it follows
that $\pi_{2}(U)$ is finite. Note that $\pi_{2}(U)=U_{2}$. A similar argument
shows that $\pi_{1}(U)=U_{1}$ is finite. But
$U\subseteq\pi_{1}(U)\times\pi_{2}(U)$ and since the latter is finite, so is
the former. ∎

######  Lemma 4.

If $V,W\subseteq\Gamma$ are finitely bounded below and downward finite, then
$V+W$ is as well.

###### Proof.

Since $V$ and $W$ are finitely bounded below by hypothesis, there are subsets
$\\{g_{1},\ldots g_{n}\\}$, $\\{h_{1},\ldots,h_{m}\\}$ such that every element
of $V$ is larger than an element of the former, and every element of $W$ is
larger than an element of the latter. Hence given $v+w\in V+W$ there exists
$1\leq i\leq n$ and $1\leq j\leq m$ such that $g_{i}\leq_{Q}v$ and
$h_{j}\leq_{Q}w$. Adding these inequalities together yields
$g_{i}+h_{j}\leq_{Q}v+w$ and hence every element of $V+W$ is larger than one
of the elements of the set $\\{g_{i}+h_{j}\mid 1\leq i\leq n,1\leq j\leq
m\\}$.

Given any $u\in\Gamma$, Lemma 3 implies that the set $U=\\{(g,h)\mid g\in
V,h\in W,g+h\leq_{Q}u\\}$ is finite, but the set $\\{g\in V+W\mid
g\leq_{Q}u\\}$ is the image of $U$ under the map $(g,h)\mapsto g+h$, so this
set is finite too. ∎

If $a,b\in\mathbb{Z}[[Q]]\\{\Gamma\\}$, by Lemma 3, all but finitely many of
the terms in the sum $c_{u}=\sum_{g+h=u}a_{g}b_{h}$ from Equation 1 are zero,
so the product of two elements of $\mathbb{Z}[[Q]]\\{\Gamma\\}$ exists in
$\mathbb{Z}^{\Gamma}$. We now show that this product is actually again in
$\mathbb{Z}[[Q]]\\{\Gamma\\}$.

######  Lemma 5.

Given $a,b\in\mathbb{Z}[[Q]]\\{\Gamma\\}$, we have that
$ab\in\mathbb{Z}[[Q]]\\{\Gamma\\}$.

###### Proof.

If $u\in\text{Supp}_{\Gamma}(ab)$ then there must exist
$g\in\text{Supp}_{\Gamma}(a)$ and $h\in\text{Supp}_{\Gamma}(b)$ such that
$g+h=u$, i.e.
$\text{Supp}_{\Gamma}(ab)\subseteq\text{Supp}_{\Gamma}(a)+\text{Supp}_{\Gamma}(b)$.
By Lemma 4, because $\text{Supp}_{\Gamma}(a)$ and $\text{Supp}_{\Gamma}(b)$
are finitely bounded below and downward finite,
$\text{Supp}_{\Gamma}(a)+\text{Supp}_{\Gamma}(b)$ is too. Hence by Lemma 1,
$\text{Supp}_{\Gamma}(ab)$ is also finitely bounded below and downward finite.
∎

It is easy to see that the multiplication defined by Equation 1 makes
$\mathbb{Z}[[Q]]\\{\Gamma\\}$ into a commutative ring with multiplicative
identity equal to $1=\textbf{t}^{0}$. We can now define Hilbert functions of
BDF modules.

######  Definition 8.

If $M$ is a finite-dimensionally $\Gamma$-graded $R$-module, the _Hilbert
function_ of $M$ is the function $\Gamma\to\mathbb{N}$ given by
$g\mapsto\dim_{k}(M_{g})$. The associated generating function
$\mathcal{H}_{\Gamma}(M)=\sum_{g\in\Gamma}\dim_{k}(M_{g})\textbf{t}^{g}$ is
called the _Hilbert series_ of $M$. By definition $\mathcal{H}_{\Gamma}(M)$ is
an element of $\mathbb{Z}^{\Gamma}$. Note that if $M$ is a BDF $\Gamma$-graded
$R$-module then $\mathcal{H}_{\Gamma}(M)\in\mathbb{Z}[[Q]]\\{\Gamma\\}$.

###  5.2. Tensor Product of BDF Modules

We are now in a position to define the $\Gamma$ grading on the tensor product
of two BDF $\Gamma$-graded $R$-modules and show that the result is still BDF.

######  Proposition 11.

Let $M$ and $N$ be BDF $\Gamma$-graded $R$-modules. Then $M\otimes_{R}N$ is a
$\Gamma$-graded BDF $R$-module with
$(M\otimes_{R}N)_{u}=\text{span}_{k}(x\otimes y\in M\otimes_{R}N\mid x\in
M_{g},y\in N_{h},g+h=u)$ for $u\in\Gamma$.

###### Proof.

First we show that $M\otimes_{R}N=\bigoplus_{u\in\Gamma}(M\otimes_{R}N)_{u}$.
Note that as $k$-vector spaces we have $M=\bigoplus_{g\in\Gamma}M_{g}$ and
$N=\bigoplus_{h\in\Gamma}N_{h}$. Therefore because tensor product commutes
with direct sum, we have
$M\otimes_{k}N=\bigoplus_{g,h\in\Gamma}M_{g}\otimes_{k}N_{h}=\bigoplus_{u\in\Gamma}\left[\bigoplus_{g+h=u}M_{g}\otimes_{k}N_{h}\right]$.
Therefore $M\otimes_{k}N$ is a $\Gamma$-graded $k$-vector space with
$(M\otimes_{k}N)_{u}=\bigoplus_{g+h=u}M_{g}\otimes_{k}N_{h}$. Let $L$ be the
$k$-subspace of $M\otimes_{k}N$ spanned by elements of the form $rx\otimes
y-x\otimes ry$ where $r\in R$ and $x\in M$, $y\in N$ are all homogeneous.
There is a $k$-linear map $\varphi:M\otimes_{k}N\to M\otimes_{R}N$ sending
$x\otimes y\mapsto x\otimes y$ and clearly $L$ is contained in the kernel of
$\varphi$, so we obtain a $k$-linear map
$\overline{\varphi}:(M\otimes_{k}N)/L\to M\otimes_{R}N$. Conversely we define
a map $\psi^{\prime}:M\times N\to(M\otimes_{k}N)/L$ by sending
$(x,y)\mapsto\overline{x\otimes y}$. This is $R$-balanced because given
$r=\sum_{g\in\Gamma}r_{g}\in R$, $x=\sum_{h\in\Gamma}x_{h}\in M$, and
$y=\sum_{u\in\Gamma}y_{u}\in N$, we have that

$\psi^{\prime}(rx,y)=\overline{(rx)\otimes
y}=\overline{\sum_{g,h,u\in\Gamma}r_{g}x_{h}\otimes
y_{u}}=\overline{\sum_{g,h,u\in\Gamma}x_{h}\otimes
r_{g}y_{u}}=\overline{x\otimes(ry)}=\psi^{\prime}(x,ry).$

We therefore obtain a map $\psi:M\otimes_{R}N\to(M\otimes_{k}N)/L$ which is
clearly the inverse of $\overline{\varphi}$, hence $\overline{\varphi}$ is an
isomorphism.

Now if $r\in R_{g}$, $x\in M_{h}$, and $y\in N_{u}$, then $rx\otimes y\in
M_{g+h}\otimes_{k}N_{u}$ and $x\otimes ry\in M_{g}\otimes_{k}N_{h+u}$, and
therefore $rx\otimes y-x\otimes ry\in(M\otimes_{k}N)_{g+h+u}$, hence $L$ is
spanned by homogeneous elements and is therefore a graded submodule. Thus
$(M\otimes_{k}N)/L$ is a $\Gamma$-graded $k$-vector space. Using the
$k$-linear isomorphism $(M\otimes_{k}N)/L\cong M\otimes_{R}N$ from above and
the fact that image of $(M\otimes_{k}N)_{u}$ under $\varphi$ is exactly
$(M\otimes_{R}N)_{u}=\text{span}_{k}(x\otimes y\in M\otimes_{R}N\mid x\in
M_{g},y\in N_{h},g+h=u)$, we obtain that
$M\otimes_{R}N=\bigoplus_{u\in\Gamma}(M\otimes_{R}N)_{u}$ so indeed
$M\otimes_{R}N$ is a $\Gamma$-graded $R$-module.

Now we use the Hilbert series machinery developed above to show that
$M\otimes_{R}N$ is a BDF $\Gamma$-graded $R$-module. We begin with
$M\otimes_{k}N$. By definition, for any $u\in\Gamma$,
$(M\otimes_{k}N)_{u}=\bigoplus_{g+h=u}M_{g}\otimes_{k}N_{h}$ and by Lemma 3
since $\text{Supp}_{\Gamma}(M)$ and $\text{Supp}_{\Gamma}(N)$ are finitely
bounded below and downward finite, the set
$\\{(g,h)\in\text{Supp}_{\Gamma}(M)\times\text{Supp}_{\Gamma}(N)\mid g+h=u\\}$
is finite, and since $M$ and $N$ are finite-dimensionally graded, $M_{g}$ and
$N_{h}$ are finite-dimensional vector spaces, so $(M\otimes_{k}N)_{u}$ is a
finite-dimensional vector space of dimension
$\sum_{g+h=u}\dim_{k}(M_{g})\dim_{k}(N_{h})$. Hence $M\otimes_{k}N$ is a
finite-dimensionally graded $k$-vector space with Hilbert series given by
$\mathcal{H}_{\Gamma}(M\otimes_{k}N)=\mathcal{H}_{\Gamma}(M)\mathcal{H}_{\Gamma}(N)$.
By Lemma 5 we have that
$\mathcal{H}_{\Gamma}(M\otimes_{k}N)\in\mathbb{Z}[[Q]]\\{\Gamma\\}$, and hence
$\text{Supp}_{\Gamma}(M\otimes_{k}N)$ is finitely bounded below and downward
finite. Since $M\otimes_{R}N$ is a quotient of $M\otimes_{k}N$ by a graded
subspace, it follows that $M\otimes_{R}N$ is finite-dimensionally graded too,
and that $\text{Supp}_{\Gamma}(M\otimes_{R}N)$ is finitely bounded below and
downward finite as well. ∎

######  Note 4.

A similar proof to that of Proposition 11 shows that if $R$ is a PDCF
$\Gamma$-graded $k$-algebra and $M_{1},\ldots,M_{n}$ are BDF $\Gamma$-graded
$R$-modules, then if we define
$(M_{1}\otimes_{R}\ldots\otimes_{R}M_{n})_{g}=\text{span}_{k}(x_{1}\otimes\ldots\otimes
x_{n}\mid x_{i}\in M_{h_{i}},\sum_{i=1}^{n}h_{i}=g)$ for $g\in\Gamma$, then
$M_{1}\otimes_{R}\ldots\otimes_{R}M_{n}$ is a BDF $\Gamma$-graded $R$-module.

Finally we record the fact that the tensor product of two graded maps is
graded, which will be useful later.

######  Proposition 12.

Let $M,M^{\prime},N,N^{\prime}$ be BDF $\Gamma$-graded $R$-modules, and let
$\varphi:M\to M^{\prime}$, $\psi:N\to N^{\prime}$ be graded $R$-linear maps.
Then $\varphi\otimes\psi:M\otimes_{R}N\to M^{\prime}\otimes_{R}N^{\prime}$ is
also graded.

###### Proof.

By Proposition 11, $(M\otimes_{R}N)_{u}$ is spanned by elements of the form
$x\otimes y$ where $\deg(x)+\deg(y)=u$. Given $x\otimes y\in M\otimes_{R}N$
with $x\in M_{g}$, $y\in N_{h}$ and $g+h=u$, since $\varphi$ and $\psi$ are
graded, we have that $\varphi(x)\in M^{\prime}_{g}$ and $\psi(y)\in
N^{\prime}_{h}$, and hence $[\varphi\otimes\psi](x\otimes
y)=\varphi(x)\otimes\psi(y)\in(M^{\prime}\otimes_{R}N^{\prime})_{u}$. Hence
indeed $\varphi\otimes\psi$ sends $(M\otimes_{R}N)_{u}$ to
$(M^{\prime}\otimes_{R}N^{\prime})_{u}$ as desired. ∎

##  6\. Free and Projective BDF Modules

In this section we’ll show that every projective BDF $\Gamma$-graded
$R$-module is in fact free. We begin by describing $\Gamma$-graded free
modules. To build these, we will want to take infinite direct sums of shifted
copies of $R$, but we must place restrictions on such direct sums to insure
that the resulting free module is still BDF. The following definition gives
the necessary restriction.

######  Definition 9.

Let $Q$ be a pointed submonoid of an abelian group $\Gamma$. A family of
elements $(g_{i}\mid i\in I)$ of $\Gamma$ is called a _BDF family_ if the set
$\\{g_{i}\mid i\in I\\}\subseteq\Gamma$ is downward finite and finitely
bounded below and for each $g\in\Gamma$, the set $\\{i\in I\mid g_{i}=g\\}$ is
finite. If $M$ is a BDF $\Gamma$-graded $R$-module, a family of homogenous
elements $(m_{i}\mid i\in I)$ of $M$ is called a _BDF family_ if each $m_{i}$
is nonzero and $(\deg(m_{i})\mid i\in I)$ is a BDF family of elements of
$\Gamma$.

We now show that the direct sum of a BDF family of shifted copies of $R$ is
again BDF.

######  Proposition 13.

If $(g_{i}\mid i\in I)$ is a BDF family of elements of $\Gamma$ then the free
module $F=\bigoplus_{i\in I}R(-g_{i})$ is a BDF $\Gamma$-graded $R$-module.

###### Proof.

First note that $\text{Supp}_{\Gamma}(F)=\bigcup_{i\in
I}\text{Supp}_{\Gamma}(R(-g_{i}))$. By hypothesis the set $\\{g_{i}\mid i\in
I\\}$ is finitely bounded below, say by $\\{h_{1},\ldots,h_{n}\\}$, and we
claim that $\text{Supp}_{\Gamma}(F)$ is also bounded below by this set. Then
given $u\in\text{Supp}_{\Gamma}(F)$, there exists $i\in I$ such that
$u\in\text{Supp}_{\Gamma}(R(-g_{i}))=\text{Supp}_{\Gamma}(R)+g_{i}$ and hence
we have that $u-g_{i}\in\text{Supp}_{\Gamma}(R)$ so $0\leq_{Q}u-g_{i}$, which
gives $g_{i}\leq_{Q}u$. But because $\\{g_{i}\mid i\in I\\}$ is bounded below
by $\\{h_{1}\ldots,h_{n}\\}$, then there exists $1\leq j\leq n$ such that
$h_{j}\leq_{Q}g_{i}$, and hence $h_{j}\leq_{Q}u$, as desired.

We now claim that $\text{Supp}_{\Gamma}(F)$ is downward finite. Given
$g\in\Gamma$, we have that $\\{u\in\text{Supp}_{\Gamma}(F)\mid
u\leq_{Q}g\\}=\bigcup_{i\in I}\\{u\in\text{Supp}_{\Gamma}(R(-g_{i}))\mid
u\leq_{Q}g\\}$. But if $u\in\text{Supp}_{\Gamma}(R(-g_{i}))$, then as above
$g_{i}\leq_{Q}u$ so if we also have $u\leq_{Q}g$, it follows that
$g_{i}\leq_{Q}g$. However by hypothesis there are only finitely many $g_{i}$
with $g_{i}\leq_{Q}g$, say $g_{i_{1}}\ldots,g_{i_{n}}$. Hence we have that
$\\{u\in\text{Supp}_{\Gamma}(F)\mid
u\leq_{Q}g\\}=\bigcup_{\ell=1}^{n}\\{u\in\text{Supp}_{\Gamma}(R(-g_{i_{\ell}}))\mid
u\leq_{Q}g\\}$. Since $\text{Supp}_{\Gamma}(R)$ is downward finite, by
Proposition 10, $\text{Supp}_{\Gamma}(R(-g_{i_{\ell}}))$ is also downward
finite, and hence $\\{u\in\text{Supp}_{\Gamma}(F)\mid u\leq_{Q}g\\}$ is a
finite union of finite sets.

Finally we claim that $F$ is finite dimensionally graded. By definition we
have that $F_{h}=\bigoplus_{i\in I}R(-g_{i})_{h}=\bigoplus_{i\in
I}R_{h-g_{i}}$. By definition of $Q=\text{Supp}_{\Gamma}(R)$, $R_{h-g_{j}}=0$
unless $h-g_{j}\geq_{Q}0$ which is equivalent to $g_{j}\leq_{Q}h$. Since the
set $\\{g_{i}\mid i\in I\\}$ is downward finite, there are only finitely many
such $g_{j}$. By hypothesis, for each of these $g_{j}$, there are only
finitely many $i\in I$ such that $g_{i}=g_{j}$. Therefore the direct sum
$\bigoplus_{i\in I}R_{h-g_{i}}$ is in fact a finite direct sum of finite-
dimensional $k$-vector spaces as desired. ∎

For each $i\in I$, we denote the image of $1\in R(g_{i})$ under the inclusion
map $R(g_{i})\hookrightarrow F$ by $e_{i}$ so that $(e_{i}\mid i\in I)$ is a
homogeneous basis for $F$. Note that if we forget the grading,
$F=\bigoplus_{i\in I}R(g_{i})$ is equal to the free $R$-module $R^{(I)}$.

In order to show that projective BDF modules are free, we will need to use the
fact that for every BDF $\Gamma$-graded $R$-module $M$, there exists a free
BDF $\Gamma$-graded $R$-module $F$ and a graded surjection $F\to M$ such that
the images of a homogeneous basis for $F$ are a minimal BDF $R$-spanning
family for $M$. That such a map exists for a given BDF family in $M$ is easy
to see, as the following lemma shows.

######  Lemma 6.

Given a BDF $\Gamma$-graded $R$-module $M$ and a BDF family $(x_{i}\mid i\in
I)$ of elements of $M$, there exists a BDF $\Gamma$-graded free $R$-module
$F=\bigoplus_{i\in I}R(-g_{i})$ and a graded $R$-linear map $\varphi:F\to M$
such that the image of the basis $(e_{i}\mid i\in I)$ of $F$ is $(x_{i}\mid
i\in I)$. In particular if $(x_{i}\mid i\in I)$ $R$-spans $M$ then this map is
surjective.

###### Proof.

For any $i\in I$, let $g_{i}=\deg(x_{i})$. Because $(x_{i}\mid i\in I)$ is a
BDF family, Proposition 13 implies that $F=\bigoplus_{i\in I}R(-g_{i})$ is a
$\Gamma$-graded free $R$-module. Then we can define an $R$-linear map
$\varphi:F\to M$ by sending $e_{i}\mapsto x_{i}$ which is graded because
$\deg(e_{i})=g_{i}=\deg(x_{i})$. ∎

We must now prove the existence of a minimal BDF $R$-spanning family in any
BDF $\Gamma$-graded $R$-module $M$. Unsurprisingly first we need a BDF
$\Gamma$-graded version of Nakayama’s Lemma, preceded by a necessary lemma
about the unique graded maximal ideal of $R$ and the graded residue field.

######  Lemma 7.

The set $R_{+}:=\bigoplus_{q>_{Q}0}R_{q}$ is the unique graded maximal ideal
of $R$ and $R/R_{+}\cong k$ in the category of PDCF $\Gamma$-graded
$k$-algebras ($k$ is a PDCF $\Gamma$-graded $k$-algebra as in Example 4).

###### Proof.

Define a function $(-)_{0}:R\to k$ by sending $r=\sum_{q\in Q}r_{q}\mapsto
r_{0}$. This is a $\Gamma$-graded $k$-linear map, and it respects
multiplication because if $r=\sum_{p\in Q}r_{p}$ and $s=\sum_{q\in Q}s_{q}$
then $(rs)_{0}=\sum_{p+q=0}r_{p}s_{q}$ and since $Q$ is pointed, if $p+q=0$
then $p=0$ and $q=0$, hence $(rs)_{0}=r_{0}s_{0}$. Therefore $(-)_{0}$ is a
graded surjective $k$-algebra homomorphism whose kernel is clearly $R_{+}$, so
$R_{+}$ is a graded maximal ideal. Furthermore suppose that $I=\bigoplus_{q\in
Q}R_{q}\cap I$ is a graded ideal of $R$. If $R_{0}\cap I\neq 0$ then $I$
contains a unit and so is equal to $R$. Hence if $I$ is proper, it must be
that $R_{0}\cap I=0$ and hence $I\subseteq R_{+}$. It follows that $R_{+}$ is
the unique graded maximal ideal of $R$.

∎

The previous lemma shows that we can view PDCF $\Gamma$-graded $k$-algebras
much like local rings and this analogy will be very advantageous. In
particular it gives our BDF analog of the graded version of Nakayama’s Lemma,
preceded by a lemma about the well-foundedness of downward finite sets.

######  Lemma 8.

If a subset $U$ of a poset $(P,\leq)$ is downward finite then $(U,\leq)$ is
well-founded.

###### Proof.

Suppose that $U$ is not well-founded. Then there exists a strictly infinite
descending chain $u_{1}>u_{2}>\ldots$ in $U$ and so clearly the set $\\{u\in
U\mid u\leq u_{1}\\}$ is not finite, a contradiction. ∎

######  Lemma 9 (BDF $\Gamma$-graded Nakayama’s Lemma).

If $R$ is a PDCF $\Gamma$-graded $k$-algebra, $M$ is a BDF $\Gamma$-graded
$R$-module, and $R_{+}M=M$, then $M=0$.

###### Proof.

Suppose that $M\neq 0$. Then $\text{Supp}_{\Gamma}(M)$ is non-empty and
downward finite, so it has a minimal element with respect to $\leq_{Q}$ by
Lemma 8, call it $h$. Note that
$R_{+}M=\sum_{q>_{Q}0,g\in\text{Supp}_{\Gamma}(M)}R_{q}M_{g}$, but
$R_{q}M_{g}\subseteq M_{g+q}$ and since $q>_{Q}0$ it follows that $g+q>_{Q}g$.
Hence every element of $\text{Supp}_{\Gamma}(R_{+}M)$ must be strictly larger
than some element of $\text{Supp}_{\Gamma}(M)$. In particular,
$h\notin\text{Supp}_{\Gamma}(R_{+}M)$, which shows that $R_{+}M\neq M$. ∎

We need one more technical lemma before we can discuss minimal $R$-spanning
families. It describes how tensoring a BDF module $M$ with $k\cong R/R_{+}$ as
an $R$-module is the same as quotienting $M$ by $R_{+}M$. The graded
$k$-algebra homomorphism $(-)_{0}:R\to k$ from the proof of Lemma 7 makes $k$
into a $\Gamma$-graded $R$-module with multiplication $R\times k\to k$ given
by sending $r,\gamma\mapsto r_{0}\gamma$. Then $k$ is in fact a BDF
$\Gamma$-graded $R$-module because $\text{Supp}_{\Gamma}(k)=\\{0\\}$ is
clearly finitely bounded below and downward finite.

######  Lemma 10.

For any BDF $\Gamma$-graded $R$-module $M$ there is a graded $R$-linear
isomorphism $M\otimes_{R}k\cong M/R_{+}M$.

###### Proof.

First we define a map $\varphi:M\times k\to M/R_{+}M$ which sends
$x,\gamma\mapsto\gamma\overline{x}$. This map is obviously $k$-bilinear and it
is $R$-balanced because given $r=\sum_{q\in Q}r_{q}\in R$ we have

$\varphi(rx,\gamma)=\gamma\overline{\sum_{q\in
Q}r_{q}x}\overset{(1)}{=}\gamma\overline{r_{0}x}=\gamma
r_{0}\overline{x}=\varphi(x,r_{0}\gamma)=\varphi(x,r\gamma)$

where (1) follows because $\sum_{q\in Q}r_{q}x=r_{0}x+\sum_{q>0}r_{q}x$ and
the latter term is in $R_{+}M$. Hence we obtain an $R$-linear map
$\tilde{\varphi}:M\otimes_{R}k\to M/R_{+}M$. The degree of $x\otimes\gamma$ is
$\deg(x)+\deg(\gamma)=\deg(x)=\deg(\overline{x})$ so this map is indeed
graded. On the other hand we define a map $\psi:M\to M\otimes_{R}k$ by sending
$x\mapsto x\otimes 1$ which is clearly $R$-linear and the kernel of $\psi$
contains $R_{+}M$ since given $rx$ where $r\in R_{q}$ for $q>0$ and $x\in
M_{g}$ we have $\psi(rx)=rx\otimes 1=x\otimes r\cdot 1$, but $r\cdot 1=0$
since $r\in R_{+}$, hence $\psi(rx)=0$. Since $R_{+}M$ is spanned by elements
of the same form as $rx$, it follows that $\ker\psi\subseteq R_{+}M$. Hence we
obtain a map $\overline{\psi}:M/R_{+}M\to M\otimes_{R}k$. The maps
$\tilde{\varphi}$ and $\overline{\psi}$ are obviously inverses of each other.
∎

We are now finally in a position to discuss minimal $R$-spanning families. The
following proposition describes them in detail.

######  Proposition 14.

Let $M$ be a BDF $\Gamma$-graded $R$-module, $(x_{i}\mid i\in I)$ be a BDF
family of elements of $M$ with $\deg(x_{i})=g_{i}$, $F=\bigoplus_{i\in
I}R(-g_{i})$, and $\varphi:F\to M$ be the $R$-linear map sending $e_{i}\mapsto
x_{i}$.

  1. (1)

The following are equivalent:

    1. (a)

$(\overline{x_{i}}\mid i\in I)$ $k$-spans $M/R_{+}M$.

    2. (b)

$(x_{i}\mid i\in I)$ $R$-spans $M$.

    3. (c)

$\operatorname{im}\varphi=M$.

  2. (2)

If $(\overline{x}_{i}\mid i\in I)$ is $k$-independent then
$\ker\varphi\subseteq R_{+}F$.

  3. (3)

The following are equivalent:

    1. (a)

$(\overline{x_{i}}\mid i\in I)$ is a $k$-basis of $M/R_{+}M$.

    2. (b)

$(x_{i}\mid i\in I)$ is a minimal $R$-spanning family for $M$ in the sense
that no proper subfamily $(x_{j}\mid j\in J)$ with $J\subsetneq I$ $R$-spans
$M$.

    3. (c)

$(x_{i}\mid i\in I)$ $R$-spans $M$ and $\ker\varphi\subseteq R_{+}F$.

###### Proof.

  1. (1)

(b) $\iff$ (c) is obvious. To prove (a) $\implies$ (c), suppose that
$(\overline{{}_{i}}\mid i\in I)$ $k$-spans $M/R_{+}M$. Let $C$ be the cokernel
of $\varphi$ so that we obtain an exact sequence $F\to M\to C\to 0$. Since the
functor $-\otimes_{R}k$ is right exact, we obtain an exact sequence
$F\otimes_{R}k\to M\otimes_{R}k\to C\otimes_{R}k\to 0$. We claim that
$\varphi\otimes\text{id}$ is surjective. Indeed, by the proof of Lemma 10,
under the isomorphism $M/R_{+}M\cong M\otimes_{R}k$, the element
$\overline{x_{i}}$ corresponds to $x_{i}\otimes 1$. Hence $M\otimes_{R}k$ is
$k$-spanned (and hence $R$-spanned) by the family $(x_{i}\otimes 1\mid i\in
I)$. But $[\varphi\otimes 1](e_{i}\otimes 1)=x_{i}\otimes 1$ and therefore
each of the elements of this spanning family is in the image of
$\varphi\otimes 1$ and surjectivty follows. By exactness we obtain that
$C\otimes_{R}k=0$, so Lemma 10 implies that $C/R_{+}C=0$, which gives that
$R_{+}C=C$, and by $\Gamma$-graded Nakayama’s Lemma we obtain that $C=0$,
implying that $\varphi:F\to M$ is surjective.

(b) $\implies$ (a). Now suppose that $(x_{i}\mid i\in I)$ generates $M$ as an
$R$-module. Given any $\overline{x}\in M/R_{+}M$ we can write $x=\sum_{i\in
I}r_{i}x_{i}$ where all but finitely many of the $r_{i}$’s are zero. Then
$\overline{x}=\sum_{i\in I}\overline{r_{i}x_{i}}=\sum_{i\in
I}(r_{i})_{0}\overline{x_{i}}$ where the latter equality follows since if we
write $r_{i}=\sum_{q\in Q}(r_{i})_{q}$ with $(r_{i})_{q}\in R_{q}$ then
$\overline{r_{i}x_{i}}=\overline{(r_{i})_{0}x_{i}}+\overline{\sum_{q>0}(r_{i})_{q}x_{i}}=\overline{(r_{i})_{0}x_{i}}$,
this time the final equality holding because $\sum_{q>0}(r_{i})_{q}x_{i}\in
R_{+}M$. But $(r_{i})_{0}\in k$ so indeed $(\overline{x_{i}}\mid i\in I)$
$k$-spans $M/R_{+}M$.

  2. (2)

Given $y\in\ker\varphi$, we can write $y=\sum_{j=1}^{n}r_{j}e_{i_{j}}$ for
$r_{j}\in R$. Applying $\varphi$ yields $0=\sum_{j=1}^{n}r_{j}x_{i_{j}}$ and
taking residues modulo $R_{+}M$ yields
$0=\sum_{j=1}^{n}(r_{j})_{0}\overline{x}_{i_{j}}$. Since
$(\overline{x}_{i}\mid i\in I)$ is $k$-independent, it follows that
$(r_{j})_{0}=0$ and hence that $r_{j}\in R_{+}$. Therefore $y\in R_{+}F$.

  3. (3)

(a) $\implies$ (b). Suppose that $(\overline{x_{i}}\mid i\in I)$ is a
$k$-basis of $M/R_{+}M$. Then by part (1) $(x_{i}\mid i\in I)$ generates $M$
as an $R$-module and if a subfamily $(x_{j}\mid j\in J)$ with $J\subseteq I$
$R$-generates $M$, then again by the same result, $(\overline{x_{j}}\mid j\in
J)$ is a $k$-basis of $M/R_{+}M$ which is only possible if $J=I$.

(b) $\implies$ (a). Conversely, if $(x_{i}\mid i\in I)$ minimally generates
$M$ as an $R$-module, then by the same corollary $(\overline{x_{i}}\mid i\in
I)$ $k$-spans $M/R_{+}M$. Therefore there is a subfamily
$(\overline{x_{j}}\mid j\in J)$ which is a $k$-basis of $M/R_{+}M$. Again by
part (1), $(x_{j}\mid j\in J)$ $R$-generates $M$ so by minimality $I=J$ and
hence $(\overline{x_{i}}\mid i\in I)$ is a $k$-basis of $M/R_{+}M$.

(a) $\implies$ (c) follows from parts (1) and (2) of this proposition.

(c) $\implies$ (a). That $(\overline{x}_{i}\mid i\in I)$ $k$-spans $M/R_{+}M$
follows from part (1). To show it is $k$-independent, given $\gamma_{i}\in k$
for $i\in I$ with all but finitely many of the $\gamma_{i}$’s equal to zero
such that $\sum_{i\in I}\gamma_{i}\overline{x}_{i}=0$ in $M/R_{+}M$, it
follows that $\sum_{i\in I}\gamma_{i}x_{i}\in R_{+}M$. Since $(x_{i}\mid i\in
I)$ $R$-spans $M$, we can write $\sum_{i\in I}\gamma_{i}x_{i}=\sum_{i\in
I}r_{i}x_{i}$ with $r_{i}\in R_{+}$ and all but finitely many of the $r_{i}$’s
equal to zero. Then $\sum_{i\in I}\gamma_{i}e_{i}-\sum_{i\in
I}r_{i}e_{i}\in\ker\varphi\subseteq R_{+}F$, hence $\sum_{i\in
I}\gamma_{i}e_{i}\in R_{+}F$, so we can write $\sum_{i\in
I}\gamma_{i}e_{i}=\sum_{i\in I}s_{i}e_{i}$ with $s_{i}\in R_{+}$. By
independence of the $e_{i}$’s it follows that $\gamma_{i}\in R_{+}$, and since
$R_{+}\cap k=0$, that $\gamma_{i}=0$ for each $i\in I$.

∎

######  Corollary 1.

There exists a minimal $R$-spanning family for any BDF $\Gamma$-graded
$R$-module $M$.

###### Proof.

Choose a homogeneous $k$-basis $(\overline{x_{i}}\mid i\in I)$ for $M/R_{+}M$.
Since $M/R_{+}M$ is a BDF module, for each $g\in\Gamma$, $(M/R_{+}M)_{g}$ is a
finite-dimensional vector space, hence $\\{i\in I\mid\deg(x_{i})=g\\}$ is
finite. Also $\\{g\in\Gamma\mid\exists i\in
I,\deg(x_{i})=g\\}=\text{Supp}_{\Gamma}(M/R_{+}M)$ which is finitely bounded
below and downward finite, hence $(\overline{x_{i}}\mid i\in I)$ is a BDF
family. By Proposition 14, $(x_{i}\mid i\in I)$ is a minimal $R$-spanning set
for $M$. ∎

Finally we are able to use the existence of a minimal $R$-spanning family to
show that all projective BDF modules are free.

######  Theorem 3.

If $R$ is a PDCF $\Gamma$-graded $k$-algebra and $P$ is BDF $\Gamma$-graded
$R$-module which is projective as an $R$-module then $P$ free and in fact
isomorphic to $\bigoplus_{i\in I}R(-g_{i})$ for some BDF family $(g_{i}\mid
i\in I)$.

###### Proof.

By Proposition 14 and Corollary 1, there exists a homogeneous minimal
$R$-spanning family $(z_{i}\mid i\in I)$ of $P$ which is a BDF family and a
BDF free $R$-module $F=\bigoplus_{i\in I}R(-g_{i})$ where $g_{i}=\deg(z_{i})$
and a graded surjective $R$-linear map $\varphi:F\to P$ with
$K:=\ker\varphi\subseteq R_{+}F$. So we have a short exact sequence of BDF
$\Gamma$-graded $R$-modules $0\to K\to F\to P\to 0$ which splits because $P$
is projective. Therefore we obtain $F=K\oplus P^{\prime}$ where $P^{\prime}$
is a (not necessarily graded) submodule of $F$ isomorphic to $P$. Then we have
that $K\subseteq R_{+}F=R_{+}(K\oplus P^{\prime})=R_{+}K\oplus
R_{+}P^{\prime}$. Hence, given any $x\in K$, we can write $x=y+z$ where $y\in
R_{+}K$ and $z\in R_{+}P^{\prime}$. But subtraction yields $z=x-y\in K$, and
since $z\in R_{+}P^{\prime}\subseteq P^{\prime}$, and $P^{\prime}\cap
K=\\{0\\}$, it must be that $z=0$ and $x=y\in R_{+}K$. Hence $K\subseteq
R_{+}K$. By $\Gamma$-graded Nakayama’s Lemma we have that $K=0$ and hence that
$\varphi$ is an isomorphism, so $P$ is isomorphic to $F=\bigoplus_{i\in
I}F(-g_{i})$ as desired. ∎

One consequence of Theorem 3 is that “$\Gamma$-graded projective” is the same
as “$\Gamma$-graded and projective” and “$\Gamma$-graded free” is the same as
“$\Gamma$-graded and free”, as the following corollary states precisely.

######  Corollary 2.

  1. (1)

If $P$ is a BDF $\Gamma$-graded $R$-module which is projective as an
$R$-module then $P$ is projective in the category
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$.

  2. (2)

If $F$ is a BDF $\Gamma$-graded $R$-module which is free as an $R$-module then
$F\cong\bigoplus_{i\in I}R(-g_{i})$ for some BDF family $(g_{i}\mid i\in I)$,
i.e. $F$ has a homogeneous basis.

###  6.1. Minimal Free Resolutions of BDF Modules

We conclude this section with an application of the theory of minimal spanning
sets we just developed to the existence of minimal free resolutions. These are
free resolutions which map homogeneous bases onto minimal generating sets, and
therefore, in some sense have the smallest possibly syzygy modules. They will
be of importance to us when proving the BDF Hilbert Syzygy Theorem.

######  Lemma 11.

Let $M$ be a BDF $\Gamma$-graded $R$-module and consider the following free
resolution in the category $R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$.

$\ldots\to
F_{2}\overset{\varphi_{2}}{\longrightarrow}F_{1}\overset{\varphi_{1}}{\longrightarrow}F_{0}\overset{\varphi_{0}}{\longrightarrow}M\to
0$

The following are equivalent:

  1. (1)

$\operatorname{im}\varphi_{n}\subseteq R_{+}F_{n-1}$ for all $n\geq 1$.

  2. (2)

The maps in the complex

$\ldots\to
F_{2}\otimes_{R}k\xrightarrow{\varphi_{2}\otimes\text{id}}F_{1}\otimes_{R}k\xrightarrow{\varphi_{1}\otimes\text{id}}F_{0}\otimes_{R}k$

are all $0$.

  3. (3)

For every $n\geq 0$, $\varphi_{n}$ maps every homogeneous basis of $F_{n}$
onto a minimal generating set for $\operatorname{im}\varphi_{n}$.

###### Proof.

(1) $\iff$ (2). First of all, for any any $n\geq 1$ the map
$\varphi_{n}:F_{n}\to F_{n-1}$ induces a map
$\overline{\varphi}_{n}:F_{n}/R_{+}F_{n}\to F_{n-1}/R_{+}F_{n-1}$. Let
$\psi_{n}:F_{n}\otimes_{R}k\to F_{n}/R_{+}F_{n}$ denote the isomorphism
described in Lemma 10. Then the following diagram commutes.

${F_{n}\otimes_{R}k}$${F_{n-1}\otimes_{R}k}$${F_{n}/R_{+}F_{n}}$${F_{n-1}/R_{+}F_{n-1}}$$\scriptstyle{\varphi_{n}\otimes\text{id}}$$\scriptstyle{\psi_{n}}$$\scriptstyle{\psi_{n-1}}$$\scriptstyle{\overline{\varphi}_{n}}$

Since $\psi_{n}$ and $\psi_{n-1}$ are isomorphisms, the top map is zero if and
only if the bottom map is zero. But the bottom map is zero if and only if
$\operatorname{im}\varphi_{n}\subseteq R_{+}F_{n-1}$, as desired.

(1) $\iff$ (3). Given $n\geq 0$, let $(e_{i}\mid i\in I)$ denote a homogeneous
basis for $F_{n}$ and $(x_{i}\mid i\in I)$ denote the image of this basis
under $\varphi_{n}$. Then $(x_{i}\mid i\in I)$ is an $R$-spanning family for
$\operatorname{im}\varphi_{n}$. By Proposition 14, this is a minimal
$R$-spanning family for $\operatorname{im}\varphi_{n}$ if and only if
$\ker\varphi_{n}\subseteq R_{+}F$, but because the complex is exact, we have
$\ker\varphi_{n}=\operatorname{im}\varphi_{n+1}$, so the result follows. ∎

######  Definition 10.

A free resolution satisfying the three equivalent conditions of Lemma 11 is
called _minimal_.

We now show that minimal free resolutions of BDF modules exist.

######  Proposition 15.

Every BDF $\Gamma$-graded $R$-module $M$ has a minimal free resolution in the
category $R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$.

###### Proof.

First of all, by Corollary 1 there exists a homogeneous minimal $R$-generating
set $(x_{i}\mid i\in I)$ of $M$. By Lemma 6 there exists a BDF free $R$-module
$F_{0}=\bigoplus_{i\in I}F(-g_{i})$ and a graded surjective $R$-linear map
$\varphi_{0}:F_{0}\to M$ and by Proposition 14 we have that
$\ker\varphi_{0}\subseteq R_{+}F_{0}$. Replacing $M$ with $\ker\varphi_{0}$ we
repeat this construction to recursively define an exact sequence

$\ldots\to
F_{2}\overset{\varphi_{2}}{\longrightarrow}F_{1}\overset{\varphi_{1}}{\longrightarrow}F_{0}\to
M\to 0$

with the property that
$\operatorname{im}\varphi_{n}=\ker\varphi_{n-1}\subseteq R_{+}F_{n-1}$ for all
$n\geq 1$, hence by Lemma 11 the resolution is minimal. ∎

##  7\. $\Gamma$-finite Sequences of BDF Modules and the Koszul Complex

In the following section we will prove the BDF Hilbert Syzygy Theorem which
states that every BDF module has $\Gamma$-finite free resolution, i.e. a free
resolution such that each graded piece is eventually zero. In this section we
discuss the notion of $\Gamma$-finite sequence of BDF modules and we show that
the Koszul complex associated to a BDF family of elements of $R$ is a
$\Gamma$-finite chain complex which is exact if the family is a regular
sequence. The Koszul complex will be used in the proof of the BDF Hilbert
Syzygy Theorem as well.

######  Definition 11.

A sequence $\ldots\to M_{2}\to M_{1}\to M_{0}\to 0$ of objects of
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ is called _$\Gamma$ -finite_ if
(GF1) $\bigcup_{n=0}^{\infty}\text{Supp}_{\Gamma}(M_{n})$ is finitely bounded
below and downward finite and (GF2) for all $g\in\Gamma$ the sequence of
$k$-vector spaces $\ldots\to(M_{2})_{g}\to(M_{1})_{g}\to(M_{0})_{g}\to 0$ is
eventually $0$.

###  7.1. Exterior Powers of BDF Modules

The Koszul complex requires graded exterior powers of BDF modules, which we
now describe.

######  Proposition 16.

If $M$ is a BDF $\Gamma$-graded $R$-module, the submodule $L$ of $M^{\otimes
n}$ spanned by elements of the form $x_{1}\otimes\ldots\otimes x_{n}$ with
$x_{i}=x_{j}$ for some $i\neq j$ is equal to the submodule $N$ spanned by
elements of the form $x_{1}\otimes\ldots\otimes x_{n}$ with $x_{i}$
homogeneous and $x_{i}=x_{j}$ for some $i\neq j$ together with elements of the
form $x_{1}\otimes\ldots\otimes x_{j}\otimes\ldots\otimes
x_{\ell}\otimes\ldots\otimes x_{n}+x_{1}\otimes\ldots\otimes
x_{\ell}\otimes\ldots\otimes x_{j}\otimes\ldots\otimes x_{n}$ with $x_{i}$
homogeneous.

###### Proof.

See [Haz16] Section 4.1. ∎

######  Corollary 3.

Let $M$ be a BDF $\Gamma$-graded $R$-module. The quotient
$\bigwedge^{n}M=M^{\otimes n}/L$ is a BDF $\Gamma$-graded $R$-module and for
any $g\in\Gamma$, $\left(\bigwedge^{n}M\right)_{g}$ is $k$-spanned by elements
of the form $x_{1}\wedge\ldots\wedge x_{n}$ where $x_{i}\in M_{h_{i}}$ and
$\sum_{i=1}^{n}h_{i}=g$.

###### Proof.

First of all, by Note 4, $M^{\otimes n}$ is a BDF $\Gamma$-graded $R$-module
and $(M^{\otimes n})_{g}$ is $k$-spanned by elements of the form
$x_{1}\otimes\ldots\otimes x_{n}$ where $x_{i}\in M_{h_{i}}$ and
$\sum_{i=1}^{n}h_{i}=g$. Therefore, in the quotient the elements
$x_{1}\wedge\ldots\wedge x_{k}$ $k$-span $\left(\bigwedge^{n}M\right)_{g}$. ∎

######  Proposition 17.

Let $F$ be a BDF $\Gamma$-graded $R$-module with homogeneous $R$-basis
$(e_{i}\mid i\in I)$. Fix any total order on $I$. Then
$(e_{i_{1}}\wedge\ldots\wedge e_{i_{n}}\mid i_{1}<\ldots<i_{n})$ is a
homogeneous $R$-basis for $\bigwedge^{n}F$ and
$\deg(e_{i_{1}}\wedge\ldots\wedge
e_{i_{n}})=\deg(e_{i_{1}})+\ldots+\deg(e_{i_{n}})$. Hence $\bigwedge^{n}F$ is
also a free BDF $\Gamma$-graded $R$-module.

###### Proof.

That $(e_{i_{1}}\wedge\ldots\wedge e_{i_{n}}\mid i_{1}<\ldots<i_{n})$ is an
$R$-basis for $\bigwedge^{n}F$ is well-known, and because $e_{i}$ is
homogeneous for all $i\in I$, it follows that $e_{i_{1}}\otimes\ldots\otimes
e_{i_{n}}$ is homogeneous in $F^{\otimes n}$ with
$\deg(e_{i_{1}}\otimes\ldots\otimes
e_{i_{n}})=\deg(e_{i_{1}})+\ldots+\deg(e_{i_{n}})$, and hence by definition of
the quotient that $e_{i_{1}}\wedge\ldots\wedge e_{i_{n}}$ is homogeneous in
$\bigwedge^{n}F$ and that $\deg(e_{i_{1}}\wedge\ldots\wedge
e_{i_{n}})=\deg(e_{i_{1}})+\ldots+\deg(e_{i_{n}})$. ∎

###  7.2. The Koszul Complex

######  Definition 12.

The _Koszul complex_ associated to a BDF family $(r_{n}\mid n\in\mathbb{N})$
in a PDCF $\Gamma$-graded $k$-algebra $R$, denoted by $K(r_{1},r_{2},\ldots)$,
is the following sequence of modules:

$\displaystyle\ldots\to\bigwedge^{3}F\overset{d_{3}}{\longrightarrow}\bigwedge^{2}F\overset{d_{2}}{\longrightarrow}F\overset{d_{1}}{\longrightarrow}R\to
0$

Here $F=\bigoplus_{i\in\mathbb{N}}R(-g_{i})$ for $g_{i}=\deg(r_{i})$ and we
define $d_{1}(e_{i})=r_{i}$ for all $i\in\mathbb{N}$ and $d_{n}$ is the
$R$-linear map defined on the homogeneous $R$-basis
$(e_{i_{1}}\wedge\ldots\wedge e_{i_{n}}\mid i_{1}<\ldots<i_{n})$ for
$\bigwedge^{n}F$ by

$e_{i_{1}}\wedge\ldots\wedge
e_{i_{n}}\mapsto\sum_{\ell=1}^{n}(-1)^{\ell}d_{1}(e_{i_{\ell}})\cdot
e_{i_{1}}\wedge\ldots\wedge\hat{e}_{i_{\ell}}\wedge\ldots\wedge e_{i_{n}}.$

######  Proposition 18.

The Koszul complex associated to a BDF sequence $r_{1},r_{2},\ldots$ of
elements of $R$ is a graded chain complex of BDF $\Gamma$-graded $R$-modules.

###### Proof.

By Lemma 6, $F$ is a BDF $\Gamma$-graded $R$-module and $d_{1}$ is a
$\Gamma$-graded $R$-linear map. By Corollary 3, $\bigwedge^{n}F$ is also a BDF
$\Gamma$-graded $R$-module and by Proposition 17,
$(e_{i_{1}}\wedge\ldots\wedge e_{i_{n}}\mid i_{1}<\ldots<i_{n})$ is a
homogeneous basis for $\bigwedge^{n}F$, so $d_{n}$ is well-defined. This same
Lemma also says that $e_{i_{1}}\wedge\ldots\wedge e_{i_{n}}$ has degree
$g_{i_{1}}+\ldots+g_{i_{n}}$. Furthermore,
$\deg(d_{1}(e_{i_{\ell}}))=\deg(r_{i_{\ell}})=g_{i_{\ell}}$, hence
$d_{1}(e_{i_{\ell}})\cdot
e_{i_{1}}\wedge\ldots\wedge\hat{e}_{i_{\ell}}\wedge\ldots\wedge e_{i_{n}}$ is
homogeneous of degree $g_{i_{1}}+\ldots+g_{i_{n}}$, so
$d_{n}(e_{i_{1}}\wedge\ldots\wedge e_{i_{n}})$ has degree
$g_{i_{1}}+\ldots+g_{i_{n}}$ as well, and $d_{n}$ is graded. It is well-known
that the Koszul complex is a chain complex, see e.g. [Eis13] Section 17.4. ∎

We now show that the Koszul complex of a BDF family is $\Gamma$-finite,
preceded by the following technical lemma.

######  Lemma 12.

If $r_{1},r_{2},\ldots$ is a BDF sequence of elements of $R$ with degrees
$\deg(r_{i})=g_{i}$, then for all $u\in\Gamma$, the set $\\{i\in\mathbb{N}\mid
g_{i}\leq_{Q}u\\}$ is finite.

###### Proof.

Given $u\in\Gamma$, because $Q$ is downward finite, the set $S:=\\{g\in Q\mid
g\leq_{Q}u\\}$ is finite. Then we notice that $T:=\\{i\in\mathbb{N}\mid
g_{i}\leq_{Q}u\\}=\bigcup_{g\in S}\\{i\in\mathbb{N}\mid g_{i}=g\\}$. Since
$(r_{i}\mid i\in\mathbb{N})$ is a BDF family each set $\\{i\in\mathbb{N}\mid
g_{i}=g\\}$ is finite, hence we have written $T$ as a finite union of finite
sets. ∎

######  Proposition 19.

The Koszul complex associated to a BDF sequence $r_{1},r_{2},\ldots$ of
elements of $R$ is $\Gamma$-finite.

###### Proof.

First of all, note that since $r_{1},r_{2},\ldots\in R$, we have that
$g_{1},g_{2},\ldots\in Q$ and hence $\text{Supp}_{\Gamma}(F)\subseteq Q$.
Furthermore, recall that the grading on $F^{\otimes_{R}n}$ is given by
$(F^{\otimes_{R}n})_{g}=\text{span}_{k}(x_{1}\otimes\ldots\otimes x_{n}\mid
x_{i}\in F_{h_{i}},\sum_{i=1}^{n}h_{i}=g)$. If $g\notin Q$, then at least one
$h_{i}$ must be not in $Q$, but then $F_{h_{i}}=0$, so $x_{i}=0$, which
implies $x_{1}\otimes\ldots\otimes x_{n}=0$. Hence
$\text{Supp}_{\Gamma}(F^{\otimes_{R}n})\subseteq Q$. Since $\bigwedge^{n}F$ is
a graded quotient of $F^{\otimes_{R}n}$, it follows that
$\text{Supp}_{\Gamma}(\bigwedge^{n}F)\subseteq Q$ as well. So we need only
check the $\Gamma$-finite condition on elements of $Q$.

Given $g\in\Gamma$, by Lemma 12, the set $T=\\{i\in\mathbb{N}\mid
g_{i}\leq_{Q}g\\}$ is finite. Suppose that $m>|T|$. We claim that
$\text{Supp}_{\Gamma}(\bigwedge^{m}F)\cap(-\infty,g]=\emptyset$. Indeed, given
any $h\leq_{Q}g$ with $h\in Q$ and any element $x\in(\bigwedge^{m}F)_{h}$, $x$
is a sum of elements of $\bigwedge^{m}F$ of the form $s\cdot
e_{i_{1}}\wedge\ldots\wedge e_{i_{m}}$ for $s\in R$ homogeneous. But we have
that $\deg(s),g_{i_{j}}\in Q$, and
$\deg(s)+\sum_{j=1}^{m}g_{i_{j}}=h\leq_{Q}g$, hence for each $1\leq j\leq m$
we have $i_{j}\in T$. Since $|T|=m-1$, it must be that $i_{n}=i_{\ell}$ for
some $n\neq\ell$. Therefore $e_{i_{1}}\wedge\ldots\wedge e_{i_{m}}=0$, so
$x=0$.

We have shown that for every $g\in\Gamma$ there exits some $n\in\mathbb{N}$
such that for all $m\geq n$, we have $(\bigwedge^{m}F)_{h}=0$ for all
$h\leq_{Q}g$. Furthermore
$\bigcup_{n\in\mathbb{N}}\text{Supp}_{\Gamma}(\bigwedge^{n}F)\subseteq Q$,
hence the Koszul complex is $\Gamma$-finite. ∎

Recall that a family of elements $(r_{i}\mid i\in\mathbb{N})$ of a ring $R$ is
said to be _regular_ on $R$ if the ideal generated by $r_{1},r_{2},\ldots$ is
not equal to $R$ and for all $n\in\mathbb{N}$, $r_{n+1}$ is not a zero divisor
on $R/(r_{1},\ldots,r_{n-1})$.

######  Example 9.

The sequence $(x_{i,j})$ listed in any order in the polynomial ring
$k[x_{i,j}]_{n}$ is clearly a regular sequence since quotienting by any number
of the variables just yields a smaller polynomial ring which is a domain.

######  Proposition 20.

Suppose $r_{1},r_{2},\ldots$ is a regular BDF sequence of elements of $R$. The
Koszul complex is a $\Gamma$-finite BDF free resolution of
$R/(r_{1},r_{2},\ldots)$, i.e. the following complex is exact:

$\displaystyle\ldots\to\bigwedge^{3}F\overset{d_{3}}{\longrightarrow}\bigwedge^{2}F\overset{d_{2}}{\longrightarrow}F\overset{d_{1}}{\longrightarrow}R\to
R/(r_{1},r_{2},\ldots)\to 0$

###### Proof.

First of all, the image of $d_{1}$ is $(r_{1},r_{2},\ldots)$ so the complex is
exact at the $R$ and $R/(r_{1},r_{2},\ldots)$ terms. We must now show that
$\ker d_{n}\subseteq\operatorname{im}d_{n+1}$ for all $n\in\mathbb{N}$. Since
$d_{n}$ is a graded $R$-linear map, $\ker d_{n}$ is a graded submodule hence
it suffices to show that $(\ker d_{n})_{u}\subseteq\operatorname{im}d_{n+1}$
for all $u\in\Gamma$. By Lemma 12 the set $T=\\{i\in\mathbb{N}\mid
g_{i}\leq_{Q}u\\}$ is finite. Then there exists $m\in\mathbb{N}$ is such that
for all $i\in T$, $i\leq m$. It is clear from the definition of a regular
sequence that $r_{1},\ldots,r_{m}$ is also regular on $R$. It is well-known
that the Koszul complex $K(r_{1},\ldots,r_{m})$ is then a free resolution of
$R/(r_{1},\ldots,r_{m})$ (again see [Eis13] 17.4). Explicitly, let
$F^{\prime}=\bigoplus_{i=1}^{m}R(-g_{i})$ and for $n\in\mathbb{N}$ define
$d_{n}^{\prime}:\bigwedge^{n}F^{\prime}\to\bigwedge^{n-1}F^{\prime}$ as
before. Then the following complex is exact:

$\displaystyle\ldots\to\bigwedge^{3}F^{\prime}\overset{d_{3}^{\prime}}{\longrightarrow}\bigwedge^{2}F^{\prime}\overset{d_{2}^{\prime}}{\longrightarrow}F^{\prime}\overset{d_{1}^{\prime}}{\longrightarrow}R\to
R/(r_{1},\ldots,r_{m})\to 0$

We now define a morphism of complexes $\varphi:K(r_{1},\ldots,r_{m})\to
K(r_{1},r_{2},\ldots)$, i.e. a collection of morphisms $\varphi_{n}$ such that
the following diagram commutes.

${\ldots}$${\bigwedge^{3}F^{\prime}}$${\bigwedge^{2}F^{\prime}}$${F^{\prime}}$${R}$${R/(r_{1},\ldots,r_{m})}$${0}$${\ldots}$${\bigwedge^{3}F}$${\bigwedge^{2}F}$${F}$${R}$${R/(r_{1},r_{2},\ldots)}$${0}$$\scriptstyle{d_{3}^{\prime}}$$\scriptstyle{\varphi_{3}}$$\scriptstyle{d_{2}^{\prime}}$$\scriptstyle{\varphi_{2}}$$\scriptstyle{d_{1}^{\prime}}$$\scriptstyle{\varphi_{1}}$$\scriptstyle{\varphi_{0}=\text{id}}$$\scriptstyle{d_{3}}$$\scriptstyle{d_{2}}$$\scriptstyle{d_{1}}$

In fact we define $\varphi_{1}$ to just be the inclusion and define
$\varphi_{n}=\bigwedge^{n}\varphi_{1}$. Then it’s clear that the diagram
commutes and that $\varphi_{n}$ is graded and injective because it sends the
homogeneous basis elements of $\bigwedge^{n}F^{\prime}$ to homogeneous basis
elements of $\bigwedge^{n}F$ of the same degree. Any $x\in(\ker d_{n})_{u}$
can be written as a sum of elements of the form $r\cdot
e_{i_{1}}\wedge\ldots\wedge e_{i_{n}}$ which are of degree $u$. It follows
that $g_{i_{j}}=\deg(e_{i_{j}})\leq_{Q}u$ for all $1\leq j\leq n$. Therefore
$i_{j}\in T$ so $i_{j}\leq m$. Hence $e_{i_{1}}\wedge\ldots\wedge
e_{i_{n}}\in\operatorname{im}\varphi_{n}$, hence
$x\in\operatorname{im}\varphi_{n}$ as well, say $x=\varphi_{n}(y)$. Applying
$d_{n}$ yields
$0=d_{n}(x)=d_{n}(\varphi_{n}(y))=\varphi_{n-1}(d^{\prime}_{n}(y))$, but since
$\varphi_{n-1}$ is injective, we have that $d^{\prime}_{n}(y)=0$ so $y\in\ker
d^{\prime}_{n}$. By exactness of the top sequence there exists
$z\in\bigwedge^{n+1}F^{\prime}$ such that $d_{n+1}^{\prime}(z)=y$. Therefore
$x=\varphi_{n}(y)=\varphi_{n}(d_{n+1}^{\prime}(z))=d_{n+1}(\varphi_{n+1}(z))$,
so $x\in\operatorname{im}\varphi_{n+1}$ as desired. ∎

##  8\. The BDF Hilbert Syzygy Theorem

In this section we will prove the BDF $\Gamma$-graded analog of the Hilbert
Syzygy Theorem, which says that every BDF module has a BDF $\Gamma$-finite
free resolution. This will require a BDF $\Gamma$-graded $\operatorname{Tor}$
functor which we now describe.

###  8.1. The $\Gamma$-graded $\operatorname{Tor}$ Functor

If $M$ is a BDF $\Gamma$-graded $R$-module, by Proposition 15 there exists a
BDF $\Gamma$-graded free resolution $F_{\bullet}$ of $M$ and tensoring
$F_{\bullet}\to 0$ with another BDF module $N$ gives a chain complex
$F_{\bullet}\otimes_{R}N\to 0$ whose terms are BDF $\Gamma$-graded $R$-modules
by Proposition 11 and whose maps are graded by Proposition 12. Therefore the
kernels and images of these maps are also BDF $\Gamma$-graded $R$-modules, so
the homology modules $\operatorname{Tor}^{R}_{i}(M,N):=H_{i}(F_{\bullet}\to
0)$ of this complex are BDF $\Gamma$-graded $R$-modules as well. It is well-
known that if $F^{\prime}_{\bullet}$ is another free resolution of $M$, then
$H_{i}(F_{\bullet}\to 0)$ and $H_{i}(F^{\prime}_{\bullet}\to 0)$ are
canonically isomorphic as ungraded $R$-modules as are
$\operatorname{Tor}^{R}_{i}(M,N)$ and $\operatorname{Tor}^{R}_{i}(N,M)$ (see
[Wei94] Section 2.7). However, we will need to know that these isomorphisms
are graded. The former is standard, but we check the latter carefully since we
could not find a reference.

######  Proposition 21.

If $M$ and $N$ are BDF $\Gamma$-graded $R$-modules, there is a graded
isomorphism
$\operatorname{Tor}^{R}_{i}(M,N)\cong\operatorname{Tor}^{R}_{i}(N,M)$.

###### Proof.

We just follow the proof for the ungraded case in [Wei94] and show that the
isomorphism constructed there is $\Gamma$-graded. Let $\ldots
F_{1}\overset{\varphi_{1}}{\to}F_{0}\overset{\varphi_{0}}{\to}M\to 0$ and
$\ldots G_{1}\overset{\psi_{1}}{\to}G_{0}\overset{\psi_{0}}{\to}N\to 0$ be BDF
$\Gamma$-graded free resolutions of $M$ and $N$ respectively. We form the
double complex in Figure 1.

${\vdots}$${\vdots}$${\vdots}$${\vdots}$${\ldots}$${F_{2}\otimes_{R}G_{2}}$${F_{1}\otimes_{R}G_{2}}$${F_{0}\otimes_{R}G_{2}}$${M\otimes_{R}G_{2}}$${0}$${\ldots}$${F_{2}\otimes_{R}G_{1}}$${F_{1}\otimes_{R}G_{1}}$${F_{0}\otimes_{R}G_{1}}$${M\otimes_{R}G_{1}}$${0}$${\ldots}$${F_{2}\otimes_{R}G_{0}}$${F_{1}\otimes_{R}G_{0}}$${F_{0}\otimes_{R}G_{0}}$${M\otimes_{R}G_{0}}$${0}$${\ldots}$${F_{2}\otimes_{R}N}$${F_{1}\otimes_{R}N}$${F_{0}\otimes_{R}N}$${0}$${0}$${0}$${0}$${0}$${0}$$\scriptstyle{\varphi_{2}\otimes\text{id}_{G_{2}}}$$\scriptstyle{\text{id}_{F_{2}}\otimes\psi_{2}}$$\scriptstyle{\varphi_{1}\otimes\text{id}_{G_{2}}}$$\scriptstyle{\text{id}_{F_{1}}\otimes\psi_{2}}$$\scriptstyle{\varphi_{0}\otimes\text{id}_{G_{2}}}$$\scriptstyle{\text{id}_{F_{0}}\otimes\psi_{2}}$$\scriptstyle{\text{id}_{M}\otimes\psi_{2}}$$\scriptstyle{\varphi_{2}\otimes\text{id}_{G_{1}}}$$\scriptstyle{\text{id}_{F_{2}}\otimes\psi_{1}}$$\scriptstyle{\varphi_{1}\otimes\text{id}_{G_{1}}}$$\scriptstyle{\text{id}_{F_{1}}\otimes\psi_{1}}$$\scriptstyle{\varphi_{0}\otimes\text{id}_{G_{1}}}$$\scriptstyle{\text{id}_{F_{0}}\otimes\psi_{1}}$$\scriptstyle{\text{id}_{M}\otimes\psi_{1}}$$\scriptstyle{\varphi_{2}\otimes\text{id}_{G_{0}}}$$\scriptstyle{\text{id}_{F_{2}}\otimes\psi_{0}}$$\scriptstyle{\varphi_{1}\otimes\text{id}_{G_{0}}}$$\scriptstyle{\text{id}_{F_{1}}\otimes\psi_{0}}$$\scriptstyle{\varphi_{0}\otimes\text{id}_{G_{0}}}$$\scriptstyle{\text{id}_{F_{0}}\otimes\psi_{0}}$$\scriptstyle{\varphi_{2}\otimes\text{id}_{N}}$$\scriptstyle{\varphi_{1}\otimes\text{id}_{N}}$
Figure 1. The Tensor Product of Two Free Resolutions

The homology of the bottom row is $\operatorname{Tor}^{R}_{i}(M,N)$ and that
of the right-most column is $\operatorname{Tor}^{R}_{i}(N,M)$. Furthermore all
rows except the bottom one are exact because $G_{n}$ is free, and hence flat,
for all $n\in\mathbb{N}$ and similarly all columns except the right-most are
also exact. The proof in the ungraded case proceeds as follows. Given
$\overline{x}\in\operatorname{Tor}^{R}_{i}(M,N)$ using exactness of the rows
and columns as well as commutativity of the diagram, one constructs a “zigzag”
sequence of elements $x=x_{i,-1}\in F_{i}\otimes N$, $x_{i,0}\in
F_{i}\otimes_{R}G_{0}$, $x_{i-1,0}\in F_{i-1}\otimes_{R}G_{0}$, $x_{i-1,1}\in
F_{i-1}\otimes_{R}G_{1}$, $x_{i-2,1}\in F_{i-2}\otimes_{R}G_{1}$, …$x_{0,i}\in
F_{0}\otimes_{R}G_{i}$, $x_{-1,i}\in M\otimes_{R}G_{i}$ such that if
$j+\ell=i$, then
$[\text{id}_{F_{j}}\otimes\psi_{\ell}](x_{j,\ell})=x_{j,\ell-1}$ and
$[\varphi_{j}\otimes\text{id}_{G_{\ell}}](x_{j,\ell})=x_{j-1,\ell}$. In other
words the elements of the zigzag sequence map to each other under the maps in
the zigzag. Then the isomorphism
$\operatorname{Tor}^{R}_{i}(M,N)\to\operatorname{Tor}^{R}_{i}(N,M)$ is defined
by $\overline{x}\mapsto\overline{x_{-1,i}}$, and one checks that this is well-
defined.

In the $\Gamma$-graded case, we take
$\overline{x}\in\operatorname{Tor}^{R}_{i}(M,N)_{u}=(\ker\varphi_{i}\otimes\text{id}_{N})_{u}/(\operatorname{im}\varphi_{i+1}\otimes\text{id}_{N})_{u}$
so that $x\in(\ker\varphi_{i}\otimes\text{id}_{N})_{u}$. Then because all maps
in this complex are graded (Proposition 12), Proposition 5 allows us to choose
all terns of the zigzag sequence to be homogeneous of degree $u$. In
particular $x_{-1,i}$ is graded of degree $u$, hence
$\overline{x_{-1,i}}\in\operatorname{Tor}^{R}_{i}(N,M)_{u}$ so indeed our
isomorphism is graded. ∎

###  8.2. Existence of $\Gamma$-finite free resolutions

######  Lemma 13.

Suppose that $\ldots\to M_{1}\to M_{0}\to 0$ and $\ldots\to N_{1}\to N_{0}\to
0$ are $\Gamma$-graded sequences of BDF $\Gamma$-graded $R$-modules. Then the
double complex $M_{\bullet}\otimes_{R}N_{\bullet}$ pictured in Figure 2 is
$\Gamma$-finite, meaning that (DGF1) the union of the $\Gamma$-supports is
finitely bounded below and downward finite, and (DGF2) for any $g\in\Gamma$
there exists $n\in\mathbb{N}$ such that $(M_{i}\otimes_{R}N_{j})_{g}=0$ for
either $i\geq n$ or $j\geq n$.

###### Proof.

Let $U=\bigcup_{i=1}^{\infty}\text{Supp}_{\Gamma}(M_{i})$ and
$V=\bigcup_{j=1}^{\infty}\text{Supp}_{\Gamma}(N_{j})$. Recall from Proposition
11 that $(M_{i}\otimes_{R}N_{j})_{g}=\text{span}_{k}(x\otimes
y\mid\deg(x)+\deg(y)=g)$. Therefore
$\text{Supp}_{\Gamma}(M_{i}\otimes_{R}N_{j})\subseteq\text{Supp}_{\Gamma}(M_{i})+\text{Supp}_{\Gamma}(N_{j})\subseteq
U+V$. Since this is true for all $i,j$, it follows that
$\bigcup_{i,j=1}^{\infty}\left[\text{Supp}_{\Gamma}(M_{i}\otimes_{R}N_{j})\right]\subseteq
U+V$. But by definition of a $\Gamma$-finite sequence, $U$ and $V$ are
finitely bounded below and downward finite, so by Lemma 4, $U+V$ is as well.
Therefore
$\bigcup_{i,j=1}^{\infty}\left[\text{Supp}_{\Gamma}(M_{i}\otimes_{R}N_{j})\right]$
is contained in a finitely bounded below and downward finite subset of
$\Gamma$, hence by Lemma 1, the double complex satisfies (DGF1).

Given $g\in\Gamma$, because $U$ and $V$ are finitely bounded below and
downward finite by hypothesis, by Lemma 3 the sets $W_{1}=\\{u\in U\mid\exists
v\in V,u+v=g\\}$ and $W_{2}=\\{v\in V\mid\exists u\in U,u+v=g\\}$ are finite.
Hence because $M_{\bullet}$ and $N_{\bullet}$ are $\Gamma$-finite, there
exists some $n\in\mathbb{N}$ such that if $i,j\geq n$ then
$(M_{i})_{u}=(N_{j})_{v}=0$ for all $u\in W_{1}$ and $v\in W_{2}$. But we have
that $(M_{i}\otimes_{R}N_{j})_{g}=\text{span}_{k}(x\otimes y\mid
x\in(M_{i})_{u},y\in(M_{j})_{v},u+v=g)$ so supposing that either $i\geq n$ or
$j\neq n$, if either $u\notin U$ or $v\notin V$ then $x=0$ or $y=0$ so
$x\otimes y=0$ and if $u\in U$ and $v\in V$ then $u+v=g$ so $u\in W_{1}$ and
$v\in W_{2}$ hence either $x=0$ or $y=0$ so $x\otimes y=0$ anyway. Hence if
either $i\geq n$ or $j\geq n$ we have that $(M_{i}\otimes_{R}N_{j})_{g}=0$, so
the double complex satisfies (DGF2). ∎

${\vdots}$${\vdots}$${\vdots}$${\ldots}$${M_{2}\otimes_{R}N_{2}}$${M_{1}\otimes_{R}N_{2}}$${M_{0}\otimes_{R}N_{2}}$${0}$${\ldots}$${M_{2}\otimes_{R}N_{1}}$${M_{1}\otimes_{R}N_{1}}$${M_{0}\otimes_{R}N_{1}}$${0}$${\ldots}$${M_{2}\otimes_{R}N_{0}}$${M_{1}\otimes_{R}N_{0}}$${M_{0}\otimes_{R}N_{0}}$${0}$${0}$${0}$${0}$
Figure 2. The Tensor Product of two Sequences

######  Corollary 4.

Let $\ldots\to M_{1}\to M_{0}\to 0$ be a $\Gamma$-finite chain complex
consisting of BDF $\Gamma$-graded $R$-modules and let $N$ also be a BDF
$\Gamma$-graded $R$-module. Then $\ldots\to M_{1}\otimes_{R}N\to
M_{0}\otimes_{R}N\to 0$ is also $\Gamma$-finite.

###### Proof.

This follows by setting $N_{0}=N$ and $N_{i}=0$ for $i\geq 1$ in Lemma 13. ∎

######  Lemma 14.

If $M$ is a BDF $\Gamma$-graded $R$-module and $g\in\Gamma$, then
$\mathcal{H}_{\Gamma}(M(-g))=\textbf{t}^{g}\mathcal{H}_{\Gamma}(M)$.

###### Proof.

We compute

$\displaystyle\mathcal{H}_{\Gamma}(M(-g))$
$\displaystyle=\sum_{h\in\Gamma}\dim(M(-g)_{h})\textbf{t}^{h}=\sum_{h\in\Gamma}\dim(M_{h-g})\textbf{t}^{h}\overset{(1)}{=}\sum_{h\in\Gamma}\dim(M_{h})\textbf{t}^{h+g}$
$\displaystyle\overset{(2)}{=}\textbf{t}^{g}\sum_{h\in\Gamma}\dim(M_{h})\textbf{t}^{h}=\textbf{t}^{g}\mathcal{H}_{\Gamma}(M).$

Here (1) holds by the change of variables $h\mapsto h+g$ and (2) follows from
the definition of the product in $\mathbb{Z}[[Q]]\\{\Gamma\\}$. ∎

######  Lemma 15.

If $(g_{i}\mid i\in I)$ is a BDF family of elements of $\Gamma$ and
$F=\bigoplus_{i\in I}R(-g_{i})$, then $F\otimes_{R}k\cong\bigoplus_{i\in
I}k(-g_{i})$. As a consequence,
$\mathcal{H}_{\Gamma}(F\otimes_{R}k)=\sum_{g\in\Gamma}\\#\\{i\in I\mid
g_{i}=g\\}\textbf{t}^{g}$.

###### Proof.

We have that $F\otimes_{R}k\cong\bigoplus_{i\in
I}\left(R(-g_{i})\otimes_{R}k\right)\cong\bigoplus_{i\in I}k(-g_{i})$, where
the first isomorphism follows because tensor product commutes with direct sum,
and the second follows because for any $\Gamma$-graded $R$-module $M$ and any
$g\in\Gamma$ we have $R(-g)\otimes_{R}M\cong M(-g)$ (see [Haz16] Section 1.2.6
Equation 1.22). Since $\mathcal{H}_{\Gamma}(k)=1$, it follows from Lemma 14
that $\mathcal{H}_{\Gamma}(k(-g_{i}))=\textbf{t}^{g_{i}}$. Therefore counting
dimensions at each $g\in\Gamma$ yields
$\mathcal{H}_{\Gamma}(F\otimes_{R}k)=\sum_{g\in\Gamma}\\#\\{i\in I\mid
g_{i}=g\\}\textbf{t}^{g}$. ∎

######  Lemma 16.

Let $(g_{i}\mid i\in I)$ be a BDF family of elements of $\Gamma$ and let
$F=\bigoplus_{i\in I}R(-g_{i})$ the associated BDF $\Gamma$-graded free
module. Then $F_{g}=\bigoplus_{i\in I,g_{i}\leq_{Q}g}R(-g_{i})_{g}$. In
particular $F_{g}=0$ if and only if $\\{i\in I\mid g_{i}\leq_{Q}g\\}$ is
empty.

###### Proof.

By definition $F_{g}=\bigoplus_{i\in I}R(-g_{i})_{g}=\bigoplus_{i\in
I}R_{g-g_{i}}$ and $R_{g-g_{i}}\neq 0$ if and only if $0\leq_{Q}g-g_{i}$ which
is equivalent to $g_{i}\leq_{Q}g$. Therefore the direct sum can be taken over
the set $\\{i\in I\mid g_{i}\leq_{Q}g\\}$, so we obtain the first desired
result. If $\\{i\in I\mid g_{i}\leq_{Q}g\\}$ is empty then obviously $F_{g}=0$
since it is equal to a direct sum over the empty set. Conversely, if there
exists $i\in I$ with $g_{i}\leq_{Q}g$ then $R(-g_{i})_{g}\neq 0$ so
$F_{g}=\bigoplus_{i\in I,g_{i}\leq_{Q}g}R(-g_{i})_{g}\neq 0$. ∎

######  Lemma 17.

Let $F$ be a free BDF $\Gamma$-graded $R$-module and let $g\in\Gamma$. Then
$F_{g}=0$ if and only if $[F\otimes_{R}k]_{\leq g}=0$ (where we define
$M_{\leq g}:=\bigoplus_{h\leq_{Q}g}M_{h}$ for any $\Gamma$-graded $R$-module
$M$).

###### Proof.

By Lemma 16 we have that $F_{g}=0$ if and only if $\\{i\in I\mid
g_{i}\leq_{Q}g\\}=\emptyset$. But by Lemma 15, this is true if and only if
$\mathcal{H}_{\Gamma}(F\otimes_{R}k)_{h}=0$ for all $h\leq_{Q}g$ which is true
if and only if $[F\otimes_{R}k]_{\leq g}=0$. ∎

######  Theorem 4 (BDF Hilbert Syzygy Theorem).

Let $R$ be an PDCF $\Gamma$-graded $k$-algebra. Then for any BDF
$\Gamma$-graded $R$-module $M$ with the property that $R_{+}$ is generated by
a homogenous regular sequence, there exists a BDF $\Gamma$-finite free
resolution of $M$.

###### Proof.

If $(r_{i}\mid i\in\mathbb{N})$ is a homogeneous regular sequence on $R$ which
generates $R_{+}$ then by Proposition 20, the Koszul complex
$K=K(r_{1},r_{2},\ldots)$ is a $\Gamma$-finite free resolution of
$R/R_{+}\cong k$. Tensoring the Koszul complex with $M$ we get a BDF
$\Gamma$-graded complex $K_{\bullet}\otimes_{R}M$ which is $\Gamma$-finite by
Corollary 4, and the $i$th homology of which is
$\operatorname{Tor}^{R}_{i}(k,M)$. By Proposition 15, there exists a minimal
BDF $\Gamma$-graded free resolution $F_{\bullet}$ of $M$. Tensoring with $k$
gives a complex $F_{\bullet}\otimes_{R}k$ the $i$th homology of which is
$\operatorname{Tor}^{R}_{i}(M,k)$. But by Proposition 21, we have that
$\operatorname{Tor}^{R}_{i}(M,k)_{g}\cong\operatorname{Tor}^{R}_{i}(k,M)_{g}$
for all $g\in\Gamma$, hence the former is zero if and only if the latter is.
Since $K_{\bullet}\otimes_{R}M$ is $\Gamma$-finite, for all $g\in\Gamma$ there
exists $n\in\mathbb{N}$ such that $[K_{i}\otimes_{R}M]_{g}=0$, and hence
$\operatorname{Tor}^{R}_{i}(k,M)=0$, and hence
$\operatorname{Tor}^{R}_{i}(M,k)_{g}=0$ for all $i\geq n$.

Because the resolution $F_{\bullet}$ of $M$ is minimal, the differentials of
the complex $F_{\bullet}\otimes_{R}k$ are all $0$ by Lemma 11, and hence
$\operatorname{Tor}^{R}_{i}(M,k)_{g}=(F_{i}\otimes_{R}k)_{g}$. But by Lemma
17, $(F_{i}\otimes_{R}k)_{g}=0$ if and only if $(F_{i})_{g}=0$, therefore
$F_{\bullet}$ is a $\Gamma$-finite free resolution of $M$ as desired. ∎

##  9\. Projective $\Gamma$-Dimension of a BDF Module

Although we will not need it in the sequel, we give an application of the
previous section to defining the graded projective dimension of a BDF module
and we show that it is equal to a certain graded torsion dimension, just as in
the ungraded finitely generated case (see [Eis13] Corollary 19.5).

Let $M$ be a BDF $\Gamma$-graded $R$-module. The _projective
$\Gamma$-dimension_ of $M$ is the function
$\text{pd}_{R}^{\Gamma}(M):\Gamma\to\mathbb{Z}_{\geq 0}\cup\\{\infty\\}$ which
sends $g\in\Gamma$ to the minimum $n$ such that there exists a BDF projective
resolution $P_{\bullet}$ of $M$ with $(P_{n})_{g}\neq 0$ but $(P_{i})_{g}=0$
for all $i>n$. The _torsion $\Gamma$-dimension_ of $M$ is the function
$\text{td}_{R}^{\Gamma}(M):\Gamma\to\mathbb{Z}_{\geq 0}\cup\\{\infty\\}$ which
sends $g\in\Gamma$ to the minimum $n$ such that
$\operatorname{Tor}^{R}_{n}(k,M)_{\leq g}\neq 0$ but
$\operatorname{Tor}^{R}_{i}(k,M)_{\leq g}=0$ for all $i>n$. In [MS05] the
$i$th graded Betti number of $M$ in degree $g$ is defined to be
$\beta_{i,g}(M)=\dim_{k}\operatorname{Tor}^{R}_{i}(k,M)_{g}$, so in this
language the torsion dimension $\text{td}_{R}^{\Gamma}(M)(g)$ is just the
first $n$ such that all Betti numbers $\beta_{n,h}(M)$ for $h\leq_{Q}g$
vanish.

######  Proposition 22.

If $M$ is a BDF $\Gamma$-graded $R$-module, then for each $g\in\Gamma$, we
have $\text{pd}_{R}^{\Gamma}(M)(g)=\text{td}_{R}^{\Gamma}(M)(g)$.

###### Proof.

First we show that
$\text{pd}_{R}^{\Gamma}(M)(g)\geq\text{td}_{R}^{\Gamma}(M)(g)$. This is
trivially true if $\text{pd}_{R}^{\Gamma}(M)(g)=\infty$, so suppose that
$\text{pd}_{R}^{\Gamma}(M)(g)=n\in\mathbb{Z}_{\geq 0}$. Then there exits a BDF
free resolution $F_{\bullet}$ of $M$ such that $(F_{i})_{g}=0$ for all $i>n$.
Tensoring with $k$ yields a complex $F_{\bullet}\otimes_{R}k$ which, by Lemma
17, has the property that $(F_{i}\otimes_{R}k)_{\leq g}=0$ for all $i>n$ so
$\operatorname{Tor}^{R}_{i}(k,M)_{\leq g}=0$ for all $i>n$.

Next we show that
$\text{pd}_{R}^{\Gamma}(M)(g)\leq\text{td}_{R}^{\Gamma}(M)(g)$. Again this is
trivial if $\text{td}_{R}^{\Gamma}(M)(g)=\infty$ so suppose that
$\text{td}_{R}^{\Gamma}(M)(g)=n\in\mathbb{Z}_{\geq 0}$. By Proposition 15,
there exists a minimal BDF $\Gamma$-graded free resolution $F_{\bullet}$ of
$M$. Tensoring with $k$ gives a complex $F_{\bullet}\otimes_{R}k$ whose
differentials are all $0$. It follows that
$\operatorname{Tor}^{R}_{i}(k,M)_{g}=(F_{i}\otimes_{R}k)_{g}$. Note that by
Lemma 17, $(F_{i}\otimes_{R}k)_{\leq g}=0$ if and only if $(F_{i})_{g}=0$, and
for $i>n$ we have that the former is $0$ so the latter is too, implying that
$\text{pd}_{R}^{\Gamma}(M)(g)\leq n$. ∎

##  10\. $K$-series of Graded Modules

The rest of the paper will be dedicated to defining the Grothendieck group of
our abelian category $R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ and proving
it is isomorphic to $\mathbb{Z}[[Q]\\{\Gamma\\}$. This isomorphism is obtained
by computing the $K$-series of a BDF module, which is a slightly altered
version of the Hilbert series. In this section we will introduce $K$-series
and, by turning the set of such into a linear topological module, show that a
free BDF module is determined by both its Hilbert series and its $K$-series

###  10.1. Definition of $K$-series

######  Lemma 18.

Let $R$ be a PDCF $\Gamma$-graded $k$-algebra. Then
$\mathcal{H}_{\Gamma}(R)\in\mathbb{Z}[[Q]]\\{\Gamma\\}$ is a unit and
$\mathcal{H}_{\Gamma}(R)^{-1}\in\mathbb{Z}[[Q]]$.

###### Proof.

Let $\mathcal{H}_{\Gamma}(R)=a=\sum_{g\in Q}a_{g}\textbf{t}^{g}$, and note
that we may take the sum over $Q$ instead of $\Gamma$ by definition of $Q$.
Now because $R_{0}=k$, it follows that $a_{0}=\dim_{k}(R_{0})=1$. Since $Q$ is
well-founded by hypothesis, we can use well-founded recursion to define an
element $b=\sum_{g\in Q}a_{g}\textbf{t}^{g}\in\mathbb{Z}[[Q]]\\{\Gamma\\}$
which will be the inverse of $a$. Let $b_{0}=1$. Then suppose that for some
$g\in Q$ we have defined $b_{h}$ for all $h\in Q$ with $h<_{Q}g$. Then we let

$b_{g}=-\sum_{\begin{subarray}{c}u,v\in Q\smallsetminus\\{0\\}\\\
u+v=g\end{subarray}}b_{u}a_{v},$

which makes sense because if $u,v\in Q\smallsetminus\\{0\\}$ and $u+v=g$ then
$u<_{Q}g$ hence $b_{u}$ is already defined by induction.

Having defined $b$, we let $c=ab$ and we note first that
$c_{0}=\sum_{\begin{subarray}{c}u,v\in Q\\\
u+v=0\end{subarray}}a_{u}b_{v}=a_{0}b_{0}=1$ because $Q$ is pointed by
hypothesis. For $h\in Q$ with $h\neq 0$, we have

$c_{h}=\sum_{\begin{subarray}{c}u,v\in Q\\\
u+v=h\end{subarray}}a_{u}b_{v}=b_{h}a_{0}+\sum_{\begin{subarray}{c}u,v\in
Q\smallsetminus\\{0\\}\\\ u+v=h\end{subarray}}b_{u}a_{v}=0.$

Therefore $c=1$ as desired. From the definition of $b$ it is clear that
$\mathcal{H}_{\Gamma}(R)^{-1}\in\mathbb{Z}[[Q]]$ ($\mathbb{Z}[[Q]]$ denotes
the Laurent series which are supported on $Q$). ∎

######  Definition 13.

The _$K$ -series_ of a $\Gamma$-graded $R$-module $M$ is defined to be
$\mathcal{K}_{\Gamma}(M)=\mathcal{H}_{\Gamma}(M)\cdot\mathcal{H}_{\Gamma}(R)^{-1}$.

An important property of $K$-series is that they are additive, as described in
the following result.

######  Proposition 23.

Hilbert series and $K$-series are additive i.e. if $0\to L\to M\to N\to 0$ is
an exact sequence of BDF $R$-modules then
$\mathcal{H}_{\Gamma}(L)-\mathcal{H}_{\Gamma}(M)+\mathcal{H}_{\Gamma}(N)=0$
and
$\mathcal{K}_{\Gamma}(L)-\mathcal{K}_{\Gamma}(M)+\mathcal{K}_{\Gamma}(N)=0$.

###### Proof.

If $0\to L\to M\to N\to 0$ is an exact sequence of BDF $R$-modules then
because kernels and images of graded module homomorphisms are graded
submodules, it follows that $0\to L_{g}\to M_{g}\to N_{g}\to 0$ is an exact
sequence of $k$-vector spaces for all $g\in\Gamma$. This implies that
$\dim_{k}(L_{g})-\dim_{k}(M_{g})+\dim_{k}(N_{g})=0$, which is exactly the
coefficient of $\textbf{t}^{g}$ in
$\mathcal{H}_{\Gamma}(L)-\mathcal{H}_{\Gamma}(M)+\mathcal{H}_{\Gamma}(N)$.
Hence the latter is $0$. Multiplying by $\mathcal{H}_{\Gamma}(R)^{-1}$ gives
the final desired result. ∎

###  10.2. Linear Topological Modules

Here we give a brief summary of the basics of linear topological modules that
we will need to define a topology on $\mathbb{Z}[[Q]]\\{\Gamma\\}$. A
_directed filtration_ of an $R$-module $M$ consists of a non-empty directed
set $D$ (i.e. a set with a reflexive and transitive binary operation where
every pair of elements has an upper bound) together with an order-reversing
function $F:D\to\text{Sub}_{R}(M)$ (where $\text{Sub}_{R}(M)$ denotes the set
of $R$-submodules of $M$ and is a poset under inclusion). By definition, if
$d,e\in D$ then there exists $t\in D$ with $d\leq t$ and $e\leq t$, and hence
$F_{d}\supseteq F_{t}$ and $F_{e}\supseteq F_{t}$, implying that
$F_{t}\subseteq F_{d}\cap F_{e}$. A _filtered $R$-module_ is an $R$-module
with a directed filtration on it.

Recall that a _topological $R$-module_ is an $R$-module $M$ with a topology on
it such that addition $M\times M\to M$ which sends $m,n\mapsto m+n$, and
scalar multiplication $R\times M\to M$ which sends $r,m\mapsto rm$ is
continuous (where $R$ is given the discrete topology).

######  Proposition 24.

Let $(F,D)$ be a directed filtration of an $R$-module $M$.

  1. (1)

The set $B=\\{a+F_{d}\mid a\in\Gamma,d\in D\\}$ is a base for a topology on
$M$ which we call the _$F$ -adic_ topology.

  2. (2)

The $F$-adic topology makes $M$ into a topological $R$-module.

###### Proof.

  1. (1)

Clearly $B$ covers $M$ since $D$ is non-empty and for any $x\in M$ and any
$d\in D$, we have that $x\in x+F_{d}$. Next, given $x+F_{d}$ and $y+F_{e}$
with $x,y\in M$ and $d,e\in D$, if $z\in(x+F_{d})\cap(y+F_{e})$ then by
hypothesis there exists $t\in D$ such that $F_{t}\subseteq F_{d}\cap F_{e}$.
Clearly then we have $z+F_{t}\subseteq z+F_{d}=x+F_{d}$ and similarly
$z+F_{t}\subseteq y+F_{e}$, hence $z+F_{t}\subseteq(x+F_{d})\cap(y+F_{e})$ as
desired.

  2. (2)

We first show that addition $M\times M\to M$ is continuous. Suppose that
$x+y\in z+F_{d}$. Then $x+F_{d}$ and $y+F_{d}$ are neighborhoods of $x$ and
$y$ respectively and $(x+F_{d})+(y+F_{d})=x+y+F_{d}=z+F_{d}$, so
$(x+F_{d})\times(y+F_{d})$ is a neighborhood of $(x,y)$ in $M\times M$ which
maps into $z+F_{d}$ under $+$, hence indeed addition is continuous. Finally to
show that scalar multiplication is continuous, suppose $rx\in z+F_{d}$. Then
we have that $r(x+F_{d})=rx+F_{d}=z+F_{d}$. Hence $\\{r\\}\times(x+F_{d})$ is
a neighborhood of $(r,x)$ which maps into $z+F_{d}$ under scalar
multiplication as desired.

∎

######  Definition 14.

A topological $R$-module $M$ is called a _linear topological $R$-module_ if
there exists a directed filtration of $M$ whose associated topology agrees
with that on $M$.

We now discuss convergence of series in a linear topological module. Let
$(F,D)$ be a directed filtration of an $R$-module $M$. A net $(x_{i}\mid i\in
I)$ in $M$ (i.e. a family of elements of $M$ indexed by a directed set $I$)
converges to $x\in M$ if and only if for all $d\in D$ there exists $i\in I$
such that if $j\geq i$, then $x_{j}-x\in F_{d}$. If $(x_{a}\mid a\in A)$ is a
family of elements of $M$, by definition the series $\sum_{a\in
A}^{\infty}x_{a}$ converges to $x$ if its net of partial sums $(\sum_{b\in
B}x_{b}\mid B\in\text{Fin}(A))$ does, where $\text{Fin}(A)$ denotes the set of
finite subsets of $A$ which is directed by containment. Explicitly this means
for all $d\in D$ there exists a finite set $B\subseteq A$ such that if
$B\subseteq C\subseteq A$ and $C$ is finite then $x-\sum_{c\in C}x_{c}\in
F_{d}$.

Later we will need to know that series in our linear topological modules
converge to at most one point. The following result gives us a method of
checking when this is true.

######  Proposition 25.

Let $(F,D)$ be a directed filtration of an $R$-module $M$. Then the $F$-adic
topology on $M$ is Hausdorff if and only if every net $(x_{i}\mid i\in I)$ in
$M$ converges to at most one point.

###### Proof.

This is true for general topological spaces, see [Kel75] Chapter 2, Theorem 3.
∎

Furthermore, for linear topological modules, the Hausdorff condition can be
checked easily simply by taking the intersection of the defining filtration.

######  Proposition 26.

If $(F,D)$ is a directed filtration of an $R$-module $M$ then the $F$-adic
topology on $M$ is Hausdorff if and only if $\bigcap_{d\in D}F_{d}=0$.

###### Proof.

See [Mat89] Section 8. ∎

###  10.3. The Topology of $\mathbb{Z}[[Q]]\\{\Gamma\\}$

We now apply the theory of linear topological modules that we just developed
to our study of $K$-series. First we note that there is an injective ring
homomorphism $\mathbb{Z}[[Q]]\to\mathbb{Z}[[Q]]\\{\Gamma\\}$ which makes
$\mathbb{Z}[[Q]]\\{\Gamma\\}$ into a $\mathbb{Z}[[Q]]$ module. Now, given any
$g\in\Gamma$, define $Z_{g}$ to be the set of series
$a\in\mathbb{Z}[[Q]]\\{\Gamma\\}$ with
$\text{Supp}_{\Gamma}(a)\cap(-\infty,g]=\emptyset$ where
$(-\infty,g]=\\{h\in\Gamma\mid h\leq_{Q}g\\}$.

######  Proposition 27.

$Z_{g}$ is a $\mathbb{Z}[[Q]]$ submodule of $\mathbb{Z}[[Q]]\\{\Gamma\\}$.

###### Proof.

It is obvious that $Z_{g}$ is a subgroup of $\mathbb{Z}[[Q]]\\{\Gamma\\}$.
Given $a\in Z_{g}$ and $b\in\mathbb{Z}[[Q]]$, for any $u\in\Gamma$ with
$u\leq_{Q}g$ by definition we have that $(ab)_{u}=\sum_{g+h=u}a_{g}b_{h}$. If
$h\notin Q$ then $b_{h}=0$ so $a_{g}b_{h}=0$. But if $h\in Q$ then $g+h=u$
implies that $g\leq_{Q}u\leq_{Q}g$ and hence $a_{g}=0$. Therefore $(ab)_{u}=0$
and $ab\in Z_{g}$. ∎

Given any subset $U$ of $\Gamma$, define $(-\infty,U]=\\{g\in\Gamma\mid\exists
u\in U,g\leq_{Q}u\\}$. Then let $\mathcal{P}_{\text{fin}}(\Gamma)$ be the set
of finite subsets $U\subseteq\Gamma$ with the relation $U\leq U^{\prime}$ if
and only if for all $u\in U$ there exists $u^{\prime}\in U^{\prime}$ such that
$u\leq_{Q}u^{\prime}$. For each $U\subseteq\Gamma$, define $Z_{U}$ to be the
subset consisting of series $f\in\mathbb{Z}[[Q]]\\{\Gamma\\}$ with
$\text{Supp}_{\Gamma}(f)\cap(-\infty,U]=\emptyset$. Note that for any subset
$U\subseteq\Gamma$ we have that $Z_{U}=\bigcap_{u\in U}Z_{u}$, hence $Z_{U}$
is a $\mathbb{Z}[[Q]]$-submodule of $\mathbb{Z}[[Q]]\\{\Gamma\\}$ for all
$U\in\mathcal{P}_{\text{fin}}(\Gamma)$.

######  Proposition 28.

The set $\mathcal{P}_{\text{fin}}(\Gamma)$ is directed and the function
$\mathcal{P}_{\text{fin}}(\Gamma)\to\text{Sub}_{\mathbb{Z}[[Q]]}(\mathbb{Z}[[Q]]\\{\Gamma\\})$
sending $U\mapsto Z_{U}$ is a directed filtration.

###### Proof.

The relation is clearly reflexive and transitive, and for any such $U$ and
$U^{\prime}$, $U\cup U^{\prime}$ is an upper bound for $U$ and $U^{\prime}$
which is also finite. If $U\leq U^{\prime}$, then clearly
$(\infty,U]\subseteq(-\infty,U^{\prime}]$, so for any $f\in Z_{U^{\prime}}$ we
have that $\text{Supp}_{\Gamma}(f)\cap(-\infty,U^{\prime}]=\emptyset$ so
certainly $\text{Supp}_{\Gamma}(f)\cap(-\infty,U]=\emptyset$ and hence $f\in
Z_{U}$. Hence the map is order-reversing. ∎

######  Proposition 29.

We have that $\bigcap_{U\in\mathcal{P}_{\text{fin}}(\Gamma)}Z_{U}=0$. As a
result the $Z$-adic topology on $\mathbb{Z}[[Q]]\\{\Gamma\\}$ is Hausdorff and
nets in $\mathbb{Z}[[Q]]\\{\Gamma\\}$ converge to at most one point.

###### Proof.

Suppose that $f\in\bigcap_{U\in\mathcal{P}_{\text{fin}}(\Gamma)}Z_{U}$. Then
given any $g\in\Gamma$, we have that $f\in Z_{g}$ so in particular $f_{g}=0$.
Since $g$ was arbitrary, it follows that $f=0$. The desired conclusions follow
from Propositions 25 and 26. ∎

Our main application of this topology is to compute the $K$-series of a free
BDF module. We begin with two technical lemmas regarding the multiplication of
convergent series by elements of the ring, and then prove the desired resut
over the course of the next three lemmas.

######  Lemma 19.

If $(F,D)$ is a directed filtration of an $R$-module $M$ then for any $r\in
R$, multiplication by $r$ gives a continuous function $M\to M$.

###### Proof.

Multiplication by $r$ is equal to the composition of the continuous functions
$M\to R\times M$ given by $m\mapsto(r,m)$ and scalar multiplication $R\times
M\to M$. ∎

######  Lemma 20.

If $(F,D)$ is a directed filtration of an $R$-module $M$ and a series
$\sum_{a\in A}^{\infty}x_{a}$ in $M$ converges to $x$ then the series
$\sum_{a\in A}^{\infty}rx_{a}$ converges to $rx$.

###### Proof.

By Lemma 19, multiplication by $r$ is a continuous function $M\to M$, and
continuous functions preserve convergence of nets, hence the net $(r\sum_{b\in
B}x_{b}\mid B\in\text{Fin}(A))$ converges to $rx$, but because each $B$ is a
finite set, we obtain $r\sum_{b\in B}x_{b}=\sum_{b\in B}rx_{b}$, so this is
actually the net of partial sums of the series $\sum_{a\in A}^{\infty}rx_{a}$.
∎

######  Lemma 21.

If $(g_{i}\mid i\in I)$ is a BDF family of elements of $\Gamma$ then
$\mathcal{K}_{\Gamma}(\bigoplus_{i\in I}R(-g_{i}))=\sum_{i\in
I}\mathcal{K}_{\Gamma}(R(-g_{i}))$, i.e. the RHS converges to the LHS in the
$Z$-adic topology.

###### Proof.

Let $F=\bigoplus_{i\in I}R(-g_{i})$. We first show that
$\mathcal{H}_{\Gamma}(F)=\sum_{i\in I}\mathcal{H}_{\Gamma}(R(-g_{i}))$. Let
$U\subseteq\Gamma$ be a finite set. Because $(g_{i}\mid i\in I)$ is a BDF
family, the set $\\{g_{i}\mid i\in I\\}$ is downward finite, hence for each
$u\in U$ the set $\\{g_{i}\mid g_{i}\leq_{Q}u\\}$ is finite and furthermore
for each $g_{i}$ in this set, the set $\\{j\in I\mid g_{j}=g_{i}\\}$ is also
finite. Since there are only finitely many elements of $U$, the set $B=\\{i\in
I\mid\exists g_{i}\in(-\infty,U]\\}$ is finite. If $C\subseteq I$ is finite
and $B\subseteq C$, we claim that $\mathcal{H}_{\Gamma}(F)-\sum_{i\in
C}\mathcal{H}_{\Gamma}(R(-g_{i}))\in Z_{U}$. First note that because $C$ is
finite, we have that $\sum_{i\in
C}\mathcal{H}_{\Gamma}(R(-g_{i}))=\mathcal{H}_{\Gamma}(\bigoplus_{i\in
C}R(-g_{i}))$.

Then if $g\in(-\infty,U]$, by Lemma 16, $\left(\bigoplus_{i\in
I}R(-g_{i})\right)_{g}=\bigoplus_{i\in I,g_{i}\leq_{Q}g}R(-g_{i})_{g}$, and
similarly $\left(\bigoplus_{i\in C}R(-g_{i})\right)_{g}=\bigoplus_{i\in
C,g_{i}\leq_{Q}g}R(-g_{i})_{g}$. But if $g_{i}\leq_{Q}g$ then $i\in B\subseteq
C$, hence these two direct sums are taken over the same set, i.e.
$\left(\bigoplus_{i\in I}R(-g_{i})\right)_{g}=\left(\bigoplus_{i\in
C}R(-g_{i})\right)_{g}$. In particular their dimensions are the same. Since
$g$ was an arbitrary element of $(-\infty,U]$, the Hilbert series
$\mathcal{H}_{\Gamma}(F)$ and $\mathcal{H}_{\Gamma}(\bigoplus_{i\in
C}R(-g_{i}))$ agree on $(-\infty,U]$, so indeed their difference is in
$Z_{U}$. It follows that $\mathcal{H}_{\Gamma}(F)=\sum_{i\in
I}\mathcal{H}_{\Gamma}(R(-g_{i}))$ as claimed.

Finally, using the fact that $\mathcal{H}_{\Gamma}(R)^{-1}\in\mathbb{Z}[[Q]]$
and Lemma 20 in step (1), we can compute the desired result:

$\displaystyle\mathcal{K}_{\Gamma}\left(\bigoplus_{i\in I}R(-g_{i})\right)$
$\displaystyle=\mathcal{H}_{\Gamma}(R)^{-1}\mathcal{H}_{\Gamma}\left(\bigoplus_{i\in
I}R(-g_{i})\right)=\mathcal{H}_{\Gamma}(R)^{-1}\sum_{i\in
I}\mathcal{H}_{\Gamma}(R(-g_{i}))$ $\displaystyle\overset{(1)}{=}\sum_{i\in
I}\mathcal{H}_{\Gamma}(R)^{-1}\mathcal{H}_{\Gamma}(R(-g_{i}))=\sum_{i\in
I}\mathcal{K}_{\Gamma}(R(-g_{i})).$

∎

######  Lemma 22.

Let $M$ be a BDF $\Gamma$-graded $R$-module, and $g\in\Gamma$. Then
$\mathcal{K}_{\Gamma}(M(-g))=\textbf{t}^{g}\mathcal{K}_{\Gamma}(M)$ and in
particular we have $\mathcal{K}_{\Gamma}(R(-g))=\textbf{t}^{g}$.

###### Proof.

By Lemma 14 we have that
$\mathcal{H}_{\Gamma}(M(-g))=\textbf{t}^{g}\mathcal{H}_{\Gamma}(M)$, so we
compute:

$\mathcal{K}_{\Gamma}(M(-g))=\mathcal{H}_{\Gamma}(M(-g))\mathcal{H}_{\Gamma}(R)^{-1}=\textbf{t}^{g}\mathcal{H}_{\Gamma}(M)\mathcal{H}_{\Gamma}(R)^{-1}=\textbf{t}^{g}\mathcal{K}_{\Gamma}(M)$

In particular this implies that
$\mathcal{K}_{\Gamma}(R(-g))=\textbf{t}^{g}\mathcal{K}_{\Gamma}(R)=\textbf{t}^{g}\mathcal{H}_{\Gamma}(R)\mathcal{H}_{\Gamma}(R)^{-1}=\textbf{t}^{g}$.
∎

######  Lemma 23.

If $(g_{i}\mid i\in I)$ is a BDF family of elements of $\Gamma$ then
$\sum_{i\in I}\textbf{t}^{g_{i}}=\sum_{g\in\Gamma}\\#\\{i\in I\mid
g_{i}=g\\}\textbf{t}^{g}$.

###### Proof.

Let $f_{g}=\\#\\{i\in I\mid g_{i}=g\\}$ so that $f=\sum_{g\in\Gamma}\\#\\{i\in
I\mid g_{i}=g\\}\textbf{t}^{g}$. Given any finite set $U\subseteq\Gamma$,
because $(g_{i}\mid i\in I)$ is a BDF family, the set $B=\\{i\in I\mid
g_{i}\in(-\infty,U]\\}$ is finite. If $C\subseteq\Gamma$ is finite and
$B\subseteq C$ then we have that $f-\sum_{i\in C}\textbf{t}^{g_{i}}\in Z_{U}$
because for any $g\in(-\infty,U]$, we have that $f_{g}$ is the size of the set
$\\{i\in I\mid g_{i}=g\\}$, but this set is contained in $B$ and hence in $C$,
so there are exactly $f_{g}$ terms of the sum $\sum_{i\in
C}\textbf{t}^{g_{i}}$ equal to $\textbf{t}^{g}$. ∎

We are now able to compute the $K$-series of a free BDF module.

######  Proposition 30.

If $(g_{i}\mid i\in I)$ is a BDF family of elements of $\Gamma$ then
$\mathcal{K}_{\Gamma}(\bigoplus_{i\in I}R(-g_{i}))=\sum_{g\in\Gamma}\\#\\{i\in
I\mid g_{i}=g\\}\textbf{t}^{g}$.

###### Proof.

We compute:

$\mathcal{K}_{\Gamma}(\bigoplus_{i\in I}R(-g_{i}))\overset{(1)}{=}\sum_{i\in
I}\mathcal{K}_{\Gamma}(R(-g_{i}))\overset{(2)}{=}\sum_{i\in
I}\textbf{t}^{g_{i}}\overset{(3)}{=}\sum_{g\in\Gamma}\\#\\{i\in I\mid
g_{i}=g\\}\textbf{t}^{g}$

Here (1) follows from Lemma 21, (2) follows from Lemma 22, and (3) follows
from Lemma 23. ∎

As a consequence, we obtain that free BDF modules are determined by their
$K$-series.

######  Corollary 5.

If $F$ and $F^{\prime}$ are free BDF $R$-modules then the following are
equivalent:

  1. (1)

$\mathcal{H}_{\Gamma}(F)=\mathcal{H}_{\Gamma}(F^{\prime})$

  2. (2)

$\mathcal{K}_{\Gamma}(F)=\mathcal{K}_{\Gamma}(F^{\prime})$

  3. (3)

$F\cong F^{\prime}$

###### Proof.

One obtains (1) $\iff$ (2) by multiplying by $\mathcal{H}_{\Gamma}(R)$ or
$\mathcal{H}_{\Gamma}(R)^{-1}$. Also (3) $\implies$ (1) is obvious. We now
show $(2)\implies(3)$. Given BDF $\Gamma$-graded free modules
$F=\bigoplus_{i\in I}R(-g_{i})$ and $F^{\prime}=\bigoplus_{j\in J}R(-h_{j})$
with $\mathcal{K}_{\Gamma}(F)=\mathcal{K}_{\Gamma}(F^{\prime})$, by
Proposition 30, we have that
$\mathcal{K}_{\Gamma}(F)=\sum_{g\in\Gamma}\\#\\{i\in I\mid
g_{i}=g\\}\textbf{t}^{g}$ and
$\mathcal{K}_{\Gamma}(F^{\prime})=\sum_{g\in\Gamma}\\#\\{j\in J\mid
h_{j}=g\\}\textbf{t}^{g}$, and hence for each $g\in\Gamma$, we have
$\\#\\{i\in I\mid g_{i}=g\\}=\\#\\{j\in J\mid h_{j}=g\\}$, so we obtain a
bijection between the standard bases of $F$ and $F^{\prime}$, yielding $F\cong
F^{\prime}$. ∎

##  11\. The Grothendieck Group of a Graded Ring

For the category of finitely generated ungraded or graded modules over a fixed
ring there are two typical ways to define the Grothendieck group of a
commutative ring $R$ which agree in favorable circumstances, for example when
the ring is a Noetherian regular local ring, or a polynomial ring in finitely
many variables. One the one hand, $K_{0}(R)$ (or $K_{\Gamma,0}(R)$ in the
graded case) is defined to be the Grothendieck group of the commutative monoid
of isomorphism classes of finitely generated projective (graded) $R$-modules
with addition given by $[P]+[Q]=[P\oplus Q]$. On the other hand, $G_{0}(R)$
(or $G_{\Gamma,0}(R)$ in the graded case) is defined to be the quotient of the
free abelian group on the set of isomorphism classes of all finitely generated
(graded) $R$-modules modulo the subgroup generated by expressions of the form
$[L]-[M]+[N]$ when there exists a short exact sequence $0\to L\to M\to N\to
0$. (See [Wei13] for details). In this section we give analogs of these
definitions for BDF $\Gamma$-graded modules and show that they agree for PDCF
$\Gamma$-graded $k$-algebras whose ideal maximal graded ideal $R_{+}$ is
generated by a regular sequence.

###  11.1. The Group $K^{\text{BDF}}_{\Gamma,0}(R)$ of a PDCF $\Gamma$-Graded
$k$-Algebra

We begin by recalling the Grothendieck group of a commutative monoid. If $Q$
is a commutative monoid, then there exists an abelian group $K$ (unique up to
a unique isomorphism) and a monoid homomorphism $i:Q\to K$ which satisfies the
following universal property: For all abelian groups $A$ and all monoid
homomorphisms $\varphi:Q\to A$ there exists a unique group homomorphism
$\tilde{\varphi}:K\to A$ such that $\varphi=\tilde{\varphi}\circ i$. Then $K$
is called the _Grothendieck group_ of the commutative monoid $Q$.

One construction of the Grothendieck group of $Q$ is to define an equivalence
relation $\sim$ on $Q\times Q$ by letting $(q_{1},q_{2})\sim(p_{1},p_{2})$ if
and only if there exists $g\in Q$ such that $q_{1}+p_{2}+g=p_{1}+q_{2}+g$. The
equivalence class of $(q,p)\in Q\times Q$ is denoted by $[q,p]$, and $K$ is
defined to be the set of equivalence classes $Q\times Q/\sim$ together with
the binary operation
$[q,p]+[q^{\prime},p^{\prime}]=[q+q^{\prime},p+p^{\prime}]$.

We will need the following proposition which describes when the Grothendieck
group of a commutative monoid is trivial.

######  Proposition 31.

Let $Q$ be a commutative monoid. Then its Grothendieck group $K$ is trivial if
and only if for all $p,q\in Q$ there exists $g\in Q$ such that $p+g=q+g$.

###### Proof.

Since the Grothendieck group of $Q$ is unique up to a unique isomorphism we
may assume that $K=Q\times Q/\sim$, and we note that the identity of $K$ is
$[0,0]$. We have that $K$ is trivial if and only if for all $p,q\in Q$,
$(p,q)\sim(0,0)$ which is true if and only if for all $p,q\in Q$ there exists
$g\in Q$ such that $p+g=q+g$. ∎

We define $R\text{-}\textbf{PMod}_{\Gamma}^{\text{BDF}}$ to be the category of
projective BDF $\Gamma$-graded $R$-modules. By Theorem 3, every such module is
actually free. Now define $K^{\text{BDF}}_{\Gamma,0}(R)$ to be the
Grothendieck group of the commutative monoid consisting of the set of
isomorphism classes of objects in
$R\text{-}\textbf{PMod}_{\Gamma}^{\text{BDF}}$, together with the operation
$[M]+[N]=[M\oplus N]$. We note that by Proposition 31 this group is not
trivial because $R$ and $R\oplus R$ are both free BDF $\Gamma$-graded
$R$-modules, if there existed an object $M$ of
$R\text{-}\textbf{PMod}_{\Gamma}^{\text{BDF}}$ with $R\oplus M\cong R\oplus
R\oplus M$ as $R$-modules then we would have $R_{0}\oplus M_{0}\cong
R_{0}\oplus R_{0}\oplus M_{0}$ as $k$-vector spaces which is impossible as
$R_{0}$ and $M_{0}$ are both finite-dimensional and $\dim_{k}(R_{0})=1$.
Recall that the famous Eilenberg swindle shows that the Grothendieck group of
the monoid consisting of all projective $R$-modules is trivial. We have
avoided this by considering only finite-dimensionally graded modules.

######  Note 5.

The category $R\text{-}\textbf{PMod}_{\Gamma}^{\text{BDF}}$ is split exact
(which can be embedded in the abelian category
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$), and therefore we could have
equivalently defined $K^{\text{BDF}}_{\Gamma,0}(R)$ to be the free abelian
group on the set of isomorphism classes of elements of
$R\text{-}\textbf{PMod}_{\Gamma}^{\text{BDF}}$ modulo the subgroup generated
by sums of the form $[L]-[M]+[N]$ whenever there is a short exact sequence
$0\to L\to M\to N\to 0$ (see [Wei13] Example 7.1.2).

######  Proposition 32.

Every element of $K^{\text{BDF}}_{\Gamma,0}(R)$ can be written as
$[F]-[F^{\prime}]$ where $F$ and $F^{\prime}$ are free BDF $R$-modules.

###### Proof.

By definition every element of $K^{\text{BDF}}_{\Gamma,0}(R)$ is of the form
$[F_{1}]+\ldots+[F_{n}]-[F^{\prime}_{1}]-\ldots-[F^{\prime}_{m}]$ for $F_{i}$
and $F^{\prime}_{j}$ free BDF $R$-modules. By definition of addition in the
Grothendieck group, this element is equal to $[F]-[F^{\prime}]$ where
$F=F_{1}\oplus\ldots\oplus F_{n}$ and
$F^{\prime}=F^{\prime}_{1}\oplus\ldots\oplus F^{\prime}_{m}$. Again $F$ and
$F^{\prime}$ are free BDF $R$-modules, as desired. ∎

######  Theorem 5.

Let $R$ be a PDCF $\Gamma$-graded $k$-algebra. There is an isomorphism
$K^{\text{BDF}}_{\Gamma,0}(R)\overset{\sim}{\to}\mathbb{Z}[[Q]]\\{\Gamma\\}$
given by sending $[F]\mapsto\mathcal{K}_{\Gamma}(F)$.

###### Proof.

By the universal property of Grothendieck groups, to define a group
homomorphism $K^{\text{BDF}}_{\Gamma,0}(R)\to\mathbb{Z}[[Q]]\\{\Gamma\\}$ it
suffices to define a map from the commutative monoid of isomorphism classes of
free $\Gamma$-graded BDF $R$-modules to $\mathbb{Z}[[Q]]\\{\Gamma\\}$ that
respects direct sum, and by Proposition 23 we have
$\mathcal{K}_{\Gamma}(F\oplus
F^{\prime})=\mathcal{K}_{\Gamma}(F)+\mathcal{K}_{\Gamma}(F^{\prime})$ so such
a map does indeed exist. To show injectivity, we note that if
$[F]-[F^{\prime}]\mapsto 0$ then
$\mathcal{K}_{\Gamma}(F)=\mathcal{K}_{\Gamma}(F^{\prime})$ hence by Corollary
5, $F\cong F^{\prime}$ and so $[F]=[F^{\prime}]$ in
$K^{\text{BDF}}_{\Gamma,0}(R)$.

For surjectivity, let
$a=\sum_{g\in\Gamma}a_{g}\textbf{t}^{g}\in\mathbb{Z}[[Q]]\\{\Gamma\\}$ be such
that $a_{g}\geq 0$ for all $g\in\Gamma$ and define $I$ to be the subset of
$\text{Supp}_{\Gamma}(a)\times\mathbb{N}$ containing the elements $(g,n)$ such
that $n\leq a_{g}$. Define a function $h:I\to\Gamma$ by $(g,n)\mapsto
g:=h_{(g,n)}$, so that we get a family $(h_{i}\mid i\in I)$ of elements of
$\Gamma$. We claim that this is a BDF family. Indeed, for each $g\in\Gamma$,
we have $\\#\\{i\in I\mid h_{i}=g\\}=a_{g}$ and the set $\\{h_{i}\mid i\in
I\\}$ is exactly $\text{Supp}_{\Gamma}(a)$ which is finitely bounded below and
downward finite by definition of $\mathbb{Z}[[Q]]\\{\Gamma\\}$. Therefore by
Proposition 13, $F:=\bigoplus_{i\in I}R(-h_{i})$ is a BDF $\Gamma$-graded
$R$-module.

Then under our map
$K^{\text{BDF}}_{\Gamma,0}(R)\to\mathbb{Z}[[Q]]\\{\Gamma\\}$, we have that
$[F]\mapsto\mathcal{K}_{\Gamma}(F)$ and the latter is equal to
$\sum_{g\in\Gamma}\\#\\{i\in I\mid h_{i}=g\\}\textbf{t}^{g}$ by Proposition
30. Since $a=\sum_{g\in\Gamma}\\#\\{i\in I\mid h_{i}=g\\}$ as well, we have
$\mathcal{K}_{\Gamma}(F)=a$.

Surjectivity follows since every element of $\mathbb{Z}[[Q]]\\{\Gamma\\}$ can
be written as $a-a^{\prime}$ where $a_{g},a^{\prime}_{g}\geq 0$. ∎

###  11.2. The Group $G^{\text{BDF}}_{\Gamma,0}(R)$ of a Graded Ring

Define $G^{\text{BDF}}_{\Gamma,0}(R)$ to be the free abelian group on the set
of isomorphism classes of objects in
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$, modulo the subgroup generated
by expressions of the form $[L]-[M]+[N]$ whenever there is a short exact
sequence $0\to L\to M\to N\to 0$. We now prove a series of lemmas which
culminate in the result that $K^{\text{BDF}}_{\Gamma,0}(R)$ and
$G^{\text{BDF}}_{\Gamma,0}(R)$ are isomorphic. We begin by defining the map
that will become this isomorphism.

######  Lemma 24.

Let $R$ be a PDCF $k$-algebra. There exists a group homomorphism
$\Phi:K^{\text{BDF}}_{\Gamma,0}(R)\to G^{\text{BDF}}_{\Gamma,0}(R)$ which
sends the isomorphism class $[F]$ in $K^{\text{BDF}}_{\Gamma,0}(R)$ of a free
module $F$ to the residue of $[F]$ in the quotient
$G^{\text{BDF}}_{\Gamma,0}(R)$.

###### Proof.

There is a function $\varphi$ from the set of isomorphism classes of objects
in $R\text{-}\textbf{PMod}_{\Gamma}^{\text{BDF}}$ to
$G^{\text{BDF}}_{\Gamma,0}(R)$ which sends an isomorphism class $[F]$ to its
residue in the quotient $G^{\text{BDF}}_{\Gamma,0}(R)$. This is well-defined
because $R\text{-}\textbf{PMod}_{\Gamma}^{\text{BDF}}$ is a subcategory of
$R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$, so if modules are isomorphic in
the former, they are also isomorphic in the latter. Furthermore, it is a
monoid homomorphism because given free modules $F$ and $F^{\prime}$ in
$R\text{-}\textbf{PMod}_{\Gamma}^{\text{BDF}}$, we have that
$\varphi([F]+[F^{\prime}])=\varphi([F\oplus F^{\prime}])=[F\oplus F^{\prime}]$
and since there is an exact sequence $0\to F\to F\oplus F^{\prime}\to
F^{\prime}\to 0$ in $R\text{-}\textbf{Mod}_{\Gamma}^{\text{BDF}}$ we have that
$[F\oplus F^{\prime}]=[F]+[F^{\prime}]$ in $G^{\text{BDF}}_{\Gamma,0}(R)$, and
the latter is equal to $\varphi([F])+\varphi([F^{\prime}])$. Therefore the
universal property of the Grothendieck group of a monoid gives the desired
group homomorphism $K^{\text{BDF}}_{\Gamma,0}(R)\to
G^{\text{BDF}}_{\Gamma,0}(R)$. ∎

Next we give a technical lemma, followed by a lemma which allows us to express
classes of modules $[M]\in G^{\text{BDF}}_{\Gamma,0}(R)$ in terms of free
modules using $\Gamma$-finite free resolutions.

######  Lemma 25.