# An arithmetic valuative criterion
for proper maps of tame algebraic stacks

Giulio Bresciani  and  Angelo Vistoli  Scuola Normale Superiore
Piazza dei Cavalieri 7
56126 Pisa
Italy<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

The valuative criterion for proper maps of schemes has many applications in
arithmetic, e.g. specializing $\mathbb{Q}_{p}$-points to
$\mathbb{F}_{p}$-points. For algebraic stacks, the usual valuative criterion
for proper maps is ill-suited for these kind of arguments, since it only gives
a specialization point defined over an extension of the residue field, e.g. a
$\mathbb{Q}_{p}$-point will specialize to an $\mathbb{F}_{p^{n}}$-point for
some $n$. We give a new valuative criterion for proper maps of tame stacks
which solves this problem and is well-suited for arithmetic applications. As a
consequence, we prove that the Lang-Nishimura theorem holds for tame stacks.

The second author was partially supported by research funds from Scuola
Normale Superiore, project SNS19_B_VISTOLI, and by PRIN project “Derived and
underived algebraic stacks and applications”. The paper is based upon work
partially supported by the Swedish Research Council under grant no. 2016-06596
while the second author was in residence at Institut Mittag-Leffler in
Djursholm

##  1\. Introduction

The well known and extremely useful valuative criterion for properness says,
in particular, that if $X\longrightarrow Y$ is a proper morphism of schemes,
$R$ is a DVR with quotient field $K$ and residue field $k$, and we have a
commutative diagram

${\operatorname{Spec}K}$${X}$${\operatorname{Spec}R}$${S}$

there exists a unique lifting $\operatorname{Spec}R\longrightarrow X$ of
$\operatorname{Spec}R\longrightarrow Y$ extending
$\operatorname{Spec}K\longrightarrow X$. This has many arithmetic
applications: most of them use that the statement above ensures the existence
of a lifting $\operatorname{Spec}k\longrightarrow X$ of the composite
$\operatorname{Spec}k\subseteq\operatorname{Spec}R\longrightarrow Y$.

If $X$ and $Y$ are algebraic stack, and $X\longrightarrow Y$ is a proper
morphism, then this fails, even in very simple examples, unless
$X\longrightarrow Y$ is representable. The correct general statement is that
there exists a local extension of DVR $R\longrightarrow R^{\prime}$, such that
if we denote by $K^{\prime}$ the fraction field of $R^{\prime}$, the composite
$\operatorname{Spec}R^{\prime}\longrightarrow\operatorname{Spec}R\longrightarrow
Y$ has a lifting $\operatorname{Spec}R^{\prime}\longrightarrow X$ extending
the composite
$\operatorname{Spec}K^{\prime}\longrightarrow\operatorname{Spec}K\longrightarrow
X$ (see for example [Sta21, Tag 0CLY]). For arithmetic applications this is
problematic, because the extension $R\subseteq R^{\prime}$ will typically
induce a nontrivial extension of residue fields, so it does not imply that
$\operatorname{Spec}k\longrightarrow Y$ lifts to
$\operatorname{Spec}k\longrightarrow X$, as in the case of schemes.

When $X$ and $Y$ are Deligne–Mumford stacks over a field of characteristic $0$
a substitute was found by the first author in [Bre21, Theorem 1]. In this note
we we extend this, in a somewhat more precise form, to positive and mixed
characteristic. In this context the correct generality is that of _tame
stacks_ , in the sense of [AOV08]. Tame stacks are algebraic stacks with
finite inertia, such that the automorphism group scheme of any object over a
field is linearly reductive. In characteristic $0$ they coincide with
Deligne–Mumford stacks with finite inertia, but in positive and mixed
characteristic there are Deligne–Mumford stacks with finite inertia that are
not tame, and tame stacks that are not Deligne–Mumford.

In our version the role of the DVR $R^{\prime}$ above is played by a _root
stack_ $\sqrt[n]{\operatorname{Spec}R}$; this is not a scheme, but a tame
stack with a map
$\sqrt[n]{\operatorname{Spec}R}\longrightarrow\operatorname{Spec}R$, which is
an isomorphism above $\operatorname{Spec}K\subseteq\operatorname{Spec}R$ (see
the discussion at the beginning of §3). The statement of our main theorem 3.1
is that if $X\longrightarrow Y$ is a proper morphism of tame algebraic stacks
and we have a commutative diagram as above, there exists a unique positive
integer $n$ and a unique representable lifting
$\sqrt[n]{\operatorname{Spec}R}\longrightarrow X$ of the composite
$\sqrt[n]{\operatorname{Spec}R}\longrightarrow\operatorname{Spec}R\longrightarrow
Y$ extending $\operatorname{Spec}K\longrightarrow X$. The key point for
arithmetic applications is that the closed point
$\operatorname{Spec}k\longrightarrow\operatorname{Spec}R$ lifts to
$\operatorname{Spec}k\longrightarrow\sqrt[n]{\operatorname{Spec}R}$. This
statement is much harder to prove in arbitrary characteristic than in
characteristic $0$.

Besides the original application to Grothendieck’s section conjecture in
[Bre21], this valuative criterion has been applied in [BVb] to give new proofs
and stronger versions of the genericity theorem for essential dimension.

Recall that the Lang-Nishimura theorem states that the property of having a
rational point is a birational invariant of smooth proper varieties. Another
consequence of our version of the valuative criterion is that the Lang-
Nishimura theorem generalizes to tame stacks, see 4.1. Our version of the
Lang-Nishimura theorem has an immediate corollary, which we find surprising:
if $\mathcal{M}$ is a smooth tame stack which is generically a scheme and
$\overline{M}\to M$ is a resolution of singularities of the coarse moduli
space $\mathcal{M}\to M$, then a rational point of $M(k)$ lifts to
$\mathcal{M}$ if and only if it lifts to $\overline{M}$. This gives a hint of
the applications of the Lang–Nishimura theorem to fields of moduli, which will
be the subject of the forthcoming papers [BVa, Bre].

##  2\. Notations and conventions

We will follow the conventions of [Knu71] and [LMB00]; so the diagonals of
algebraic spaces and algebraic stacks will be separated and of finite type. In
particular, every algebraic space will be _decent_ , in the sense of [Sta21,
Definition 03I8].

We will follow the terminology of [AOV08]: a _tame stack_ is an algebraic
stack $X$ with finite inertia, such that its geometric points have linearly
reductive automorphism group. This is equivalent to requiring that $X$ is
étale locally over its moduli space a quotient by a finite, linearly reductive
group scheme [AOV08, Theorem 3.2].

More generally, a morphism $f\colon X\longrightarrow Y$ of algebraic stacks is
_tame_ if the relative inertia group stack $I_{X/Y}\longrightarrow X$, defined
as in [Sta21, Section 050P], is finite and has linearly reductive geometric
fibers. See [AOV11, §3].

Using [Sta21, Lemma 0CPK] one can easily prove the following.

######  Proposition 2.1.

Let $f\colon X\longrightarrow Y$ be a morphism of algebraic stacks. The
following conditions are equivalent.

  1. (1)

$f$ is tame.

  2. (2)

If $Z\longrightarrow Y$ is a morphism, and $Z$ is a scheme, then
$Z\times_{Y}X$ is a tame stack.

  3. (3)

If $Z\longrightarrow Y$ is a morphism, and $Z$ a tame stack, then
$Z\times_{Y}X$ is also tame.

Furthermore, if $X$ is tame, then the morphism $f$ is also tame.

##  3\. The valuative criterion

A basic example of tame stacks is _root stacks_ (see [AGV08, Appendix B2]). We
will need this in the following situation. Let $R$ be a DVR with uniformizing
parameter $\pi$ and residue field
$k\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle def}}{=}}}R/(\pi)$. If
$n$ is a positive integer, we will denote by $\sqrt[n]{\operatorname{Spec}R}$
the $n^{\text{th}}$ root of the Cartier divisor
$\operatorname{Spec}k\subseteq\operatorname{Spec}R$. It is a stack over
$\operatorname{Spec}R$, such that given a morphism $\phi\colon
T\longrightarrow\operatorname{Spec}R$, the groupoid of liftings
$T\longrightarrow\sqrt[n]{\operatorname{Spec}R}$ is equivalent to the groupoid
whose objects are triples $(L,s,\alpha)$, where $L$ is an invertible sheaf on
$T$, $s\in L(T)$ is a global section of $L$, and $\alpha$ is an isomorphism
$L^{\otimes n}\simeq\mathcal{O}_{T}$, such that $\alpha(s^{\otimes
n})=\phi^{\sharp}(\pi)$. Alternatively, $\sqrt[n]{\operatorname{Spec}R}$ can
be described as the quotient stack
$[\operatorname{Spec}R[t]/(t^{n}-\pi)/\boldsymbol{\mu}_{n}]$, where the action
of $\boldsymbol{\mu}_{n}$ on $\operatorname{Spec}R[t]/(t^{n}-\pi)$ is by
multiplication on $t$. The morphism
$\rho\colon\sqrt[n]{\operatorname{Spec}R}\longrightarrow\operatorname{Spec}R$
is an isomorphism outside of
$\operatorname{Spec}k\subseteq\operatorname{Spec}R$, while the reduced fiber
$\rho^{-1}(\operatorname{Spec}k)_{\mathrm{red}}$ is non-canonically isomorphic
to the classifying stack $\mathcal{B}_{k}\boldsymbol{\mu}_{n}$. In particular
the embedding $\operatorname{Spec}k\longrightarrow\operatorname{Spec}R$ lifts
to a morphism
$\operatorname{Spec}k\longrightarrow\sqrt[n]{\operatorname{Spec}R}$.

The following is our version of the valuative criterion for properness.

######  Theorem 3.1.

Let $f\colon X\longrightarrow Y$ be a tame, proper morphism of algebraic
stacks, $R$ a DVR with quotient field $K$. Suppose that we have a
$2$-commutative square

${\operatorname{Spec}K}$${X}$${\operatorname{Spec}R}$${Y}$$\scriptstyle{f}$

Then there exists a unique positive integer $n$ and a representable lifting
$\sqrt[n]{\operatorname{Spec}R}\longrightarrow X$ of the given morphism
$\operatorname{Spec}R\longrightarrow Y$, making the diagram

${\operatorname{Spec}K\
}$${X}$${\sqrt[n]{\operatorname{Spec}R}}$${\operatorname{Spec}R}$${Y}$

$2$-commutative. Furthermore, the lifting is unique up to a unique
isomorphism.

######  Corollary 3.2.

In the situation above, if $k$ is the residue field of $R$, the composite
$\operatorname{Spec}k\subseteq\operatorname{Spec}R\longrightarrow Y$ has a
lifting $\operatorname{Spec}k\longrightarrow X$.

As one would expect, these statements fail without the tameness hypothesis,
even when $Y$ is a scheme and $X$ is a separated Deligne–Mumford stack.

######  Example 3.3.

Let $p$ be a prime, $R$ a DVR whose fraction field $K$ has characteristic $0$
and contains a $p$-th root of $1$, denoted by $\zeta_{p}$, while its residue
field $k$ has characteristic $p$ and is not perfect. An example would be the
localization of $\mathbb{Z}[\zeta_{p}][t]$ at a prime ideal of height 1
containing $p$.

Choose an element $a\in R^{*}$ whose image in $k$ is not a $p$-th power, and
set $R^{\prime}\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}R(\sqrt[p]{a})$. Then $R^{\prime}$ is a DVR, since the
$R^{\prime}\otimes_{R}k=k(\sqrt[p]{\overline{a}})$ is a field (here
$\overline{a}$ is the class of $a$ in $k$). Write $K^{\prime}=K(\sqrt[p]{a})$
for its fraction field and $k^{\prime}=k(\sqrt[p]{\overline{a}})$ for its
residue field.

Call $C_{p}$ the cyclic group of order $p$ generated by $\zeta_{p}\in K^{*}$.
The extension $K^{\prime}/K$ is Galois with cyclic Galois group $C_{p}$ acting
by $\sqrt[p]{a}\mapsto\zeta_{p}\sqrt[p]{a}$. The action of $C_{p}$ on
$K^{\prime}$ naturally extends to $R^{\prime}$.

Let $X$ be the quotient stack $[\operatorname{Spec}R^{\prime}/C_{p}]$; this is
a separated Deligne–Mumford stack, but it is not tame. Since
$(R^{\prime})^{C_{p}}=R$ the moduli space of $X$ is $\operatorname{Spec}R$,
and we have we have a natural map $X\longrightarrow\operatorname{Spec}R$,
which is an isomorphism over
$\operatorname{Spec}K\subseteq\operatorname{Spec}R$. Since $k^{\prime}/k$ is
purely inseparable, then $X_{k}(k)$ is empty: such a $k$-rational point would
correspond to a $C_{p}$-torsor $\operatorname{Spec}A\to\operatorname{Spec}k$
with an equivariant morphism
$\operatorname{Spec}A\to\operatorname{Spec}k^{\prime}$ and thus an embedding
of $k^{\prime}$ in the étale $k$-algebra $A$, which is clearly absurd. In
particular, there is no map $\sqrt[n]{\operatorname{Spec}R}\to X$ for any $n$.

######  Definition 3.4.

In the situation of Theorem 3.1, we call the integer $n$ the _loop index_ of
the morphism $\operatorname{Spec}K\to X$ at the place associated with
$R\subseteq K$. If the loop index is $1$, we say that $\operatorname{Spec}K\to
X$ is _untangled_.

######  Lemma 3.5.

Let $R\subseteq R^{\prime}$ be an extension of DVRs with ramification index
$e$, and let $K\subseteq K^{\prime}$ be the fraction fields of $R$ and
$R^{\prime}$ respectively. If $X$ is a tame stack proper over $R$ and
$\operatorname{Spec}K\to X$ is a morphism with loop index $n$, the composite
$\operatorname{Spec}K^{\prime}\to\operatorname{Spec}K\to X$ has loop index
equal to $n/\gcd(n,e)$.

###### Proof.

Write $m\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}n/\gcd(n,e)$; the statement follows from the fact that there is a
natural representable morphism
$\sqrt[m]{\operatorname{Spec}R^{\prime}}\to\sqrt[n]{\operatorname{Spec}R}$
inducing the given morphism
$\operatorname{Spec}K^{\prime}\longrightarrow\operatorname{Spec}K$. ∎

We spend the rest of this section proving Theorem 3.1. Given a DVR $R$ and
$\pi\in R$ a uniformizing parameter, write
$R^{(n)}\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}R[t]/(t^{n}-\pi)$,
$K^{(n)}\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}K[t]/(t^{n}-\pi)$. We have
$\sqrt[n]{\operatorname{Spec}R}=[\operatorname{Spec}R^{(n)}/\boldsymbol{\mu}_{n}]$.

######  Lemma 3.6.

Let $R$ be a DVR, $m,n$ integers. A morphism
$\sqrt[m]{\operatorname{Spec}R}\to\sqrt[n]{\operatorname{Spec}R}$ over $R$
exists if and only if $n|m$, and in this case it is unique up to equivalence.

###### Proof.

This follows from the fact that a section
$\operatorname{Spec}R^{(m)}\to\sqrt[n]{\operatorname{Spec}R}$ exists if and
only if $n|m$, and in this case it is unique up to equivalence. ∎

######  Lemma 3.7.

Let $R$ be a DVR, $D\subset\operatorname{Spec}R$ the divisor corresponding to
the closed point, $m,n_{i},r_{i}$ for $i=1,\dots,m$ positive integers, with
$n_{i}\geq 2$ for every $i$. The fibered product

$X=\prod_{i=1}^{m}\sqrt[n_{i}]{\operatorname{Spec}R,r_{i}D}.$

is normal if and only if $m=1$ and $r_{1}=1$.

###### Proof.

Let $V_{n,r}$ be the scheme $\operatorname{Spec}R[t,s]_{s}/(t^{n}s-\pi^{r})$,
there is an action of $\mathbb{G}_{m}$ on $V_{n,r}$ given by
$(\lambda,t,s)\mapsto(\lambda t,\lambda^{-n}s)$ and
$\sqrt[n]{\operatorname{Spec}R,rD}\simeq[V_{n,r}/\mathbb{G}_{m}]$, see [AGV08,
Appendix B]. Consider the fibered product

$Y=\prod_{i}V_{n_{i},r_{i}}=R[t_{1},s_{1},\dots,t_{m},s_{m}]/(t_{i}^{n_{i}}s_{i}-\pi^{r_{i}}).$

The prime ideal $\mathfrak{p}=(t_{1},\dots,t_{m},\pi)$ is the generic point of
the special fiber and has height $1$. Since
$V_{r,n}\to\sqrt[n]{\operatorname{Spec}R,rD}$ is smooth, then $Y\to X$ is
smooth and hence $X$ is normal if and only if $Y$ is normal at $\mathfrak{p}$.

Now consider the prime ideal $\mathfrak{p}_{0}=(t_{1},\dots,t_{m},\pi)\subset
R[t_{1},s_{1},\dots,t_{m},s_{m}]$, $\mathfrak{p}$ and $\mathfrak{p}_{0}$ have
equal residue fields and there is a natural surjective linear map
$\mathfrak{p}_{0}/\mathfrak{p}_{0}^{2}\to\mathfrak{p}/\mathfrak{p}^{2}$. We
have that $\mathfrak{p}_{0}/\mathfrak{p}_{0}^{2}$ has dimension $m+1$
generated by the classes of $[t_{i}],[\pi]$. If $m=1$ and $r_{1}=1$, then
$\mathfrak{p}_{0}/\mathfrak{p}_{0}^{2}$ has dimension $2$ and $[\pi]$ is in
the kernel of
$\mathfrak{p}_{0}/\mathfrak{p}_{0}^{2}\to\mathfrak{p}/\mathfrak{p}^{2}$, hence
$Y$ is normal at $\mathfrak{p}$.

On the other hand, assume that $Y$ is normal at $\mathfrak{p}$, so that
$\mathfrak{p}/\mathfrak{p}^{2}$ has dimension $1$. Since $n_{i}\geq 2$ for
every $i$, the kernel of
$\mathfrak{p}_{0}/\mathfrak{p}_{0}^{2}\to\mathfrak{p}/\mathfrak{p}^{2}$ is
generated by the classes $[\pi^{r_{i}}]$ and hence has dimension $1$ if
$r_{i}=1$ for some $i$, and dimension $0$ otherwise. Since
$\mathfrak{p}_{0}/\mathfrak{p}_{0}^{2}$ has dimension $m+1$ and
$\mathfrak{p}/\mathfrak{p}^{2}$ has dimension $1$, this implies that $m=1$ and
$r_{1}=1$. ∎

######  Lemma 3.8.

Let $A$ be a Dedekind domain with fraction field $K$,
$D\subset\operatorname{Spec}A$ an effective, reduced divisor. Let
$f:X\longrightarrow\sqrt[n]{(\operatorname{Spec}A,D)}$ be a representable,
proper morphism. Every generic section $\operatorname{Spec}K\to X$ of $f$
extends uniquely to a global section $\sqrt[n]{(\operatorname{Spec}A,D)}\to
X$.

###### Proof.

Let $Y\subset X$ be the schematic image of a generic section
$\operatorname{Spec}K\to X$, we want to prove that
$Y\longrightarrow\sqrt[n]{(\operatorname{Spec}A,D)}$ is an isomorphism. Since
the problem is local, we may assume that $A=R$ is a DVR and $D$ is either
empty or the closed point. If $D$ is empty, then
$\sqrt[n]{\operatorname{Spec}R,D}=\operatorname{Spec}R$ and this is simply the
valuative criterion of properness. Suppose that $D$ is the closed point.
Consider the flat morphism
$\operatorname{Spec}R^{(n)}\longrightarrow\sqrt[n]{\operatorname{Spec}R}$ and
write
$X^{\prime}=X\times_{\sqrt[n]{\operatorname{Spec}R}}\operatorname{Spec}R^{(n)}$,
$Y^{\prime}=Y\times_{\sqrt[n]{\operatorname{Spec}R}}\operatorname{Spec}R^{(n)}$.
Thanks to [Sta21, Lemma 0CMK] we have that $Y^{\prime}\subset X^{\prime}$ is
the schematic image of the induced generic section
$\operatorname{Spec}K^{(n)}\to X^{\prime}$. By the valuative criterion of
properness, there is a section $\operatorname{Spec}R^{(n)}\to X^{\prime}$
which is a closed immersion since $X^{\prime}$ is representable, this implies
that $Y^{\prime}\to\operatorname{Spec}R^{(n)}$ is an isomorphism. It follows
that $Y\longrightarrow\sqrt[n]{\operatorname{Spec}R}$ is an isomorphism, too.
∎

######  Lemma 3.9.

Let $R$ be a DVR, $n,m$ positive integers. Assume that $n$ is prime with the
residue characteristic of $R$. Consider the $\mu_{n}$-torsor
$\operatorname{Spec}R^{(n)}\to\sqrt[n]{\operatorname{Spec}R}$. There exists a
unique way of extending the action of $\mu_{n}$ to
$\sqrt[m]{\operatorname{Spec}R^{(n)}}$, and the quotient
$[\sqrt[m]{\operatorname{Spec}R^{(n)}}/\mu_{n}]$ is isomorphic to
$\sqrt[mn]{\operatorname{Spec}R}$.

###### Proof.

We have a natural action
$\rho:\sqrt[m]{\operatorname{Spec}R^{(n)}}\times_{R}\mu_{n}\to\sqrt[m]{\operatorname{Spec}R^{(n)}}$
induced by the action on $\operatorname{Spec}R^{(n)}$. The action $\rho$ gives
a structure of $\mu_{n}$-torsor to the natural morphism
$\sqrt[m]{\operatorname{Spec}R^{(n)}}\to\sqrt[mn]{\operatorname{Spec}R}$. Let
$\eta:\sqrt[m]{\operatorname{Spec}R^{(n)}}\times_{R}\mu_{n}\to\sqrt[m]{\operatorname{Spec}R^{(n)}}$
be an action such that the diagram

${\sqrt[m]{\operatorname{Spec}R^{(n)}}\times_{R}\mu_{n}}$${\sqrt[m]{\operatorname{Spec}R^{(n)}}}$${\operatorname{Spec}R^{(n)}\times_{R}\mu_{n}}$${\operatorname{Spec}R^{(n)}}$$\scriptstyle{\eta}$

is $2$-commutative, we want to show that $\rho$ and $\eta$ are equivalent.

Let $D\subset\operatorname{Spec}R^{(n)}\times_{R}\mu_{n}$ be the pullback of
the closed point of $\operatorname{Spec}R^{(n)}$, since $\mu_{n}$ is finite
étale over $R$ we have that $D$ is a reduced divisor. Since $\mu_{n}$ is
étale, the natural morphism
$\sqrt[m]{\operatorname{Spec}R^{(n)}\times_{R}\mu_{n},D}\to\sqrt[m]{\operatorname{Spec}R^{(n)}}\times_{R}\mu_{n}$
is an isomorphism. The scheme $\operatorname{Spec}R^{(n)}\times_{R}\mu_{n}$ is
finite étale over $\operatorname{Spec}R^{(n)}$, hence it is a disjoint union
of Dedekind domains, and
$\sqrt[m]{\operatorname{Spec}R^{(n)}}\times_{R}\mu_{n}=\sqrt[m]{\operatorname{Spec}R^{(n)}\times_{R}\mu_{n},D}$
is a disjoint union of root stacks over Dedekind domains.

The stack $\mathop{\underline{\mathrm{Isom}}}\nolimits(\rho,\eta)$ has a
proper, representable morphism

$\mathop{\underline{\mathrm{Isom}}}\nolimits(\rho,\eta)\to\sqrt[m]{\operatorname{Spec}R^{(n)}}\times_{R}\mu_{n},$

and for every connected component of
$\sqrt[m]{\operatorname{Spec}R^{(n)}}\times_{R}\mu_{n}$ there is a generic
section. By Lemma 3.8, these generic sections extend to global sections, hence
$\eta\simeq\rho$. ∎

######  Corollary 3.10.

Let $R$ be a DVR, $n,m$ positive integers. Assume that $n$ is prime with the
residue characteristic of $R$. Let $X\to\sqrt[n]{\operatorname{Spec}R}$ be a
morphism, and assume that the base change of $X$ to
$\operatorname{Spec}R^{(n)}$ is isomorphic to
$\sqrt[m]{\operatorname{Spec}R^{(n)}}$. Then
$X\simeq\sqrt[mn]{\operatorname{Spec}R}$.

######  Lemma 3.11.

Let $R^{\prime}/R$ be a local, quasi-finite étale extension of DVRs and $X$ a
tame stack over $R$,
$X^{\prime}\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}X_{R^{\prime}}$. If
$X^{\prime}\simeq\sqrt[n]{\operatorname{Spec}R^{\prime}}$, then
$X\simeq\sqrt[n]{\operatorname{Spec}R}$.

###### Proof.

Let $K$ be the residue field of $R$, clearly we have that
$X_{K}\to\operatorname{Spec}K$ is an isomorphism. Write
$A=\operatorname{Spec}R^{\prime}\otimes_{R}R^{\prime}$, since $R^{\prime}$ is
quasi-finite étale over $R$ then $A$ is a product of Dedekind domains with a
finite number of closed points. Let $D\subset\operatorname{Spec}A$ be the
effective, reduced divisor of all closed points and
$S\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}\sqrt[n]{\operatorname{Spec}R^{\prime}}\times_{\sqrt[n]{\operatorname{Spec}R}}\sqrt[n]{\operatorname{Spec}R^{\prime}}$,
it is easy to see that $S\simeq\sqrt[n]{(\operatorname{Spec}A,D)}$.

Let $\phi:\sqrt[n]{\operatorname{Spec}R^{\prime}}\simeq X^{\prime}\to X$ be
the composite, and consider the two projections
$p_{1},p_{2}:S\to\sqrt[n]{\operatorname{Spec}R^{\prime}}$. Since
$X_{R^{\prime}}$ is separated, then $X$ is separated, too, and hence
$\mathop{\underline{\mathrm{Isom}}}\nolimits(p_{1}^{*}\phi,p_{2}^{*}\phi)$ is
an algebraic stack with a proper, representable morphism to $S$. There is a
generic section
$S_{K}\to\mathop{\underline{\mathrm{Isom}}}\nolimits(p_{1}^{*}\phi,p_{2}^{*}\phi)$
which extends to a global section thanks to Lemma 3.8, this gives descent data
for a morphism $f:\sqrt[n]{\operatorname{Spec}R}\to X$ (the cocycle condition
can be checked on the generic point, where it is obvious). Since the base
change to $R^{\prime}$ of $f$ is an isomorphism, we have that $f$ is an
isomorphism, too. ∎

######  Proposition 3.12.

Let $X$ be a normal, tame stack of finite type over a DVR $R$, and assume that
there is a generic section $\operatorname{Spec}K\to X$ which is an open,
scheme-theoretically dense embedding. Then
$X\simeq\sqrt[n]{\operatorname{Spec}R}$ for some $n$.

###### Proof.

Since $X$ is of finite type over $R$, there exists a DVR $R_{0}\subset R$
which is the localization of a $\mathbb{Z}$-algebra of finite type and a stack
$X_{0}/R_{0}$ such that $X\simeq X_{0,R}$. Furthermore, we may assume that the
uniformizing parameter of $R_{0}$ maps to a uniformizing parameter of $R$, so
that
$\sqrt[n]{\operatorname{Spec}R_{0}}\times_{R_{0}}\operatorname{Spec}R\simeq\sqrt[n]{\operatorname{Spec}R}$.
Up to replacing $R,X$ with $R_{0},X_{0}$ we may assume that $R$ is Nagata. Let
$k$ be the residue field of $R$ and $p$ its characteristic.

By [AOV08, Theorem 3.2], there exists a DVR $R^{\prime}$ quasi-finite and
étale over $R$ and a finite, flat, linearly reductive group scheme
$G/R^{\prime}$ with an action on a scheme $U$ finite over $R^{\prime}$ such
that $X_{R^{\prime}}\simeq[U/G]$. Up to enlarging $R^{\prime}$, by [AOV08,
Lemma 2.20] there exists a diagonalizable flat, closed subgroup $\Delta\subset
G$ such that $H\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}G/\Delta$ is constant and tame. We may furthermore assume that the
degree of $\Delta$ is a power of $p$. Thanks to Lemma 3.11, we may assume
$R^{\prime}=R$.

Case 1. $X$ is tame and Deligne-Mumford. Since $\Delta_{k}$ is connected and
$X$ is Deligne-Mumford and generically a scheme, the action of $\Delta$ is
free (because otherwise $X$ would have ramified inertia), hence up to
replacing $U$ with $U/\Delta$ we may assume that $G$ is constant and tame.
Since $X$ is normal and $G$ is constant and tame, then $U$ is normal, too. If
$u\in U$ is a geometric point, the stabilizer $G_{u}$ acts faithfully on the
tangent space, hence the automorphism groups of the points of $X$ are cyclic
and tame. By [Ryd11, Lemma 8.5] and Lemma 3.7, since $X$ is normal we have
$X\simeq\sqrt[n]{\operatorname{Spec}R}$ for some $n$.

Case 2. $X$ is tame. Let $V\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}U/\Delta$ and
$Y_{0}\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle def}}{=}}}[V/H]$, we
have that $Y_{0}$ is Deligne-Mumford and there is a natural birational
morphism $X\to Y_{0}$ whose relative inertia is diagonalizable. Let $Y\to
Y_{0}$ be the normalization, it is finite over $Y_{0}$ since $R$ is Nagata and
since $X$ is normal the morphism $X\to Y_{0}$ lifts to a morphism $X\to Y$. By
case $1$, there exists an $n$ prime with $p$ and an isomorphism
$Y\simeq\sqrt[n]{\operatorname{Spec}R}$. Consider the morphism
$\operatorname{Spec}R^{(n)}\to\sqrt[n]{\operatorname{Spec}R}$, it is a
$\mu_{n}$-torsor and hence finite étale since $n$ is prime with $p$, it
follows that the base change
$X\times_{\sqrt[n]{\operatorname{Spec}R}}\operatorname{Spec}R^{(n)}$ is normal
with diagonalizable inertia. By [Ryd11, Lemma 8.5] and Lemma 3.7, we have
$X\times_{\sqrt[n]{\operatorname{Spec}R}}\operatorname{Spec}R^{(n)}\simeq\sqrt[m]{\operatorname{Spec}R^{(n)}}$
for some integer $m$, hence $X\simeq\sqrt[mn]{\operatorname{Spec}R}$ thanks to
Corollary 3.10. ∎

###### Proof of Theorem 3.1.

By base change, we may assume that $Y=\operatorname{Spec}R$ and that $X$ is a
tame stack proper over $R$. With an argument similar to the one in the proof
of Proposition 3.12, we may reduce to the case in which $R$ is Nagata.

By [Ryd11, Theorem B], we may assume that $\operatorname{Spec}K\to X$ is an
open, scheme theoretically dense embedding. Since $R$ is Nagata, the
normalization $\overline{X}$ is finite and representable over $X$. By
Proposition 3.12 we have $\overline{X}=\sqrt[n]{\operatorname{Spec}R}$, hence
an extension exists. If $m$ is another integer with a representable extension
$\sqrt[m]{\operatorname{Spec}R}\to X$, it factors through
$\overline{X}=\sqrt[n]{\operatorname{Spec}R}$ since
$\sqrt[m]{\operatorname{Spec}R}$ is normal by Lemma 3.7. We conclude the proof
of Theorem 3.1 by Lemma 3.6.

∎

##  4\. The Lang–Nishimura theorem

Here is our version of the Lang–Nishimura theorem for tame stacks.

######  Theorem 4.1.

Let $S$ be a scheme and $X\dashrightarrow Y$ a rational map of algebraic
stacks over $S$, with $X$ locally noetherian and integral and $Y$ tame and
proper over $S$. Let $k$ be a field,
$s\colon\operatorname{Spec}k\longrightarrow S$ a morphism. Assume that $s$
lifts to a regular point $\operatorname{Spec}k\longrightarrow X$; then it also
lifts to a morphism $\operatorname{Spec}k\longrightarrow Y$.

In the standard version of the Lang–Nishimura theorem (see for example [Poo17,
Theorem 3.6.11]), which is a standard tool in arithmetic geometry, $X$ and $Y$
are schemes, and $S=\operatorname{Spec}k$. In the applications that we have in
mind, the additional flexibility of having a base scheme is important.

###### Proof.

According to [LMB00, Théorème 6.3] we can find a smooth morphism
$U\longrightarrow X$ with a lifting $\operatorname{Spec}k\longrightarrow U$ of
$\operatorname{Spec}k\longrightarrow X$; hence we can replace $X$ by $U$, and
assume that $X$ is scheme. Furthermore, if $x$ denotes the image of
$\operatorname{Spec}k\longrightarrow X$ and $k(x)$ its residue field, we have
a factorization
$\operatorname{Spec}k\longrightarrow\operatorname{Spec}k(x)\longrightarrow X$,
and we may assume $k=k(x)$. If $x$ has height $0$, then $\operatorname{Spec}k$
dominates $X$, and the composite $\operatorname{Spec}k\longrightarrow
X\dashrightarrow Y$ is well defined.

Otherwise, call $U\subseteq X$ the open subset where $f$ is defined. By [BVb,
Lemma 4.3] there exists a DVR $R$ with residue field $k=k(x)$ and a morphism
$\operatorname{Spec}R\longrightarrow X$ that maps the generic point
$\operatorname{Spec}K$ of $\operatorname{Spec}R$ into $U$. Thus we get a
morphism $\operatorname{Spec}K\longrightarrow U$, and we apply Corollary 3.2
to the diagram

${\operatorname{Spec}K}$${U}$${Y}$${\operatorname{Spec}R}$${X}$${S\
}$$\scriptstyle{f}$

thus getting the desired morphism $\operatorname{Spec}k\longrightarrow Y$. ∎

The Lang–Nishimura theorem fails for non-tame separated stacks. Let us give
two examples, one in mixed characteristic, the other in positive
characteristic.

######  Example 4.2.

Let $X\longrightarrow\operatorname{Spec}R$ be the stack constructed in 3.3, it
is a non-tame regular Deligne-Mumford stack. Let $k$ be the residue field of
$R$. There is a rational map $\operatorname{Spec}R\dashrightarrow X$ and
$\operatorname{Spec}R$ has a $k$ rational point, but $X$ has no $k$-rational
points.

######  Example 4.3.

Let $C_{0}$ be a smooth, projective curve of positive genus over a finite
field $F$ of characteristic $p$ with $C_{0}(F)\neq\emptyset$. Let $a$ be an
indeterminate, write $k\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}F(a)$ and $C\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}C_{0,k}$; since $C_{0}$ has positive genus $C(k)=C_{0}(F)$ is
finite. Let $f\in k(C)$ be a rational function such that each rational point
is a zero of $f$ (this can be easily found using Riemann-Roch). Consider the
ramified cover $D\to C$ given by the equation

$t^{p}-f^{p-1}t=a\,;$

in other words, $D$ is the smooth projective curve associated with the field
extension $k(C)[t]/(t^{p}-f^{p-1}t-a)$. Let $c\in C(k)$ be a rational point,
and write $R_{c}\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}\mathcal{O}_{C,c}[t]/(t^{p}-f^{p-1}t-a)$, it is a normal domain: if
$\overline{R}_{c}$ is the normalization, both $R_{c}\otimes
k(C)\to\overline{R}_{c}\otimes k(C)$ and $R_{c}\otimes
k\to\overline{R}_{c}\otimes k$ are isomorphisms for degree reasons since
$R_{c}\otimes k=k[t]/(t^{p}-a)$ is a field of degree $p$ over $k$. Hence,
$R_{c}$ is a DVR with residue field
$k^{\prime}\mathrel{\smash{\overset{\mathrm{\scriptscriptstyle
def}}{=}}}k[t]/(t^{p}-a)$. It follows that $D$ has no $k$-rational points.

The cyclic group $C_{p}$ acts on $D$ by $t\mapsto t+f$, the field extension
$k(D)/k(C)$ is a cyclic Galois cover and $C$ is the quotient scheme $D/C_{p}$.
Let $X$ be the quotient stack $[D/C_{p}]$, there is a natural birational
morphism $X\to C=D/C_{p}$. A rational point $\operatorname{Spec}k\to X$
corresponds to a $C_{p}$-torsor $\operatorname{Spec}A\to\operatorname{Spec}k$
with an equivariant morphism $\operatorname{Spec}A\to D$: since the fibers of
$D\to C$ over rational points are isomorphic to
$\operatorname{Spec}k^{\prime}$, a rational point of $X$ gives an embedding of
$k^{\prime}$ in an étale algebra $A$, which is clearly absurd. It follows that
$X$ is a proper Deligne-Mumford stack over $k$ with $X(k)=\emptyset$ and a
birational map $C\dashrightarrow X$.

As a consequence of Theorem 4.1, we can decide whether a residual gerbe of a
tame stack is neutral or not by looking at a resolution of singularities of
the coarse moduli space. We find this rather surprising.

######  Corollary 4.4.

Let $X$ be a locally noetherian, regular and integral tame stack with coarse
moduli space $X\to M$, and $\overline{M}\longrightarrow M$ a proper birational
morphism, with $\overline{M}$ integral and regular. Assume that there is a
lifting $\operatorname{Spec}k(M)\longrightarrow X$ of the generic point
$\operatorname{Spec}k(M)\longrightarrow M$.

If $k$ is a field and $m\colon\operatorname{Spec}k\to M$ a morphism, then $m$
lifts to a morphism $\operatorname{Spec}k\longrightarrow X$ if and only if it
lifts to a morphism $\operatorname{Spec}k\longrightarrow\overline{M}$.∎

So, for example, if $M$ is regular all morphisms
$\operatorname{Spec}k\longrightarrow M$ lift to $X$, and all residual gerbes
are neutral.

## References

  * [AGV08] Dan Abramovich, Tom Graber, and Angelo Vistoli, _Gromov-Witten theory of Deligne-Mumford stacks_ , Amer. J. Math. 130 (2008), no. 5, 1337–1398.
  * [AOV08] Dan Abramovich, Martin Olsson, and Angelo Vistoli, _Tame stacks in positive characteristic_ , Ann. Inst. Fourier (Grenoble) 58 (2008), no. 4, 1057–1091.
  * [AOV11] by same author, _Twisted stable maps to tame Artin stacks_ , J. Algebraic Geom. 20 (2011), no. 3, 399–477.
  * [Bre] Giulio Bresciani, _The arithmetic of tame quotient singularities in dimension 2_ , in preparation.
  * [Bre21] Giulio Bresciani, _On the birational section conjecture with strong birationality assumptions_ , arXiv:2108.13397, 2021.
  * [BVa] Giulio Bresciani and Angelo Vistoli, _Fields of moduli and the arithmetic of tame quotient singularities_ , in preparation.
  * [BVb] by same author, _The genericity theorem for the essential dimension of tame stacks_ , arXiv:2111.01117. To appear in Pure and Applied Mathematics Quarterly.
  * [Knu71] Donald Knutson, _Algebraic spaces_ , Lecture Notes in Mathematics, Vol. 203, Springer-Verlag, Berlin-New York, 1971.
  * [LMB00] Gérard Laumon and Laurent Moret-Bailly, _Champs algébriques_ , Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics, vol. 39, Springer-Verlag, Berlin, 2000.
  * [Poo17] Bjorn Poonen, _Rational points on varieties_ , Graduate Studies in Mathematics, vol. 186, American Mathematical Society, Providence, RI, 2017.
  * [Ryd11] David Rydh, _Compactification of tame Deligne–Mumford stacks_ , https://people.kth.se/~dary/tamecompactification20110517.pdf, 2011.
  * [Sta21] The Stacks Project Authors, _Stacks project_ , http://stacks.math.columbia.edu, 2021.


# Upper bounds on two Hilbert coefficients

Le Xuan Dung1, Juan Elias2 and Le Tuan Hoa3 1Faculty of Natural Science, Hong
Duc University
565 Quang Trung, Dong Ve, Thanh Hoa, Vietnam  2Facultat de Matemàtiques i
Informàtica, Universitat de Barcelona, Gran Via 585, 08007 Barcelona, Spain
3Institute of Mathematics, VAST, 18 Hoang Quoc Viet, 10307 Hanoi, Viet Nam
<EMAIL_ADDRESS><EMAIL_ADDRESS>(L. X. Dung<EMAIL_ADDRESS>(J.
Elias<EMAIL_ADDRESS>(L. T. Hoa)

###### Abstract.

New upper bounds on the first and the second Hilbert coefficients of a Cohen-
Macaulay module over a local ring are given. Characterizations are provided
for some upper bounds to be attained. The characterizations are given in terms
of Hilbert series as well as in terms of the Castelnuovo-Mumford regularity of
the associated graded module.

###### Key words and phrases:

Hilbert function, Hilbert coefficient, Cohen-Macaulay, Local cohomology

###### 1991 Mathematics Subject Classification:

13D40, 13H10, 13D45

##  1\. Introduction

Let $(A,\mathfrak{m})$ be a Noetherian local ring with an infinite residue
field $A/\mathfrak{m}$ and $I$ an $\mathfrak{m}$-primary ideal. Let $M$ be a
finitely generated $A$-module of dimension $d$. Then the Hilbert-Samuel
function $H^{1}_{I,M}(n):=\ell_{A}(M/I^{n+1}M)$ agrees with a polynomial, so-
called Hilbert-Samuel polynomial, $H\\!P^{1}_{I,M}(n)$ for all $n\gg 0$. If we
write

$H\\!P^{1}_{I,M}(n)=e_{0}(I,M)\binom{n+d}{d}-e_{1}(I,M)\binom{n+d-1}{d-1}+\cdots+(-1)^{d}e_{d}(I,M),$

then the integers $e_{0}(I,M),...,,e_{0}(I,M)$ are called the Hilbert
coefficients of $M$ with respect to $I$.

There are intensive studies on finding bounds on several Hilbert coefficients.
We refer the interested readers to the book [15] for main developments and
rich references. Our aim is to give new bounds on the first two Hilbert
coefficients in terms of $e_{0}(I,M)$ for Cohen-Macaulay modules. Recall that
$e_{0}(I,M),e_{1}(I,M),e_{2}(I,M)\geq 0$ and that $e_{3}(I,M)$ could be
negative, see [11] and [15].

In this paper we establish two upper bounds on $e_{1}(I,M)$. The first one
(Proposition 3.1) holds for Cohen-Macaulay modules and is a slight improvement
of the bound given by Rossi and Valla for filtrations in [15, Proposition 2.8
and Proposition 2.10]. Moreover, in the case of dimension one, we can give
conditions for this bound to be attained (see Proposition 3.5 and Proposition
4.2). In difference to the approach in [15, Section 2.2], we use here local
cohomology modules of the associated graded modules $G_{I}(M)$. As a
consequence, we can show that if $IM\subseteq\mathfrak{m}^{b}M$ for some
$b\geq 2$, then $e_{1}(I,M)\leq\binom{e_{0}(I,M)-b}{2}$, see Proposition 3.2.
In the ring case, this result was given by Elias [3, Proposition 2.5 and
Remark 2.6] under an additional condition, which was then removed by
Hanumanthu and Huneke in [6, Corollary 3.7]. This part can be also seen as a
preparation for our study on the $e_{2}(I,M)$ in Section 5.

The second upper bound on $e_{1}(I,M)$ (Theorem 4.6) only holds for $M=A$ and
is based on the bound given by Elias [4] in the dimension one case. When
$b\geq 2$, this new bound provides a much better bound than the above
mentioned bound $\binom{e_{0}(I,A)-b}{2}$, see Remark 4.7.

For $e_{2}(I,M)$ we can give a new upper bound in terms of $e_{0}(I,M)$ and
$b$, see Theorem 5.1. As we know that the first upper bound on $e_{2}(I,M)$
was given by Rhodes, see [13, Proposition 6.1(iv)] (also see [10, Corollary
4.2] for modules over a semi-local ring). In [5, Theorem 2.3] a much better
bound is given. However, all bounds on $e_{2}(I,M)$ in [13, 10, 5] involve
$e_{1}(I,M)$, while our bound only depend on $e_{0}(I,M)$ (and the largest
number $b$ such that $IM\subseteq\mathfrak{m}^{b}M$). Moreover, we can
characterize when this bound is attained (Theorem 5.3). Like Proposition 3.5,
the conditions in Theorem 5.3 are given in terms of Hilbert series as well as
in terms of the Castelnuovo-Mumford regularity of the associated graded module
$G_{I}(M)$. In the case $M=A$ and $b\geq 2$, using Theorem 4.6 one can get a
better bound than the one in Theorem 5.1, see Theorem 5.9.

We now give a brief content of the paper. In Section 2 we recall some basic
notions and give some estimations on the Hilbert function of $G_{I}(M)$ and of
$\overline{G_{I}(M)}$. In Section 3 we give two bounds on $e_{1}(I,M)$ (see
Propositions 3.1 and 3.2) and characterize when the first bound in
Propositions 3.1 is attained, provided $\dim M=1$. In Section 4 we restrict to
the case $M=A$. Here we give further structures of $I$ and $A$ such that the
first bound in Propositions 3.1 is attained, see Proposition 4.2. Then we
prove an essentially new bound on $e_{1}(I)$ (Theorem 4.6). Main known upper
bounds on $e_{1}(I)$ of an $\mathfrak{m}$-primary ideal $I$ of an one-
dimensional Cohen-Macaulay ring $(A,\mathfrak{m})$ such that
$I\varsubsetneq\mathfrak{m}^{2}$ are summarized in Remark 4.8. In the last
Section 5, we prove the new bounds on $e_{2}(I,M)$ (Theorems 5.1 and 5.9), and
give equivalent conditions for the bound in Theorem 5.1 to be attained
(Theorem 5.3).

##  2\. Preliminaries

Let $R=\oplus_{n\geq 0}R_{n}$ be a Noetherian standard graded ring over a
local Artinian ring $(R_{0},\mathfrak{m}_{0})$. Let $E$ be a finitely
generated graded module of dimension $d$. The function
$H_{E}(n):=\ell_{R_{0}}(E_{n})$ is called Hilbert function of $E$. For all
$n\gg 0$, it agrees with the so-called Hilbert polynomial denoted by
$H\\!P_{E}(t)$, that is polynomial of degree $d-1$. The number

$pn(E):=\min\\{n|\ H_{E}(t)=H\\!P_{E}(t)\ \text{for all}\ t\geq n\\},$

is called the postulation number of $H_{E}$.

If we denote by $R_{+}:=\oplus_{n>0}R_{n}$ the irrelevant ideal of $R$ then we
set

$a_{i}(E):=\sup\\{n|\ H^{i}_{R_{+}}(E)_{n}\neq 0\\},$

$0\leq i\leq d$. The Castelnuovo-Mumford regularity of $E$ is the number:

$\operatorname{reg}(E):=\max\\{a_{i}(E)+i|\ 0\leq i\leq d\\}.$

Let $(A,\mathfrak{m})$ be a Noetherian local ring with an infinite residue
field $A/\mathfrak{m}$ and $I$ an $\mathfrak{m}$-primary ideal. Let $M$ be a
finitely generated $A$-module of dimension $d$. The associated graded module
of $I$ with respect to $M$ is the standard $A/I$-algebra

$G_{I}(M):=\oplus_{n\geq 0}I^{n}M/I^{n+1}M,$

There is another way to define the Hilbert coefficients $e_{i}(I,M)$ already
defined in the introduction. We recall here this approach from [15, Section
1.3]. The function

$H_{I,M}(n):=H_{G_{I}(M)}(n)=\ell_{A}(I^{n}M/I^{n+1}M)$

is called the Hilbert function of $M$. The Hilbert polynomial of $M$ is
$H\\!P_{I,M}=H\\!P_{G_{I}(M)}$.

By the Hilbert-Serre theorem, the Hilbert series

(2.1) $P_{I,M}(z):=\sum_{n\geq 0}H_{I,M}(n)z^{n},$

is a rational function of $z$, that means one can find a polynomial
$Q_{I,M}(z)\in\mathbb{Z}[z]$ such that $Q_{I,M}(1)\neq 0$ and

$P_{I,M}(z)=\frac{Q_{I,M}(z)}{(1-z)^{d}}.$

If we set for every $i\geq 0$

(2.2) $e_{i}(I,M)=\frac{Q_{I,M}^{(i)}(1)}{i!},$

where $Q_{I,M}^{(i)}$ denotes the $i$-th derivation of $Q_{I,M}$, then for all
$0\leq i\leq d$, this value of $e_{i}(I,M)$ agrees with the one defined in the
introduction. Moreover, unlike the definition in the introduction, using (2.2)
we can talk about the Hilbert coefficients $e_{i}(I,M)$ with $i>d$. This
simple observation is useful in the study of the second Hilbert coefficient,
where we can reduce the case of dimension two to dimension one.

Together with Hilbert series, the power series

$P^{1}_{I,M}(z):=\sum_{n\geq
0}\ell(M/I^{n+1}M)z^{n}=\frac{P_{I,M}(z)}{(1-z)^{d+1}}$

is also often used; this is called Hilbert-Samuel series.

In the sequel we use the following notations

$pn(I,M):=pn(G_{I}(M))=\min\\{n|\ H\\!P_{I,M}(t)=H_{I,M}(t)\ \text{for all}\
t\geq n\\},$

and if $M=A$ then we write $H_{I}:=H_{I,A}$, $H\\!P_{I}:=H\\!P_{I,A}$,
$pn(I):=pn(I,A)$, $e_{i}(I):=e_{i}(I,A)$ and so on.

Recall that an element $x\in I$ is called $M$-superficial (of order one) for
$I$, if there exists a non-negative integer $c$ such that

$(I^{n+1}M:x)\cap I^{c}M=I^{n}M,$

for all $n\geq c$. When $M=A$ we simply say that $x$ is a superficial element
for $I$. This is equivalent to the condition that the initial form

$x^{*}\in G(I):=\oplus_{n\geq 0}I^{n}/I^{n+1}$

has degree one and it is a filter-regular element on the associated graded
module $G_{I}(M)$ which means

$[0:_{G_{I}(M)}x^{*}]_{m}=0\ \text{for all}\ m\gg 0.$

(See, e.g., the equivalence of (1) and (5) in [15, Theorem 1.2].) A sequence
$x_{1},...,x_{s}\in I$ is called $M$-superficial sequence for $I$ if $x_{i}$
is an $M/(x_{1},...,x_{i-1})M$-superficial element, for $I$ for $i=1,...,s$.

The following result is now standard and useful for proceeding by induction.

######  Lemma 2.1.

(See, e.g., [14, Proposition 1.2]) Let $x\in I$ be an $M$-superficial element
for $I$. Then

  * (i)

$\dim(M/xM)=d-1,$

  * (ii)

$e_{j}(I,M/xM)=e_{j}(I,M)$ for every $j=0,...,d-2,$

  * (iii)

$e_{d-1}(I,M/xM)=e_{d-1}(I,M)+(-1)^{d-1}\ell(0:x),$

  * (iv)

There exists an integer $n_{0}$ such that for every $n\geq n_{0}-1$ we have

$e_{d}(I,M/xM)=e_{d}(I,M)+(-1)^{d}\left[\sum_{i=0}^{n}\ell(I^{n+1}M:x/I^{n}M)-(n+1)\ell(0:x)\right],$

  * (v)

$x^{*}$ is a regular element on $G_{I}(M)$ if only if
$P_{I,M}(z)=P^{1}_{I,M/xM}(z)=\frac{P_{I,M/xM}(z)}{1-z}$ if only if $x$ is
$M$-regular and $e_{d}(I,M)=e_{d}(I,M/xM)$.

We would like to comment that in the above statements (iv) and (v),
$e_{d}(I,M/xM)$ is the one defined by (2.2).

The Ratliff-Rush closure of an ideal introduced in [12] plays an important
role in the study of Hilbert functions, see e.g. [14, 15]. It is defined by

(2.3) $\widetilde{I^{n}M}=\bigcup_{k\geq 1}I^{n+k}M:I^{k}=I^{n+l}:I^{l}\
\text{for some}\ l\gg 0.$

Using this notion, we can compute the zero-th local cohomology module of
$G_{I}(M)$ with respect to $G_{+}:=\oplus_{n\geq 1}I^{n}/I^{n+1}$ as follows
(see, e.g., [14, p. 26]):

(2.4) $[H^{0}_{G_{+}}(G_{I})M))]_{n}=\frac{\widetilde{I^{n+1}M}\cap
I^{n}M}{I^{n+1}M}.$

We set $\overline{G_{I}(M)}=G_{I}(M)/H^{0}_{G_{+}}(G_{I}(M))$.

######  Lemma 2.2.

Let $M$ be an one-dimensional $A$-module. Let $b$ be a positive integer such
that $IM\subseteq\mathfrak{m}^{b}M$. Then

  * (i)

([1, Lemma 2.5]) $\ell(\overline{G_{I}(M)}_{0})\geq b$.

  * (ii)

If $e_{0}(I,M)\neq e_{0}(\mathfrak{m}^{b},M)$, then
$\ell(\overline{G_{I}(M)}_{0})\geq b+1$.

###### Proof.

(i) is [1, Lemma 2.5]. It is based on the the fact
$\ell(\overline{G_{I}(M)}_{0})=\ell(E)$, where $E=M/\widetilde{IM}$, and the
strict inclusions:

(2.5) $E\supsetneq\mathfrak{m}E\supsetneq\cdots\supsetneq\mathfrak{m}^{b}E.$

(ii) Assume that $\mathfrak{m}^{b}E=0$. Then
$\mathfrak{m}^{b}M\subseteq\widetilde{IM}$. By (2.3), it implies that

$I^{l+1}M\subseteq I^{l}(\mathfrak{m}^{b}M)\subseteq I^{l+1}M,$

for some $l\gg 0$. Hence $I^{l+1}M=\mathfrak{m}^{b}I^{l}M$. Let $c$ be an
integer such that $\mathfrak{m}^{bc}\subseteq I^{l}$, then for all $n>0$, it
yields

$I^{l+n}M=(\mathfrak{m}^{b})^{n}I^{l}M\supseteq(\mathfrak{m}^{b})^{n+c}M.$

Hence

$\begin{array}[]{ll}(n+l)e_{0}(I,M)-e_{1}(I,M)&=\ell(M/I^{l+n}M)\\\
&\leq\ell(M/(\mathfrak{m}^{b})^{n+c}M)=(n+c)e_{0}(\mathfrak{m}^{b},M)-e_{1}(\mathfrak{m}^{b},M),\end{array}$

for all $n\gg 0$. This implies $e_{0}(I,M)\leq e_{0}(\mathfrak{m}^{b},M)$,
whence $e_{0}(I,M)=e_{0}(\mathfrak{m}^{b},M)$, a contradiction to the
assumption. So, we must have $\mathfrak{m}^{b}E\neq 0$. From (2.5) we then get
$\ell(\overline{G_{I}(M)}_{0})=\ell(E)\geq b+1$, as required. ∎

######  Lemma 2.3.

Let $R$ be a Noetherian standard graded ring over a local Artinian ring and
$E$ an one-dimensional Cohen-Macaulay graded $R$-module. Let
$\Delta:=\Delta(E)$ be the maximal generating degree of $E$. Then

  * (i)

$H_{E}(\Delta)<H_{E}(\Delta+1)<\cdots<H_{E}(pn(E))$,

  * (ii)

$H_{E}(n)\geq(n-\Delta)+H_{E}(\Delta)$ for all $\Delta+1\leq n\leq pn(E).$

###### Proof.

Let $z\in R_{1}$ be an $E$-regular element. Since $E$ is a Cohen-Macaulay
module, $pn(E)=pn(E/zE)-1$. Note that $\Delta(E/zE)\leq\Delta=\Delta(E)$ and
since $\dim E/zE=0$, $H_{E/zE}(t)\geq 1$ for all $\Delta(E/zE)\leq t\leq
pn(E/zE)-1=pn(E)$. Hence the statements follow from the following equality

$H_{E}(n)=H_{E}(\Delta)+\sum_{\Delta+1\leq i\leq n}H_{E/zE}(i).$

∎

The following result is a slight improvement of [15, Proposition 2.7] and the
remark after it.

######  Lemma 2.4.

Let $M$ be an one-dimensional Cohen-Macaulay $A$-module such that
$IM\subseteq\mathfrak{m}^{b}M$. Then

  * (i)

$a_{1}(G_{I}(M))=pn(I,M)-1>a_{0}(G_{I}(M))$ and
$\operatorname{reg}(G_{I}(M))=pn(I,M)$,

  * (ii)

$H_{I,M}(n)\geq n+b+\ell(H^{0}_{G_{+}}(G_{I}(M))_{n})$ for all $0\leq n\leq
pn(I,M)$,

  * (iii)

([15, Proposition 2.7(2)]) $pn(I,M)\leq e_{0}(I,M)-b$. If the equality holds,
then $\ell(\overline{G_{I}(M)}_{n})=n+b$ for all $0\leq n\leq pn(I,M)$.

If, in addition, $e_{0}(I,M)\neq e_{0}(\mathfrak{m}^{b},M)$, then we further
have:

  * (iv)

$H_{I,M}(n)\geq n+b+1+\ell(H^{0}_{G_{+}}(G_{I}(M))_{n})$ for all $0\leq n\leq
pn(I,M)$,

  * (v)

([15, Remark (b) after Proposition 2.7])  $pn(I,M)\leq e_{0}(I,M)-b-1$. If the
equality holds, then $\ell(\overline{G_{I}(M)}_{n})=n+b+1$ for all $0\leq
n\leq pn(I,M)$.

###### Proof.

(i) By [7, Theorem 5.2], $a_{0}(G_{I}(M))<a_{1}(G_{I}(M))$. From the
Grothendieck-Serre formula

$H_{I,M}(n)-H\\!P_{I,M}(n)=H_{G_{I,M}}(n)-H\\!P_{G_{I,M}}(n)=\\\
=\ell(H^{0}_{G_{+}}(G_{I}(M))_{n})-\ell(H^{1}_{G_{+}}(G_{I}(M))_{n}),$

it follows that $pn(G_{I}(M))=a_{1}(G_{I}(M))+1$.

(ii) From the short exact sequence

$0\longrightarrow H^{0}_{G_{+}}(G_{I}(M))\longrightarrow
G_{I}(M)\longrightarrow\overline{G_{I}(M)}:=G_{I}(M)/H^{0}_{G_{+}}(G_{I}(M))\longrightarrow
0,$

we get

$H_{I,M}(n)=\ell(\overline{G_{I}(M)}_{n})+\ell(H^{0}_{G_{+}}(G_{I}(M))_{n}).$

From (i) it follows that

$pn(I,M)=a_{1}(G_{I}(M))+1=pn(\overline{G_{I}(M)}).$

Note that $\Delta(G_{I}(M))=0$, and by Lemma 2.2(i),
$\ell(\overline{G_{I}(M)}_{0})\geq b$. Since $\overline{G_{I}(M)}$ is a Cohen-
Macaulay module, by Lemma 2.3, we have

(2.6) $\displaystyle H_{I,M}(n)$ $\displaystyle=$
$\displaystyle\ell(H^{0}_{G_{+}}(G_{I}(M))_{n})+H_{\overline{G_{I}(M)}}(n)$
$\displaystyle\geq$
$\displaystyle\ell(H^{0}_{G_{+}}(G_{I}(M))_{n})+H_{\overline{G_{I}(M)}}(0)+n$
(2.7) $\displaystyle\geq$
$\displaystyle\ell(H^{0}_{G_{+}}(G_{I}(M))_{n})+b+n,$

for all $0\leq n\leq pn(\overline{G_{I}(M)})=pn(G_{I}(M))=pn(I,M)$.

(iii) Since $H_{I,M}(n)\leq e_{0}(I,M)$ (see, e.g., the remark after Lemma 2.1
in [15]), from (ii) we immediately get

$p\leq e_{0}(I,M)-b-\ell(H^{0}_{G_{+}}(G_{I}(M))_{p})\leq e_{0}(I,M)-b,$

where $p:=pn(I,M)$. If the equality holds, then from (2.6) and (2.7) we must
have $H_{\overline{G_{I}(M)}}(0)=b$ and $H_{\overline{G_{I}(M)}}(p)=p+b$. From
Lemma 2.3 we then get

$\ell(\overline{G_{I}(M)}_{n})=H_{\overline{G_{I}(M)}}(n)=n+b$

for all $0\leq n\leq p$.

(iv) and (v) If, in addition, $e_{0}(I,M)\neq e_{0}(\mathfrak{m}^{b},M)$, then
using Lemma 2.3(ii), instead of (2.7), we get a little bit stronger
inequality:

$H_{I,M}(n)\geq\ell(H^{0}_{G_{+}}(G_{I}(M))_{n})+b+1+n,$

which then implies (iv) and (v). ∎

##  3\. The first Hilbert coefficient of a module

In this section we always assume that $M$ is a Cohen-Macaulay module over a
local ring $(A,\mathfrak{m})$ and $I$ is an $\mathfrak{m}$-primary ideal. We
start with a slight improvement of [15, Proposition 2.8 and Proposition 2.10].
Its proof is also a modification of the one of [15, Proposition 2.8]. Note
that [15, Proposition 2.8 and Proposition 2.10] is formulated for an arbitrary
filtered module (not necessarily Cohen-Macaulay).

######  Proposition 3.1.

Let $M$ be a Cohen-Macaulay module and of dimension $d\geq 1$. Let $b$ be a
positive integer such that $IM\subseteq\mathfrak{m}^{b}M.$ Then

(3.1) $e_{1}(I,M)\leq\binom{e_{0}(I,M)-b+1}{2}+b-\ell(M/IM).$

If $d=1$ and the equality in (3.1) holds, then we have

  * (i)

$a_{0}(G_{I}(M))\leq 0$,

  * (ii)

Either $\operatorname{reg}(G_{I}(M))=pn(I,M)=e_{0}(I,M)-b$ or
$e_{0}(I,M)\in\\{b,b+1\\}$,

  * (iii)

$H_{I,M}(n)=b+n$ for all $1\leq n\leq pn(I,M)-1$.

If $d=1$ and $e_{0}(I,M)>e_{0}(\mathfrak{m}^{b},M)$, then

(3.2) $e_{1}(I,M)\leq\binom{e_{0}(I,M)-b}{2}+b+1-\ell(M/IM).$

If the equality in (3.2) holds, then we have

  * (i’)

$a_{0}(G_{I}(M))\leq 0$,

  * (ii’)

Either $\operatorname{reg}(G_{I}(M))=pn(I,M)=e_{0}(I,M)-b-1$ or
$e_{0}(I,M)\in\\{b+1,b+2\\}$,

  * (iii’)

$H_{I,M}(n)=n+b+1$ for all $1\leq n\leq pn(I,M)-1$.

###### Proof.

For simplicity, set $e_{0}:=e_{0}(I,M),\ e_{1}:=e_{1}(I,M)$ and $p:=pn(I,M)$.
By standard technique (using Lemma 2.1) we may assume that $d=1$. We have

$e_{1}=\sum_{i=0}^{p-1}(e_{0}-H_{I,M}(i))=e_{0}-H_{I,M}(0)+\sum_{i=1}^{p-1}(e_{0}-H_{I,M}(i)).$

Using Lemma 2.4(ii) and (iii), we get

(3.3) $\displaystyle e_{1}$ $\displaystyle\leq$ $\displaystyle
e_{0}-H_{I,M}(0)+\sum_{i=1}^{p-1}(e_{0}-i-b-\ell(H^{0}_{G_{+}}(G_{I}(M))_{i})$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{e_{0}-b-1}(e_{0}-i-b-\ell(H^{0}_{G_{+}}(G_{I}(M))_{i})+e_{0}-\ell(M/IM)$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{e_{0}-b-1}(e_{0}-i-b)+e_{0}-\ell(M/IM)$
$\displaystyle=$ $\displaystyle\binom{e_{0}-b+1}{2}+b-\ell(M/IM).$

If $e_{1}=\binom{e_{0}-b+1}{2}+b-\ell(M/IM)$, then from (3.3) and (3) we must
have:

  * (a)

$H^{0}_{G_{+}}(G_{I}(M))_{i}=0$ for all $1\leq i\leq e_{0}-b-1$,

  * (b)

$H_{I,M}(i)=b+i$ for all $1\leq i\leq e_{0}-b-1$,

  * (c)

$p=e_{0}-b$ if $e_{0}-b\geq 2$.

Since $p\leq e_{0}-b$ by Lemma 2.4(iii), (b) implies (iii). By Lemma 2.4(i),
$p-1>a_{0}(G_{I}(M))$. Hence (a) implies (i). Since
$a_{1}(G_{I}(M))>a_{0}(G_{I}(M))$ (by Lemma 2.4(i)),
$\operatorname{reg}(G_{I}(M))=a_{1}(G_{I}(M))+1$. Using again Lemma 2.4(i), we
get $\operatorname{reg}(G_{I}(M))=p$. Then (c) implies (ii).

Finally, if $e_{0}(I,M)>e_{0}(\mathfrak{m}^{b},M)$, then by Lemma 2.4(v),
$p\leq e_{0}-b-1$. Hence as above, we get

(3.5) $\displaystyle e_{1}$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{e_{0}-b-2}(e_{0}-i-b-1-\ell(H^{0}_{G_{+}}(G_{I}(M))_{i})+e_{0}-\ell(M/IM)$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{e_{0}-b-2}(e_{0}-i-b-1)+e_{0}-\ell(M/IM)$
$\displaystyle=$ $\displaystyle\binom{e_{0}-b}{2}+b+1-\ell(M/IM).$

The proof of (i’), (ii’) and (iii’) is similar to that of (i), (ii) and (iii),
where (3.5) and (3) are used. ∎

Assume that $A$ is a Cohen-Macaulay ring. Elias [3, Proposition 2.5] showed
that if $I\subseteq\mathfrak{m}^{b}$ for some $b\geq 2$, then under an
additional condition we have:

$e_{1}(I)\leq\binom{e_{0}(I)-b}{2}.$

Using integral closures of an ideal, K. Hanumanthu and C. Huneke were able to
remove that additional condition (see [6, Corollary 3.7]). We can now extend
this result to the case of modules.

######  Proposition 3.2.

Assume that $M$ is a Cohen-Macaulay $A$-module of positive dimension and $I$
is an $\mathfrak{m}$-primary ideal of $(A,\mathfrak{m})$ such that
$IM\subseteq\mathfrak{m}^{b}M$ for some $b\geq 2$. Then

$e_{1}(I,M)\leq\binom{e_{0}(I,M)-b}{2}.$

###### Proof.

Using standard technique we may assume that $d=1$. If
$e_{0}(I,M)>e_{0}(\mathfrak{m}^{b},M)$, then the statement follows from
Proposition 3.1 (3.2), since

$\ell(M/IM)>\ell(M/\mathfrak{m}^{b}M)\geq b.$

Assume now that $e_{0}(I,M)=e_{0}(\mathfrak{m}^{b},M)$. For $n\gg 0$, we have

$\begin{array}[]{ll}e_{0}(I,M)(n+1)-e_{1}(I,M)=\ell(M/I^{n+1}M)&\geq\ell(M/(\mathfrak{m}^{b})^{n+1}M)\\\
\\\ &=e_{0}(\mathfrak{m}^{b},M)(n+1)-e_{1}(\mathfrak{m}^{b},M).\end{array}$

Hence $e_{1}(I,M)\leq e_{1}(\mathfrak{m}^{b},M)$. Note that for $n\gg 0$,

$\ell(M/(\mathfrak{m}^{b})^{n+1}M)=\ell(M/(\mathfrak{m}^{(n+1)b}M)=e_{0}(\mathfrak{m},M)(n+1)b-e_{1}(\mathfrak{m},M).$

This implies $e_{0}(\mathfrak{m}^{b},M)=be_{0}(\mathfrak{m},M)$ and
$e_{1}(\mathfrak{m}^{b},M)=e_{1}(\mathfrak{m},M)$. Applying Proposition 3.1
(3.1) to the case $b=1$ we get

$e_{1}(\mathfrak{m},M)\leq\binom{e_{0}(\mathfrak{m},M)}{2}.$

(Of course, this inequality is known in [9].) If $e_{0}(\mathfrak{m},M)=1$,
then the above inequality give $e_{1}(I,M)\leq e_{1}(\mathfrak{m},M)=0$ and
the statement trivially holds. Assume $e_{0}:=e_{0}(\mathfrak{m},M)\geq 2$.
Since $b\geq 2$, we have $be_{0}-b\geq e_{0}$. Hence

$e_{1}(I,M)\leq
e_{1}(\mathfrak{m}^{b},M)=e_{1}(\mathfrak{m},M)\leq\binom{e_{0}}{2}\leq\binom{be_{0}-b}{2}=\binom{e_{0}(I,M)-b}{2}.$

∎

######  Remark 3.3.

Let us examine when the bound in Proposition 3.1 (3.1) holds in the case $\dim
M=1$. Let $b$ be the largest positive integer such that
$IM\subseteq\mathfrak{m}^{b}M$. Note that
$b\leq\ell(M/\mathfrak{m}^{b}M)\leq\ell(M/IM)\leq e_{0}(I,M)$, and
$e_{0}(I,M)\geq e_{0}(\mathfrak{m}^{b},M)=be_{0}(\mathfrak{m},M)$. If
$e_{0}(I,M)=b$, then $\ell(M/\mathfrak{m}M)=1,\ e_{0}(\mathfrak{m},M)=1$, and
from the inequality Proposition 3.1 (3.1) we get $e_{1}(I,M)\leq 0$, whence
$e_{1}(I,M)=0$. Then Proposition 3.1(3.1) becomes an equality. Replacing $A$
by $A/\text{Ann}(M)$, one can now conclude that $e_{0}(I,M)=b$ if and only if
$M\cong A$, $A$ is a regular ring and $I=(x^{b})$, where $\mathfrak{m}=(x)$.
Hence we can exclude this case in further investigation.

######  Lemma 3.4.

Let $M$ be a Cohen-Macaulay module of positive dimension $d$ and $I$ an
$\mathfrak{m}$-primary ideal. Let $b$ be the largest positive integer such
that $IM\subseteq\mathfrak{m}^{b}M$. Assume that $e_{0}(I,M)>b$ and

$e_{1}(I,M)=\binom{e_{0}(I,M)-b+1}{2}+b-\ell(M/IM).$

Then $b=1$.

###### Proof.

Assume that $b\geq 2$. First assume that $d=1$. We have

$\begin{array}[]{ll}e_{1}(I,M)&=\binom{e_{0}(I,M)-b+1}{2}+b-\ell(M/IM)\\\ \\\
&=\binom{e_{0}(I,M)-b}{2}+e_{0}(I,M)-\ell(M/IM).\end{array}$

Since $e_{0}(I,M)\geq\ell(M/IM)$, the above equality together with the
inequality in Proposition 3.2 imply that

(3.7) $e_{1}(I,M)=\binom{e_{0}(I,M)-b}{2},$

and $e_{0}(I,M)=\ell(M/IM)$. Since $M$ is an one-dimensional Cohen-Macaulay
module, $e_{0}(I,M)=\ell(M/xM)$ for some $x\in I$. This implies $IM=xM$, i.e.
we can assume that $I$ is a parameter ideal. Then $e_{1}(I,M)=0$, and by
(3.7), $e_{0}(I,M)\leq b+1$. By the assumption, we get $e_{0}(I,M)=b+1$.

Since $b+1=e_{0}(I,M)\geq e_{0}(\mathfrak{m}^{b},M)=be_{0}(\mathfrak{m},M)$
and $b\geq 2$, we can conclude that $e_{0}(\mathfrak{m},M)=1$. Let
$y\in\mathfrak{m}$ such that $e_{0}(\mathfrak{m},M)=\ell(M/yM)$. Note that
$\ell(M/yM)\geq\ell(M/\mathfrak{m}M)=\mu(M)$. Hence, then we must have
$yM=\mathfrak{m}M$ and $M$ is generated by one element, say $M=Au$. Replacing
$A$ by $A/\text{Ann}(M)$, we may assume that $M=A$. Then $A$ is a regular
ring, $\mathfrak{m}=(y)$, see Remark 3.3. Since $x\in\mathfrak{m}^{b}$ and $b$
is the largest number satisfying this property, it implies that $x=ry^{b}$ for
some unit $r$. But then $e_{0}(I,M)=e_{0}(x,A)=b$, a contradiction. Hence, the
assumption $b\geq 2$ is wrong and then $b=1$.

Now assume that $d\geq 2$. Let $x_{1},...,x_{d-1}$ be an $M$-superficial
sequence for $I$. Let $N=M/(x_{1},...,x_{d-1})M$. Then $\dim N=1$ and

$e_{1}(I,N)=\binom{e_{0}(I,N)-b+1}{2}+b-\ell(N/IN).$

Since $IN\subseteq\mathfrak{m}^{b}N$, we must have $b=1$. ∎

Below are some characterizations for the equality in (3.1).

######  Proposition 3.5.

Let $M$ be an one-dimensional Cohen-Macaulay $A$-module and $I$ an
$\mathfrak{m}$-primary ideal. Let $b$ be the largest positive integer such
that $IM\subseteq\mathfrak{m}^{b}M$. Assume that $e_{0}(I,M)\geq b+2$. Then
the following conditions are equivalent:

  * (i)

$e_{1}(I,M)=\binom{e_{0}(I,M)-b+1}{2}+b-\ell(M/IM),$

  * (ii)

$P_{I,M}(z)=\frac{\ell(M/IM)+(b+1-\ell(M/IM))z+\sum_{i=2}^{e_{0}(I,M)-b}z^{i}}{1-z},$

  * (iii)

$a_{0}(G_{I}(M))\leq 0$ and $\operatorname{reg}(G_{I}(M))=e_{0}(I,M)-b$,

  * (iv)

$\operatorname{reg}(G_{I}(M))=\binom{e_{0}(I,M)-b+2}{2}+b-e_{1}(I,M)-\ell(M/IM)-1$.

If one of the above conditions is satisfied, then $b=1$ and
$e_{0}(I,M)=e_{0}(\mathfrak{m},M)$.

###### Proof.

For simplicity, in this proof we set $e_{0}:=e_{0}(I,M),\ e_{1}:=e_{1}(I,M)$
and $p:=pn(I,M)$.

(ii) $\Longrightarrow$ (i) is immediate from (2.2).

(i) $\Longrightarrow$ (ii) Assume that
$e_{1}=\binom{e_{0}-b+1}{2}+1-\ell(M/IM)$. Since $e_{0}\geq b+2$, by
Proposition 3.1(ii) and (iii), $p=e_{0}-b$, $H_{I,M}(n)=i+b$ for all $1\leq
i\leq e_{0}-b-1$ and $H_{I,M}(n)=e_{0}$ for all $n\geq e_{0}-b$. Substituting
these values into the definition (2.1) of the Hilbert series we then get (ii).

(i) $\Longrightarrow$ (iii) By Proposition 3.1(i), $a_{0}(G_{I}(M))\leq 0$.
Since $e_{0}\geq b+2$, by Proposition 3.1(ii),
$\operatorname{reg}(G_{I}(M))=e_{0}-b$.

(iii) $\Longrightarrow$ (ii) By Lemma 2.4(i), we have
$p=\operatorname{reg}(G_{I}(M))=e_{0}-b$. Since $a_{0}(G_{I}(M))\leq 0$,

$\ell(I^{t}M/I^{t+1}M)=H_{G_{I}(M)}(t)=H_{\overline{G_{I}(M)}}(t)\ \text{for
all}\ t\geq 1.$

On the other hand, by Lemma 2.4(iii),

$H_{\overline{G_{I}(M)}}(t)=\begin{cases}t+b&\text{if}\ 0\leq t\leq
e_{0}-b-1,\\\ e_{0}&\text{if}\ t\geq e_{0}-b.\end{cases}$

Hence

$\begin{array}[]{ll}P_{I,M}(z)&=\ell(M/IM)+\sum_{t=1}^{e_{0}-b-1}(t+b)z^{t}+\sum_{t\geq
e_{0}-b}e_{0}z^{t}\\\ \\\
&=\frac{\ell(M/IM)+(b+1-\ell(M/IM))z+\sum_{i=2}^{e_{0}-b}z^{i}}{1-z}.\end{array}$

(i) $\Longrightarrow$ (iv) Using (i) $\Leftrightarrow$ (iii), we have

$\begin{array}[]{ll}\operatorname{reg}(G_{I}(M))&=e_{0}-b\\\ \\\
&=\binom{e_{0}-b+2}{2}-\binom{e_{0}-b+1}{2}+e_{1}+\ell(M/IM)-b-e_{1}-\ell(M/IM)+b-1\\\
\\\ &=\binom{e_{0}-b+2}{2}+b-e_{1}-\ell(M/IM)-1.\end{array}$

(iv) $\Longrightarrow$ (i) By [1, Proposition 2.1],
$\operatorname{reg}(G_{I}(M))\leq e_{0}-b$. Therefore

$\begin{array}[]{ll}\operatorname{reg}(G_{I}(M))&\leq e_{0}-b\\\ \\\
&=\binom{e_{0}-b+2}{2}-\binom{e_{0}-b+1}{2}+e_{1}+\ell(M/IM)-b-e_{1}-\ell(M/IM)+b-1\\\
\\\ &\leq\binom{e_{0}-b+2}{2}+b-e_{1}-\ell(M/IM)-1\ \ \text{(by Proposition
\ref{H1})}.\end{array}$

By virtue of (iv), this implies $e_{1}=\binom{e_{0}}{2}+1-\ell(M/IM)$.

Finally assume (i). By Lemma 3.4, $b=1$. Further, since $e_{0}-1\geq 2$, we
have $\binom{e_{0}-1}{2}+1\leq\binom{e_{0}}{2}$. Hence, if $e_{0}\neq
e_{0}(\mathfrak{m},M)$, by virtue of Proposition 3.1(3.2), we cannot have (i),
a contradiction. ∎

######  Example 3.6.

Let $A=k[[t^{2},t^{3}]]$ and $M=\mathfrak{m}$. Then
$e_{0}(\mathfrak{m},M)=e_{0}(\mathfrak{m},A)=2$, while
$e_{1}(\mathfrak{m},A)=1$ and $e_{1}(\mathfrak{m},M)=0$. Hence, the condition
(i) of Proposition 3.5 is satisfied for both pairs $(\mathfrak{m},A)$ and
$(\mathfrak{m},M)$. Note that $\ell(\mathfrak{m}^{n}/\mathfrak{m}^{n+1})=2$
for all $n\geq 1$. So, $pn(\mathfrak{m},A)=1$ and $pn(\mathfrak{m},M)=0$. The
ring $G_{\mathfrak{m}}(A)$ and the module $G_{I}(M)$ are Cohen-Macaulay, but
$\operatorname{reg}(G_{\mathfrak{m}}(A))=1=e_{0}(\mathfrak{m},A)-1$, while
$\operatorname{reg}(G_{\mathfrak{m}}(M))=0<e_{0}(\mathfrak{m},M)-1=1$. This
shows that no of the conditions (ii), (iii) and (iv) in Proposition 3.5 holds.
So the condition $e_{0}(I,M)\geq b+2$ in Proposition 3.5 cannot be omitted.

Using (iv) and (v) of Lemma 2.4 and (i’), (ii’) and (iii’) of Proposition 3.1,
similar arguments of the proof of Proposition 3.5 give:

######  Proposition 3.7.

Let $M$ be an one-dimensional Cohen-Macaulay $A$-module and $I$ an
$\mathfrak{m}$-primary ideal such that $I\subseteq\mathfrak{m}^{b}$,
$e_{0}(I,M)>e_{0}(\mathfrak{m}^{b},M)$ and $e_{0}(I,M)\geq b+3$, where $b$ is
a positive integer. Then the following conditions are equivalent:

  * (i)

$e_{1}=\binom{e_{0}(I,M)-b}{2}+b+1-\ell(M/IM)$,

  * (ii)

$P_{I,M}(z)=\frac{\ell(M/IM)+(b+2-\ell(M/IM))z+\sum_{i=2}^{e_{0}(I,M)-b-1}z^{i}}{1-z}$,

  * (iii)

$a_{0}(G_{I}(M))\leq 0$ and $\operatorname{reg}(G_{I}(M))=e_{0}(I,M)-b-1$,

  * (iv)

$\operatorname{reg}(G_{I}(M))=\binom{e_{0}(I,M)-b+1}{2}+b-e_{1}(I,M)-\ell(M/IM)$.

##  4\. The first Hilbert function of an $\mathfrak{m}$-primary ideal

In this section we consider the case $M=A$, that is we study the first Hilbert
coefficient of an $\mathfrak{m}$-primary ideal $I$ of a Cohen-Macaulay local
ring $(A,\mathfrak{m})$. If $b\geq 2$, see Theorem 4.6 below. If $b=1$, then
the Rossi-Valla bound in the statement (i) of the following lemma is clearly
much better.

######  Lemma 4.1.

Let $(A,\mathfrak{m})$ be a $d$-dimensional Cohen-Macaulay ring and $I$ an
$\mathfrak{m}$-primary ideal. Then

  * (i)

([14, Theorem 3.2])

(4.1) $e_{1}(I)\leq\binom{e_{0}(I)}{2}-\binom{\mu(I)-d}{2}-\ell(A/I)+1,$

where $\mu(I)$ denotes the number of generators of $I$.

  * (ii)

(A partial case of [14, Theorem 3.2])  If $d=1$, then we also have

$e_{1}(I)\leq\binom{e_{0}(I)}{2}-\binom{\mu(\tilde{I})-1}{2}-\ell(A/\tilde{I})+1.$

###### Proof.

There is an unclear step in the proof of [14, Theorem 3.2] in the case $d=1$:
from the context, $\lambda$ in [14, (8)] should be $\ell(A/\tilde{I})$, see at
the beginning of [14, Section 3]. Therefore we give here a correction of this
part. So, we may assume that $d=1$ and we need to show

(4.2) $e_{1}(I)\leq\binom{e_{0}(I)}{2}-\binom{\mu(I)-1}{2}-\ell(A/I)+1.$

If $\mu(I)=1$, then $I$ is a parameter ideal, which implies $e_{1}(I)=0$ and
the inequality holds true. Now let $\mu(I)\geq 2$. By [14, Theorem 3.1],

(4.3) $e_{1}(I)\leq\binom{e_{0}(I)}{2}-\binom{g-1}{2}-\ell(A/\tilde{I})+1,$

where

$g=\ell(\tilde{I}/\widetilde{I^{2}})+\sum_{i\geq
2}\ell\left(\frac{\widetilde{I^{i+1}}}{I\widetilde{I^{i}}+\widetilde{I^{i+2}}}\right).$

At the end of the proof of [14, Theorem 3.1], it is shown that

$\begin{array}[]{ll}g&\geq\ell(\tilde{I}/\widetilde{I^{2}}+I\mathfrak{m})+\ell(\widetilde{I^{2}}+I\mathfrak{m}/I\mathfrak{m})\\\
\\\ &=\ell(\tilde{I}/I\mathfrak{m})=\ell(\tilde{I}/I)+\mu(I).\end{array}$

Set $\tilde{l}:=\ell(\tilde{I}/I)$. Then we get

(4.4) $\displaystyle\binom{g-1}{2}+\ell(A/\tilde{I})$ $\displaystyle\geq$
$\displaystyle\binom{\mu(I)-1+\tilde{l}}{2}+\ell(A/\tilde{I})$
$\displaystyle=$
$\displaystyle\binom{\mu(I)-1}{2}+\frac{(2\mu(I)-3)\tilde{l}+\tilde{l}^{2}}{2}+\ell(A/\tilde{I})$
$\displaystyle\geq$
$\displaystyle\binom{\mu(I)-1}{2}+\frac{\tilde{l}(\tilde{l}+1)}{2}+\ell(A/\tilde{I})\
\text{(since}\ \mu(I)\geq 2).$

If $\tilde{l}=0$, then $\ell(A/\tilde{I})=\ell(A/I)$. If $\tilde{l}\geq 1$,
then

(4.5)
$\frac{\tilde{l}(\tilde{l}+1)}{2}+\ell(A/\tilde{I})\geq\tilde{l}+\ell(A/\tilde{I})=\ell(A/I).$

In both cases, from (4.4) we get

(4.6) $\binom{g-1}{2}+\ell(A/\tilde{I})\geq\binom{\mu(I)-1}{2}+\ell(A/I).$

Combining this with (4.3) we immediately get (4.2). ∎

Using the Rossi-Valla bound (4.1) we can immediately see that if $I$ satisfies
the condition (i) of Proposition 3.5 (with $M=A$), then $\mu(I)\leq 2$.
However, if $I$ is a parameter ideal, then $e_{1}(I)=0$, while
$\binom{e_{0}(I)}{2}+1-\ell(A/I)=\binom{e_{0}(I)}{2}+1-e_{0}(I)\geq 1$,
provided $e_{0}(I)\geq 3$. This contradicts the condition (i). So, $\mu(I)=2$.
Below are more information on the structure of $I$ and $A$ itself, when $I$
satisfies the condition (i) of Proposition 3.5, or equivalently, when the
Rossi-Valla bound (4.1) is attained, provided $\mu=2$.

######  Proposition 4.2.

Let $(A,\mathfrak{m})$ be an one-dimensional Cohen-Macaulay ring and $I$ an
$\mathfrak{m}$-primary ideal such that $e_{0}(I)\geq 3$ and
$e_{1}(I)=\binom{e_{0}(I)}{2}+1-\ell(A/I)$. Then we have

  * (i)

$\tilde{I}=\mathfrak{m}$, $I^{2}=\mathfrak{m}I$ and $\mu(I)=2$.

  * (ii)

$\mu(\mathfrak{m})\in\\{2,3\\}$, and

  * (iii)

If $\mu(\mathfrak{m})=2$, then $I=\mathfrak{m}$ and $G(\mathfrak{m})$ is a
Cohen-Macaulay ring.

  * (iv)

If $\mu(\mathfrak{m})=3$, then $I^{n}=\mathfrak{m}^{n}$ for all $n\geq 2$ and
$\ell(A/I)=2$. In this case $\operatorname{depth}G(I)=0$.

###### Proof.

We set $e_{0}:=e_{0}(I)$. (i) $\mu(I)=2$ was shown above. By Proposition
3.1(ii), $b=1$ and $pn(I)=e_{0}(I)-1$. Hence, by Lemma 2.4(iii),
$\ell(\overline{G(I)_{0}})=1$. By (2.4), we then get
$\ell(A/\tilde{I})=\ell(\overline{G(I)_{0}})=1$. Since
$\tilde{I}\subseteq\mathfrak{m}$, we must have $\tilde{I}=\mathfrak{m}$. By
Proposition 3.1(iii), $\ell(I/I^{2})=2$. Since
$2=\mu(I)=\ell(I/\mathfrak{m}I)\leq\ell(I/I^{2})=2$, we get
$I^{2}=\mathfrak{m}I$.

(ii) Since $\mu(I)=2$, the equality in (4.1) also holds for $I$. From (4.4),
(4.5) and (4.6) we must have $\ell(\tilde{I}/I)\leq 1$. Since
$\tilde{I}=\mathfrak{m}$, we get $\ell(A/I)\leq 2$, and by Lemma 4.1(ii), we
now have

$\binom{e_{0}}{2}+1-\ell(A/I)=e_{1}\leq\binom{e_{0}}{2}-\binom{\mu(\mathfrak{m})-1}{2}.$

This implies $\binom{\mu(\mathfrak{m})-1}{2}\leq 1$, whence
$\mu(\mathfrak{m})\leq 3$. By Proposition 3.5,
$e_{0}(\mathfrak{m})=e_{0}(I)\geq 3$. Hence $\mu(\mathfrak{m})\geq 2$.

Assume that $a$ is an element in a minimal basis of $I$. We first show that
$a\not\in\mathfrak{m}^{2}$. Assume by contrary, that $a\in\mathfrak{m}^{2}$.
Since $\tilde{I}=\mathfrak{m}$, we get

$a\in\mathfrak{m}^{2}\cap I=(\tilde{I})^{2}\cap
I\subseteq\widetilde{I^{2}}\cap I.$

By Proposition 3.5(iii) and (2.4), we get

$0=H^{0}_{G(I)_{+}}(G(I))_{1}\cong\frac{\widetilde{I^{2}}\cap I}{I^{2}},$

which implies

(4.7) $\widetilde{I^{2}}\cap I=I^{2}.$

Hence $a\in I^{2}$, a contradiction.

The condition that any element in a minimal basis of $I$ does not belong to
$\mathfrak{m}^{2}$ implies that the images of $a_{1},a_{2}$ of a minimal basis
of $I$ in $\mathfrak{m}/\mathfrak{m}^{2}$ are linearly independent. This means
that $\\{a_{1},a_{2}\\}$ is a part of minimal basis of $\mathfrak{m}$.

(iii) If $\mu(\mathfrak{m})=2$ then $I=\mathfrak{m}$ and
$H^{0}_{G_{+}}(G(I))_{0}=0$. Since $a_{0}(G(I))\leq 0$ (by Proposition
3.5(iii)), $H^{0}_{G_{+}}(G(I))=0$, and $G(\mathfrak{m})=G(I)$ is a Cohen-
Macaulay ring.

(iv) Assume now that $\mu(\mathfrak{m})=3$. As shown above $\ell(A/I)\leq 2$.
So we must have $\ell(A/I)=2$. Assume that $\mathfrak{m}=(a_{1},a_{2},a_{3})$,
where $\\{a_{1},a_{2}\\}$ is a minimal basis of $I$. Moreover, we may assume
that both elements $a_{1},a_{2}$ are non-zero divisors of $A$. Since
$\tilde{I}=\mathfrak{m}$, by (4.7), we have

$a_{1}a_{3}\in I\cap(\tilde{I})^{2}\subseteq
I\cap\widetilde{I^{2}}=I^{2}=(a_{1},a_{2})^{2}.$

Hence

(4.8) $a_{1}a_{3}=y_{1}a_{1}^{2}+y_{2}a_{1}a_{2}+y_{3}a_{2}^{2},$

for some $y_{1},y_{2},y_{3}\in A$. Replacing $a_{3}$ by
$a_{3}-y_{1}a_{1}-y_{2}a_{2}$ in the above relation, we may assume that

$a_{1}a_{3}=y_{3}a_{2}^{2}.$

Analogously, we can find $z_{1},z_{2},z_{3}\in A$ such that

(4.9) $a_{2}a_{3}=z_{1}a_{1}^{2}+z_{2}a_{1}a_{2}+z_{3}a_{2}^{2}.$

Then

$\begin{array}[]{ll}a_{2}a_{3}^{2}&=z_{1}a_{1}^{2}a_{3}+z_{2}a_{1}a_{2}a_{3}+z_{3}a_{2}^{2}a_{3}\\\
&=z_{1}a_{1}y_{3}a_{2}^{2}+z_{2}y_{3}a_{2}^{3}+z_{3}a_{2}^{2}a_{3}.\end{array}$

Since $a_{2}$ is a non-zero divisor, this implies

$a_{3}^{2}=z_{1}y_{3}a_{1}a_{2}+z_{2}y_{3}a_{2}^{2}+z_{3}a_{2}a_{3}\in I^{2}.$

Together with (4.8) and (4.9), this shows that $I^{2}=\mathfrak{m}^{2}$, and
by induction $I^{n}=\mathfrak{m}^{n}$ for all $n\geq 2$. In this case, by
(2.4),
$H^{0}_{G_{+}}(G(I))_{0}\cong\frac{\tilde{I}}{I}=\frac{\mathfrak{m}}{I}\neq
0$, $\operatorname{depth}G(I)=0$. ∎

######  Remark 4.3.

The Cohen-Macaulayness of $G(\mathfrak{m})$ in (iii) of the above proposition
is known long time ago, see, e.g. [16, p. 19].

If $\mu(I)>2$, then the Rossi-Valla bound (4.1) is much better. An ideal, for
which the Rossi-Valla bound (4.1) is attained, may have an arbitrary number of
generators. For an example, take $I=\mathfrak{m}$ in
$A=k[[t^{a},t^{a+1},...,t^{2a-1}]],a\geq 3$. Then $e_{0}(\mathfrak{m})=a$ and

$e_{1}(\mathfrak{m})=a-1=\binom{e_{0}(\mathfrak{m})}{2}-\binom{\mu(\mathfrak{m})-1}{2}.$

If $I=\mathfrak{m}$ the Rossi-Valla bound (4.1) is Elias’ bound given in [2,
Theorem 1.6]. In [5, Theorem 3.1], there is a characterization in terms of
Hilbert series for an one-dimensional Cohen-Macaulay ring such that the Elias’
bound is attained. See also [14, Proposition 3.3] for a shorter proof.

######  Example 4.4.

Let $a\geq 3$ and $A=k[[t^{a},t^{a+1},t^{a^{2}-a-1}]]$ and
$I=(t^{a},t^{a+1})$. Then

$\ell(A/I)=2,\ e_{0}(I)=e_{0}(\mathfrak{m})=a,\
e_{1}(I)=e_{1}(\mathfrak{m})=\binom{a}{2}-1=\binom{e_{0}(I)}{2}+1-\ell(A/I).$

This is the situation in (iv) of the above proposition. Note that
$G(\mathfrak{m})$ is a Cohen-Macaulay ring only in the case $a=3$. This was
indicated in [16, p. 19] in the case $a=3$ and in [2, Proposition 4.6(2)] for
$a\geq 4$.

We now give a new bound on $e_{1}(I)$ for an $\mathfrak{m}$-primary ideal
$I\subseteq\mathfrak{m}^{b}$ and $b\geq 2$. It is in fact a correction of the
bound given in [4, Proposition 1.1]. The following result was stated for any
dimension $d\geq 1$, but the proof there is valid only for $d=1$, because in
general one cannot find an element $x\in\mathfrak{m}$ such that it is
simultaneously superficial for both $\mathfrak{m}$ and $I$.

######  Lemma 4.5.

[4, Proposition 1.1] Let $I\subseteq\mathfrak{m}^{b}$ be an
$\mathfrak{m}$-primary ideal of an one-dimensional Cohen-Macaulay ring $A$.
Then

$e_{1}(I)\leq(e_{0}(\mathfrak{m})-1)(e_{0}(I)-be_{0}(\mathfrak{m}))+e_{1}(\mathfrak{m}).$

Modifying the bound in the above lemma, we can give a new bound on $e_{1}(I)$
for any dimension.

######  Theorem 4.6.

Let $A$ be a Cohen-Macaulay ring of dimension $d\geq 1$. Let
$I\subseteq\mathfrak{m}^{b}$ be an $\mathfrak{m}$-primary ideal, where $b\geq
1$. Then

$e_{1}(I)\leq\frac{1}{2b-1}\binom{e_{0}(I)-b+1}{2}-\binom{\mu(m)-d}{2}.$

###### Proof.

First consider the case $d=1$. By Lemma 4.5,

$e_{1}(I)\leq(e_{0}-1)(e_{0}(I)-be_{0})+e_{1}(\mathfrak{m}),$

where we set $e_{i}:=e_{i}(\mathfrak{m})$. By [2, Theorem 1.6],

$e_{1}\leq\binom{e_{0}}{2}-\binom{\mu(m)-1}{2}.$

Hence

$\begin{array}[]{ll}e_{1}(I)&\leq(e_{0}-1)(e_{0}(I)-be_{0})+\frac{e_{0}(e_{0}-1)}{2}-\binom{\mu(m)-1}{2}\\\
\\\
&=e_{0}^{2}(-b+\frac{1}{2})+e_{0}(e_{0}(I)+b-\frac{1}{2})-e_{0}(I)-\binom{\mu(m)-1}{2}.\end{array}$

The function

$f(t)=(-b+\frac{1}{2})t^{2}+(e_{0}(I)+b-\frac{1}{2})t-e_{0}(I)$

reaches its maximum at $t_{0}=\frac{e_{0}(I)+b-\frac{1}{2}}{2(b-\frac{1}{2})}$
and

$f(t_{0})=\frac{(2e_{0}(I)-2b+1)^{2}}{8(2b-1)}=\frac{1}{2b-1}\left\\{\binom{e_{0}(I)-b+1}{2}+\frac{1}{8}\right\\}.$

Note that $\lfloor\frac{m+\alpha}{n}\rfloor=\lfloor\frac{m}{n}\rfloor$ for any
integers $n\geq 1$, $m$ and a real number $0\leq\alpha<1$. Hence

$\begin{array}[]{ll}e_{1}(I)&\leq\lfloor
f(t_{0})\rfloor-\binom{\mu(m)-1}{2}\\\ \\\
&=\lfloor\frac{1}{2b-1}\binom{e_{0}(I)-b+1}{2}\rfloor-\binom{\mu(m)-1}{2}\\\
\\\
&\leq\frac{1}{2b-1}\binom{e_{0}(I)-b+1}{2}-\binom{\mu(m)-1}{2}.\end{array}$

Now let $d\geq 2$. Let $x\in I$ be a superficial element. Then
$e_{0}(I/x)=e_{0}(I),\ e_{1}(I/x)=e_{1}(I)$,
$I/x\subseteq(\mathfrak{m}/x)^{b}$ and
$\mu(\mathfrak{m}/x)\geq\mu(\mathfrak{m})-1$. Hence, the conclusion follows by
induction on the dimension. ∎

######  Remark 4.7.

Let $b\geq 2$. Then Theorem 4.6 gives

(4.10)
$e_{1}(I)\leq\lfloor\frac{1}{3}\binom{e_{0}(I)-b+1}{2}\rfloor-\binom{\mu(m)-d}{2}.$

It is easy to check that

$\lfloor\frac{1}{3}\binom{e_{0}(I)-b+1}{2}\rfloor\leq\binom{e_{0}(I)-b}{2}.$

This give another proof of Corollary 3.2 in the case $M=A$. It is clearly
better than the bound of Corollary 3.2 if $\mathfrak{m}$ is generated by at
least $d+2$ elements. If $e_{0}(I)\geq b+5$, then from (4.10) we get a much
better bound:

$e_{1}(I)\leq\frac{1}{2}\binom{e_{0}(I)-b}{2}-\binom{\mu(m)-d}{2}.$

######  Remark 4.8.

We now give a brief account of upper bounds on $e_{1}(I)$ of an
$\mathfrak{m}$-primary ideal $I$ of an one-dimensional Cohen-Macaulay ring $A$
such that $I\subsetneq\mathfrak{m}^{2}$.

  * (i)

The first Rossi-Valla bound (4.1):

$e_{1}(I)\leq\binom{e_{0}(I)}{2}-\binom{\mu(I)-1}{2}-\ell(A/I)+1.$

  * (ii)

The second Rossi-Valla bound (see Lemma 4.1(ii)):

$e_{1}(I)\leq\binom{e_{0}(I)}{2}-\binom{\mu(\tilde{I})-1}{2}-\ell(A/\tilde{I})+1.$

  * (iii)

Elias’ bound (see Proposition 4.5):

$e_{1}(I)\leq(e_{0}(\mathfrak{m})-1)(e_{1}(I)-be_{0}(\mathfrak{m}))+e_{1}(\mathfrak{m}).$

  * (iv)

The Hanumanthu-Huneke bound [6, Corollary 2.9]: Under the additional condition
that $A$ is an analytically unramified local domain with algebraically closed
residue field, we have

$e_{1}(I)\leq\binom{e_{0}(I)-\ell(A/\bar{I})+1}{2},$

where $\bar{I}$ denotes the integral closure of $I$.

  * (v)

The case $b=1$ of Theorem 4.6

$e_{1}(I)\leq\binom{e_{0}(I)}{2}-\binom{\mu(m)-1}{2}.$

One can give examples to show that these bounds are independent. Note that the
bounds in (i) and (v) can be lifted to higher dimensions, while we could not
do the same for the other bounds.

##  5\. The second Hilbert coefficient

Rhodes [13, Proposition 6.1(iv)] proved that
$e_{2}(I,M)\leq\binom{e_{1}(I,M)}{2}$. Combining with the bound in Proposition
3.1, we get $e_{2}(I,M)<\frac{1}{8}e_{0}(I,M)^{4}$. In the case
$I=\mathfrak{m}$ and $M=A$, there is a much better bound given in [5, Theorem
2.3]. The bound also involves $e_{1}(\mathfrak{m})$ and some rather technical
invariants. As a consequence, it was shown there that
$e_{2}(\mathfrak{m})\leq\binom{e_{1}(\mathfrak{m})}{2}-\binom{\mu(\mathfrak{m})-d}{2}$,
which is of course better than Rhodes’ bound in the case $I=\mathfrak{m}$.
Applying known bounds on $e_{1}(\mathfrak{m})$ to the bound in [5, Theorem
2.3], one can show that
$e_{2}(\mathfrak{m})<\frac{2}{3}e_{0}(\mathfrak{m})^{3}$.

The aim of this section is to give a new bound on $e_{2}(I,M)$ in terms of
$e_{0}(I,M)$, which is less than $\frac{1}{6}e_{0}(I,M)^{3}$, and to
characterize when this bound is attained. In the case $M=A$, after finding
some relationships between the reduction number and the Hilbert coefficients,
using Theorem 4.6 we can give a better bound for a large class of $I$, see
Theorem 5.9.

######  Theorem 5.1.

Let $M$ be a Cohen-Macaulay module of $\dim(M)=d\geq 2$ over
$(A,\mathfrak{m})$. Let $I$ be an $\mathfrak{m}$-primary ideal such that
$IM\subseteq\mathfrak{m}^{b}M$, where $b$ is a positive integer. Then

$e_{2}(I,M)\leq\binom{e_{0}(I,M)-b+1}{3}.$

###### Proof.

By standard technique we may assume that $d=2$.

Let $x\in I\backslash I^{2}$ be an $M$-superficial element for $I$. Let
$N:=M/xM$. By Lemma 2.1(ii) and (iii), $e_{i}(I,M)=e_{i}(I,N)$ for $i=0,1$.
For short, we write $p:=pn(I,N)$ and $e_{0}:=e_{0}(I,M)=e_{0}(I,N)$. Then

(5.1)
$P_{I,N}(z)=\frac{H_{I,N}(0)+\sum_{i=0}^{p-1}(H_{I,N}(i)-H_{I,N}(i-1))z^{i}+(e_{0}-H_{I,N}(p-1))z^{p}}{1-z}.$

By (2.2), we have

(5.2) $\displaystyle e_{2}(I,N)$ $\displaystyle=$
$\displaystyle\frac{\sum_{i=0}^{p-1}(i-1)i(H_{I,N}(i)-H_{I,N}(i-1))+p(p-1)(e_{0}-H_{I,N}(p-1))}{2!}$
$\displaystyle=$
$\displaystyle-\sum_{i=1}^{p-1}iH_{I,N}(i)+\frac{p(p-1)}{2}e_{0}$
$\displaystyle=$ $\displaystyle\sum_{i=1}^{p-1}i(e_{0}-H_{I,N}(i))$
$\displaystyle\leq$ $\displaystyle\sum_{i=1}^{e_{0}-b-1}i(e_{0}-H_{I,N}(i))\ \
\text{(by Lemma \ref{I6}(iii)) }$ (5.3) $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{e_{0}-b-1}i(e_{0}-i-b-\ell(H^{0}_{G(I)_{+}}(G_{I}(N))_{i})\
\ \text{(by Lemma \ref{I6}(ii))}$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{e_{0}-b-1}i(e_{0}-i-b)=\binom{e_{0}-b+1}{3}.$

Since $M$ is a Cohen-Macaulay module, by Lemma 2.1(iv),

(5.4)
$e_{2}(I,N)=e_{2}(I,M)+\sum_{i=0}^{n}\ell\left(\frac{I^{i+1}M:x}{I^{i}M}\right)\geq
e_{2}(I,M).$

Hence the inequality (5.3) gives $e_{2}(I,M)\leq\binom{e_{0}-b+1}{3}$. ∎

######  Remark 5.2.

Assume that $IM\subseteq\mathfrak{m}^{b}M$. If $e_{0}(I,M)\leq b+1$, then by
the above theorem, we get $e_{2}(I,M)\leq 0$. From the famous result of Narita
[11] on the non negativity of the second Hilbert coefficient (see [15,
Proposition 3.1] for a short proof in the module case), this implies
$e_{2}(I,M)=0$. Hence we can omit this case when dealing with the border case
of the above theorem. The following result say that if the above bound is
attained, then $b=1$ and $I$ satisfies the conditions of Proposition 3.5.

######  Theorem 5.3.

Let $M$ be a Cohen-Macaulay module of $\dim(M)=d\geq 2$ over
$(A,\mathfrak{m})$ and $I$ an $\mathfrak{m}$-primary ideal. Let $b$ be the
largest integer such that $IM\subseteq\mathfrak{m}^{b}M$. Assume that
$e_{0}(I,M)\geq b+2$. The following conditions are equivalent:

  * (i)

$e_{2}(I,M)=\binom{e_{0}(I,M)-b+1}{3},$

  * (ii)

$P_{I,M}(z)=\textstyle{\frac{\ell(M/IM)+(1+b-\ell(M/IM))z+\sum_{i=2}^{e_{0}(I,M)-b}z^{i}}{(1-z)^{d}}},$

  * (iii)

$\operatorname{depth}(G_{I}(M))\geq d-1$ and
$e_{1}(I,M)=\binom{e_{0}(I,M)-b+1}{2}+b-\ell(M/IM)$,

  * (iv)

$\operatorname{depth}(G_{I}(M))\geq d-1$,
$\operatorname{reg}(G_{I}(M))=e_{0}(I)-b$ and $a_{d-1}(G_{I}(M))\leq 1-d$,

  * (v)

$\operatorname{depth}(G_{I}(M))\geq d-1$ and
$\operatorname{reg}(G_{I}(M))=\binom{e_{0}(I,M)-b+2}{2}+b-e_{1}(I,M)-\ell(M/IM)-1$.

If one of the above conditions holds, then $b=1$.

###### Proof.

For simplicity, we set $e_{i}:=e_{i}(I,M),\ i\in\\{0,1,2\\}$ and $G:=G(I)$.
First, let $d=2$.

By (2.2) it is clear that (ii) implies (i). Assume (i), i.e.
$e_{2}=\binom{e_{0}-b+1}{3}$. Let $x\in I\setminus I^{2}$ be an
$M$-superficial element for $I$. Let $N:=M/xM$. By Lemma 2.1,
$e_{i}(I,N)=e_{i}(I,M)=e_{i}$ for $i=0,1$, and by (5.4), $e_{2}(I,M)\leq
e_{2}(I,N)$. Since $e_{2}(I,N)\leq\binom{e_{0}-b+1}{3}$ (see (5.3)), we must
have $e_{2}(I,N)=e_{2}(I,M)=\binom{e_{0}-b+1}{3}$. By Lemma 2.1(v), the
initial form $x^{*}\in I/I^{2}$ is a regular element on $G_{I}(M)$. This means
$\operatorname{depth}G_{I}(M)>0$. Note that $G_{I}(N)\cong
G_{I}(M)/x^{*}G_{I}(M)$.

Moreover, since $e_{0}\geq b+2$, using (5.2) and (5.3) we also have
$p=e_{0}-b$, where $p:=pn(I,N)$, and $H_{I,N}(n)=n+b$ for all $1\leq n\leq p$.
By (5.1) we then get

(5.5)
$P_{I,N}(z)=\frac{\ell(N/IN)+(1+b-\ell(N/IN))z+\sum_{i=2}^{e_{0}-b}z^{i}}{1-z}.$

Therefore, using Lemma 2.1(v) again, we get (ii). Thus (i)
$\Longleftrightarrow$ (ii) and they imply $\operatorname{depth}G_{I}(M)>0$.

(ii) $\Longrightarrow$ (iii) The first part $\operatorname{depth}G_{I}(M)>0$
was just shown, while the second part immediately follows from (2.2).

(ii) $\Longrightarrow$ (iv) and (v) The first part
$\operatorname{depth}G_{I}(M)>0$ was shown above. Since $x^{*}$ is a regular
element on $G_{I}(M)$, by Lemma 2.1(v), it implies that (5.5) holds. This
means $(I,N)$ satisfies the condition (ii) of Proposition 3.5. By the
conditions (iii) and (iv) of Proposition 3.5, we get

$\begin{array}[]{ll}\operatorname{reg}(G_{I}(N))&=\binom{e_{0}-b+2}{2}+b-e_{1}-\ell(N/IN)-1,\\\
\operatorname{reg}(G_{I}(N))&=e_{0}-b\ \text{and}\ a_{0}(G_{I}(N))\leq
0.\end{array}$

Note that
$\operatorname{reg}G_{I}(M)=\operatorname{reg}G_{I}(M)/x^{*}G_{I}(M)=\operatorname{reg}G_{I}(N)$
and $\ell(M/IM)=\ell(N/IN)$. Hence

$\begin{array}[]{ll}\operatorname{reg}(G_{I}(M))&=\binom{e_{0}-b+2}{2}+b-e_{1}-\ell(M/IM)-1,\\\
\operatorname{reg}(G_{I}(M))&=e_{0}-b.\end{array}$

Thus (v) is proved. Further, since $a_{0}(G_{I}(N))\leq 0$, from the exact
sequence

$0=H^{0}_{G_{+}}(G_{I}(N))_{n}\cong
H^{0}_{G_{+}}(G_{I}(M)/x^{*}G_{I}(M))_{n}\rightarrow
H^{1}_{G_{+}}(G_{I}(M))_{n-1}\rightarrow H^{1}_{G_{+}}(G_{I}(M))_{n},$

we get inclusions

$H^{1}_{G_{+}}(G_{I}(M))_{n-1}\hookrightarrow H^{1}_{G_{+}}(G_{I}(M))_{n},$

for all $n\geq 1$. This implies that $H^{1}_{G_{+}}(G_{I}(M))_{n}=0$ for all
$n\geq 0$, or equivalently, $a_{1}(G_{I}(M))\leq-1$. Summing up, (ii) also
implies (iv).

If one of the conditions (iii), (iv) and (v) is fulfilled, then one of the
conditions (i), (iii) or (iv) in Proposition 3.5 holds for the pair $(I,N)$.
Hence by Proposition 3.5(ii)

$\begin{array}[]{ll}P_{I,N}(z)&=\frac{\ell(N/IN)+(b+1-\ell(N/IN))z+\sum_{i=2}^{e_{0}-b}z^{i}}{(1-z)}\\\
&=\frac{\ell(M/IM)+(b+1-\ell(M/IM))z+\sum_{i=2}^{e_{0}-b}z^{i}}{(1-z)}.\end{array}$

Using Lemma 2.1(v), we then get (ii). The proof of the case $d=2$ is
completed.

Assume now $d>2$. Then (ii) $\Longrightarrow$ (i) follows from (2.2).

Assume (i). Let $x\in I\setminus I^{2}$ be an $M$-superficial element for $I$
and $N:=M/xM$. Then $\dim N=d-1$ and the pair $(I,N)$ satisfies the condition
(i). By induction hypothesis, $\operatorname{depth}G_{I}(N)\geq d-2$. Using
Sally’s descent (see [8, Lemma 2.2] or [15, Lemma 1.4]), we can deduce that
$\operatorname{depth}G_{I}(M)\geq d-1$. This implies that $x^{*}$ is regular
on $G_{I}(M)$. By Lemma 2.1(v), (ii) follows. Further, we have
$\operatorname{reg}(G_{I}(M))=\operatorname{reg}(G_{I}(N))$. Using the exact
sequence

$H^{d-2}_{G_{+}}(G_{I}(N))_{n}\cong
H^{d-2}_{G_{+}}(G_{I}(M)/x^{*}G_{I}(M))_{n}\rightarrow
H^{d-1}_{G_{+}}(G_{I}(M))_{n-1}\rightarrow H^{d-1}_{G_{+}}(G_{I}(M))_{n},$

one can see that $a_{d-2}(G_{I}(N))\leq 2-d$ implies $a_{d-1}(G_{I}(M))\leq
1-d$. Since $(I,N)$ satisfies the condition (iii), (iv), (v), we then get that
also $(I,M)$ satisfies these conditions.

Conversely, assume that $\operatorname{depth}G_{I}(M)\geq d-1$. Then, by
Sally’s descent, we get $\operatorname{depth}(G_{I}(N))\geq d-2$ and $x^{*}$
is regular on $G_{I}(M)$. Hence, we have the following exact sequence

$0\rightarrow H^{d-2}_{G_{+}}(G_{I}(N))_{n}\cong
H^{d-2}_{G_{+}}(G_{I}(M)/x^{*}G_{I}(M))_{n}\rightarrow
H^{d-1}_{G_{+}}(G_{I}(M))_{n-1}.$

From this one can see that $a_{d-1}(G_{I}(M))\leq 1-d$ implies
$a_{d-2}(G_{I}(N))\leq 2-d$. Since $e_{i}(I,M)=e_{i}(I,N)$ for all $i\leq 2$,
if $(I,M)$ satisfies one of the conditions (iii), (iv) and (v), then the same
condition holds for $(I,N)$. Therefore, $(i)$ holds for $(I,N)$, whence also
holds for $(I,M)$.

Finally, if one of conditions (i)…(v) is satisfied, then from the condition
(iv) we see that $(I,M)$ satisfies the condition in Lemma 3.4. Hence $b=1$. ∎

######  Example 5.4.

Using Example 4.4, we can see that the pair $(I,M)$ satisfies the conditions
of Theorem 5.3, where

$I=(t^{a},t^{a+1},u_{1},...,u_{d-1})\subset
A=k[[t^{a},t^{a+1},t^{a^{2}-a-1},u_{1},...,u_{d-1}]],$

($a\geq 3$, $d\geq 2$) and $M=A$.

The above theorem says that if $e_{0}(I,M)\geq b+2$ and $b\geq 2$, then the
inequality in Theorem 5.1 is strict. For the case $M=A$, using the bound of
Theorem 4.6, we can give a better bound in the case $b\geq 2$. We need some
more preparation.

Recall that the ideal $J\subseteq I$ is called an $M$-reduction of $I$ if
$I^{n+1}M=JI^{n}M$ for all $n\gg 0$. The number:

$r_{J}(I,M)=\min\\{n\geq 0|\ I^{n+1}M=JI^{n}M\\}$

is called the $M$-reduction number of $I$ with respect to $J$. An
$M$-reduction of $I$ is called minimal if it does not strictly contain another
$M$-reduction of $I$. The number

$r(I,M):=\min\\{r_{J}(I,M)|\ J\ \text{is a minimal \ }M\text{-reduction of \
}I\\}$

is called the $M$-reduction number  of $I$. The above definitions of
reductions and reduction numbers remain valid for any ideal $I$ of a
Noetherian ring $R$ and any finitely generated $R$-module $M$.

######  Remark 5.5.

We recall here some facts on reductions.

  * (i)

A minimal $M$-reduction of $I$ is generated by exactly $d$ elements.

  * (ii)

A minimal $M$-reduction of $I$ can be generated by a maximal $M$-superficial
sequence for $I$.

Below are some relationships between the reduction number and Hilbert
coefficients.

######  Lemma 5.6.

Let $M$ be an one-dimensional Cohen-Macaulay module and $I$ an
$\mathfrak{m}$-primary ideal such that $IM\subseteq\mathfrak{m}^{b}M$ for some
positive integer $b$. Then

$r(I,M)\leq e_{0}(I,M)-b.$

###### Proof.

Assume that $x\in I$ is an $M$-superficial element for $I$ such that
$r(I,M)=r_{(x)}(I,M)$. Then $r(I,M)=r_{(x^{*})}(G_{+},G_{I}(M))$. By [17,
Proposition 3.2],

$r_{(x^{*})}(G_{+},G_{I}(M))\leq\operatorname{reg}(G_{I}(M)).$

By Lemma 2.4(i) and (iii), $\operatorname{reg}(G_{I}(M))=pn(I,M)\leq
e_{0}(I,M)-b$. Hence $r(I,M)\leq e_{0}(I,M)-b$. ∎

######  Lemma 5.7.

Let $M$ be an one-dimensional Cohen-Macaulay module and $I$ an
$\mathfrak{m}$-primary ideal. Then

$e_{2}(I,M)\leq(r^{\prime}(I,M)-1)e_{1}(I,M),$

where we set $r^{\prime}(I,M):=\max\\{1,r(I,M)\\}$.

###### Proof.

Assume that $x\in I$ is an $M$-superficial element for $I$ such that
$r:=r(I,M)=r_{(x)}(I,M)$. Set $r^{\prime}:=\max\\{1,r\\}$. By [15, Lemmas 2.1
and 2.2]

$e_{1}(I,M)=\sum_{j=0}^{r-1}\ell(I^{j+1}M/xI^{j}M).$

Hence

$\begin{array}[]{ll}e_{2}(I,M)&=\sum_{j=1}^{r-1}j\ell(I^{j+1}M/xI^{j}M)\\\ \\\
&\leq(r^{\prime}-1)\sum_{j=0}^{r-1}\ell(I^{j+1}M/xI^{j}M)=(r^{\prime}-1)e_{1}(I,M).\end{array}$

∎

Using the above two lemmas, we can give a new bound on $e_{2}(I,M)$.

######  Proposition 5.8.

Let $M$ be a Cohen-Macaulay module of dimension $d\geq 2$ and $I$ an
$\mathfrak{m}$-primary ideal such that $IM\subseteq\mathfrak{m}^{b}M$ for some
positive integer $b$. Assume that $e_{0}(I,M)\geq b+1$. Then

$e_{2}(I,M)\leq(e_{0}(I,M)-b-1)e_{1}(I,M).$

###### Proof.

By standard technique, we only need to consider the case $d=2$. Let $x\in I$
be an $M$-superficial element for $I$. Set $N=M/xM$. Then $N$ is an one-
dimensional Cohen-Macaulay module. By the assumption, $e_{0}(I,M)-b\geq 1$.
Hence, by Lemma 5.6, $r^{\prime}(I,M)\leq e_{0}(I,M)-b$. Applying Lemma 5.7 to
$N$, by Lemma 2.1(ii) and (iii), we get

$e_{2}(I,N)\leq(r^{\prime}(I,M)-1)e_{1}(I,M)\\\
\leq(e_{0}(I,N)-b-1)e_{1}(I,N)=(e_{0}(I,M)-b-1)e_{1}(I,M).$

By (5.4), $e_{2}(I,M)\leq e_{2}(I,N)$. Hence
$e_{2}(I,M)\leq(e_{0}(I,M)-b-1)e_{1}(I,M).$ ∎

Combining the above result with Theorem 4.6 we get the following bound which
is clearly better the bound of Theorem 5.1 in the case $M=A$ and $b\geq 2$.

######  Theorem 5.9.

Let $I$ be an $\mathfrak{m}$-primary ideal of a Cohen-Macaulay ring
$(A,\mathfrak{m})$ of dimension $d\geq 2$ and such that
$I\subseteq\mathfrak{m}^{b}$ for some positive integer $b$. Assume that
$e_{0}(I,M)\geq b+1$. Then

$e_{2}(I)\leq\frac{3}{2b-1}\binom{e_{0}(I)-b+1}{3}-(e_{0}(I)-b-1)\binom{\mu(\mathfrak{m})-d}{2}.$

###### Proof.

We may assume that $d=2$. For simplicity we set $e_{i}:=e_{i}(I),\ i=0,1,2$.
By Theorem 4.6,

$e_{1}\leq\frac{1}{2b-1}\binom{e_{0}-b+1}{2}-\binom{\mu(\mathfrak{m})-2}{2}.$

Hence, by Proposition 5.8,

$\begin{array}[]{ll}e_{2}&\leq(e_{0}-b-1)e_{1}\\\ \\\
&\leq(e_{0}-b-1)\left\\{\frac{1}{2b-1}\binom{e_{0}-b+1}{2}-\binom{\mu(\mathfrak{m})-2}{2}\right\\}\\\
\\\
&=\frac{3}{2b-1}\binom{e_{0}-b+1}{3}-(e_{0}-b-1)\binom{\mu(\mathfrak{m})-2}{2}.\end{array}$

∎

Acknowledgement. This work is the result of discussions during various stays
of all three authors at Vietnam Institute for Advanced Study in Mathematics
(VIASM). The authors would like to thank VIASM for the financial support and
generous hospitality. The second author is partially supported by
PID2019-104844GB-I00, while the third author is partially supported by
NCXS02.01/22-23 of Vietnam Academy of Sciences and Technology.

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