††footnotetext: 2020 Mathematics Subject Classification. Primary: 13N15,
14R20. Secondary: 14C20, 14R05, 14R25.

# Rigidity of graded integral domains
and of their Veronese subrings

Daniel Daigle  Department of Mathematics and Statistics
University of Ottawa
Ottawa, Canada K1N 6N5<EMAIL_ADDRESS>
###### Abstract.

A ring $R$ is said to be rigid if the only locally nilpotent derivation
$D:R\to R$ is the zero derivation. Let $B=\bigoplus_{n\in\mathbb{Z}}B_{n}$ be
a $\mathbb{Z}$-graded commutative integral domain of characteristic $0$. For
each $d\in\mathbb{N}\setminus\\{0\\}$, consider the Veronese subring
$B^{(d)}=\bigoplus_{n\in d\mathbb{Z}}B_{n}$ of $B$. We study the properties of
the set

$X(B)=\linebreak[3]\big{\\{}\,d\in\mathbb{N}\setminus\\{0\\}\,\mid\,\text{$B^{(d)}$
is non-rigid}\,\big{\\}}$

and give some applications. One of the results states that if $B$ is a
finitely generated normal domain then $X(B)=\bigcup_{d\in
M(B)}\mathscr{I}_{d}$ where $\mathscr{I}_{d}=\\{d,2d,3d,\dots\\}$ and $M(B)$
is a finite primitive subset of $\mathbb{N}\setminus\\{0\\}$ that depends on
$B$. (A subset $Y$ of $\mathbb{N}\setminus\\{0\\}$ is primitive if the
conditions $a,b\in Y$ and $a\mid b$ imply $a=b$.) We study analogous questions
in the more general context of domains graded by abelian groups.

##  1\. Introduction

Let $B$ be a ring (by which we mean a commutative, associative, unital ring).
A derivation $D:B\to B$ is locally nilpotent if, for each $b\in B$, there
exists $n>0$ such that $D^{n}(b)=0$. We write $\operatorname{\text{\rm
LND}}(B)$ for the set of locally nilpotent derivations $D:B\to B$. The ring
$B$ is said to be rigid if $\operatorname{\text{\rm LND}}(B)=\\{0\\}$.

A graded ring is said to be rigid if it is rigid as a non-graded ring.

Let $B=\bigoplus_{n\in\mathbb{Z}}B_{n}$ be a $\mathbb{Z}$-graded ring.
Following [CD22], we say that an element $f$ of $B$ is cylindrical if it is
nonzero and homogeneous of nonzero degree, and $B_{(f)}$ is a polynomial ring
in one variable over a subring. Here, $B_{(f)}$ denotes the degree-zero
subring of the graded ring $B_{f}=S^{-1}B$, where
$S=\\{1,f,f^{2},f^{3},\dots\\}$. Also recall that if
$d\in\mathbb{N}\setminus\\{0\\}$ then the Veronese subring $B^{(d)}$ of $B$ is
the graded subring of $B$ defined by $B^{(d)}=\bigoplus_{n\in
d\mathbb{Z}}B_{n}$. The following is part (1) of Thm 1.2 of [CD22], with minor
edits:

######  1.1 Theorem.

Let ${\rm\bf k}$ be a field of characteristic $0$ and
$B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ a $\mathbb{Z}$-graded affine ${\rm\bf
k}$-domain that satisfies one of:

  * (i)

the transcendence degree of $B$ over $B_{0}$ is at least $2$;

  * (ii)

there exist $i,j$ such that $i<0<j$, $B_{i}\neq 0$ and $B_{j}\neq 0$.

Then $B$ has a cylindrical element if and only if there exists a positive
integer $d$ such that $B^{(d)}$ is non-rigid.

It was shown in [CD22] that this result generalizes Theorem 0.6 of [TKZ13].
The result also raises the question whether something can be said about the
set

$X(B)=\big{\\{}\,d\in\mathbb{N}\setminus\\{0\\}\,\mid\,\text{$B^{(d)}$ is non-
rigid}\,\big{\\}},$

where $B$ is a $\mathbb{Z}$-graded domain of characteristic $0$. One objective
of this article is to study the properties of $X(B)$. However, we shall not
study cylindrical elements; we quoted Thm 1.1 mainly because it motivates the
study of $X(B)$, and also because it suggests that the dichotomy “$X(B)$ is
empty or nonempty” may be of some significance. Note that the geometric
interpretation of Thm 1.1 is discussed in [CD22].

The article is organized as follows.

Section 2 gathers some definitions and facts needed in this paper.

Given a domain $B=\bigoplus_{i\in G}B_{i}$ graded by an abelian group $G$, and
given a subgroup $H$ of $G$, one defines the graded subring
$B^{(H)}=\bigoplus_{i\in H}B_{i}$ of $B$, which generalizes the notion of
Veronese subring. Sections 3–6 study the rigidity of $B$ and of these subrings
$B^{(H)}$, under the assumption that $G/H$ is torsion (it turns out that very
little can be said without this assumption). In fact Sections 3–6 study the
set $\mathscr{X}(B)$ (see Notation 6.1), which is the generalization of $X(B)$
in the context of $G$-graded rings. For the sake of brevity, we shall not say
more (in this Introduction) about the results of these four sections.

Section 7 focuses on $\mathbb{Z}$-gradings, and studies $X(B)$ in that
context. Several results in this section are corollaries of results from
Sections 3–6, but we also obtain some results that go beyond what is done in
the general case. Let us describe some of our results. Let ${\rm\bf k}$ denote
an arbitrary field of characteristic $0$. For each
$d\in\mathbb{N}\setminus\\{0\\}$, let $\mathscr{I}_{d}=\\{d,2d,3d,\dots\\}$.
Let $\preceq$ be the partial order on $\mathbb{N}\setminus\\{0\\}$ defined by
declaring that $d_{1}\preceq d_{2}$ $\Leftrightarrow$
$\mathscr{I}_{d_{1}}\subseteq\mathscr{I}_{d_{2}}$ $\Leftrightarrow$ $d_{2}\mid
d_{1}$. If $B$ is a $\mathbb{Z}$-graded domain of characteristic $0$, let
$M(B)$ be the set of maximal elements of the poset $(X(B),\preceq)$.

Cor. 7.2 asserts that if $B$ is a non-rigid $\mathbb{Z}$-graded affine
${\rm\bf k}$-domain then $B^{(d)}$ is non-rigid for every
$d\in\mathbb{N}\setminus\\{0\\}$. (It seems that this fact was not previously
known; this is perhaps surprising, given the simplicity of the statement.)

If $B=\bigoplus_{n\in\mathbb{Z}}B_{n}$ is a $\mathbb{Z}$-graded domain,
$e(B)=\gcd\big{\\{}\,n\in\mathbb{Z}\,\mid\,B_{n}\neq 0\,\big{\\}}$ is called
the saturation index of $B$. Let $Z$ temporarily denote the set of all height
$1$ homogeneous prime ideals of $B$, and define
$\bar{e}(B)=\operatorname{\text{\rm
lcm}}\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in Z\,\big{\\}}$ if
$Z\neq\varnothing$, and $\bar{e}(B)=e(B)$ if $Z=\varnothing$. We call
$\bar{e}(B)$ the codimension $1$ saturation index of $B$. We have
$e(B),\bar{e}(B)\in\mathbb{N}$ and $e(B)\mid\bar{e}(B)$. We say that $B$ is
saturated in codimension $1$ if $\bar{e}(B)=e(B)$.111The fact that $B$ is
saturated in codimension $1$ if and only if $\bar{e}(B)=e(B)$ is a result
(Lemma 7.6), not a definition, but there is no harm in using it as a
definition in this Introduction.

Let $B$ be a $\mathbb{Z}$-graded affine ${\rm\bf k}$-domain. Part (a) of Cor.
7.14 states that $X(B)=\bigcup_{d\in M(B)}\mathscr{I}_{d}$. Assuming that $B$
is normal, Part (b) gives some information about $M(B)$; it asserts that every
element $d$ of $M(B)$ satisfies $d\mid\bar{e}(B)$ and $d\odiv e(B)=d$ (the
binary operation $\odiv$ is defined in Notation 7.9).

Cor. 7.15 states that if $B$ is a $\mathbb{Z}$-graded normal affine ${\rm\bf
k}$-domain which is saturated in codimension $1$, then $X(B)=\varnothing$ or
$X(B)=\mathbb{N}\setminus\\{0\\}$ (i.e., either $B^{(d)}$ is rigid for all
$d\in\mathbb{N}\setminus\\{0\\}$ or $B^{(d)}$ is non-rigid for all
$d\in\mathbb{N}\setminus\\{0\\}$). This significantly improves part (2) of Thm
1.2 of [CD22], and it is shown in [CD22] that [CD22, Thm 1.2(2)] generalizes
[TKZ13, Cor. 3.2].

In view of Cor. 7.14(a), it is clear that the problem of describing $X(B)$
reduces to that of describing $M(B)$. So it is interesting to ask which
subsets of $\mathbb{N}\setminus\\{0\\}$ are realizable (we call a subset $Y$
of $\mathbb{N}\setminus\\{0\\}$ realizable if there exists a
$\mathbb{Z}$-graded normal affine $\mathbb{C}$-domain $B$ satisfying
$M(B)=Y$). Prop. 7.24 shows that every realizable set is finite and primitive,
but we don’t know if every finite primitive subset of
$\mathbb{N}\setminus\\{0\\}$ is realizable. (A subset $Y$ of
$\mathbb{N}\setminus\\{0\\}$ is primitive if the conditions $a,b\in Y$ and
$a\mid b$ imply $a=b$.)

Section 8 determines the set $M(B)$ for certain Pham-Brieskorn rings. The
results imply, in particular, that for each $n\in\mathbb{N}$ there is a
realizable set of cardinality $n$.

Sections 9 and 10 are not aimed specifically at studying $X(B)$, but are still
concerned with the same general subject, i.e., rigidity of graded domains. Thm
9.6 states that if $B=\bigoplus_{n\in\mathbb{N}}B_{n}$ is an
$\mathbb{N}$-graded normal affine ${\rm\bf k}$-domain whose prime ideal
$B_{+}=\bigoplus_{n>0}B_{n}$ has height at least $2$, and if $x$ is a
homogeneous prime element of $B$ of degree $d>0$, then non-rigidity of
$B_{(x)}$ implies non-rigidity of $(B/xB)^{(d)}$. Prop. 10.5 considers the
polynomial ring $R={\rm\bf k}[X_{1},\dots,X_{n}]$ ($n\geq 2$) endowed with an
$\mathbb{N}$-grading such that each $X_{i}$ is homogeneous and $X_{1}$ has
positive degree; it asserts that if $f$ is a homogeneous prime element of
$R=\bigoplus_{i\in\mathbb{N}}R_{i}$ such that $f\notin R_{0}[X_{1}]$, then
rigidity of $R/fR$ implies rigidity of $R/(f-c)R$ for every $c\in{\rm\bf k}$.

Section 11 returns to rings graded by general abelian groups. Before
describing its main result, we note that the roles played by the numbers
$e(B)$ and $\bar{e}(B)$ in $\mathbb{Z}$-graded rings are played by the
subgroups $\operatorname{\mathbb{G}}(B)$ and
$\operatorname{\bar{}\mathbb{G}}(B)$ of $G$ in the context of $G$-graded rings
(see paragraphs 2.2 and 4.1 for definitions). These groups satisfy
$\operatorname{\bar{}\mathbb{G}}(B)\subseteq\operatorname{\mathbb{G}}(B)\subseteq
G$ and are usually easy to compute. Thm 11.4 implies that if $B$ is a
$G$-graded normal affine ${\rm\bf k}$-domain then $\operatorname{\text{\rm
trdeg}}(B:\operatorname{\text{\rm ML}}(B))\geq\operatorname{\text{\rm
rank}}\big{(}\operatorname{\mathbb{G}}(B)/\operatorname{\bar{}\mathbb{G}}(B)\big{)}$,
where $\operatorname{\text{\rm ML}}(B)$ denotes the intersection of the
kernels of all elements of $\operatorname{\text{\rm LND}}(B)$. It follows that
if $\operatorname{\mathbb{G}}(B)/\operatorname{\bar{}\mathbb{G}}(B)$ is an
infinite group then $B$ is non-rigid. This is one of several results in the
article that show that $\operatorname{\bar{}\mathbb{G}}(B)$ (or $\bar{e}(B)$
in the case of $\mathbb{Z}$-gradings) plays a role in the study of locally
nilpotent derivations of graded domains.

##  2\. Preliminaries

This section gathers the definitions and facts that we need in this article.
All results are known, but we provide proofs when we are unable to give a
reference.

######  2.1.

We use “$\setminus$” for set difference, “$\subset$” for strict inclusion and
“$\subseteq$” for general inclusion. We follow the convention that
$0\in\mathbb{N}$.

All rings and algebras are assumed to be associative, commutative and unital.
If $A$ is a ring then $A^{*}$ denotes the set of units of $A$. If $B$ is an
algebra over a ring $A$, the notation $B=A^{[n]}$ (where $n\in\mathbb{N}$)
means that $B$ is isomorphic as an $A$-algebra to a polynomial ring in $n$
variables over $A$. If $L/K$ is a field extension then $L=K^{(n)}$ means that
$L$ is a purely transcendental extension of $K$ of transcendence degree $n$.

The word “domain” means “integral domain”. We write $\operatorname{\text{\rm
Frac}}A$ for the field of fractions of a domain $A$. If ${\rm\bf k}$ is a
field, then a ${\rm\bf k}$-domain is a domain that is also a ${\rm\bf
k}$-algebra; an affine ${\rm\bf k}$-domain (or ${\rm\bf k}$-affine domain) is
a domain which is a finitely generated ${\rm\bf k}$-algebra. If $A\subseteq B$
are domains then the transcendence degree of $B$ over $A$ is denoted
$\operatorname{\text{\rm trdeg}}_{A}(B)$ or $\operatorname{\text{\rm
trdeg}}(B:A)$.

If $A$ is a ring, we write $\operatorname{\text{\rm
Spec}}^{1}(A)=\big{\\{}\,{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}A\,\mid\,\operatorname{\text{\rm ht}}{\mathfrak{p}}=1\,\big{\\}}$.

If $S$ is a subset of a group $G$, then $\langle S\rangle$ denotes the
subgroup of $G$ generated by $S$.

######  2.2.

Let $G$ be an abelian group (with additive notation). A $G$-grading of a ring
$B$ is a family $\big{(}B_{i}\big{)}_{i\in G}$ of subgroups of $(B,+)$
satisfying $B=\bigoplus_{i\in G}B_{i}$ and $B_{i}B_{j}\subseteq B_{i+j}$ for
all $i,j\in G$. The phrase “let $B=\bigoplus_{i\in G}B_{i}$ be a $G$-graded
ring” means that we are considering the ring $B$ together with the $G$-grading
$\big{(}B_{i}\big{)}_{i\in G}$.

Let $B=\bigoplus_{i\in G}B_{i}$ be a $G$-graded ring.

  1. (1)

The subgroup of $G$ generated by $\big{\\{}\,i\in G\,\mid\,B_{i}\neq
0\,\big{\\}}$ is denoted $\operatorname{\mathbb{G}}(B)$.

  2. (2)

An element of $B$ is homogeneous if it belongs to $\bigcup_{i\in G}B_{i}$. If
$x$ is a nonzero homogeneous element, the degree of $x$, $\deg(x)$, is the
unique $i\in G$ such that $x\in B_{i}$. The degree of a non-homogeneous
element is not defined.

  3. (3)

$B_{0}$ is a subring of $B$ and is called the degree-$0$ subring of $B$.

  4. (4)

Given a homogeneous element $f$ of $B$, $B_{(f)}$ denotes the degree-$0$
subring of the $G$-graded ring $B_{f}=S^{-1}B$ where
$S=\\{1,f,f^{2},\dots\\}$. Given a homogeneous prime ideal ${\mathfrak{p}}$ of
$B$, $B_{({\mathfrak{p}})}$ denotes the degree-$0$ subring of the $G$-graded
ring $S^{-1}B$ where $S$ is the set of homogeneous elements of
$B\setminus{\mathfrak{p}}$.

  5. (5)

If $H$ is a subgroup of $G$, define $B^{(H)}=\bigoplus_{i\in H}B_{i}$ and note
that $B^{(H)}$ is a graded subring of $B$.

  6. (6)

If $G=\mathbb{Z}$ and $B_{i}=0$ for all $i<0$, we say that $B$ is
$\mathbb{N}$-graded. In this case we write $B=\bigoplus_{i\in\mathbb{N}}B_{i}$
and define $B_{+}=\bigoplus_{i>0}B_{i}$, which is an ideal of $B$.

######  2.3 Lemma.

Let $B=\bigoplus_{i\in G}B_{i}$ be a domain graded by an abelian group $G$. If
$\operatorname{\mathbb{G}}(B)$ is torsion then
$\operatorname{\mathbb{G}}(B)=\big{\\{}\,i\in G\,\mid\,B_{i}\neq
0\,\big{\\}}$.

###### Proof.

Since $B$ is a domain, the set $M=\big{\\{}\,i\in G\,\mid\,B_{i}\neq
0\,\big{\\}}$ is a nonempty subset of $\operatorname{\mathbb{G}}(B)$ and is
closed under addition. If $\operatorname{\mathbb{G}}(B)$ is torsion then $M$
is a group and hence $M=\langle M\rangle=\operatorname{\mathbb{G}}(B)$. ∎

######  2.4 Theorem (Thm 1.1 of [GY83]).

Let $G$ be a finitely generated abelian group and $A$ a $G$-graded ring. The
following are equivalent:

  1. (a)

$A$ is a noetherian ring;

  2. (b)

the ring $A_{0}$ is noetherian and the $A_{0}$-algebra $A$ is finitely
generated;

  3. (c)

every homogeneous ideal of $A$ is finitely generated.

######  2.5 Corollary.

Let $A$ be a noetherian ring graded by an abelian group $G$. If $H$ is a
finitely generated subgroup of $G$ then $A^{(H)}$ is a noetherian ring.

###### Proof.

By Thm 2.4, it suffices to show that every homogeneous ideal of $A^{(H)}$ is
finitely generated. Let $I$ be a homogeneous ideal of $A^{(H)}$. Since $A$ is
noetherian, $IA$ is a finitely generated ideal of $A$. So there exists a
finite subset $\\{a_{1},\dots,a_{n}\\}$ of $I$ such that
$IA=\sum_{i=1}^{n}a_{i}A$. Moreover, we can choose $a_{1},\dots,a_{n}$ to be
homogeneous; so $a_{i}\in A_{h_{i}}$ and $h_{i}\in H$ for all $i=1,\dots,n$.
We claim that $\\{a_{1},\dots,a_{n}\\}$ generates $I$. To see this, it
suffices to show that each homogeneous element of $I$ belongs to
$\sum_{i=1}^{n}a_{i}A^{(H)}$. Consider $a\in I\cap A_{h}$, where $h\in H$.
Then there exist $b_{1},\dots,b_{n}\in A$ such that
$a=\sum_{i=1}^{n}a_{i}b_{i}$; moreover, we can choose $b_{1},\dots,b_{n}$ to
be homogeneous ($b_{i}\in A_{g_{i}}$, $g_{i}\in G$) and such that
$a_{i}b_{i}\in A_{h}$ for all $i=1,\dots,n$. If $i$ is such that
$a_{i}b_{i}\neq 0$ then $A_{h}\ni a_{i}b_{i}\in A_{h_{i}}A_{g_{i}}\subseteq
A_{h_{i}+g_{i}}$ implies that $h_{i}+g_{i}=h$, so $g_{i}\in H$ and hence
$b_{i}\in A^{(H)}$. So $a\in\sum_{i=1}^{n}a_{i}A^{(H)}$, which shows that
$\\{a_{1},\dots,a_{n}\\}$ generates $I$. Thus, every homogeneous ideal of
$A^{(H)}$ is finitely generated. ∎

######  2.6 Lemma (Lemma 2.2 of [CD22]).

Let $G$ be an abelian group, $B=\bigoplus_{i\in G}B_{i}$ a $G$-graded ring,
and $R$ a subring of $B_{0}$. If $B$ is finitely generated as an $R$-algebra
then so is $B^{(H)}$ for every subgroup $H$ of $G$.

######  2.7 Lemma.

Let $A$ be a domain graded by an abelian group $G$, and let $H$ be a subgroup
of $G$. Then $A\cap\operatorname{\text{\rm Frac}}(A^{(H)})=A^{(H)}$. Moreover,
if $A$ is normal then so is $A^{(H)}$.

###### Proof.

Write $A=\bigoplus_{g\in G}A_{g}$. Given $x=\sum_{g\in G}x_{g}\in A$ (where
$x_{g}\in A_{g}$ for all $g\in G$), define $S(x)=\big{\\{}\,g\in
G\,\mid\,x_{g}\neq 0\,\big{\\}}$. We claim that

(1) if $x\in A\cap\operatorname{\text{\rm Frac}}(A^{(H)})$ and $x\neq 0$ then
$S(x)\cap H\neq\varnothing$.

Indeed, write $x=u/v$ where $u,v\in A^{(H)}\setminus\\{0\\}$. Then $u\neq 0$
implies $S(u)\neq\varnothing$ and $xv=u$ implies
$S(u)\subseteq\big{\\{}\,i+j\,\mid\,i\in S(x),\ j\in S(v)\,\big{\\}}$, so
there exist $i\in S(x)$ and $j\in S(v)$ such that $i+j\in S(u)\subseteq H$;
since $j\in H$, it follows that $i\in H$, so $S(x)\cap H\neq\varnothing$. This
proves (1).

Now consider $x\in A\cap\operatorname{\text{\rm Frac}}(A^{(H)})$. We can write
$x=x_{H}+x^{\prime}$ with $x_{H},x^{\prime}\in A$, $S(x_{H})\subseteq H$ and
$S(x^{\prime})\cap H=\varnothing$. Since $x$ and $x_{H}$ belong to
$A\cap\operatorname{\text{\rm Frac}}(A^{(H)})$, we have $x^{\prime}=x-x_{H}\in
A\cap\operatorname{\text{\rm Frac}}(A^{(H)})$. Since $S(x^{\prime})\cap
H=\varnothing$, (1) implies that $x^{\prime}=0$, so $x=x_{H}\in A^{(H)}$. This
shows that $A\cap\operatorname{\text{\rm Frac}}(A^{(H)})=A^{(H)}$. If $A$ is
normal then $A^{(H)}$ is an intersection of two normal domains and hence is
normal. ∎

######  2.8 Lemma.

Let $B$ be a domain graded by an abelian group $G$, let $H$ be a subgroup of
$G$, and let $S$ be a multiplicative subset of $B^{(H)}\setminus\\{0\\}$ such
that each element of $S$ is homogeneous. Then

$S^{-1}\big{(}B^{(H)}\big{)}=\big{(}S^{-1}B\big{)}^{(H)}\,.$

###### Proof.

Note that $S^{-1}B$ is a $G$-graded domain. Consider $\frac{b}{s}\in S^{-1}B$
where $b$ is a homogeneous element of $B\setminus\\{0\\}$ and $s\in S$. Since
$\deg\big{(}\frac{b}{s}\big{)}=\deg(b)-\deg(s)$ and $\deg(s)\in H$, we have
$\deg(b)\in H$ $\Leftrightarrow$ $\deg\big{(}\frac{b}{s}\big{)}\in H$, i.e.,
$\frac{b}{s}\in S^{-1}\big{(}B^{(H)}\big{)}$ $\Leftrightarrow$
$\frac{b}{s}\in\big{(}S^{-1}B\big{)}^{(H)}$. ∎

######  2.9 Lemma.

Let $B$ be a ring graded by an abelian group $G$. If $G$ is torsion-free then
the following hold.

  1. (a)

If ${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}B$ and $I$ is the ideal of
$B$ generated by all homogeneous elements of ${\mathfrak{p}}$, then
$I\in\operatorname{\text{\rm Spec}}B$.

  2. (b)

If $B$ is a domain, ${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}^{1}(B)$
and ${\mathfrak{p}}$ contains a nonzero homogeneous element of $B$, then
${\mathfrak{p}}$ is a homogeneous ideal.

###### Proof.

(a) Since $G$ is torsion-free, there exists an order relation $\leq$ on $G$
such that $(G,\leq)$ is a totally ordered abelian group (cf. [AF88, Prop.
1.1.7]); we choose such an order relation. Proceeding by contradiction,
suppose that $x,y\in B\setminus I$ are such that $xy\in I$. Write
$x=\sum_{i\in G}x_{i}$ and $y=\sum_{i\in G}y_{i}$ with $x_{i},y_{i}\in B_{i}$
for all $i$. The sets $S_{x}=\big{\\{}\,i\in G\,\mid\,x_{i}\notin
I\,\big{\\}}$ and $S_{y}=\big{\\{}\,i\in G\,\mid\,y_{i}\notin I\,\big{\\}}$
are nonempty and finite, so we may define $m_{x}=\max S_{x}$ and $m_{y}=\max
S_{y}$. Let $x^{\prime}=\sum_{i\leq m_{x}}x_{i}$ and $y^{\prime}=\sum_{i\leq
m_{y}}y_{i}$; then $x\equiv x^{\prime}$ and $y\equiv y^{\prime}\pmod{I}$, so
$x^{\prime}y^{\prime}\in I$. Moreover, if we write
$x^{\prime}y^{\prime}=\sum_{i\in G}z_{i}$ with $z_{i}\in B_{i}$ for all $i$,
then $x_{m_{x}}y_{m_{y}}=z_{m_{x}+m_{y}}\in I$, so
$x_{m_{x}}y_{m_{y}}\in{\mathfrak{p}}$, so $x_{m_{x}}\in{\mathfrak{p}}$ or
$y_{m_{y}}\in{\mathfrak{p}}$, so $x_{m_{x}}\in I$ or $y_{m_{y}}\in I$, a
contradiction. This proves (a).

To prove (b), consider $I$ as in part (a) and observe that ${\mathfrak{p}}=I$.
∎

######  2.10 Lemma.

Let $B$ be a noetherian normal domain graded by a finitely generated abelian
group $G$. Let $H$ be a subgroup of $G$ such that $G/H$ is torsion.

  1. (a)

$\operatorname{\text{\rm ht}}(J)=\operatorname{\text{\rm ht}}(J\cap B^{(H)})$
for every ideal $J$ of $B$.

  2. (b)

The map $f:\operatorname{\text{\rm Spec}}B\to\operatorname{\text{\rm
Spec}}B^{(H)}$, $f({\mathfrak{P}})={\mathfrak{P}}\cap B^{(H)}$, is surjective
and satisfies $f^{-1}\big{(}\operatorname{\text{\rm
Spec}}^{1}B^{(H)}\big{)}=\operatorname{\text{\rm Spec}}^{1}B$.

  3. (c)

If $G$ is torsion-free then $f^{-1}(Z^{\prime})=Z$, where $Z$ (resp.
$Z^{\prime}$) denotes the set of height $1$ homogeneous prime ideals of $B$
(resp. $B^{(H)}$). Moreover, $f|_{Z}:Z\to Z^{\prime}$ is bijective.

###### Proof.

By Lemmas 2.5 and 2.7, $B^{(H)}$ is a noetherian normal domain; moreover, $B$
is integral over $B^{(H)}$. By [Mat80, (5.E)], the Going-Down Theorem holds
for $B^{(H)}\subseteq B$; so assertion (a) follows from [Mat80, (13.C)], and
(b) follows from (a). Now assume that $G$ is torsion-free. If
${\mathfrak{P}}\in\operatorname{\text{\rm Spec}}^{1}B$ and
${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}^{1}B^{(H)}$ are such that
$f({\mathfrak{P}})={\mathfrak{p}}$, then Lemma 2.9(b) implies that
${\mathfrak{P}}$ is homogeneous if and only if ${\mathfrak{p}}$ is
homogeneous. So $f^{-1}(Z^{\prime})=Z$. It is clear that $f|_{Z}:Z\to
Z^{\prime}$ is surjective. Consider ${\mathfrak{p}}\in Z^{\prime}$ and
${\mathfrak{P}}_{1},{\mathfrak{P}}_{2}\in Z$ such that
$f({\mathfrak{P}}_{1})={\mathfrak{p}}=f({\mathfrak{P}}_{2})$. If $x$ is a
nonzero homogeneous element of ${\mathfrak{P}}_{1}$ then we may choose $d\geq
1$ such that $x^{d}\in B^{(H)}$; then
$x^{d}\in{\mathfrak{p}}\subseteq{\mathfrak{P}}_{2}$ and hence
$x\in{\mathfrak{P}}_{2}$, showing that
${\mathfrak{P}}_{1}\subseteq{\mathfrak{P}}_{2}$ and hence that
${\mathfrak{P}}_{1}={\mathfrak{P}}_{2}$. So $f|_{Z}$ is bijective and (c) is
proved. ∎

######  2.11 Lemma.

Let $R=\bigoplus_{i\in G}R_{i}$ be a domain graded by an abelian group $G$,
let $S$ be the set of all nonzero homogeneous elements of $R$ and consider the
$G$-graded domain $\mathcal{R}=S^{-1}R=\bigoplus_{i\in G}\mathcal{R}_{i}$. If
$\operatorname{\mathbb{G}}(R)\cong\mathbb{Z}^{r}$ where $r\in\mathbb{N}$, then
$\mathcal{R}=\mathcal{R}_{0}[t_{1}^{\pm 1},\dots,t_{r}^{\pm 1}]$ where
$\mathcal{R}_{0}$ is a field that contains $R_{0}$ and $t_{1},\dots,t_{r}$ are
nonzero homogeneous elements of $\mathcal{R}$ that are algebraically
independent over $\mathcal{R}_{0}$.

###### Proof.

We may assume that $r\neq 0$. Observe that
$\operatorname{\mathbb{G}}(\mathcal{R})=\operatorname{\mathbb{G}}(R)\cong\mathbb{Z}^{r}$.
Since every nonzero homogeneous element of $\mathcal{R}$ is a unit, we have
$\operatorname{\mathbb{G}}(\mathcal{R})=\big{\\{}\,i\in
G\,\mid\,\mathcal{R}_{i}\neq 0\,\big{\\}}$, so we can choose nonzero
homogeneous elements $t_{1},\dots,t_{r}$ of $\mathcal{R}$ such that
$(\deg(t_{j}))_{j=1}^{r}$ is a basis of the free $\mathbb{Z}$-module
$\operatorname{\mathbb{G}}(\mathcal{R})$. The reader can check that
$\mathcal{R}_{0}$ and $t_{1},\dots,t_{r}$ have the desired properties. ∎

## Derivations

The notion of locally nilpotent derivation is defined in the introduction, and
so is the notation $\operatorname{\text{\rm LND}}(B)$ for any ring $B$. A ring
$B$ is said to be rigid if $\operatorname{\text{\rm LND}}(B)=\\{0\\}$.

Let $A\subseteq B$ be domains. We say that $A$ is factorially closed in $B$ if
the conditions $x,y\in B\setminus\\{0\\}$ and $xy\in A$ imply that $x,y\in A$.
Clearly, if $A$ is factorially closed in $B$ then $B^{*}=A^{*}$ and
$B\cap\operatorname{\text{\rm Frac}}(A)=A$.

It is well known that if $B$ is a domain of characteristic $0$,
$D\in\operatorname{\text{\rm LND}}(B)$ and $A=\ker(D)$, then (i) $A$ is
factorially closed in $B$; (ii) if $a\in A$ then the derivation $aD:B\to B$ is
locally nilpotent; (iii) if $S$ is a multiplicative set of $A$ then the
derivation $S^{-1}D:S^{-1}B\to S^{-1}B$ is locally nilpotent and
$\ker(S^{-1}D)=S^{-1}A$.

The following is well known and easy to prove:

######  2.12 Lemma.

Let $S$ be a multiplicative set of a domain $R$ of characteristic $0$ and let
$D:S^{-1}R\to S^{-1}R$ be a derivation. Suppose that $R$ is finitely generated
as an algebra over the ring $R\cap\ker(D)$.

  1. (a)

There exists $s\in S$ such that the derivation $sD:S^{-1}R\to S^{-1}R$ maps
$R$ into itself. Consequently, $(sD)|_{R}:R\to R$ is a derivation of $R$.

  2. (b)

If $D$ is locally nilpotent then so are $sD$ and $(sD)|_{R}$.

Part (a) of the following fact appeared in [NG67, Section 1] and [Wri81, Prop.
2.1]; part (b) easily follows from (a).

######  2.13 Lemma.

Let $B$ be a $\mathbb{Q}$-algebra, $D\in\operatorname{\text{\rm LND}}(B)$ and
$A=\ker(D)$.

  1. (a)

If $s\in B$ satisfies $Ds=1$ then $B=A[s]=A^{[1]}$ and $D=\frac{d}{ds}:A[s]\to
A[s]$.

  2. (b)

Let $t\in B$ be such that $D(t)\neq 0$ and $D^{2}(t)=0$, and let $a=D(t)\in
A\setminus\\{0\\}$. Then $B_{a}=A_{a}[t]=(A_{a})^{[1]}$.

The following is a slightly improved222The result in [Vas69] assumes that
$\mathbb{Q}\subseteq A$. version of the Theorem of Vasconcelos [Vas69]:

######  2.14 Lemma (Lemma 2.8 of [CD22]).

Let $A\subseteq B$ be domains of characteristic $0$ such that $B$ is integral
over $A$. If $\delta:A\to A$ is a locally nilpotent derivation and $D:B\to B$
is a derivation that extends $\delta$, then $D$ is locally nilpotent.

######  2.15 Definition.

Let $B=\bigoplus_{i\in G}B_{i}$ be a ring graded by an abelian group $G$.

  1. (1)

A derivation $D:B\to B$ is homogeneous if there exists $d\in G$ such that
$D(B_{i})\subseteq B_{i+d}$ holds for all $i\in G$; if $D$ is homogeneous and
nonzero then $d$ is unique, and is called the degree of $D$.

  2. (2)

$\operatorname{\text{\rm HLND}}(B)=\big{\\{}\,D\in\operatorname{\text{\rm
LND}}(B)\,\mid\,\text{$D$ is homogeneous}\,\big{\\}}$

  3. (3)

$\operatorname{\text{\rm
KHLND}}(B)=\big{\\{}\,\ker(D)\,\mid\,D\in\operatorname{\text{\rm
HLND}}(B)\text{ and }D\neq 0\,\big{\\}}$

  4. (4)

The $G$-graded ring $B$ is said to be rigid if $\operatorname{\text{\rm
LND}}(B)=\\{0\\}$, i.e., if it is rigid as a non-graded ring.

Graded rings $B$ satisfying $\operatorname{\text{\rm HLND}}(B)=\\{0\\}$ and
$\operatorname{\text{\rm LND}}(B)\neq\\{0\\}$ do exist, and are not rigid; see
for instance Propositions 6.5 and 6.6 of [DFMJ17]. The following well-known
fact states that the conditions $\operatorname{\text{\rm HLND}}(B)=\\{0\\}$
and $\operatorname{\text{\rm LND}}(B)=\\{0\\}$ are equivalent under certain
assumptions on $B$ and $G$.

######  2.16 Lemma (Lemma 2.7 of [CD22]).

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ an affine ${\rm\bf
k}$-domain graded by a torsion-free abelian group $G$. Then,

$B$ is non-rigid if and only if $\operatorname{\text{\rm
HLND}}(B)\neq\\{0\\}$.

##  3\. From derivations of $B$ to derivations of $B^{(H)}$

The main result of the section is Thm 3.2. For its proof, we need:

######  3.1 Lemma.

Let $\mathcal{B}$ be a domain of characteristic $0$ graded by an abelian group
$G$ and suppose that $\mathcal{B}=R[t]=R^{[1]}$ where $R$ is a graded subring
of $\mathcal{B}$ and $t$ is homogeneous. For any subgroup $H$ of $G$ such that
$G/H$ is torsion, we have $R^{(H)}\in\operatorname{\text{\rm
KHLND}}\big{(}\mathcal{B}^{(H)}\big{)}$.

###### Proof.

Note that $I=\big{\\{}\,m\in\mathbb{Z}\,\mid\,m\deg(t)\in
H+\operatorname{\mathbb{G}}(R)\,\big{\\}}$ is a nonzero ideal of $\mathbb{Z}$,
because $G/H$ is torsion. Let $n>0$ be such that $I=n\mathbb{Z}$. We claim
that

(2) there exists $i\in G$ such that $R_{i}\neq 0$ and $i+n\deg(t)\in H$.

Indeed, we have $n\deg(t)\in\operatorname{\mathbb{G}}(R)+H$, so we may choose
$g\in\operatorname{\mathbb{G}}(R)$ and $h\in H$ such that $n\deg(t)=g+h$.
Since $M=\big{\\{}\,i\in G\,\mid\,R_{i}\neq 0\,\big{\\}}$ is closed under
addition and the subgroup of $G$ generated by $M$ is
$\operatorname{\mathbb{G}}(R)$, we have $g=i_{1}-i_{2}$ for some
$i_{1},i_{2}\in M$. Since $G/H$ is torsion, there exists
$m\in\mathbb{N}\setminus\\{0\\}$ such that $mi_{1}\in H$. Define
$i=(m-1)i_{1}+i_{2}$, then $i\in M$ (so $R_{i}\neq 0$) and
$i+n\deg(t)=(m-1)i_{1}+i_{2}+g+h=mi_{1}+h\in H$, which proves (2).

Choose $i$ as in (2), choose $\rho\in R_{i}\setminus\\{0\\}$, and choose
$d\in\mathbb{N}\setminus\\{0\\}$ such that $di\in H$. Define $\xi=\rho^{d}$
and $\tau=\rho t^{n}$; these are nonzero homogeneous elements of $R^{(H)}$ and
$\mathcal{B}^{(H)}$ respectively, and $\tau$ is transcendental over $R^{(H)}$.
We claim that

(3)
$R^{(H)}[\tau]\subseteq\mathcal{B}^{(H)}\subseteq\big{(}R^{(H)}\big{)}_{\xi}[\tau].$

The first part of (3) is clear. To prove the second part, note that each
element of $\mathcal{B}^{(H)}$ is a finite sum of elements $rt^{m}$ where $r$
is a homogeneous element of $R\setminus\\{0\\}$, $m\in\mathbb{N}$, and
$\deg(rt^{m})\in H$. For such an element $rt^{m}$, we have
$\deg(r)+m\deg(t)\in H$ and $\deg(r)\in\operatorname{\mathbb{G}}(R)$, so $m\in
I=n\mathbb{Z}$ and hence $m=nk$ for some $k\in\mathbb{N}$. Then
$rt^{m}=r(t^{n})^{k}=r\rho^{-k}\tau^{k}$. Choose $\ell\in\mathbb{N}$ such that
$\ell d-k\geq 0$; then

(4) $\textstyle rt^{m}=\frac{r\rho^{\ell d-k}}{\xi^{\ell}}\tau^{k}\,.$

Since $\deg(rt^{m}),\deg(\xi),\deg(\tau)\in H$, (4) gives $\deg(r\rho^{\ell
d-k})\in H$, so $r\rho^{\ell d-k}\in R^{(H)}$ and hence
$rt^{m}\in\big{(}R^{(H)}\big{)}_{\xi}[\tau]$ by (4) again. This proves
$\mathcal{B}^{(H)}\subseteq\big{(}R^{(H)}\big{)}_{\xi}[\tau]$ and completes
the proof of (3).

Now (3) implies that $\big{(}\mathcal{B}^{(H)}\big{)}_{\xi}=S[\tau]=S^{[1]}$
where $S=\big{(}R^{(H)}\big{)}_{\xi}$, so we may consider the derivative
$\frac{d}{d\tau}:S[\tau]\to S[\tau]$. Let
$\Delta=\xi\frac{d}{d\tau}:(\mathcal{B}^{(H)})_{\xi}\to(\mathcal{B}^{(H)})_{\xi}$;
then $\Delta\in\operatorname{\text{\rm
HLND}}\big{(}(\mathcal{B}^{(H)})_{\xi}\big{)}$ and $\ker\Delta=S$. We claim
that $\Delta(\mathcal{B}^{(H)})\subseteq\mathcal{B}^{(H)}$. To see this,
consider the same element $rt^{m}$ as before and let us check that
$\Delta(rt^{m})\in\mathcal{B}^{(H)}$. If $m=0$ then $\deg(r)=\deg(rt^{m})\in
H$, so $rt^{m}=r\in R^{(H)}$ and hence $\Delta(rt^{m})=0\in\mathcal{B}^{(H)}$.
If $m>0$ then $k>0$, so (4) gives

$\textstyle(\mathcal{B}^{(H)})_{\xi}\ni\Delta(rt^{m})=\xi\frac{d}{d\tau}\big{(}\frac{r\rho^{\ell
d-k}}{\xi^{\ell}}\tau^{k}\big{)}=\xi\cdot\frac{r\rho^{\ell
d-k}}{\xi^{\ell}}\cdot k\tau^{k-1}=kr\rho^{d-1}t^{n(k-1)}\in\mathcal{B}.$

Since $\mathcal{B}\cap(\mathcal{B}^{(H)})_{\xi}=\mathcal{B}^{(H)}$ by Lemma
2.7, we get $\Delta(rt^{m})\in\mathcal{B}^{(H)}$. Since every element of
$\mathcal{B}^{(H)}$ is a finite sum of elements $rt^{m}$ of this type, we
obtain $\Delta(\mathcal{B}^{(H)})\subseteq\mathcal{B}^{(H)}$. Let
$D:\mathcal{B}^{(H)}\to\mathcal{B}^{(H)}$ be the restriction of $\Delta$. Then
$D\in\operatorname{\text{\rm HLND}}\big{(}\mathcal{B}^{(H)}\big{)}$ and $\ker
D=\mathcal{B}^{(H)}\cap\ker\Delta=\mathcal{B}^{(H)}\cap\big{(}R^{(H)}\big{)}_{\xi}$.
Since $R$ is factorially closed in $\mathcal{B}=R^{[1]}$, it follows that
$R^{(H)}$ is factorially closed in $\mathcal{B}^{(H)}$, so
$\mathcal{B}^{(H)}\cap\big{(}R^{(H)}\big{)}_{\xi}=R^{(H)}$ and hence $\ker
D=R^{(H)}$. We have $D(\tau)=\xi\frac{d}{d\tau}(\tau)=\xi\neq 0$, so $D\neq 0$
and hence $R^{(H)}\in\operatorname{\text{\rm
KHLND}}\big{(}\mathcal{B}^{(H)}\big{)}$. ∎

######  3.2 Theorem.

Let ${\rm\bf k}$ be a field of characteristic $0$, let $B$ be an affine
${\rm\bf k}$-domain graded by an abelian group $G$, and let $H$ be a subgroup
of $G$ such that $G/H$ is torsion.

  1. (a)

For each $A\in\operatorname{\text{\rm KHLND}}(B)$, we have
$A^{(H)}\in\operatorname{\text{\rm KHLND}}(B^{(H)})$.

  2. (b)

The map $\operatorname{\text{\rm KHLND}}(B)\to\operatorname{\text{\rm
KHLND}}(B^{(H)})$, $A\mapsto A^{(H)}$, is injective.333This map is not
necessarily surjective. Section 8 gives examples where
$\operatorname{\text{\rm KHLND}}(B)=\varnothing$ and $\operatorname{\text{\rm
KHLND}}(B^{(H)})\neq\varnothing$.

  3. (c)

If $G$ is torsion-free and $B$ is non-rigid then $B^{(H)}$ is non-rigid.

###### Proof.

(a) Let $A\in\operatorname{\text{\rm KHLND}}(B)$. Choose
$D\in\operatorname{\text{\rm HLND}}(B)\setminus\\{0\\}$ such that $\ker D=A$,
and choose a homogeneous element $t\in B$ such that $D(t)\neq 0$ and
$D^{2}(t)=0$. Consider the homogeneous element $a=D(t)$ of $A\setminus\\{0\\}$
and let $\mathcal{B}=B_{a}$ and $R=A_{a}$; then $\mathcal{B}=R[t]=R^{[1]}$ by
Lemma 2.13(b), and $R^{(H)}\in\operatorname{\text{\rm
KHLND}}\big{(}\mathcal{B}^{(H)}\big{)}$ by Lemma 3.1. Choose
$\Delta\in\operatorname{\text{\rm HLND}}(\mathcal{B}^{(H)})$ such that
$\ker(\Delta)=R^{(H)}$.

Since $G/H$ is torsion, there exists $n\in\mathbb{N}\setminus\\{0\\}$ such
that $\deg(a^{n})\in H$; let $\alpha=a^{n}$ and observe that
$\mathcal{B}=B_{\alpha}$. By Lemma 2.8, we have
$\mathcal{B}^{(H)}=\big{(}B_{\alpha}\big{)}^{(H)}=\big{(}B^{(H)}\big{)}_{\alpha}$,
so $\Delta\in\operatorname{\text{\rm HLND}}\big{(}(B^{(H)})_{\alpha}\big{)}$.
Since (by Lemma 2.6) $B^{(H)}$ is ${\rm\bf k}$-affine, Lemma 2.12 implies that
there exists $\ell\in\mathbb{N}$ such that
$\alpha^{\ell}\Delta:(B^{(H)})_{\alpha}\to(B^{(H)})_{\alpha}$ maps $B^{(H)}$
into itself. Let $\delta:B^{(H)}\to B^{(H)}$ be the restriction of
$\alpha^{\ell}\Delta$; then $\delta\in\operatorname{\text{\rm HLND}}(B^{(H)})$
and $\ker(\delta)=B^{(H)}\cap\ker(\Delta)=B^{(H)}\cap R^{(H)}$. To finish the
proof of assertion (a), it suffices to verify that

(5) $B^{(H)}\cap R^{(H)}=A^{(H)}\quad\text{and}\quad A^{(H)}\neq B^{(H)}.$

The fact that $A$ is factorially closed in $B$ implies that $B\cap R=B\cap
A_{a}=A$, so

$A^{(H)}\subseteq B^{(H)}\cap R^{(H)}\subseteq B^{(H)}\cap B\cap R=B^{(H)}\cap
A=A^{(H)},$

which proves the first part of (5). To prove the second part, recall that
$D(t)\neq 0$. Since $G/H$ is torsion, we can pick
$m\in\mathbb{N}\setminus\\{0\\}$ such that $t^{m}\in B^{(H)}$. Since
$D(t^{m})=mt^{m-1}D(t)\neq 0$, we have $t^{m}\notin A$, so $t^{m}\notin
A^{(H)}$ and hence $A^{(H)}\neq B^{(H)}$. So (5) is true and assertion (a) is
proved.

(b) Suppose that $A_{1},A_{2}\in\operatorname{\text{\rm KHLND}}(B)$ are such
that $A_{1}^{(H)}=A_{2}^{(H)}$. Let $a\in A_{1}$. Since $G/H$ is torsion, we
can pick $m\in\mathbb{N}\setminus\\{0\\}$ such that $a^{m}\in A_{1}^{(H)}$.
Then $a^{m}\in A_{2}^{(H)}\subseteq A_{2}$, so $a\in A_{2}$ since $A_{2}$ is
factorially closed in $B$. This shows that $A_{1}\subseteq A_{2}$, so
$A_{1}=A_{2}$ follows by symmetry. This proves (b).

(c) We have $\operatorname{\text{\rm HLND}}(B)\neq\\{0\\}$ by Lemma 2.16, so
this follows from (a). ∎

######  3.3 Remark.

Part (a) of the Theorem implies in particular that

if $\operatorname{\text{\rm HLND}}(B)\neq\\{0\\}$ then
$\operatorname{\text{\rm HLND}}(B^{(H)})\neq\\{0\\}$.

######  3.4 Remark.

The Theorem is false if we do not assume that $G/H$ is torsion. For instance,
consider the polynomial ring $\mathbb{C}[X,Y]$ with the
$\mathbb{Z}^{2}$-grading defined by declaring that $X$ and $Y$ are homogeneous
of degrees $(1,0)$ and $(0,1)$, respectively. Let $B$ be the
$\mathbb{Z}^{2}$-graded subalgebra $\mathbb{C}[X^{2},X^{3},Y]$ of
$\mathbb{C}[X,Y]$. Then $0\neq\frac{\partial}{\partial
Y}\in\operatorname{\text{\rm HLND}}(B)$ but if $H$ is the subgroup of
$\mathbb{Z}^{2}$ generated by $(1,0)$ then $B^{(H)}=\mathbb{C}[X^{2},X^{3}]$
is rigid.

######  3.5 Remark.

If $B$ is a ring of characteristic $0$ (graded or not), the Makar-Limanov
invariant of $B$ (denoted $\operatorname{\text{\rm ML}}(B)$) is the
intersection of the kernels of all elements of $\operatorname{\text{\rm
LND}}(B)$. If $B$ is graded, define the homogeneous Makar-Limanov invariant of
$B$ (denoted $\operatorname{\text{\rm HML}}(B)$) to be the intersection of the
kernels of all elements of $\operatorname{\text{\rm HLND}}(B)$. Note that if
$B$ is graded then $\operatorname{\text{\rm
ML}}(B)\subseteq\operatorname{\text{\rm HML}}(B)$.

Thm 3.2 implies that if ${\rm\bf k}$ is a field of characteristic $0$ and $B$
an affine ${\rm\bf k}$-domain graded by an abelian group $G$, then
$\operatorname{\text{\rm HML}}(B^{(H)})\subseteq\operatorname{\text{\rm
HML}}(B)^{(H)}$ for every subgroup $H$ of $G$ such that $G/H$ is torsion.

##  4\. The group $\operatorname{\bar{}\mathbb{G}}(B)$

This section is devoted to defining the subgroup
$\operatorname{\bar{}\mathbb{G}}(B)$ of $\operatorname{\mathbb{G}}(B)$ and
proving some of its properties. We will see that
$\operatorname{\bar{}\mathbb{G}}(B)$ is related to locally nilpotent
derivations in several ways. This idea is first encountered in Prop. 4.7, and
is further developed in subsequent sections.

Recall that $\operatorname{\text{\rm
Spec}}^{1}(A)=\big{\\{}\,{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}A\,\mid\,\operatorname{\text{\rm ht}}{\mathfrak{p}}=1\,\big{\\}}$ for
any ring $A$.

######  4.1 Definition.

Let $B=\bigoplus_{i\in G}B_{i}$ be a domain graded by an abelian group $G$.
Given ${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}B$, the set
$\mathbb{M}(B,{\mathfrak{p}})=\big{\\{}\,i\in
G\,\mid\,B_{i}\nsubseteq{\mathfrak{p}}\,\big{\\}}$ is a submonoid of
$\operatorname{\mathbb{G}}(B)$. The subgroup of $\operatorname{\mathbb{G}}(B)$
generated by $\mathbb{M}(B,{\mathfrak{p}})$ is denoted
$\overline{\mathbb{M}}(B,{\mathfrak{p}})$. We define the subgroup
$\operatorname{\bar{}\mathbb{G}}(B)$ of $\operatorname{\mathbb{G}}(B)$ by

$\operatorname{\bar{}\mathbb{G}}(B)=\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}\overline{\mathbb{M}}(B,{\mathfrak{p}})$

if $\operatorname{\text{\rm Spec}}^{1}(B)\neq\varnothing$, and
$\operatorname{\bar{}\mathbb{G}}(B)=\operatorname{\mathbb{G}}(B)$ if
$\operatorname{\text{\rm Spec}}^{1}(B)=\varnothing$. We say that $B$ is
saturated in codimension $1$ if
$\operatorname{\bar{}\mathbb{G}}(B)=\operatorname{\mathbb{G}}(B)$.

######  4.2 Lemma.

Let $B=\bigoplus_{i\in G}B_{i}$ be a domain graded by an abelian group $G$. If
$\operatorname{\mathbb{G}}(B)$ is torsion and $\operatorname{\text{\rm
Spec}}^{1}(B)\neq\varnothing$ then

$\operatorname{\bar{}\mathbb{G}}(B)=\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}\\!\\!{\mathbb{M}}(B,{\mathfrak{p}})\ =\ \big{\\{}\,i\in
G\,\mid\,\text{$B_{i}\nsubseteq{\mathfrak{p}}$ for every
${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}^{1}(B)$}\,\big{\\}}.$

###### Proof.

Since every submonoid of a torsion group is a group, we have
$\mathbb{M}(B,{\mathfrak{p}})=\overline{\mathbb{M}}(B,{\mathfrak{p}})$ for
each ${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}^{1}(B)$. The conclusion
follows. ∎

######  4.3 Lemma.

Let $B=\bigoplus_{i\in G}B_{i}$ be a domain graded by an abelian group $G$. If
${\mathfrak{p}}$ is a homogeneous prime ideal of $B$ then
$\overline{\mathbb{M}}(B,{\mathfrak{p}})=\operatorname{\mathbb{G}}(B/{\mathfrak{p}})$.

###### Proof.

Clearly, $\operatorname{\mathbb{G}}(B/{\mathfrak{p}})$ is the subgroup of $G$
generated by $\big{\\{}\,i\in G\,\mid\,(B/{\mathfrak{p}})_{i}\neq
0\,\big{\\}}=\big{\\{}\,i\in
G\,\mid\,B_{i}\nsubseteq{\mathfrak{p}}\,\big{\\}}=\mathbb{M}(B,{\mathfrak{p}})$,
so
$\operatorname{\mathbb{G}}(B/{\mathfrak{p}})=\overline{\mathbb{M}}(B,{\mathfrak{p}})$.
∎

The following Proposition explains how to compute the groups
$\overline{\mathbb{M}}(B,{\mathfrak{p}})$.

######  4.4 Proposition.

Let $G$ be an abelian group and let $B=\bigoplus_{i\in G}B_{i}$ be a
$G$-graded domain which is finitely generated as a $B_{0}$-algebra.

  1. (a)

Let $S$ be a finite generating set of the $B_{0}$-algebra $B$ such that each
element of $S$ is nonzero and homogeneous. Let
${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}(B)$ and define

$D_{\mathfrak{p}}=\big{\\{}\,\deg(x)\,\mid\,x\in
S\setminus{\mathfrak{p}}\,\big{\\}}.$

Then ${\mathbb{M}}(B,{\mathfrak{p}})$ is the submonoid of $G$ generated by
$D_{\mathfrak{p}}$, and $\overline{\mathbb{M}}(B,{\mathfrak{p}})$ is the
subgroup of $G$ generated by $D_{\mathfrak{p}}$.

  2. (b)

The set
$\big{\\{}\,\overline{\mathbb{M}}(B,{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}(B)\,\big{\\}}$ is finite.

  3. (c)

The set
$\big{\\{}\,\operatorname{\mathbb{G}}(B/{\mathfrak{p}})\,\mid\,\text{${\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}(B)$ and ${\mathfrak{p}}$ is homogeneous}\,\big{\\}}$ is finite.

###### Proof.

It is straightforward to see that if ${\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}(B)$ then ${\mathbb{M}}(B,{\mathfrak{p}})$ is the submonoid of $G$
generated by $D_{\mathfrak{p}}$ (we leave this to the reader). This implies
that $\overline{\mathbb{M}}(B,{\mathfrak{p}})$ is the subgroup of $G$
generated by $D_{\mathfrak{p}}$, so (a) follows. Since
$D=\big{\\{}\,\deg(x)\,\mid\,x\in S\,\big{\\}}$ is a finite set, and since
$\big{\\{}\,D_{\mathfrak{p}}\,\mid\,{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}(B)\,\big{\\}}$ is a collection of subsets of $D$, we see that
$\big{\\{}\,D_{\mathfrak{p}}\,\mid\,{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}(B)\,\big{\\}}$ is a finite set; so (b) follows from (a). In view of
Lemma 4.3, (c) follows from (b). ∎

######  4.5 Remark.

If $B$ is a noetherian domain graded by a finitely generated abelian group,
then Thm 2.4 implies that $B$ is finitely generated as a $B_{0}$-algebra, so
the hypothesis of Proposition 4.4 is satisfied.

In the case where $B$ is a domain graded by a torsion-free abelian group, we
have the following description of $\operatorname{\bar{}\mathbb{G}}(B)$.

######  4.6 Proposition.

Let $B=\bigoplus_{i\in G}B_{i}$ be a domain graded by a torsion-free abelian
group $G$, and let $Z$ denote the set of all height $1$ homogeneous prime
ideals of $B$. If $Z\neq\varnothing$ then
$\operatorname{\bar{}\mathbb{G}}(B)=\bigcap_{{\mathfrak{p}}\in
Z}\operatorname{\mathbb{G}}(B/{\mathfrak{p}})$, and if $Z=\varnothing$ then
$\operatorname{\bar{}\mathbb{G}}(B)=\operatorname{\mathbb{G}}(B)$.

###### Proof.

It suffices to show that

(6)
$\overline{\mathbb{M}}(B,{\mathfrak{p}})=\begin{cases}\operatorname{\mathbb{G}}(B/{\mathfrak{p}})&\text{if
${\mathfrak{p}}\in Z$,}\\\ \operatorname{\mathbb{G}}(B)&\text{if
${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}^{1}(B)\setminus
Z$.}\end{cases}$

If ${\mathfrak{p}}\in Z$ then
$\operatorname{\mathbb{G}}(B/{\mathfrak{p}})=\overline{\mathbb{M}}(B,{\mathfrak{p}})$
by Lemma 4.3. If ${\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)\setminus Z$ then Lemma 2.9(b) (together with the fact that $G$
is torsion-free) implies that $B_{i}\cap{\mathfrak{p}}=\\{0\\}$ for all $i\in
G$, so $B_{i}\nsubseteq{\mathfrak{p}}$ for all $i\in G$ such that $B_{i}\neq
0$, so $\overline{\mathbb{M}}(B,{\mathfrak{p}})=\operatorname{\mathbb{G}}(B)$.
∎

######  4.7 Proposition.

Let $B$ be a domain of characteristic $0$ graded by an abelian group $G$.

  1. (a)

For each $D\in\operatorname{\text{\rm HLND}}(B)\setminus\\{0\\}$, there exists
a height $1$ homogeneous prime ideal ${\mathfrak{p}}$ of $B$ satisfying:

    1. (i)

some homogeneous element $t\in{\mathfrak{p}}$ satisfies $D(t)\neq 0$ and
$D^{2}(t)=0$;

    2. (ii)

$\operatorname{\mathbb{G}}(B/{\mathfrak{p}})=\operatorname{\mathbb{G}}(\ker
D)$.

  2. (b)

$\operatorname{\bar{}\mathbb{G}}(B)\subseteq\operatorname{\mathbb{G}}(\ker
D)\subseteq\operatorname{\mathbb{G}}(B)$ for all $D\in\operatorname{\text{\rm
HLND}}(B)$.

  3. (c)

If $B$ is saturated in codimension $1$ then $\operatorname{\mathbb{G}}(\ker
D)=\operatorname{\mathbb{G}}(B)$ for all $D\in\operatorname{\text{\rm
HLND}}(B)$.

###### Proof.

(a) Let $D\in\operatorname{\text{\rm HLND}}(B)\setminus\\{0\\}$ and let
$A=\ker D$. Choose a homogeneous $t\in B$ such that $Dt\in A\setminus\\{0\\}$.
Let $S$ be the set of homogeneous elements of $A\setminus\\{0\\}$. Note that
$S^{-1}B$ and $S^{-1}A$ are $G$-graded $\mathbb{Q}$-domains,
$S^{-1}D\in\operatorname{\text{\rm HLND}}(S^{-1}B)\setminus\\{0\\}$ and
$\ker(S^{-1}D)=S^{-1}A$. Since $(S^{-1}D)(t)$ is a unit of $S^{-1}A$, we have
$S^{-1}B=(S^{-1}A)[t]=(S^{-1}A)^{[1]}$ by Lemma 2.13. The condition
$S^{-1}B=(S^{-1}A)[t]=(S^{-1}A)^{[1]}$ implies that ${\mathfrak{P}}=tS^{-1}B$
is a height $1$ prime ideal of $S^{-1}B$. By basic properties of localization,
it follows that ${\mathfrak{p}}=B\cap{\mathfrak{P}}$ is a height $1$ prime
ideal of $B$. It is clear that ${\mathfrak{P}}$ is homogeneous, so
${\mathfrak{p}}$ is homogeneous. We have
$\operatorname{\mathbb{G}}(B/{\mathfrak{p}})=\operatorname{\mathbb{G}}(S^{-1}(B/{\mathfrak{p}}))=\operatorname{\mathbb{G}}(S^{-1}B/tS^{-1}B)=\operatorname{\mathbb{G}}(S^{-1}A)=\operatorname{\mathbb{G}}(A)$
and $t\in{\mathfrak{p}}$, so (a) is proved.

(b) Let $D\in\operatorname{\text{\rm HLND}}(B)$. It is clear that
$\operatorname{\mathbb{G}}(\ker D)\subseteq\operatorname{\mathbb{G}}(B)$, so
it’s enough to prove that
$\operatorname{\bar{}\mathbb{G}}(B)\subseteq\operatorname{\mathbb{G}}(\ker
D)$. If $D=0$ then the claim is obvious. If $D\neq 0$ then part (a) implies
that there exists a height $1$ homogeneous prime ideal ${\mathfrak{p}}$ of $B$
such that $\operatorname{\mathbb{G}}(\ker
D)=\operatorname{\mathbb{G}}(B/{\mathfrak{p}})$. We have
$\operatorname{\mathbb{G}}(B/{\mathfrak{p}})=\overline{\mathbb{M}}(B,{\mathfrak{p}})$
by Lemma 4.3, and
$\overline{\mathbb{M}}(B,{\mathfrak{p}})\supseteq\operatorname{\bar{}\mathbb{G}}(B)$
by Def. 4.1, so (b) follows.

(c) The assumption means that
$\operatorname{\bar{}\mathbb{G}}(B)=\operatorname{\mathbb{G}}(B)$, so this
follows from (b). ∎

We find it intriguing that $\operatorname{\bar{}\mathbb{G}}(B)$ is related to
locally nilpotent derivations (as demonstrated by Prop. 4.7). We will see in
Section 5 that $\operatorname{\bar{}\mathbb{G}}(B)$ plays a role in the
process of extending derivations of $B^{(H)}$ to derivations of $B$. The group
$\operatorname{\bar{}\mathbb{G}}(B)$ also appears in the statement of Thm 6.3,
and Thm 11.4 will reveal yet another facet of the relation between
$\operatorname{\bar{}\mathbb{G}}(B)$ and $\operatorname{\text{\rm LND}}(B)$.

##  5\. From derivations of $B^{(H)}$ to derivations of $B$

The following is Notation 3.3 of [CD22]:

######  5.1 Notation.

We write $(A,B)\in\operatorname{\text{\rm EXT}}$ as an abbreviation for:

> $A$ is a ring, $B$ is an $A$-algebra, and for every derivation $\delta:A\to
> A$ there exists a unique derivation $D:B\to B$ that makes the following
> diagram commute:
>
>
> $\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{D}$$\textstyle{B}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\textstyle{A\,.\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

######  5.2 Example.

If $L/K$ is a separable algebraic field extension, then
$(K,L)\in\operatorname{\text{\rm EXT}}$ (see [ZS75], Corollary $2^{\prime}$,
p. 125).

This section revolves around the following question: Given a $G$-graded domain
$B$ of characteristic $0$ and a subgroup $H$ of $G$ such that $G/H$ is
torsion, when do we have $(B^{(H)},B)\in\operatorname{\text{\rm EXT}}$? In
other words, when is it the case that every derivation $\delta:B^{(H)}\to
B^{(H)}$ extends uniquely to a derivation $D:B\to B$?

A priori, “$(B^{(H)},B)\in\operatorname{\text{\rm EXT}}$” is a condition on
general derivations of $B$ and $B^{(H)}$. To relate this to locally nilpotent
or homogeneous derivations, we need:

######  5.3 Lemma.

Let $B$ be a domain of characteristic $0$ graded by an abelian group $G$, and
let $H$ be a subgroup of $G$ such that $G/H$ is torsion. Assume that
$(B^{(H)},B)\in\operatorname{\text{\rm EXT}}$.

  1. (a)

Suppose that $\delta:B^{(H)}\to B^{(H)}$ is a derivation and that $D:B\to B$
is the unique derivation of $B$ that extends it. If $\delta$ is locally
nilpotent (resp. homogeneous) then so is $D$.

  2. (b)

If $B^{(H)}$ is non-rigid then so is $B$.

  3. (c)

If $\operatorname{\text{\rm HLND}}(B^{(H)})\neq\\{0\\}$ then
$\operatorname{\text{\rm HLND}}(B)\neq\\{0\\}$.

###### Proof.

(a) If $\delta$ is locally nilpotent then, since $B$ is integral over
$B^{(H)}$, $D$ is locally nilpotent by Lemma 2.14. If $\delta$ is homogeneous
and $b\in B\setminus\\{0\\}$ is a homogeneous element then, for a suitable
$n\geq 1$, we have $b^{n}\in B^{(H)}$, so
$nb^{n-1}D(b)=D(b^{n})=\delta(b^{n})$ is homogeneous; since $nb^{n-1}$ is
homogeneous and nonzero, it follows that $D(b)$ is homogeneous, which shows
that $D$ is homogeneous. This proves (a). Assertions (b) and (c) easily follow
from (a). ∎

The next two results are Lemmas 3.5 and 3.6 of [CD22], respectively.

######  5.4 Lemma.

Let $B$ be a noetherian normal domain and $A$ a subring of $B$. Suppose that
$(\operatorname{\text{\rm Frac}}A,\operatorname{\text{\rm
Frac}}B)\in\operatorname{\text{\rm EXT}}$ and that there exists a family
$(f_{i})_{i\in I}$ of elements of $A\setminus\\{0\\}$ satisfying:

  * •

$(A_{f_{i}},B_{f_{i}})\in\operatorname{\text{\rm EXT}}$ for every $i\in I$;

  * •

no height $1$ prime ideal of $B$ contains all $f_{i}$.

Then $(A,B)\in\operatorname{\text{\rm EXT}}$.

######  5.5 Lemma.

Let $A$ be a ring, $A[\mathbf{X}]=A[X_{1},\dots,X_{n}]=A^{[n]}$,
$f_{1},\dots,f_{n}\in A[\mathbf{X}]$, $B=A[\mathbf{X}]/(f_{1},\dots,f_{n})$,
and $\pi:A[\mathbf{X}]\to B$ the canonical homomorphism of the quotient ring.
Let $P\in A[\mathbf{X}]$ be the determinant of the Jacobian matrix
$\frac{\partial(f_{1},\dots,f_{n})}{\partial(X_{1},\dots,X_{n})}$. If $\pi(P)$
is a unit of $B$, then $(A,B)\in\operatorname{\text{\rm EXT}}$.

We shall now establish a sequence of results about rings graded by finite
abelian groups. All these results will be superseded by the main result of
this section, Thm 5.11.

######  5.6 Lemma.

Let $B=\bigoplus_{i\in G}B_{i}$ be a $\mathbb{Q}$-domain graded by a finite
abelian group $G$. Suppose that there exist homogeneous units
$x_{1},\dots,x_{n}\in B^{*}$ such that $\langle\deg
x_{1}\rangle\oplus\cdots\oplus\langle\deg
x_{n}\rangle=\operatorname{\mathbb{G}}(B)$. Then
$(B_{0},B)\in\operatorname{\text{\rm EXT}}$.

###### Proof.

We may assume that $\deg(x_{i})\neq 0$ for all $i$. For each $i=1,\dots,n$,
let $e_{i}\geq 2$ be the order of $\deg(x_{i})$ in
$\operatorname{\mathbb{G}}(B)$. Let
$L=\big{\\{}\,({\ell_{1}},\dots,{\ell_{n}})\in\mathbb{N}^{n}\,\mid\,\text{$0\leq\ell_{i}<e_{i}$
for all $i=1,\dots,n$}\,\big{\\}}$ and observe that the map
$({\ell_{1}},\dots,{\ell_{n}})\mapsto\deg(x_{1}^{\ell_{1}}\cdots
x_{n}^{\ell_{n}})$ from $L$ to $\operatorname{\mathbb{G}}(B)$ is bijective.
This implies that if $b$ is a nonzero homogeneous element of $B$ then there is
a unique $({\ell_{1}},\dots,{\ell_{n}})\in L$ such that
$\deg(b)=\deg(x_{1}^{\ell_{1}}\cdots x_{n}^{\ell_{n}})$; since
$x_{1},\dots,x_{n}$ are homogeneous units of $B$, it follows that
$b/(x_{1}^{\ell_{1}}\cdots x_{n}^{\ell_{n}})\in B_{0}$, so
$b=ax_{1}^{\ell_{1}}\cdots x_{n}^{\ell_{n}}$ for some $a\in B_{0}$. This shows
that $B$ is a free $B_{0}$-module with basis
$\mathcal{B}=\big{\\{}\,x_{1}^{\ell_{1}}\cdots
x_{n}^{\ell_{n}}\,\mid\,({\ell_{1}},\dots,{\ell_{n}})\in L\,\big{\\}}$.

Consider the ideal
$I=(X_{1}^{e_{1}}-x_{1}^{e_{1}},\dots,X_{n}^{e_{n}}-x_{n}^{e_{n}})$ of the
polynomial ring $B_{0}[X_{1},\dots,X_{n}]$ and the surjective homomorphism of
$B_{0}$-algebras $\varphi:B_{0}[X_{1},\dots,X_{n}]\to B$ given by
$\varphi(X_{i})=x_{i}$ ($1\leq i\leq n$). Clearly, $I\subseteq\ker\varphi$. If
$F\in\ker\varphi$ then the division algorithm shows that there exists $f\in
B_{0}[X_{1},\dots,X_{n}]$ such that $F\equiv f\pmod{I}$ and
$\deg_{X_{i}}(f)<e_{i}$ for all $i=1,\dots,n$; then $f\in\ker\varphi$, so
$f(x_{1},\dots,x_{n})=0$, so the fact that $\mathcal{B}$ is linearly
independent over $B_{0}$ implies that $f=0$. This shows that $\ker\varphi=I$,
so

$B\cong
B_{0}[X_{1},\dots,X_{n}]/(X_{1}^{e_{1}}-x_{1}^{e_{1}},\dots,X_{n}^{e_{n}}-x_{n}^{e_{n}}).$

Let $P\in B_{0}[X_{1},\dots,X_{n}]$ be the determinant of the Jacobian matrix
$\frac{\partial(X_{1}^{e_{1}}-x_{1}^{e_{1}},\dots,X_{n}^{e_{n}}-x_{n}^{e_{n}})}{\partial(X_{1},\dots,X_{n})}$.
Since $\mathbb{Q}\subseteq B$,
$P(x_{1},\dots,x_{n})=\prod_{i=1}^{n}(e_{i}x_{i}^{e_{i}-1})$ is a unit of $B$;
so $(B_{0},B)\in\operatorname{\text{\rm EXT}}$ by Lemma 5.5. ∎

######  5.7 Definition.

Let $B$ be a domain graded by a finite abelian group $G$. An element $x$ of
$B$ is admissible if there exist nonzero homogeneous elements
$x_{1},\dots,x_{n}$ of $B$ such that $x=x_{1}x_{2}\cdots x_{n}$ and
$\operatorname{\mathbb{G}}(B)=\langle\deg(x_{1})\rangle\oplus\cdots\oplus\langle\deg(x_{n})\rangle$.

######  5.8 Lemma.

Let $B=\bigoplus_{i\in G}B_{i}$ be a $\mathbb{Q}$-domain graded by a finite
abelian group $G$. If $x$ is an admissible element of $B$ and $m\geq 1$ is
such that $x^{m}\in B_{0}$, then
$((B_{0})_{x^{m}},B_{x^{m}})\in\operatorname{\text{\rm EXT}}$.

###### Proof.

There exist nonzero homogeneous elements $x_{1},\dots,x_{n}$ of $B$ such that
$x=x_{1}x_{2}\cdots x_{n}$ and
$\operatorname{\mathbb{G}}(B)=\langle\deg(x_{1})\rangle\oplus\cdots\oplus\langle\deg(x_{n})\rangle$.
Consider the $G$-graded domain $\mathcal{B}=B_{x^{m}}$ and note that
$x_{1},\dots,x_{n}$ are homogeneous units of $\mathcal{B}$ such that
$\langle\deg x_{1}\rangle\oplus\cdots\oplus\langle\deg
x_{n}\rangle=\operatorname{\mathbb{G}}(B)=\operatorname{\mathbb{G}}(\mathcal{B})$.
By Lemma 5.6, we have $(\mathcal{B}_{0},\mathcal{B})\in\operatorname{\text{\rm
EXT}}$. Since (by Lemma 2.8 with $H=0$) $\mathcal{B}_{0}=(B_{0})_{x^{m}}$, we
are done. ∎

See Def. 4.1 for the concept of saturation in codimension $1$.

######  5.9 Lemma.

Let $B=\bigoplus_{i\in G}B_{i}$ be a domain graded by a finite abelian group
$G$. The following are equivalent:

  1. (i)

$B$ is saturated in codimension $1$;

  2. (ii)

no height $1$ prime ideal of $B$ contains all admissible elements of $B$.

###### Proof.

If $\operatorname{\text{\rm Spec}}^{1}(B)=\varnothing$ then both (i) and (ii)
are true, so (i)$\Leftrightarrow$(ii). From now-on, assume that
$\operatorname{\text{\rm Spec}}^{1}(B)\neq\varnothing$. Note that
$\operatorname{\bar{}\mathbb{G}}(B)=\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}{\mathbb{M}}(B,{\mathfrak{p}})$ by Lemma 4.2.

Suppose that (i) holds. Then
$\operatorname{\mathbb{G}}(B)=\operatorname{\bar{}\mathbb{G}}(B)=\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}{\mathbb{M}}(B,{\mathfrak{p}})$. Consider
${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}^{1}(B)$. Then
$\operatorname{\mathbb{G}}(B)=\mathbb{M}(B,{\mathfrak{p}})$. Choose
$d_{1},\dots,d_{n}\in\operatorname{\mathbb{G}}(B)$ such that
$\operatorname{\mathbb{G}}(B)=\langle d_{1}\rangle\oplus\cdots\oplus\langle
d_{n}\rangle$. Then $d_{1},\dots,d_{n}\in\mathbb{M}(B,{\mathfrak{p}})$, so for
each $j\in\\{1,\dots,n\\}$ we have $B_{d_{j}}\nsubseteq{\mathfrak{p}}$ and
hence we can choose $x_{j}\in B_{d_{j}}\setminus{\mathfrak{p}}$. Then
$x=x_{1}\cdots x_{n}$ is an admissible element of $B$ such that
$x\notin{\mathfrak{p}}$. So (ii) holds.

Conversely, suppose that (ii) holds. To prove (i), it suffices to show that if
$i\in\operatorname{\mathbb{G}}(B)$ and
${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}^{1}(B)$, then
$B_{i}\nsubseteq{\mathfrak{p}}$. So consider
$i\in\operatorname{\mathbb{G}}(B)$ and
${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}^{1}(B)$. By (ii), some
admissible element $x$ of $B$ satisfies $x\notin{\mathfrak{p}}$. There exist
nonzero homogeneous elements $x_{1},\dots,x_{n}$ of $B$ such that
$x=x_{1}\cdots x_{n}$ and
$\operatorname{\mathbb{G}}(B)=\langle\deg(x_{1})\rangle\oplus\cdots\oplus\langle\deg(x_{n})\rangle$.
For some $j_{1},\dots,j_{n}\in\mathbb{N}$, we have
$i=j_{1}\deg(x_{1})+\cdots+j_{n}\deg(x_{n})$, so $x_{1}^{j_{1}}\cdots
x_{n}^{j_{n}}\in B_{i}$. Since $x\notin{\mathfrak{p}}$, we have
$x_{1}^{j_{1}}\cdots x_{n}^{j_{n}}\notin{\mathfrak{p}}$ and hence
$B_{i}\nsubseteq{\mathfrak{p}}$. So (i) holds. ∎

######  5.10 Proposition.

Let $B=\bigoplus_{i\in G}B_{i}$ be a noetherian normal $\mathbb{Q}$-domain
graded by a finite abelian group $G$. If $B$ is saturated in codimension $1$
then $(B_{0},B)\in\operatorname{\text{\rm EXT}}$.

###### Proof.

Let $d$ be a positive integer such that $di=0$ for all $i\in G$, and consider
the subset $Y=\big{\\{}\,x^{d}\,\mid\,\text{$x$ is an admissible element of
$B$}\,\big{\\}}$ of $B_{0}$. Since $\operatorname{\text{\rm
Frac}}(B)/\operatorname{\text{\rm Frac}}(B_{0})$ is an algebraic extension of
fields of characteristic $0$, we have $(\operatorname{\text{\rm
Frac}}(B_{0}),\operatorname{\text{\rm Frac}}(B))\in\operatorname{\text{\rm
EXT}}$. By Lemma 5.8, $((B_{0})_{y},B_{y})\in\operatorname{\text{\rm EXT}}$
for every $y\in Y$. By Lemma 5.9, no height $1$ prime ideal of $B$ contains
$Y$. So Lemma 5.4 implies that $(B_{0},B)\in\operatorname{\text{\rm EXT}}$. ∎

######  5.11 Theorem.

Let $B=\bigoplus_{i\in G}B_{i}$ be a noetherian normal $\mathbb{Q}$-domain
graded by a finitely generated abelian group $G$. For every subgroup $H$ of
$G$ such that $G/H$ is torsion,

$\big{(}B^{(H)},B^{(H+\operatorname{\bar{}\mathbb{G}}(B))}\big{)}\in\operatorname{\text{\rm
EXT}}.$

###### Proof.

Consider the group $X=H+\operatorname{\bar{}\mathbb{G}}(B)$, the finite group
$Y=X/H$, and the canonical homomorphism of the quotient group $\pi:X\to Y$.
Let $A=B^{(X)}$. Since $A$ has an $X$-grading, it also has a $Y$-grading,
namely $A=\bigoplus_{y\in Y}A_{y}$ where for each $y\in Y$ we define
$A_{y}=\bigoplus_{i\in X_{y}}B_{i}$ (with $X_{y}=\big{\\{}\,i\in
X\,\mid\,\pi(i)=y\,\big{\\}}$). We always regard $A$ as being graded by the
finite group $Y$. Since $A_{0}=B^{(H)}$, it suffices to show that
$(A_{0},A)\in\operatorname{\text{\rm EXT}}$. By Cor. 2.5 and Lemma 2.7, $A$ is
a noetherian normal $\mathbb{Q}$-domain; so, by Prop. 5.10, it suffices to
show that $A$ is saturated in codimension $1$.444The $X$-graded ring $B^{(X)}$
is not necessarily saturated in codimension $1$, but we will show that the
$Y$-graded ring $A$ is.

Consider the inclusion $A\subseteq B$; by Lemma 2.10(b),
${\mathfrak{p}}\mapsto{\mathfrak{p}}\cap A$ is a surjective map from
$\operatorname{\text{\rm Spec}}^{1}(B)$ to $\operatorname{\text{\rm
Spec}}^{1}(A)$. In particular, $\operatorname{\text{\rm
Spec}}^{1}(B)\neq\varnothing$ $\Leftrightarrow$ $\operatorname{\text{\rm
Spec}}^{1}(A)\neq\varnothing$. If $\operatorname{\text{\rm
Spec}}^{1}(A)=\varnothing$ then (by Def. 4.1) $A$ is saturated in codimension
$1$ and we are done. So we may assume that $\operatorname{\text{\rm
Spec}}^{1}(A)\neq\varnothing$. It follows that $\operatorname{\text{\rm
Spec}}^{1}(B)\neq\varnothing$,
$\operatorname{\bar{}\mathbb{G}}(B)={\textstyle\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}\overline{\mathbb{M}}(B,{\mathfrak{p}})}$ and
$\operatorname{\bar{}\mathbb{G}}(A)={\textstyle\bigcap_{{\mathfrak{q}}\in\operatorname{\text{\rm
Spec}}^{1}(A)}\overline{\mathbb{M}}(A,{\mathfrak{q}})}$.

Consider ${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}(B)$. We claim that

(7) $\pi\big{(}\overline{\mathbb{M}}(B,{\mathfrak{p}})\cap
X\big{)}=\pi\big{(}{\mathbb{M}}(B,{\mathfrak{p}})\cap X\big{)}.$

Indeed, let $\xi\in\overline{\mathbb{M}}(B,{\mathfrak{p}})\cap X$. Then
$\xi=i-j$ for some $i,j\in{\mathbb{M}}(B,{\mathfrak{p}})$. Since $G/H$ is
torsion, there exists $m\in\mathbb{N}$ such that $(m+1)j\in H$. Then
${\mathbb{M}}(B,{\mathfrak{p}})\ni i+mj=\xi+(m+1)j\in X$, so
$i+mj\in{\mathbb{M}}(B,{\mathfrak{p}})\cap X$ and
$\pi(\xi)=\pi(\xi+(m+1)j)=\pi(i+mj)\in\pi({\mathbb{M}}(B,{\mathfrak{p}})\cap
X)$. This proves (7).

For each $i\in{\mathbb{M}}(B,{\mathfrak{p}})\cap X$, we have
$B_{i}\nsubseteq{\mathfrak{p}}$, so $A_{\pi(i)}\nsubseteq{\mathfrak{p}}$
because $B_{i}\subseteq A_{\pi(i)}$, so
$A_{\pi(i)}\nsubseteq{\mathfrak{p}}\cap A$ and hence
$\pi(i)\in\mathbb{M}(A,{\mathfrak{p}}\cap A)$. This shows that
$\pi\big{(}{\mathbb{M}}(B,{\mathfrak{p}})\cap
X\big{)}\subseteq{\mathbb{M}}(A,{\mathfrak{p}}\cap A)$. In view of (7), and
since ${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}(B)$ is arbitrary, this
gives

(8) $\pi\big{(}\overline{\mathbb{M}}(B,{\mathfrak{p}})\cap
X\big{)}\subseteq\overline{\mathbb{M}}(A,{\mathfrak{p}}\cap A)\quad\text{for
every ${\mathfrak{p}}\in\operatorname{\text{\rm Spec}}(B)$.}$

Keeping in mind that $\operatorname{\text{\rm Spec}}^{1}(B)\neq\varnothing$,
we obtain

(9) $\displaystyle Y$
$\displaystyle=\pi(H+\operatorname{\bar{}\mathbb{G}}(B))=\pi(\operatorname{\bar{}\mathbb{G}}(B))=\pi\bigg{(}{\textstyle\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}\overline{\mathbb{M}}(B,{\mathfrak{p}})\bigg{)}}$
$\displaystyle=\pi\bigg{(}{\textstyle\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}\big{(}\overline{\mathbb{M}}(B,{\mathfrak{p}})\cap
X\big{)}\bigg{)}}$
$\displaystyle\subseteq{\textstyle\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}\pi\big{(}\overline{\mathbb{M}}(B,{\mathfrak{p}})\cap
X\big{)}}\subseteq{\textstyle\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}\overline{\mathbb{M}}(A,{\mathfrak{p}}\cap A)},$

where the last inclusion follows from (8). Since
${\mathfrak{p}}\mapsto{\mathfrak{p}}\cap A$ is a surjective map from
$\operatorname{\text{\rm Spec}}^{1}(B)$ to $\operatorname{\text{\rm
Spec}}^{1}(A)$, we have

${\textstyle\bigcap_{{\mathfrak{p}}\in\operatorname{\text{\rm
Spec}}^{1}(B)}\overline{\mathbb{M}}(A,{\mathfrak{p}}\cap
A)}={\textstyle\bigcap_{{\mathfrak{q}}\in\operatorname{\text{\rm
Spec}}^{1}(A)}\overline{\mathbb{M}}(A,{\mathfrak{q}})}=\operatorname{\bar{}\mathbb{G}}(A),$

so $Y\subseteq\operatorname{\bar{}\mathbb{G}}(A)$ by (9). We have
$\operatorname{\bar{}\mathbb{G}}(A)\subseteq\operatorname{\mathbb{G}}(A)\subseteq
Y$ by definition, so we obtain
$\operatorname{\bar{}\mathbb{G}}(A)=\operatorname{\mathbb{G}}(A)$, i.e., $A$
is saturated in codimension $1$. This completes the proof. ∎

######  5.12 Corollary.

Let $B=\bigoplus_{i\in G}B_{i}$ be a noetherian normal $\mathbb{Q}$-domain
graded by a finitely generated abelian group $G$. If $B$ is saturated in
codimension $1$ then $(B^{(H)},B)\in\operatorname{\text{\rm EXT}}$ for every
subgroup $H$ of $G$ such that $G/H$ is torsion.

###### Proof.

If $B$ is saturated in codimension $1$ then
$B^{(H+\operatorname{\bar{}\mathbb{G}}(B))}=B^{(H+\operatorname{\mathbb{G}}(B))}=B$,
so
$(B^{(H)},B)=\big{(}B^{(H)},B^{(H+\operatorname{\bar{}\mathbb{G}}(B))}\big{)}\in\operatorname{\text{\rm
EXT}}$ by Thm 5.11. ∎

##  6\. The set $\mathscr{X}(B)$

######  6.1 Notation.

Given an abelian group $G$, let $\mathscr{T}(G)$ be the set of all subgroups
$H$ of $G$ such that $G/H$ is torsion. Observe that if $H\in\mathscr{T}(G)$
then $\mathscr{T}(H)\subseteq\mathscr{T}(G)$. If $B$ is a $G$-graded domain of
characteristic $0$, let

$\mathscr{X}(B)=\big{\\{}\,H\in\mathscr{T}(G)\,\mid\,\operatorname{\text{\rm
HLND}}(B^{(H)})\neq\\{0\\}\,\big{\\}}$

and let $\mathscr{M}(B)$ be the set of maximal elements of the poset
$(\mathscr{X}(B),\subseteq)$.

######  6.2 Remark.

Let $B$ be a domain of characteristic $0$ graded by an abelian group $G$.

  1. (a)

$\mathscr{X}(B)\subseteq\big{\\{}\,H\in\mathscr{T}(G)\,\mid\,\text{$B^{(H)}$
is non-rigid}\,\big{\\}}$

  2. (b)

If $\operatorname{\mathbb{G}}(B)$ is torsion-free and $B$ is finitely
generated as an algebra over a field, then
$\mathscr{X}(B)=\big{\\{}\,H\in\mathscr{T}(G)\,\mid\,\text{$B^{(H)}$ is non-
rigid}\,\big{\\}}$ by Lemma 2.16.

Combining the main results of Sections 3 and 5, we obtain:

######  6.3 Theorem.

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ an affine ${\rm\bf
k}$-domain graded by a finitely generated abelian group $G$.

  1. (a)

$\mathscr{X}(B)=\bigcup_{H\in\mathscr{M}(B)}\mathscr{T}(H)$

  2. (b)

If $B$ is normal then each element $H$ of $\mathscr{M}(B)$ satisfies
$H\supseteq\operatorname{\bar{}\mathbb{G}}(B)$.

###### Proof.

Let us abbreviate $\mathscr{X}(B)$ and $\mathscr{M}(B)$ to $\mathscr{X}$ and
$\mathscr{M}$ respectively. We may assume that $\mathscr{X}\neq\varnothing$,
otherwise the claim holds trivially.

(a) Let $H\in\mathscr{X}$. By Lemma 2.6, $\mathcal{B}=B^{(H)}$ is an affine
${\rm\bf k}$-domain graded by $H$. Since $\operatorname{\text{\rm
HLND}}(\mathcal{B})\neq\\{0\\}$ (because $H\in\mathscr{X}$), Thm 3.2 implies
that $\operatorname{\text{\rm HLND}}(\mathcal{B}^{(H^{\prime})})\neq\\{0\\}$
for all $H^{\prime}\in\mathscr{T}(H)$; since
$\mathcal{B}^{(H^{\prime})}=B^{(H^{\prime})}$, we get $\operatorname{\text{\rm
HLND}}(B^{(H^{\prime})})\neq\\{0\\}$ for all $H^{\prime}\in\mathscr{T}(H)$.
This shows that if $H\in\mathscr{X}$ then
$\mathscr{T}(H)\subseteq\mathscr{X}$. So
$\mathscr{X}=\bigcup_{H\in\mathscr{X}}\mathscr{T}(H)$.

Since $G$ is finitely generated, the poset $(\mathscr{X},\subseteq)$ satisfies
the ascending chain condition. This implies that for each
$H_{1}\in\mathscr{X}$, there exists $H_{2}\in\mathscr{M}$ such that
$H_{1}\subseteq H_{2}$ and hence
$\mathscr{T}(H_{1})\subseteq\mathscr{T}(H_{2})$. Thus,
$\mathscr{X}=\bigcup_{H\in\mathscr{X}}\mathscr{T}(H)=\bigcup_{H\in\mathscr{M}}\mathscr{T}(H)$,
which proves (a).

(b) Assume that $B$ is normal and consider $H\in\mathscr{M}$. If
$H^{\prime}=H+\operatorname{\bar{}\mathbb{G}}(B)$ then
$(B^{(H)},B^{(H^{\prime})})\in\operatorname{\text{\rm EXT}}$ by Thm 5.11. We
have $\operatorname{\text{\rm HLND}}(B^{(H)})\neq\\{0\\}$ (because
$H\in\mathscr{X}$), so Lemma 5.3 implies that $\operatorname{\text{\rm
HLND}}(B^{(H^{\prime})})\neq\\{0\\}$. So we have $\mathscr{M}\ni H\subseteq
H^{\prime}\in\mathscr{X}$, which implies $H=H^{\prime}$ and hence
$H\supseteq\operatorname{\bar{}\mathbb{G}}(B)$. This proves (b). ∎

######  6.4 Corollary.

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ a normal affine
${\rm\bf k}$-domain graded by a finitely generated abelian group $G$. If $B$
is saturated in codimension $1$ then $\mathscr{X}(B)=\varnothing$ or
$\mathscr{X}(B)=\mathscr{T}(G)$.

###### Proof.

Suppose that $\mathscr{X}(B)\neq\varnothing$. Then
$\mathscr{M}(B)\neq\varnothing$ by Thm 6.3(a). Consider $H\in\mathscr{M}(B)$.
We have $H\supseteq\operatorname{\bar{}\mathbb{G}}(B)$ by Thm 6.3(b), and
$\operatorname{\bar{}\mathbb{G}}(B)=\operatorname{\mathbb{G}}(B)$ because $B$
is saturated in codimension $1$, so $H\supseteq\operatorname{\mathbb{G}}(B)$
and consequently $B^{(H)}=B$. Since $H\in\mathscr{X}(B)$ we have
$\operatorname{\text{\rm HLND}}(B^{(H)})\neq\\{0\\}$ and hence
$\operatorname{\text{\rm HLND}}(B)\neq\\{0\\}$. It then follows from Thm 3.2
that $\operatorname{\text{\rm HLND}}(B^{(H^{\prime})})\neq\\{0\\}$ for all
$H^{\prime}\in\mathscr{T}(G)$, i.e., $\mathscr{X}(B)=\mathscr{T}(G)$. ∎

##  7\. $\mathbb{Z}$-gradings

In the first part of this section (up to Cor. 7.16), we use the general
results of the previous sections to derive consequences for
$\mathbb{Z}$-graded rings, adding new pieces of information where possible.
The material covered from Lemma 7.17 to the end of the section goes beyond the
results of the previous sections.

The roles played by the groups $\operatorname{\mathbb{G}}(B)$ and
$\operatorname{\bar{}\mathbb{G}}(B)$ in the previous sections are played by
the natural numbers $e(B)$ and $\bar{e}(B)$ in the context of
$\mathbb{Z}$-gradings. These numbers are defined in Notation 7.3, and their
relation to $\operatorname{\mathbb{G}}(B)$ and
$\operatorname{\bar{}\mathbb{G}}(B)$ is given in Lemma 7.6.

######  7.1 Notation.

If $B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ is a $\mathbb{Z}$-graded ring and $d$
is a positive integer then we define $B^{(d)}=\bigoplus_{i\in
d\mathbb{Z}}B_{i}$. (In other words, $B^{(d)}=B^{(H)}$ with $H=d\mathbb{Z}$.)
This is called the $d$-th Veronese subring of $B$.

It is obvious that Thm 3.2 has the following consequence:

######  7.2 Corollary.

Let ${\rm\bf k}$ be a field of characteristic $0$, let $B$ be a
$\mathbb{Z}$-graded affine ${\rm\bf k}$-domain, and let $d$ be a positive
integer.

  1. (a)

For each $A\in\operatorname{\text{\rm KHLND}}(B)$, we have
$A^{(d)}\in\operatorname{\text{\rm KHLND}}(B^{(d)})$.

  2. (b)

The map $\operatorname{\text{\rm KHLND}}(B)\to\operatorname{\text{\rm
KHLND}}(B^{(d)})$, $A\mapsto A^{(d)}$, is injective.

  3. (c)

If $B$ is non-rigid then $B^{(d)}$ is non-rigid.

If $S$ is a nonempty subset of $\mathbb{Z}$ then $\operatorname{\text{\rm
lcm}}(S)$ is defined to be the nonnegative generator of the ideal
$\bigcap_{a\in S}a\mathbb{Z}$ of $\mathbb{Z}$. In particular, if $S$ is an
infinite subset of $\mathbb{Z}$ then $\operatorname{\text{\rm lcm}}(S)=0$.

######  7.3 Notation.

Let $B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ be a $\mathbb{Z}$-graded domain.

  1. (1)

We define $e(B)=\gcd\big{\\{}\,n\in\mathbb{Z}\,\mid\,B_{n}\neq 0\,\big{\\}}$.
The number $e(B)$ is called the saturation index of $B$.

  2. (2)

Let $Z$ temporarily denote the set of all height $1$ homogeneous prime ideals
of $B$. We define $\bar{e}(B)=\operatorname{\text{\rm
lcm}}\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in Z\,\big{\\}}$ if
$Z\neq\varnothing$, and $\bar{e}(B)=e(B)$ if $Z=\varnothing$. We call
$\bar{e}(B)$ the codimension $1$ saturation index of $B$.

The following example shows that it is possible to have $\bar{e}(B)=0\neq
e(B)$. We will see in Lemma 7.17 that the rings that satisfy that condition
are very special.

######  7.4 Example.

The ring $B=\mathbb{C}[U,V,X,Y]/(UY-VX)$ is a normal domain, and is
$\mathbb{N}$-graded by declaring that $U,V,X,Y$ are homogeneous of degrees
$0,0,1,1$. Clearly, $e(B)=1$. Define $Z$ as in Notation 7.3. The prime ideal
$B_{+}=\bigoplus_{i>0}B_{i}$ belongs to $Z$ and $e(B/B_{+})=0$, so
$0\in\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in Z\,\big{\\}}$ and
hence $\bar{e}(B)=0$.

######  7.5 Remark.

Let $B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ be a $\mathbb{Z}$-graded domain, and
let $Z$ denote the set of all height $1$ homogeneous prime ideals of $B$.

  1. (1)

We have $e(B)\in\mathbb{N}$ and $\operatorname{\mathbb{G}}(B)=e(B)\mathbb{Z}$.
Moreover, $e(B)=0$ if and only if the grading is trivial (i.e., $B=B_{0}$).

  2. (2)

We have $\bar{e}(B)\in\mathbb{N}$ and $e(B)\mid\bar{e}(B)$. Moreover,
$\bar{e}(B)=0$ if and only if one of the following holds:

     * •

$e(B)=0$;

     * •

$e(B/{\mathfrak{p}})=0$ for some ${\mathfrak{p}}\in Z$;

     * •

the set $\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in Z\,\big{\\}}$
is infinite.

  3. (3)

If $B$ is noetherian then
$\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in Z\,\big{\\}}$ is a
finite set. Indeed, Prop. 4.4 and Rem. 4.5 imply that
$\big{\\{}\,\operatorname{\mathbb{G}}(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in
Z\,\big{\\}}$ is a finite set. By part (1) we have
$\operatorname{\mathbb{G}}(B/{\mathfrak{p}})=e(B/{\mathfrak{p}})\mathbb{Z}$
for all ${\mathfrak{p}}\in Z$, so
$\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in Z\,\big{\\}}$ is
finite.

######  7.6 Lemma.

Let $B$ be a $\mathbb{Z}$-graded domain.

  1. (a)

$\operatorname{\mathbb{G}}(B)=e(B)\mathbb{Z}$ and
$\operatorname{\bar{}\mathbb{G}}(B)=\bar{e}(B)\mathbb{Z}$.

  2. (b)

$B$ is saturated in codimension $1$ if and only if $\bar{e}(B)=e(B)$.

###### Proof.

We already noted that $\operatorname{\mathbb{G}}(B)=e(B)\mathbb{Z}$. Let $Z$
denote the set of all height $1$ homogeneous prime ideals of $B$. If
$Z=\varnothing$ then Prop. 4.6 gives
$\operatorname{\bar{}\mathbb{G}}(B)=\operatorname{\mathbb{G}}(B)$ and the
definition of $\bar{e}(B)$ gives $\bar{e}(B)=e(B)$, so
$\bar{e}(B)\mathbb{Z}=e(B)\mathbb{Z}=\operatorname{\mathbb{G}}(B)=\operatorname{\bar{}\mathbb{G}}(B)$,
which proves (a) in the case $Z=\varnothing$. If $Z\neq\varnothing$ then we
claim that the following equalities hold:

$\operatorname{\bar{}\mathbb{G}}(B)=\bigcap_{{\mathfrak{p}}\in
Z}\operatorname{\mathbb{G}}(B/{\mathfrak{p}})=\bigcap_{{\mathfrak{p}}\in
Z}\bigg{(}e(B/{\mathfrak{p}})\mathbb{Z}\bigg{)}=\operatorname{\text{\rm
lcm}}\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in
Z\,\big{\\}}\mathbb{Z}=\bar{e}(B)\mathbb{Z}\,.$

Indeed, the first equality is Prop. 4.6, the second one follows from the fact
that
$\operatorname{\mathbb{G}}(B/{\mathfrak{p}})=e(B/{\mathfrak{p}})\mathbb{Z}$
for each ${\mathfrak{p}}\in Z$, the third one is trivial and the last one is
the definition of $\bar{e}(B)$. This proves (a). Assertion (b) follows from
(a) and Def. 4.1. ∎

######  7.7 Remark.

In [CD22], saturation in codimension $1$ is only defined for
$\mathbb{Z}$-graded domains, and the definition is as in part (b) of Lemma
7.6. So our notion of saturation in codimension $1$ (Def. 4.1) generalizes
that of [CD22].

In view of Lemma 7.6, it is clear that Prop. 4.7 has the following
consequence:

######  7.8 Corollary.

Let $B$ be a $\mathbb{Z}$-graded domain of characteristic $0$.

  1. (a)

For each $D\in\operatorname{\text{\rm HLND}}(B)\setminus\\{0\\}$, there exists
a height $1$ homogeneous prime ideal ${\mathfrak{p}}$ of $B$ satisfying:

    1. (i)

some homogeneous element $t\in{\mathfrak{p}}$ satisfies $D(t)\neq 0$ and
$D^{2}(t)=0$;

    2. (ii)

$e(B/{\mathfrak{p}})=e(\ker D)$.

  2. (b)

$e(B)\mid e(\ker D)\mid\bar{e}(B)$ for all $D\in\operatorname{\text{\rm
HLND}}(B)$.

  3. (c)

If $B$ is saturated in codimension $1$ then $e(\ker D)=e(B)$ for all
$D\in\operatorname{\text{\rm HLND}}(B)$.

Note that Corollaries 4.2–4.4 of [DFMJ17] are special cases of Cor. 7.8(c),
which is itself a special case of Prop. 4.7(c).

######  7.9 Notation.

Given a prime number $p$ and any $m\in\mathbb{Z}\setminus\\{0\\}$, we set

$v_{p}(m)=\max\big{\\{}\,i\in\mathbb{N}\,\mid\,\text{$p^{i}$ divides
$m$}\,\big{\\}}.$

We also define $v_{p}(0)=\infty$, so $v_{p}(0)>v_{p}(m)$ for all
$m\in\mathbb{Z}\setminus\\{0\\}$.

Given $m\in\mathbb{N}\setminus\\{0\\}$ and $n\in\mathbb{N}$, we define the
element $m\odiv n$ of $\mathbb{N}\setminus\\{0\\}$ by

$\textstyle m\odiv n=\prod\limits_{p\in P(m,n)}p^{v_{p}(m)}$

where $P(m,n)$ is the set of prime numbers $p$ such that $v_{p}(m)>v_{p}(n)$.

######  7.10 Lemma.

The operation $\odiv$ has the following properties.

  1. (a)

If $m\in\mathbb{N}\setminus\\{0\\}$ then $m\odiv 1=m$ and $m\odiv 0=1$.

  2. (b)

If $m\in\mathbb{N}\setminus\\{0\\}$ and $n\in\mathbb{N}$ then

     * •

$m\odiv n=1\Leftrightarrow m\mid n$,

     * •

$(m\odiv n)\mid m\mid\operatorname{\text{\rm lcm}}(m\odiv n,n)$.

  3. (c)

If $B$ is a $\mathbb{Z}$-graded domain then $B^{(m\odiv e(B))}=B^{(m)}$ for
all $m\in\mathbb{N}\setminus\\{0\\}$.

###### Proof.

Assertions (a) and (b) are easy. We prove (c). We have $(m\odiv e(B))\mid m$
by (b), so $B^{(m\odiv e(B))}\supseteq B^{(m)}$. To prove the reverse
inclusion, consider $x\in B^{(m\odiv e(B))}\setminus\\{0\\}$ homogeneous of
degree $d$. Then $0\neq B_{d}\subseteq B^{(m\odiv e(B))}$, so $d$ is a
multiple of both $m\odiv e(B)$ and $e(B)$, i.e., $\operatorname{\text{\rm
lcm}}(m\odiv e(B),e(B))\mid d$. Since $m\mid\operatorname{\text{\rm
lcm}}(m\odiv e(B),e(B))$ by (b), we get $m\mid d$, so $x\in B^{(m)}$. ∎

The main result of Section 5 has the following consequence:

######  7.11 Corollary.

Let $B$ be a $\mathbb{Z}$-graded noetherian normal $\mathbb{Q}$-domain. Let
$d\in\mathbb{N}\setminus\\{0\\}$ and define $d^{\prime}=\gcd(d,\bar{e}(B))$
and $d^{\prime\prime}=d^{\prime}\odiv e(B)$. Then

$\big{(}B^{(d)},B^{(d^{\prime})}\big{)}\in\operatorname{\text{\rm
EXT}}\quad\text{and}\quad\big{(}B^{(d)},B^{(d^{\prime\prime})}\big{)}\in\operatorname{\text{\rm
EXT}}.$

###### Proof.

We have $\big{(}B^{(d)},B^{(d^{\prime})}\big{)}\in\operatorname{\text{\rm
EXT}}$ by Thm 5.11 and Lemma 7.6. It follows that
$\big{(}B^{(d)},B^{(d^{\prime\prime})}\big{)}\in\operatorname{\text{\rm
EXT}}$, because $B^{(d^{\prime\prime})}=B^{(d^{\prime})}$ by Lemma 7.10(c). ∎

We shall now apply the ideas and results of Section 6 to
$\mathbb{Z}$-gradings.

######  7.12 Notation.

For each positive integer $d$, let $\mathscr{I}_{d}=\\{d,2d,3d,\dots\\}$. Let
$\preceq$ be the partial order on $\mathbb{N}\setminus\\{0\\}$ defined by
declaring that $d_{1}\preceq d_{2}$ $\Leftrightarrow$
$\mathscr{I}_{d_{1}}\subseteq\mathscr{I}_{d_{2}}$ $\Leftrightarrow$ $d_{2}\mid
d_{1}$.

If $B$ is a $\mathbb{Z}$-graded domain of characteristic $0$, let

$X(B)=\big{\\{}\,d\in\mathbb{N}\setminus\\{0\\}\,\mid\,\text{$B^{(d)}$ is non-
rigid}\,\big{\\}}$

and let $M(B)$ be the set of maximal elements of the poset $(X(B),\preceq)$.
Note that

$B$ is rigid $\iff$ $1\notin X(B)$ $\iff$ $1\notin M(B)$.

######  7.13 Lemma.

If ${\rm\bf k}$ is a field of characteristic $0$ and $B$ a $\mathbb{Z}$-graded
affine ${\rm\bf k}$-domain, then
$\mathscr{X}(B)=\big{\\{}\,d\mathbb{Z}\,\mid\,d\in X(B)\,\big{\\}}$ and
$\mathscr{M}(B)=\big{\\{}\,d\mathbb{Z}\,\mid\,d\in M(B)\,\big{\\}}$.

###### Proof.

Follows from Rem. 6.2(b). ∎

######  7.14 Corollary.

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ a
$\mathbb{Z}$-graded affine ${\rm\bf k}$-domain.

  1. (a)

$X(B)=\bigcup_{d\in M(B)}\mathscr{I}_{d}$

  2. (b)

If $B$ is normal then each element $d$ of $M(B)$ satisfies

$d\mid\bar{e}(B)\quad\text{and}\quad d\odiv e(B)=d.$

###### Proof.

We have $\mathscr{X}(B)=\big{\\{}\,d\mathbb{Z}\,\mid\,d\in X(B)\,\big{\\}}$
and $\mathscr{M}(B)=\big{\\{}\,d\mathbb{Z}\,\mid\,d\in M(B)\,\big{\\}}$ by
Lemma 7.13, so Thm 6.3 and Lemma 7.6 imply that (a) and (b′) are true, where

  1. (b′)

If $B$ is normal then each element $d$ of $M(B)$ satisfies $d\mid\bar{e}(B)$.

It remains to check that each $d\in M(B)$ satisfies $d\odiv e(B)=d$ (we don’t
need to assume that $B$ is normal for this part). Let $d\in M(B)$ and define
$d^{\prime}=d\odiv e(B)$. We have $B^{(d)}=B^{(d^{\prime})}$ by Lemma 7.10(c),
so clearly $d^{\prime}\in X(B)$. Lemma 7.10(b) gives $d^{\prime}\mid d$, so
$d\preceq d^{\prime}$. Thus $M(B)\ni d\preceq d^{\prime}\in X(B)$, so
$d^{\prime}=d$, i.e., $d\odiv e(B)=d$. ∎

In the following result, the conclusion “$X(B)=\varnothing$ or
$X(B)=\mathbb{N}\setminus\\{0\\}$” means that all Veronese subrings of $B$
have the same rigidity status, i.e., either $B^{(d)}$ is rigid for all
$d\in\mathbb{N}\setminus\\{0\\}$ or $B^{(d)}$ is non-rigid for all
$d\in\mathbb{N}\setminus\\{0\\}$. So Cor. 7.15 is stronger than part (2) of
Thm 1.2 of [CD22].

######  7.15 Corollary.

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ a
$\mathbb{Z}$-graded normal affine ${\rm\bf k}$-domain. If $B$ is saturated in
codimension $1$ then $X(B)=\varnothing$ or $X(B)=\mathbb{N}\setminus\\{0\\}$.

###### Proof.

Cor. 6.4 implies that $\mathscr{X}(B)=\varnothing$ or
$\mathscr{X}(B)=\mathscr{T}(\mathbb{Z})$, so the claim follows from Lemma
7.13. ∎

######  7.16 Corollary.

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ a
$\mathbb{Z}$-graded normal affine ${\rm\bf k}$-domain.

  1. (a)

For every $d\in\mathbb{N}\setminus\\{0\\}$, the following are equivalent:

    1. (i)

$d\in X(B)$

    2. (ii)

$\mathscr{I}_{d}\subseteq X(B)$

    3. (iii)

$\gcd(d,\bar{e}(B))\in X(B)$

    4. (iv)

$d\odiv e(B)\in X(B)$.

  2. (b)

If $\bar{e}(B)\neq 0$ and $X(B)\neq\varnothing$ then $\bar{e}(B)\in X(B)$.

###### Proof.

To prove (a), consider $d\in\mathbb{N}\setminus\\{0\\}$.

If $d\in X(B)$ then Cor. 7.14 implies that $d\in\mathscr{I}_{m}\subseteq X(B)$
for some $m\in M(B)$. Then $\mathscr{I}_{d}\subseteq\mathscr{I}_{m}\subseteq
X(B)$, so (i) implies (ii). The converse is obvious, so (i) $\Leftrightarrow$
(ii).

If $d\in X(B)$ then Cor. 7.14 implies that $d\in\mathscr{I}_{m}\subseteq X(B)$
for some $m\in M(B)$, and moreover $m\mid\bar{e}(B)$. So
$m\mid\gcd(d,\bar{e}(B))$, so $\gcd(d,\bar{e}(B))\in\mathscr{I}_{m}\subseteq
X(B)$ and hence $\gcd(d,\bar{e}(B))\in X(B)$. Conversely, if
$\gcd(d,\bar{e}(B))\in X(B)$ then define $d^{\prime}=\gcd(d,\bar{e}(B))$ and
note that $d\in\mathscr{I}_{d^{\prime}}\subseteq X(B)$ (using that (i) implies
(ii)), so $d\in X(B)$. So (i) $\Leftrightarrow$ (iii).

Since $B^{(d)}=B^{(d\odiv e(B))}$ by Lemma 7.10(c), it is clear that (i)
$\Leftrightarrow$ (iv). This proves (a).

To prove (b), suppose that $\bar{e}(B)\neq 0$ and $X(B)\neq\varnothing$. Pick
$d\in X(B)$; then $\bar{e}(B)d\in\mathscr{I}_{d}\subseteq X(B)$ by (a), so
$\gcd(\bar{e}(B)d,\bar{e}(B))\in X(B)$ again by (a), so $\bar{e}(B)\in X(B)$.
∎

## More about $\bar{e}(B)$

The ring $B$ of Ex. 7.4 satisfies $\bar{e}(B)=0\neq e(B)$. The following
result shows that the rings that satisfy that condition are very special.

######  7.17 Lemma.

Let ${\rm\bf k}$ be a field and $B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ a
$\mathbb{Z}$-graded affine ${\rm\bf k}$-domain such that $B_{j}\neq 0$ for
some $j>0$.

  1. (a)

$\bar{e}(B)=0$ if and only if $B$ is $\mathbb{N}$-graded and the prime ideal
$B_{+}=\bigoplus_{i>0}B_{i}$ has height $1$.

  2. (b)

If $B$ is normal and $\bar{e}(B)=0$, then there exist $s\in
B_{0}\setminus\\{0\\}$ and a homogeneous element $t\in B\setminus\\{0\\}$ of
positive degree such that
$B_{s}=(B_{0})_{s}[t]=\big{(}(B_{0})_{s}\big{)}^{[1]}$.

###### Proof.

(a) Let $Z$ denote the set of all height $1$ homogeneous prime ideals of $B$.

If $B$ is $\mathbb{N}$-graded and $\operatorname{\text{\rm ht}}(B_{+})=1$ then
$B_{+}\in Z$ and $e(B/B_{+})=0$, so $\bar{e}(B)=0$ by part (2) of Rem. 7.5.

Conversely, assume that $\bar{e}(B)=0$. Since $\bar{e}(B)\neq e(B)$, we have
$Z\neq\varnothing$; since $B$ is noetherian,
$\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in Z\,\big{\\}}$ is a
finite set by part (3) of Rem. 7.5; since $\operatorname{\text{\rm
lcm}}\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in
Z\,\big{\\}}=\bar{e}(B)=0$, there exists ${\mathfrak{p}}\in Z$ such that
$e(B/{\mathfrak{p}})=0$. Thus,

(10) $B_{\ell}\subseteq{\mathfrak{p}}$ for all
$\ell\in\mathbb{Z}\setminus\\{0\\}$.

Define ${\mathfrak{p}}_{0}={\mathfrak{p}}\cap B_{0}$. To prove that $B$ is
$\mathbb{N}$-graded, assume the contrary. Since $B_{j}\neq 0$ for some $j>0$,
this means that there exist $i,j\in\mathbb{Z}$ such that $i<0<j$, $B_{i}\neq
0$ and $B_{j}\neq 0$. Choose $x_{i}\in B_{i}\setminus\\{0\\}$ and $x_{j}\in
B_{j}\setminus\\{0\\}$; since $x_{i},x_{j}\in{\mathfrak{p}}$, we have
$x_{i}^{j}x_{j}^{|i|}\in{\mathfrak{p}}_{0}$, so ${\mathfrak{p}}_{0}\neq 0$. It
is easy to see that ${\rm\bf k}\subseteq B_{0}$, so Lemma 2.6 with $H=0$
implies that $B_{0}$ is ${\rm\bf k}$-affine. Since $x_{j}$ is transcendental
over $B_{0}$, we have $\operatorname{\text{\rm trdeg}}_{B_{0}}(B)>0$; since
${\mathfrak{p}}_{0}\neq 0$ and $B_{0}$ is ${\rm\bf k}$-affine, we have
$\operatorname{\text{\rm trdeg}}_{\rm\bf
k}(B_{0}/{\mathfrak{p}}_{0})<\operatorname{\text{\rm trdeg}}_{\rm\bf
k}(B_{0})\leq\operatorname{\text{\rm trdeg}}_{\rm\bf k}(B)-1$. On the other
hand, we have $B_{0}/{\mathfrak{p}}_{0}\cong B/{\mathfrak{p}}$ by (10), and
(since $\operatorname{\text{\rm ht}}{\mathfrak{p}}=1$ and $B$ is ${\rm\bf
k}$-affine) $\operatorname{\text{\rm trdeg}}_{\rm\bf
k}(B/{\mathfrak{p}})=\operatorname{\text{\rm trdeg}}_{\rm\bf k}(B)-1$, so
$\operatorname{\text{\rm trdeg}}_{\rm\bf
k}(B_{0}/{\mathfrak{p}}_{0})=\operatorname{\text{\rm trdeg}}_{\rm\bf k}(B)-1$,
a contradiction.

So $B$ is $\mathbb{N}$-graded and consequently $B_{+}=\bigoplus_{i>0}B_{i}$ is
a prime ideal of $B$. Since $B_{j}\neq 0$ for some $j>0$, we have $B_{+}\neq
0$, so $\operatorname{\text{\rm ht}}(B_{+})\geq 1$. We have
$B_{+}\subseteq{\mathfrak{p}}$ by (10), so $\operatorname{\text{\rm
ht}}(B_{+})=1$, which proves (a).

(b) By (a), $B$ is $\mathbb{N}$-graded and $\operatorname{\text{\rm
ht}}(B_{+})=1$. By the proof of (a), ${\rm\bf k}\subseteq B_{0}$ and both $B$
and $B_{0}$ are ${\rm\bf k}$-affine. Since $B/B_{+}\cong B_{0}$ and
$\operatorname{\text{\rm ht}}(B_{+})=1$, we have $\operatorname{\text{\rm
trdeg}}_{B_{0}}(B)=1$.

We first prove the case where $B_{0}$ is a field. Let
$S=\bigcup_{i\in\mathbb{N}}(B_{i}\setminus\\{0\\})$ and $\mathcal{B}=S^{-1}B$.
Then $\mathcal{B}=\bigoplus_{i\in\mathbb{Z}}\mathcal{B}_{i}$ where
$\mathcal{B}_{0}$ is a field that contains $B_{0}$. By Lemma 2.11, there
exists a nonzero homogeneous element $t\in\mathcal{B}$ of degree $e(B)$ such
that $\mathcal{B}=\mathcal{B}_{0}[t^{\pm 1}]$ is the ring of Laurent
polynomials over $\mathcal{B}_{0}$. In particular, $\operatorname{\text{\rm
trdeg}}_{\mathcal{B}_{0}}(\mathcal{B})=1$. We also have
$\operatorname{\text{\rm trdeg}}_{B_{0}}(\mathcal{B})=\operatorname{\text{\rm
trdeg}}_{B_{0}}(B)=1$, so $\mathcal{B}_{0}/B_{0}$ is an algebraic extension of
fields. Since $B$ is normal, it follows that $\mathcal{B}_{0}\subseteq B$ and
hence that $\mathcal{B}_{0}=B_{0}$. So $B\subseteq\mathcal{B}=B_{0}[t^{\pm
1}]$. Since $B$ is $\mathbb{N}$-graded, it follows that $B\subseteq B_{0}[t]$.
Since $B_{0}$ is a field, it is clear that some monic polynomial $f(t)\in
B_{0}[t]$ belongs to $B$; so $t$ is integral over $B$; since $B$ is normal and
$t\in S^{-1}B$, we get $t\in B$ and hence $B=B_{0}[t]=B_{0}^{[1]}$, as
desired.

Now consider the general case. This time, let $S=B_{0}\setminus\\{0\\}$ and
$\mathcal{B}=S^{-1}B=\bigoplus_{i\in\mathbb{N}}\mathcal{B}_{i}$, where
$\mathcal{B}_{0}=\operatorname{\text{\rm Frac}}B_{0}$. It is easy to see that
$\mathcal{B}_{+}\cap B=B_{+}$; by basic properties of localization, it follows
that $\operatorname{\text{\rm ht}}(\mathcal{B}_{+})=\operatorname{\text{\rm
ht}}(B_{+})=1$. Since $\mathcal{B}_{0}$ is a field and $\mathcal{B}$ is an
$\mathbb{N}$-graded normal affine $\mathcal{B}_{0}$-domain such that
$\operatorname{\text{\rm ht}}(\mathcal{B}_{+})=1$, the preceding paragraph
implies that $\mathcal{B}=\mathcal{B}_{0}[t]=\mathcal{B}_{0}^{[1]}$ where $t$
is homogeneous of degree $e(\mathcal{B})=e(B)>0$. Multiplying $t$ by an
element of $S$, we may arrange that $t\in B$. Since $B$ is ${\rm\bf
k}$-affine, we have $B={\rm\bf k}[f_{1}(t),\dots,f_{n}(t)]$ where
$f_{i}(t)\in\mathcal{B}_{0}[t]$ for all $i$. Clearly, there exists $s\in S$
such that all coefficients of $f_{1}(t),\dots,f_{n}(t)$ belong to
$(B_{0})_{s}$. Thus, $B_{s}=(B_{0})_{s}[t]=\big{(}(B_{0})_{s}\big{)}^{[1]}$,
which proves (b). ∎

######  7.18 Corollary.

Let ${\rm\bf k}$ be a field of characteristic $0$ and
$B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ a $\mathbb{Z}$-graded normal affine
${\rm\bf k}$-domain. If $e(B)\neq 0=\bar{e}(B)$ then there exists
$D\in\operatorname{\text{\rm HLND}}(B)\setminus\\{0\\}$ such that
$\ker(D)=B_{0}$. In particular, $B$ is non-rigid.

###### Proof.

First consider the case where $B_{j}\neq 0$ for some $j>0$. By Lemma 7.17(b),
there exist $s\in B_{0}\setminus\\{0\\}$ and a homogeneous element $t\in
B\setminus\\{0\\}$ of positive degree such that
$B_{s}=(B_{0})_{s}[t]=\big{(}(B_{0})_{s}\big{)}^{[1]}$. For each
$m\in\mathbb{N}$, $s^{m}\frac{d}{dt}:(B_{0})_{s}[t]\to(B_{0})_{s}[t]$ belongs
to $\operatorname{\text{\rm HLND}}(B_{s})\setminus\\{0\\}$. Since $B$ is
${\rm\bf k}$-affine, we can choose $m$ such that $s^{m}\frac{d}{dt}$ maps $B$
into itself; then $D=(s^{m}\frac{d}{dt})|_{B}:B\to B$ belongs to
$\operatorname{\text{\rm HLND}}(B)\setminus\\{0\\}$ and $\ker(D)=B_{0}$. This
proves the result in this case.

If no $j>0$ is such that $B_{j}\neq 0$, then let $B^{-}$ be the ring $B$ with
the opposite grading, i.e., $B^{-}=\bigoplus_{i\in\mathbb{Z}}B_{i}^{-}$ where
$B_{i}^{-}=B_{-i}$ for each $i\in\mathbb{Z}$. We have $e(B^{-})\neq
0=\bar{e}(B^{-})$, and there exists $j>0$ such that $B_{j}^{-}\neq 0$. By the
preceding paragraph, there exists $D\in\operatorname{\text{\rm
HLND}}(B^{-})\setminus\\{0\\}$ such that $\ker(D)=B_{0}^{-}$. Clearly, the
same $D$ satisfies $D\in\operatorname{\text{\rm HLND}}(B)\setminus\\{0\\}$ and
$\ker(D)=B_{0}$. ∎

######  7.19 Lemma.

Let $B$ be a $\mathbb{Z}$-graded domain and let
$d\in\mathbb{N}\setminus\\{0\\}$.

  1. (a)

$e(B^{(d)})=\operatorname{\text{\rm lcm}}(e(B),d)$

If moreover $B$ is noetherian and normal, then the following hold.

  1. (b)

$\bar{e}(B^{(d)})=\operatorname{\text{\rm lcm}}(\bar{e}(B),d)$

  2. (c)

If $\bar{e}(B)\neq 0$, then $B^{(d)}$ is saturated in codimension $1$ if and
only if $d$ is a multiple of $\bar{e}(B)\odiv e(B)$.

###### Proof.

Since $B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ is a domain, the set
$X=\big{\\{}\,n\in\mathbb{Z}\,\mid\,B_{n}\neq 0\,\big{\\}}$ is a submonoid of
$\mathbb{Z}$. Consequently, there exist $i,j\in X$ such that
$i-j=\gcd(X)=e(B)$. Since $i,j\in e(B)\mathbb{Z}$, it follows that
$\gcd(i,j)=e(B)$. Let $L=\operatorname{\text{\rm lcm}}(e(B),d)$; then
$L=re(B)$ for some $r\in\mathbb{N}$. Since $i,j\in e(B)\mathbb{Z}$, we have
$ri,rj\in re(B)\mathbb{Z}=L\mathbb{Z}\subseteq d\mathbb{Z}$, so

$ri,rj\in X\cap d\mathbb{Z}=\big{\\{}\,n\in\mathbb{Z}\,\mid\,(B^{(d)})_{n}\neq
0\,\big{\\}}\subseteq e(B^{(d)})\mathbb{Z},$

so $e(B^{(d)})$ divides the number $\gcd(ri,rj)=r\gcd(i,j)=re(B)=L$. It is
clear that $e(B^{(d)})$ is a common multiple of $e(B)$ and $d$, so $L\mid
e(B^{(d)})$, so (a) is proved.

Assume that $B$ is noetherian and normal and let $Z$ (resp. $Z^{\prime}$) be
the set of height $1$ homogeneous prime ideals of $B$ (resp. of $B^{(d)}$).
Consider the natural surjective map $f:\operatorname{\text{\rm
Spec}}B\to\operatorname{\text{\rm Spec}}B^{(d)}$,
${\mathfrak{P}}\mapsto{\mathfrak{P}}\cap B^{(d)}$. By Lemma 2.10(c),

$f^{-1}(Z^{\prime})=Z$ and $f|_{Z}:Z\to Z^{\prime}$ is bijective.

In particular, $Z=\varnothing$ $\Leftrightarrow$ $Z^{\prime}=\varnothing$. We
first prove (b) in the case where $Z=\varnothing=Z^{\prime}$. The definition
of $\bar{e}$ gives $\bar{e}(B)=e(B)$ and $\bar{e}(B^{(d)})=e(B^{(d)})$, so
$\bar{e}(B^{(d)})=e(B^{(d)})=\operatorname{\text{\rm
lcm}}(e(B),d)=\operatorname{\text{\rm lcm}}(\bar{e}(B),d)$ by part (a), so (b)
is true in this case. It remains to prove (b) in the case where $Z$ and
$Z^{\prime}$ are nonempty. If ${\mathfrak{p}}\in Z^{\prime}$ and
${\mathfrak{P}}\in Z$ are such that $f({\mathfrak{P}})={\mathfrak{p}}$ then
$B^{(d)}/{\mathfrak{p}}=(B/{\mathfrak{P}})^{(d)}$, so
$e\big{(}B^{(d)}/{\mathfrak{p}}\big{)}=e\big{(}(B/{\mathfrak{P}})^{(d)}\big{)}=\operatorname{\text{\rm
lcm}}(d,e(B/{\mathfrak{P}}))$ by (a). From this and the fact that
$f^{-1}(Z^{\prime})=Z$, it follows that

$\displaystyle\bar{e}(B^{(d)})$ $\displaystyle=\operatorname{\text{\rm
lcm}}\big{\\{}\,e(B^{(d)}/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in
Z^{\prime}\,\big{\\}}=\operatorname{\text{\rm
lcm}}\big{\\{}\,\operatorname{\text{\rm
lcm}}(d,e(B/{\mathfrak{P}}))\,\mid\,{\mathfrak{P}}\in Z\,\big{\\}}$
$\displaystyle=\operatorname{\text{\rm lcm}}\big{(}d,\bar{e}(B)\big{)},$

which completes the proof of (b).

(c) By (a) and (b) together with Lemma 7.6(b), we see that $B^{(d)}$ is
saturated in codimension $1$ if and only if $\operatorname{\text{\rm
lcm}}(d,\bar{e}(B))=\operatorname{\text{\rm lcm}}(d,e(B))$. Assuming that
$\bar{e}(B)\neq 0$, and keeping in mind that $e(B)\mid\bar{e}(B)$, it is a
straightforward exercise to check that $\operatorname{\text{\rm
lcm}}(d,\bar{e}(B))=\operatorname{\text{\rm lcm}}(d,e(B))$ is equivalent to
$d$ being a multiple of $\bar{e}(B)\odiv e(B)$. ∎

######  7.20 Corollary.

For a $\mathbb{Z}$-graded noetherian normal domain $B$ such that $e(B)\neq 0$,
the following conditions are equivalent:

  1. (i)

There exists $d\in\mathbb{N}\setminus\\{0\\}$ such that $B^{(d)}$ is saturated
in codimension $1$.

  2. (ii)

$\bar{e}(B)\neq 0$

###### Proof.

Lemma 7.19(c) shows that (ii) implies (i). For the converse, assume that
$\bar{e}(B)=0$ and let $d\in\mathbb{N}\setminus\\{0\\}$. By parts (a) and (b)
of Lemma 7.19, we have $e(B^{(d)})=\operatorname{\text{\rm lcm}}(e(B),d)\neq
0$ and $\bar{e}(B^{(d)})=\operatorname{\text{\rm
lcm}}(\bar{e}(B),d)=\operatorname{\text{\rm lcm}}(0,d)=0$, so $B^{(d)}$ is not
saturated in codimension $1$. ∎

## The subring $B_{\text{\rm sat}}$ of $B$

######  7.21 Definition.

Let $B$ be a $\mathbb{Z}$-graded domain such that $\bar{e}(B)\neq 0$. We
define the graded subring $B_{\text{\rm sat}}$ of $B$ by $B_{\text{\rm
sat}}=B^{(\bar{e}(B))}=B^{(\bar{e}(B)\odiv e(B))}$, where the last equality
follows from Lemma 7.10(c).

######  7.22 Corollary.

Let $B$ be a $\mathbb{Z}$-graded noetherian normal domain such that
$\bar{e}(B)\neq 0$.

  1. (a)

The $\mathbb{Z}$-graded ring $B_{\text{\rm sat}}$ is a noetherian normal
domain and $\bar{e}\big{(}B_{\text{\rm sat}}\big{)}=\bar{e}(B)\neq 0$.
Moreover, if ${\rm\bf k}$ is a field and $B$ is ${\rm\bf k}$-affine then
$B_{\text{\rm sat}}$ is ${\rm\bf k}$-affine.

  2. (b)

$(B_{\text{\rm sat}})_{\text{\rm sat}}=B_{\text{\rm sat}}$

  3. (c)

$B_{\text{\rm sat}}$ is saturated in codimension $1$.

  4. (d)

$B$ is saturated in codimension $1$ if and only if $B=B_{\text{\rm sat}}$.

###### Proof.

By Cor. 2.5 and Lemma 2.7, $B_{\text{\rm sat}}$ is a noetherian normal domain.
We have $\bar{e}(B_{\text{\rm
sat}})=\bar{e}\big{(}B^{(\bar{e}(B))}\big{)}=\operatorname{\text{\rm
lcm}}(\bar{e}(B),\bar{e}(B))=\bar{e}(B)\neq 0$ by Lemma 7.19(b). If $B$ is
${\rm\bf k}$-affine then so is $B_{\text{\rm sat}}$ by Lemma 2.6. This proves
(a).

Since $\bar{e}\big{(}B_{\text{\rm sat}}\big{)}=\bar{e}(B)$, we have
$(B_{\text{\rm sat}})_{\text{\rm
sat}}=\big{(}B^{(\bar{e}(B))}\big{)}^{(\bar{e}(B))}=B^{(\bar{e}(B))}=B_{\text{\rm
sat}}$, proving (b). Part (c) follows from $B_{\text{\rm
sat}}=B^{(\bar{e}(B)\odiv e(B))}$ and Lemma 7.19(c).

If $B$ is saturated in codimension $1$ then $\bar{e}(B)=e(B)$, so
$B_{\text{\rm sat}}=B^{(\bar{e}(B))}=B^{(e(B))}=B$. Conversely, if
$B_{\text{\rm sat}}=B$ then (c) implies that $B$ is saturated in codimension
$1$. This proves (d). ∎

######  7.23 Corollary.

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ a
$\mathbb{Z}$-graded normal affine ${\rm\bf k}$-domain such that
$\bar{e}(B)\neq 0$. The following are equivalent:

  1. (i)

There exists $d\in\mathbb{N}\setminus\\{0\\}$ such that $B^{(d)}$ is non-
rigid;

  2. (ii)

$B_{\text{\rm sat}}$ is non-rigid.

###### Proof.

This follows from Cor. 7.16(b). ∎

## The set $M(B)$

By Cor. 7.14, the problem of describing $X(B)$ reduces to that of describing
$M(B)$. So it is interesting to ask what can be said about $M(B)$. Let us
agree that a primitive set is a subset $Y$ of $\mathbb{N}\setminus\\{0\\}$
such that the conditions $y,y^{\prime}\in Y$ and $y\mid y^{\prime}$ imply
$y=y^{\prime}$. It is obvious that $M(B)$ is primitive, and Prop. 7.24 shows
that it is finite.

######  7.24 Proposition.

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ a
$\mathbb{Z}$-graded normal affine ${\rm\bf k}$-domain.

  1. (a)

$M(B)$ is a finite primitive set.

  2. (b)

If $e(B)=0$ then $M(B)$ is either $\varnothing$ or $\\{1\\}$.

  3. (c)

If $e(B)\neq 0$ and $\bar{e}(B)=0$ then $M(B)=\\{1\\}$.

###### Proof.

(b) If $e(B)=0$ then $B^{(d)}=B$ for all positive integers $d$, so $X(B)$ is
either $\varnothing$ or $\mathbb{N}\setminus\\{0\\}$ and consequently (b) is
true.

(c) Assume that $e(B)\neq 0$ and $\bar{e}(B)=0$. By Cor. 7.18, $B$ is non-
rigid. Then Cor. 7.14 implies that $X(B)=\mathbb{N}\setminus\\{0\\}$ and hence
that $M(B)=\\{1\\}$, which proves (c).

(a) It is obvious that $M(B)$ is a primitive set. If $\bar{e}(B)\neq 0$ then
$M(B)$ is a finite set by Cor. 7.14(b). If $\bar{e}(B)=0$ then (b) and (c)
imply that $M(B)$ is either $\varnothing$ or $\\{1\\}$, so $M(B)$ is finite in
all cases. ∎

######  7.25 Remark.

Let us say that a subset $Y$ of $\mathbb{N}\setminus\\{0\\}$ is realizable if
there exists a $\mathbb{Z}$-graded normal affine $\mathbb{C}$-domain $B$
satisfying $M(B)=Y$. Prop. 7.24 implies that every realizable set is finite
and primitive, but we don’t know if every finite primitive set is realizable.
For instance, we don’t know whether $\\{6,10\\}$ is realizable. We will learn
more about realizable sets in Section 8, by studying Pham-Brieskorn rings.

The purpose of the following trivial example is to show what kind of
information can be derived from the knowledge of the numbers $e(B)$ and
$\bar{e}(B)$.

######  7.26 Example.

Let ${\rm\bf k}$ be a field of characteristic $0$ and $B$ a
$\mathbb{Z}$-graded normal affine ${\rm\bf k}$-domain. Suppose that $e(B)=6$
and $\bar{e}(B)=60$. Since $60\odiv 6=20$, we have $B_{\text{\rm
sat}}=B^{(60)}=B^{(20)}$. The positive integers $d$ satisfying $d\mid 60$ and
$d\odiv 6=d$ are $1,4,5,20$; so $M(B)\subseteq\\{1,4,5,20\\}$ by Cor. 7.14(b).
Since $M(B)$ is primitive, it must be one of the sets
$\varnothing,\\{1\\},\\{4\\},\\{5\\},\\{20\\},\\{4,5\\}$. So Cor. 7.14(a)
implies that $X(B)$ is one of the following sets: $\varnothing,\
\mathscr{I}_{1}=\mathbb{N}\setminus\\{0\\},\ \mathscr{I}_{4},\
\mathscr{I}_{5},\ \mathscr{I}_{20},\ \mathscr{I}_{4}\cup\mathscr{I}_{5}$.

It is interesting to observe that each one of the sets
$\varnothing,\\{1\\},\\{4\\},\\{5\\},\\{20\\},\\{4,5\\}$ (mentioned in Ex.
7.26) is realizable. This follows from Prop. 8.9 (see below).

##  8\. Pham-Brieskorn rings

This section determines the set $M(B)$ for certain Pham-Brieskorn rings $B$.

######  8.1 Notation.

Given ${\mathbf{a}}=(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\\{0\\})^{n}$
(where $n\geq 3$), we define:

  1. (1)

$L=\operatorname{\text{\rm lcm}}(a_{1},\dots,a_{n})$

  2. (2)

For each $i\in\\{1,\dots,n\\}$, $J_{i}=\\{1,\dots,n\\}\setminus\\{i\\}$ and
$L_{i}=\displaystyle\operatorname{\text{\rm lcm}}\big{\\{}\,a_{j}\,\mid\,j\in
J_{i}\,\big{\\}}$.

  3. (3)

$S({\mathbf{a}})=S(a_{1},\dots,a_{n})=\big{\\{}\,j\,\mid\,1\leq j\leq n\text{
and }L_{j}\neq L\,\big{\\}}$

  4. (4)

The cotype of ${\mathbf{a}}$ is the cardinality of $S({\mathbf{a}})$. Note
that

$\operatorname{\text{\rm cotype}}({\mathbf{a}})=\operatorname{\text{\rm
cotype}}(a_{1},\dots,a_{n})\in\\{0,1,\dots,n\\}.$

######  8.2 Notation.

If ${\rm\bf k}$ is a field of characteristic $0$, $n\geq 3$ and
$a_{1},\dots,a_{n}\in\mathbb{N}\setminus\\{0\\}$, define $B_{{\rm\bf
k};a_{1},\dots,a_{n}}={\rm\bf
k}[X_{1},\dots,X_{n}]/(X_{1}^{a_{1}}+\cdots+X_{n}^{a_{n}})$. This is called a
Pham-Brieskorn ring, and it is well known that

$B_{{\rm\bf k};a_{1},\dots,a_{n}}$ is a normal domain.

Consider the $\mathbb{N}$-grading of ${\rm\bf k}[X_{1},\dots,X_{n}]$ such that
(for each $i=1,\dots,n$) $X_{i}$ is homogeneous of degree $d_{i}=L/a_{i}$,
where $L=\operatorname{\text{\rm lcm}}(a_{1},\dots,a_{n})$; then
$X_{1}^{a_{1}}+\cdots+X_{n}^{a_{n}}$ is homogeneous and consequently
$B_{{\rm\bf k};a_{1},\dots,a_{n}}$ is $\mathbb{N}$-graded. We have $B_{{\rm\bf
k};a_{1},\dots,a_{n}}={\rm\bf k}[x_{1},\dots,x_{n}]$, where (for each $i$)
$x_{i}\in B_{{\rm\bf k};a_{1},\dots,a_{n}}$ denotes the canonical image of
$X_{i}$. Clearly, $x_{i}\neq 0$ is homogeneous of degree $d_{i}$, and if
$n\geq 4$ then $x_{i}$ is a prime element of $B_{{\rm\bf
k};a_{1},\dots,a_{n}}$.

When considering Pham-Brieskorn rings, we shall use the notations $L$,
$J_{i}$, $L_{i}$, $x_{i}$ and $d_{i}$ (defined in 8.1 and 8.2) without
necessarily recalling their definitions.

######  8.3.

For each $n\geq 3$, let $\Gamma_{n}$ be the set of
$(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\\{0\\})^{n}$ such that

$\min(a_{1},\dots,a_{n})>1$ and at most one $i\in\\{1,\dots,n\\}$ satisfies
$a_{i}=2$.

It is well known that $B_{\mathbb{C};a_{1},\dots,a_{n}}$ is non-rigid whenever
$(a_{1},\dots,a_{n})\notin\Gamma_{n}$. It is conjectured that the following
statement $\mathscr{P}(n)$ is true for all $n\geq 3$:

$\mathscr{P}(n)$:  The ring $B_{\mathbb{C};a_{1},\dots,a_{n}}$ is rigid for
every $(a_{1},\dots,a_{n})\in\Gamma_{n}$.

Lemma 4 of [KZ00] shows that $\mathscr{P}(3)$ is true. For $n>3$, only special
cases of $\mathscr{P}(n)$ are known to be true.555At the time of writing, it
seems likely that $\mathscr{P}(4)$ is true. In particular, the PhD thesis
[Chi] shows that if $(a_{1},a_{2},a_{3},a_{4})$ belongs to $\Gamma_{4}$ and is
not a permutation of $(2,3,4,12m)$ or $(2,3,5,30m)$ for any
$m\in\mathbb{N}\setminus\\{0\\}$, then
$B_{\mathbb{C};a_{1},a_{2},a_{3},a_{4}}$ is rigid. Lemma 8.5 is an example of
such a special case; we will use it in the proofs of Propositions 8.8 and 8.9.

######  8.4 Remark.

It is easy to see that if $B_{\mathbb{C};a_{1},\dots,a_{n}}$ is rigid then so
is $B_{{\rm\bf k};a_{1},\dots,a_{n}}$ for any field ${\rm\bf k}$ of
characteristic $0$.

######  8.5 Lemma.

Let ${\rm\bf k}$ be a field of characteristic $0$, $n\geq 3$, and
$(a_{1},\dots,a_{n})\in\Gamma_{n}$. If

$n=3$ or $\operatorname{\text{\rm cotype}}(a_{1},\dots,a_{n})\geq n-2$,

then $B_{{\rm\bf k};a_{1},\dots,a_{n}}$ is rigid.

###### Proof.

The case $n=3$ is true by $\mathscr{P}(3)$ and Rem. 8.4. The case $n>3$ is
Cor. 4.16 of [CD20]. ∎

######  8.6 Notation.

Let $B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ and
$B^{\prime}=\bigoplus_{i\in\mathbb{Z}}B_{i}^{\prime}$ be $\mathbb{Z}$-graded
rings.

  1. (1)

We write $B\cong B^{\prime}$ to indicate that there exists an isomorphism of
rings $\varphi:B\to B^{\prime}$ satisfying $\varphi(B_{i})=B_{i}^{\prime}$ for
all $i\in\mathbb{Z}$.

  2. (2)

We write $B\overset{\text{\tiny$*$}}{\cong}B^{\prime}$ to indicate that there
exists an isomorphism of rings $\varphi:B\to B^{\prime}$ such that, for every
$x\in B$, $x$ is homogeneous in $B$ if and only if $\varphi(x)$ is homogeneous
in $B^{\prime}$ (but the degree of $x$ in $B$ is not necessarily equal to that
of $\varphi(x)$ in $B^{\prime}$). Note that if
$B\overset{\text{\tiny$*$}}{\cong}B^{\prime}$ then $e(B)$ is not necessarily
equal to $e(B^{\prime})$.

The following result computes $e(B_{{\rm\bf k};a_{1},\dots,a_{n}})$ and
$\bar{e}(B_{{\rm\bf k};a_{1},\dots,a_{n}})$, and shows that if $d$ is a
positive divisor of $\bar{e}(B_{{\rm\bf k};a_{1},\dots,a_{n}})$ then
$(B_{{\rm\bf k};a_{1},\dots,a_{n}})^{(d)}$ is again a Pham-Brieskorn ring.
More precisely, we have $(B_{{\rm\bf
k};a_{1},\dots,a_{n}})^{(d)}\overset{\text{\tiny$*$}}{\cong}B_{{\rm\bf
k};a_{1}^{\prime},\dots,a_{n}^{\prime}}$ where
$a_{1}^{\prime},\dots,a_{n}^{\prime}$ are defined in the statement of the
Lemma.

######  8.7 Lemma.

Let ${\rm\bf k}$ be a field of characteristic $0$, $n\geq 3$ and
$a_{1},\dots,a_{n}\in\mathbb{N}\setminus\\{0\\}$.

  1. (a)

The integers $\frac{L}{L_{1}},\dots,\frac{L}{L_{n}}$ are pairwise relatively
prime.

  2. (b)

$e(B_{{\rm\bf k};a_{1},\dots,a_{n}})=1$ and $\bar{e}(B_{{\rm\bf
k};a_{1},\dots,a_{n}})=\prod_{i=1}^{n}\frac{L}{L_{i}}$.

  3. (c)

$B_{{\rm\bf k};a_{1},\dots,a_{n}}$ is saturated in codimension $1$ if and only
if $\operatorname{\text{\rm cotype}}(a_{1},\dots,a_{n})=0$.

  4. (d)

Let $d$ be a positive divisor of $\prod_{i=1}^{n}\frac{L}{L_{i}}$ and define
$(a_{1}^{\prime},\dots,a_{n}^{\prime})$ by

$a_{i}^{\prime}=\gcd(a_{i},L_{i})\,{\textstyle\left[\frac{L/L_{i}}{\gcd(d,L/L_{i})}\right]},\quad
1\leq i\leq n.$

Then $(B_{{\rm\bf
k};a_{1},\dots,a_{n}})^{(d)}\overset{\text{\tiny$*$}}{\cong}B_{{\rm\bf
k};a_{1}^{\prime},\dots,a_{n}^{\prime}}$.

  5. (e)

$(B_{{\rm\bf k};a_{1},\dots,a_{n}})_{\text{\rm
sat}}\overset{\text{\tiny$*$}}{\cong}B_{{\rm\bf
k};a_{1}^{\prime},\dots,a_{n}^{\prime}}$ where
$a_{i}^{\prime}=\gcd(a_{i},L_{i})$ for all $i\in\\{1,\dots,n\\}$.

###### Proof.

Let $B=B_{{\rm\bf k};a_{1},\dots,a_{n}}$. For each $i\in\\{1,\dots,n\\}$,
$x_{i}\neq 0$ is homogeneous of degree $d_{i}$; so
$e(B)=\gcd(d_{1},\dots,d_{n})=\gcd(\frac{L}{a_{1}},\dots,\frac{L}{a_{n}})=\frac{L}{\operatorname{\text{\rm
lcm}}(a_{1},\dots,a_{n})}=1$. Define

$\hat{d}_{i}=\gcd\big{\\{}\,d_{j}\,\mid\,j\in J_{i}\,\big{\\}}\quad\text{for
each $i\in\\{1,\dots,n\\}$.}$

Then
$\hat{d}_{1}=\gcd(d_{2},\dots,d_{n})\linebreak[3]=\gcd(\frac{L}{a_{2}},\dots,\frac{L}{a_{n}})=\frac{L}{\operatorname{\text{\rm
lcm}}(a_{2},\dots,a_{n})}=\frac{L}{L_{1}}$, and by the same argument we get

$\textstyle\hat{d}_{i}=\frac{L}{L_{i}}\quad\text{for all $i\in\\{1,\dots
n\\}$.}$

If $i\neq j$ then
$\gcd(L/L_{i},L/L_{j})=\gcd(\hat{d}_{i},\hat{d}_{j})=\gcd(d_{1},\dots,d_{n})=1$,
which proves (a).

(b) We already proved that $e(B)=1$. If $Z$ denotes the set of height $1$
homogeneous prime ideals of $B$ then:

  * •

for each $i\in\\{1,\dots,n\\}$, some ${\mathfrak{p}}\in Z$ satisfies
$x_{i}\in{\mathfrak{p}}$;

  * •

for each ${\mathfrak{p}}\in Z$, at most one $i\in\\{1,\dots,n\\}$ satisfies
$x_{i}\in{\mathfrak{p}}$, and if $x_{i}\in{\mathfrak{p}}$ then
$e(B/{\mathfrak{p}})=\gcd\big{\\{}\,d_{j}\,\mid\,j\in
J_{i}\,\big{\\}}=\hat{d}_{i}=L/L_{i}$.

It follows that $\big{\\{}\,e(B/{\mathfrak{p}})\,\mid\,{\mathfrak{p}}\in
Z\,\big{\\}}=\\{1,\frac{L}{L_{1}},\dots,\frac{L}{L_{n}}\\}$, so
$\bar{e}(B)=\operatorname{\text{\rm
lcm}}\big{(}{\textstyle\frac{L}{L_{1}},\dots,\frac{L}{L_{n}}}\big{)}=\prod_{i=1}^{n}\frac{L}{L_{i}}$,
the last equality by (a). This proves (b), and (c) follows from (b).

(d) Let $d$ be a divisor of $\bar{e}(B)=\prod_{i=1}^{n}\frac{L}{L_{i}}$.
Define $\alpha_{i}=\gcd(d,\hat{d}_{i})$ for each $i\in\\{1,\dots,n\\}$. Let us
prove that

(11) $\text{for every
$i\in\\{1,\dots,n\\}$,}\qquad\gcd(d_{i},\alpha_{i})=1,\quad d\mid
d_{i}\alpha_{i}\quad\text{and}\quad a_{i}=a_{i}^{\prime}\alpha_{i}.$

Indeed, let $i\in\\{1,\dots,n\\}$. Since $\alpha_{i}\mid\hat{d}_{i}$ and
$\gcd(d_{i},\hat{d}_{i})=\gcd(d_{1},\dots,d_{n})=1$, the first part of (11) is
clear. To prove the second part, we first show that

(12) for all $j\in\\{1,\dots,n\\}$, $\frac{L}{L_{j}}$ divides
$d_{i}\hat{d}_{i}$.

To see this, we note that
$\frac{d_{i}\hat{d}_{i}}{(L/L_{j})}=\frac{(L/a_{i})(L/L_{i})}{(L/L_{j})}=\frac{L_{j}L}{a_{i}L_{i}}\in\mathbb{Z}$
because if $i=j$ then $a_{i}\mid L$ and $L_{i}\mid L_{j}$, and if $i\neq j$
then $a_{i}\mid L_{j}$ and $L_{i}\mid L$. This proves (12). By (a), (b) and
(12), it follows that $\bar{e}(B)=\prod_{j=1}^{n}\frac{L}{L_{j}}$ divides
$d_{i}\hat{d}_{i}$. Since $d\mid\bar{e}(B)$, we obtain $d\mid
d_{i}\hat{d}_{i}$, so $d$ divides
$\gcd(d_{i}d,d_{i}\hat{d}_{i})=d_{i}\gcd(d,\hat{d}_{i})=d_{i}\alpha_{i}$,
which proves the second part of (11). For the third part, note that
$a_{i}^{\prime}\alpha_{i}=\gcd(a_{i},L_{i})\,{\textstyle\left[\frac{L/L_{i}}{\gcd(d,L/L_{i})}\right]}\,\gcd(d,\hat{d}_{i})$
and $\hat{d}_{i}=\frac{L}{L_{i}}$, so
$a_{i}^{\prime}\alpha_{i}=\gcd(a_{i},L_{i})(L/L_{i})$. On the other hand,
$\frac{a_{i}L_{i}}{\gcd(a_{i},L_{i})}=\operatorname{\text{\rm
lcm}}(a_{i},L_{i})=L$, so
$a_{i}=\gcd(a_{i},L_{i})(L/L_{i})=a_{i}^{\prime}\alpha_{i}$, proving (11).

Consider a monomial $m=x_{1}^{\nu_{1}}\cdots x_{n}^{\nu_{n}}$
($\nu_{1},\dots,\nu_{n}\in\mathbb{N}$) such that $d$ divides
$\deg(m)=d_{1}\nu_{1}+\cdots+d_{n}\nu_{n}$. If $i\in\\{1,\dots,n\\}$ then
$\alpha_{i}\mid d_{i}\nu_{i}$ (because $\alpha_{i}\mid d\mid\deg(m)$ and
$\alpha_{i}\mid\hat{d}_{i}$), so $\alpha_{i}\mid\nu_{i}$ by (11). This shows
that $m\in{\rm\bf k}[x_{1}^{\alpha_{1}},\dots,x_{n}^{\alpha_{n}}]$ and hence
that $B^{(d)}\subseteq{\rm\bf
k}[x_{1}^{\alpha_{1}},\dots,x_{n}^{\alpha_{n}}]$. By (11), we have $d\mid
d_{i}\alpha_{i}$ and hence $x_{i}^{\alpha_{i}}\in B^{(d)}$ for all $i$, so
$B^{(d)}={\rm\bf k}[x_{1}^{\alpha_{1}},\dots,x_{n}^{\alpha_{n}}]$. Observing
that
$(x_{1}^{\alpha_{1}})^{a_{1}^{\prime}}+\cdots+(x_{n}^{\alpha_{n}})^{a_{n}^{\prime}}=0$
(again by (11)), one easily obtains that ${\rm\bf
k}[x_{1}^{\alpha_{1}},\dots,x_{n}^{\alpha_{n}}]\overset{\text{\tiny$*$}}{\cong}B_{{\rm\bf
k};a_{1}^{\prime},\dots,a_{n}^{\prime}}$. This proves (d), and (e) follows
from (b) and (d). ∎

######  8.8 Proposition.

Let $m,n\in\mathbb{N}$ be such that $n\geq 3$ and $0\leq m\leq n-2$. Let
$\ell_{1},\dots,\ell_{n}\geq 2$ be pairwise relatively prime integers such
that $\ell_{1},\dots,\ell_{m}$ are odd. Let
$B=B_{\mathbb{C};\ell_{1},\dots,\ell_{m},2\ell_{m+1},\dots,2\ell_{n}}$. Then

$M(B)=\\{\ell_{1},\dots,\ell_{m}\\}\cup\big{\\{}\,\ell_{j}\ell_{k}\,\mid\,m+1\leq
j<k\leq n\,\big{\\}}.$

###### Proof.

We have $B=B_{\mathbb{C};a_{1},\dots,a_{n}}$ where
$(a_{1},\dots,a_{n})=(\ell_{1},\dots,\ell_{m},2\ell_{m+1},\dots,2\ell_{n})$.
In the notation of paragraph 8.2 and Lemma 8.7, we have
$L=2\prod_{i=1}^{n}\ell_{i}$ and

$L_{i}=2\prod_{j\in J_{i}}\ell_{j}$ , $L/L_{i}=\ell_{i}$ and
$\gcd(a_{i},L_{i})=\begin{cases}1,&\text{if $i\leq m$,}\\\ 2,&\text{if
$i>m$,}\end{cases}$

for all $i\in\\{1,\dots,n\\}$. So Lemma 8.7 gives
$\bar{e}(B)=\prod_{i=1}^{n}\ell_{i}$, and implies that if $d$ is a positive
divisor of $\prod_{i=1}^{n}\ell_{i}$ then
$B^{(d)}\overset{\text{\tiny$*$}}{\cong}B_{\mathbb{C};a_{1}^{\prime},\dots,a_{n}^{\prime}}$
where $(a_{1}^{\prime},\dots,a_{n}^{\prime})$ is given by

(13)
$a_{i}^{\prime}=\gcd(a_{i},L_{i})\left[\frac{L/L_{i}}{\gcd(d,L/L_{i})}\right]=\gcd(a_{i},L_{i})\left[\frac{\ell_{i}}{\gcd(d,\ell_{i})}\right]$

for every $i\in\\{1,\dots,n\\}$.

Let
$Y=\\{\ell_{1},\dots,\ell_{m}\\}\cup\big{\\{}\,\ell_{j}\ell_{k}\,\mid\,m+1\leq
j<k\leq n\,\big{\\}}$. By (13), we find that if $i\leq m$ then
$B^{(\ell_{i})}\overset{\text{\tiny$*$}}{\cong}B_{\mathbb{C};a_{1}^{\prime},\dots,a_{n}^{\prime}}$
where $a_{i}^{\prime}=1$; and if $j,k$ are such that $m+1\leq j<k\leq n$ then
$B^{(\ell_{j}\ell_{k})}\overset{\text{\tiny$*$}}{\cong}B_{\mathbb{C};a_{1}^{\prime},\dots,a_{n}^{\prime}}$
where $a_{j}^{\prime}=2=a_{k}^{\prime}$. In other words, if $d\in Y$ then
$B^{(d)}\overset{\text{\tiny$*$}}{\cong}B_{\mathbb{C};a_{1}^{\prime},\dots,a_{n}^{\prime}}$
where $(a_{1}^{\prime},\dots,a_{n}^{\prime})\notin\Gamma_{n}$ and hence
$B^{(d)}$ is non-rigid. This shows that $Y\subseteq X(B)$. It follows from
Cor. 7.16 that $\bigcup_{y\in Y}\mathscr{I}_{y}\subseteq X(B)$.

Arguing by contradiction, assume that $\bigcup_{y\in Y}\mathscr{I}_{y}\neq
X(B)$. Then Cor. 7.14 implies that there exists $d\in
M(B)\setminus\bigcup_{y\in Y}\mathscr{I}_{y}$, and that this $d$ must satisfy
$d\mid\bar{e}(B)$, so

$d\in M(B)$, $d$ divides $\textstyle\prod_{i=1}^{n}\ell_{i}$ and no element of
$Y$ divides $d$.

So there exist positive integers
$\alpha_{1},\dots,\alpha_{n},\beta_{1},\dots,\beta_{n}$ such that

  1. (i)

$\ell_{i}=\alpha_{i}\beta_{i}$ ($1\leq i\leq n$) and
$d=\prod_{i=1}^{n}\beta_{i}$,

  2. (ii)

$\alpha_{1},\dots,\alpha_{m}>1$, and at most one $j\geq m+1$ satisfies
$\alpha_{j}=1$.

Using (13) again, we find that
$B^{(d)}\overset{\text{\tiny$*$}}{\cong}B_{\mathbb{C};a_{1}^{\prime},\dots,a_{n}^{\prime}}$
where

$(a_{1}^{\prime},\dots,a_{n}^{\prime})=(\alpha_{1},\dots,\alpha_{m},2\alpha_{m+1},\dots,2\alpha_{n}).$

Clearly, (ii) implies that $a_{i}^{\prime}\neq 1$ for all
$i\in\\{1,\dots,n\\}$. Recall that $\ell_{1},\dots,\ell_{m}$ are odd; so
$\alpha_{1},\dots,\alpha_{m}$ are odd, which together with (ii) implies that
at most one $i\in\\{1,\dots,n\\}$ satisfies $a_{i}^{\prime}=2$. This shows
that $(a_{1}^{\prime},\dots,a_{n}^{\prime})\in\Gamma_{n}$. A simple
calculation shows that $\operatorname{\text{\rm
cotype}}(a_{1}^{\prime},\dots,a_{n}^{\prime})=|\big{\\{}\,i\,\mid\,1\leq i\leq
n\text{ and }\alpha_{i}>1\,\big{\\}}|\geq n-1$, so Lemma 8.5 implies that
$B^{(d)}$ is rigid. So $d\notin X(B)$, contradicting our choice of $d$. This
shows that $X(B)=\bigcup_{y\in Y}\mathscr{I}_{y}$. Since $Y$ is a primitive
set, it follows that $M(B)=Y$. ∎

######  8.9 Proposition.

Let $Y$ be a finite primitive set whose elements are pairwise relatively
prime. Then there exists an $\mathbb{N}$-graded normal affine
$\mathbb{C}$-domain $B$ satisfying $M(B)=Y$.

###### Proof.

If $Y=\varnothing$ then let $B=B_{\mathbb{C};3,3,3}$. We have $\bar{e}(B)=1$
by Lemma 8.7(b), so Cor. 7.14 implies that $M(B)$ is either $\varnothing$ or
$\\{1\\}$. Lemma 8.5 implies that $B$ is rigid, so $1\notin M(B)$ and hence
$M(B)=\varnothing=Y$.

If $Y=\\{1\\}$ then let $B=\mathbb{C}[X,Y]=\mathbb{C}^{[2]}$ with standard
$\mathbb{N}$-grading ($X,Y$ are homogeneous of degree $1$). Then $B$ is non-
rigid, so $M(B)=\\{1\\}=Y$.

From now-on, assume that $Y$ is neither $\varnothing$ nor $\\{1\\}$. Let
$m=|Y|\geq 1$ and let $a_{1},\dots,a_{m}$ be the distinct elements of $Y$.
Then $a_{1},\dots,a_{m}$ are pairwise relatively prime and
$a_{1},\dots,a_{m}\geq 2$. Choose an integer $c\geq 3$ such that
$a_{1},\dots,a_{m},c$ are pairwise relatively prime and let
$B=B_{\mathbb{C};a_{1},\dots,a_{m},c,c}$. We leave it to the reader to show
that $M(B)=\\{a_{1},\dots,a_{m}\\}=Y$ (the proof is similar to that of Prop.
8.8, but is simpler). ∎

We mentioned in Rem. 7.25 that we don’t know whether the set $\\{6,10\\}$ is
realizable. So it is interesting to note that $\\{6\\}$, $\\{10\\}$,
$\\{15\\}$ and $\\{6,10,15\\}$ are realizable, by Propositions 8.8 and 8.9.

##  9\. Rigidity of $B_{(x)}$

The main result of this section is Theorem 9.6. We begin with some
preliminaries on ramification indices.

######  9.1 Notation.

Let $X$ be an integral scheme that is normal and noetherian. Then we may
consider the group $\operatorname{\text{\rm Div}}(X)$ of Weil divisors of $X$,
and the group $\operatorname{\text{\rm
Div}}_{\mathbb{Q}}(X)=\mathbb{Q}\otimes_{\mathbb{Z}}\operatorname{\text{\rm
Div}}(X)$ of $\mathbb{Q}$-divisors of $X$. If $f$ is a nonzero element of the
function field $K(X)$ of $X$ then the divisor of $f$ is denoted
$\operatorname{{\rm div}}_{X}(f)$; so $\operatorname{{\rm
div}}_{X}:K(X)^{*}\to\operatorname{\text{\rm Div}}(X)$ is a group
homomorphism.

######  9.2.

(This is paragraph 5.4 in [CD22], with minor edits.) Let
$B=\bigoplus_{i\in\mathbb{N}}B_{i}$ be an $\mathbb{N}$-graded noetherian
normal domain such that the prime ideal $B_{+}=\bigoplus_{i>0}B_{i}$ has
height greater than $1$. Let $X=\operatorname{\text{\rm Spec}}B$ and
$Y=\operatorname{\text{\rm Proj}}B$, which are noetherian normal integral
schemes. We shall now define a $\mathbb{Q}$-linear map $D\mapsto D^{*}$ from
$\operatorname{\text{\rm Div}}_{\mathbb{Q}}(Y)$ to $\operatorname{\text{\rm
Div}}_{\mathbb{Q}}(X)$.

Note that the function fields of $X$ and $Y$ satisfy $K(X)\supseteq K(Y)$. Let
$Y^{(1)}$ be the set of homogeneous prime ideals of $B$ of height $1$. Since
$\operatorname{\text{\rm ht}}(B_{+})>1$, each element of $Y^{(1)}$ is a point
of $Y$; in fact we have $Y^{(1)}=\big{\\{}\,y\in
Y\,\mid\,\dim\mathcal{O}_{Y,y}=1\,\big{\\}}$. For each ${\mathfrak{p}}\in
Y^{(1)}$, $B_{\mathfrak{p}}\supset B_{({\mathfrak{p}})}$ is an extension of
DVRs; let $e_{\mathfrak{p}}$ denote the ramification index of this extension.
Then $e_{\mathfrak{p}}\in\mathbb{N}\setminus\\{0\\}$. If
$v^{Y}_{\mathfrak{p}}:K(Y)^{*}\to\mathbb{Z}$ and
$v^{X}_{\mathfrak{p}}:K(X)^{*}\to\mathbb{Z}$ denote the normalized666The word
“normalized” means that the maps $v^{Y}_{\mathfrak{p}}$ and
$v^{X}_{\mathfrak{p}}$ are surjective. valuations of $B_{({\mathfrak{p}})}$
and $B_{\mathfrak{p}}$ respectively, then
$v^{X}_{\mathfrak{p}}(\xi)=e_{\mathfrak{p}}v^{Y}_{\mathfrak{p}}(\xi)$ for all
$\xi\in K(Y)^{*}$. Let $C_{\mathfrak{p}}^{Y}$ (resp. $C_{\mathfrak{p}}^{X}$)
denote the closure of $\\{{\mathfrak{p}}\\}$ in $Y$ (resp. in $X$). Then
$C_{\mathfrak{p}}^{Y}$ (resp. $C_{\mathfrak{p}}^{X}$) is a prime divisor of
$Y$ (resp. of $X$), and every prime divisor of $Y$ is a $C^{Y}_{\mathfrak{p}}$
for some ${\mathfrak{p}}\in Y^{(1)}$. We define
$(C_{\mathfrak{p}}^{Y})^{*}=e_{\mathfrak{p}}C_{\mathfrak{p}}^{X}$ for each
${\mathfrak{p}}\in Y^{(1)}$, and extend linearly to a $\mathbb{Q}$-linear map
$\operatorname{\text{\rm Div}}_{\mathbb{Q}}(Y)\to\operatorname{\text{\rm
Div}}_{\mathbb{Q}}(X)$, $D\mapsto D^{*}$. It is not hard to see that the
linear map $D\mapsto D^{*}$ has the following property:

(14) $\big{(}\operatorname{{\rm div}}_{Y}(\xi)\big{)}^{*}=\operatorname{{\rm
div}}_{X}(\xi)$ for all $\xi\in K(Y)^{*}$.

######  9.3 Lemma.

Let the assumptions and notations be as in paragraph 9.2. Let
${\mathfrak{p}}\in Y^{(1)}$ and consider the ramification index
$e_{\mathfrak{p}}$ of $B_{\mathfrak{p}}$ over $B_{({\mathfrak{p}})}$.

  1. (a)

There exists a homogeneous element $g$ of ${\mathfrak{p}}$ such that
${\mathfrak{p}}B_{\mathfrak{p}}=gB_{\mathfrak{p}}$.

  2. (b)

For any $g$ as in (a) we have $\displaystyle
e_{\mathfrak{p}}=\frac{\operatorname{\text{\rm
lcm}}(\deg(g),e(B/{\mathfrak{p}}))}{\deg(g)}$.

###### Proof.

(a) There exists $g\in{\mathfrak{p}}$ such that
${\mathfrak{p}}B_{\mathfrak{p}}=gB_{\mathfrak{p}}$. We have to show that $g$
can be chosen to be homogeneous. Note that $v^{X}_{\mathfrak{p}}(g)=1$. Write
$g=\sum_{i\in I}g_{i}$ where $I$ is a finite subset of $\mathbb{N}$ and
$g_{i}\in B_{i}\setminus\\{0\\}$ for all $i\in I$. Since $g\in{\mathfrak{p}}$,
we have $g_{i}\in{\mathfrak{p}}$ and hence $v^{X}_{\mathfrak{p}}(g_{i})\geq 1$
for all $i\in I$. If $v^{X}_{\mathfrak{p}}(g_{i})>1$ for all $i\in I$ then
$v^{X}_{\mathfrak{p}}(g)>1$, which is not the case. So there exists $i\in I$
such that $v^{X}_{\mathfrak{p}}(g_{i})=1$. Then
${\mathfrak{p}}B_{\mathfrak{p}}=g_{i}B_{\mathfrak{p}}$, which proves (a).

(b) Let $g$ be a homogeneous element of ${\mathfrak{p}}$ such that
${\mathfrak{p}}B_{\mathfrak{p}}=gB_{\mathfrak{p}}$. Let
$\varepsilon=e(B/{\mathfrak{p}})$ and $d=\frac{\operatorname{\text{\rm
lcm}}(\deg(g),\varepsilon)}{\deg(g)}$. Note that
$\deg(g^{d})\in\varepsilon\mathbb{Z}$; so there exist $i,j\in\mathbb{N}$ such
that $(B/{\mathfrak{p}})_{i}\neq 0$, $(B/{\mathfrak{p}})_{j}\neq 0$, and
$\deg(g^{d})=i-j$; consequently, there exist homogeneous elements $u,v\in
B\setminus{\mathfrak{p}}$ such that $\deg(g^{d})=\deg(v)-\deg(u)$. Then
$\xi=g^{d}u/v$ belongs to the maximal ideal ${\mathfrak{m}}$ of
$B_{({\mathfrak{p}})}$. Let us prove that ${\mathfrak{m}}=\xi
B_{({\mathfrak{p}})}$.

Consider an arbitrary nonzero element $\frac{x}{s}$ of ${\mathfrak{m}}$, where
$x,s\in B$ are homogeneous of the same degree and $s\notin{\mathfrak{p}}$.
Define $k=v^{X}_{\mathfrak{p}}\big{(}\frac{x}{s}\big{)}$. Then $k>0$, because
$\frac{x}{s}\in{\mathfrak{m}}\subseteq{\mathfrak{p}}B_{\mathfrak{p}}$. So
$\frac{x}{s}=g^{k}w$ for some $w\in B_{\mathfrak{p}}^{*}$, i.e.,
$\frac{x}{s}=g^{k}\frac{b}{c}$ where $b,c\in B\setminus{\mathfrak{p}}$. Write
$b=\sum_{i\in\mathbb{N}}b_{i}$ and $c=\sum_{i\in\mathbb{N}}c_{i}$
($b_{i},c_{i}\in B_{i}$ for all $i$). Since $c\notin{\mathfrak{p}}$, there
exists $\ell\in\mathbb{N}$ such that $c_{\ell}\in B\setminus{\mathfrak{p}}$.
The equation $xc=g^{k}bs$ gives $xc_{\ell}=g^{k}b_{m}s$ for some $m$; so
$\frac{x}{s}=g^{k}\frac{b_{m}}{c_{\ell}}$. Also,

$\textstyle
v^{X}_{\mathfrak{p}}(g^{k}b_{m}s)=v^{X}_{\mathfrak{p}}(xc_{\ell})=v^{X}_{\mathfrak{p}}(x)=v^{X}_{\mathfrak{p}}\big{(}\frac{x}{s}\big{)}=k$

implies that $b_{m}\notin{\mathfrak{p}}$. Since $c_{\ell},b_{m}\in
B\setminus{\mathfrak{p}}$, we have $\ell,m\in\varepsilon\mathbb{Z}$. Since
$\frac{x}{s}=g^{k}\frac{b_{m}}{c_{\ell}}$, we have
$k\deg(g)+m-\ell=\deg(x)-\deg(s)=0$; it follows that
$k\deg(g)\in\varepsilon\mathbb{Z}$, so $d\mid k$. Write $k=k_{0}d$ (where
$k_{0}>0$), then

$\frac{x}{s}=g^{k_{0}d}\,\frac{b_{m}}{c_{\ell}}=\left(g^{d}\frac{u}{v}\right)^{k_{0}}\frac{v^{k_{0}}b_{m}}{u^{k_{0}}c_{\ell}}=\xi^{k_{0}}\frac{v^{k_{0}}b_{m}}{u^{k_{0}}c_{\ell}}$

where $\frac{v^{k_{0}}b_{m}}{u^{k_{0}}c_{\ell}}\in B_{({\mathfrak{p}})}^{*}$.
Since $k_{0}>0$, we have $\frac{x}{s}\in\xi B_{({\mathfrak{p}})}$, which shows
that ${\mathfrak{m}}=\xi B_{({\mathfrak{p}})}$. It follows that
$e_{\mathfrak{p}}=v^{X}_{\mathfrak{p}}(\xi)=v^{X}_{\mathfrak{p}}(g^{d}u/v)=d=\frac{\operatorname{\text{\rm
lcm}}(\deg(g),\varepsilon)}{\deg(g)}$. ∎

######  9.4 Corollary.

Let the assumptions and notations be as in paragraph 9.2. If $B$ is saturated
in codimension $1$, then $e_{\mathfrak{p}}=1$ for all ${\mathfrak{p}}\in
Y^{(1)}$.

###### Proof.

The hypothesis and Lemma 7.6 imply that $\bar{e}(B)=e(B)$, so
$e(B/{\mathfrak{p}})=e(B)$ for all ${\mathfrak{p}}\in Y^{(1)}$. So the claim
follows from Lemma 9.3. ∎

The following notation is used in the proof of the next Theorem.

######  9.5 Notation.

Given an $\mathbb{N}$-graded ring $A=\bigoplus_{i\in\mathbb{N}}A_{i}$ and
$d\in\mathbb{N}\setminus\\{0\\}$, we define the $\mathbb{N}$-graded ring
$A^{\langle d\rangle}=S=\bigoplus_{i\in\mathbb{N}}S_{i}$ by setting
$S_{i}=A_{di}$ for all $i\in\mathbb{N}$. Note that $A^{\langle d\rangle}$ and
$A^{(d)}$ are the same ring, but have different gradings: if $x\in
S_{i}\setminus\\{0\\}$ then $x$ is a homogeneous element of both $A^{\langle
d\rangle}$ and $A^{(d)}$, the degree of $x$ in $A^{\langle d\rangle}$ is $i$,
and the degree of $x$ in $A^{(d)}$ is $di$. Observe that
$A^{(d)}\overset{\text{\tiny$*$}}{\cong}A^{\langle d\rangle}$, in the notation
of 8.6.

######  9.6 Theorem.

Let ${\rm\bf k}$ be a field of characteristic $0$ and
$B=\bigoplus_{n\in\mathbb{N}}B_{n}$ an $\mathbb{N}$-graded normal affine
${\rm\bf k}$-domain such that the prime ideal $B_{+}=\bigoplus_{n>0}B_{n}$ has
height at least $2$. Let $x$ be a homogeneous prime element of $B$ of degree
$d>0$. If $B_{(x)}$ is non-rigid then so is $(B/xB)^{(d)}$.

######  9.7 Remark.

As is well known, $B_{(x)}$ is non-rigid if and only if the basic open subset
$\mathbb{D}_{+}(x)=\big{\\{}\,{\mathfrak{p}}\in\operatorname{\text{\rm
Proj}}B\,\mid\,x\notin{\mathfrak{p}}\,\big{\\}}$ of $\operatorname{\text{\rm
Proj}}B$ admits a nontrivial $G_{a}$-action.

###### Proof of Thm 9.6.

Let $X=\operatorname{\text{\rm Spec}}B$ and $Y=\operatorname{\text{\rm
Proj}}B$, and let us use the notations of paragraph 9.2 (this is why we need
to assume that $\operatorname{\text{\rm ht}}(B_{+})\geq 2$). Let
${\mathfrak{p}}=xB\in Y^{(1)}$ and consider the valuations
$v_{\mathfrak{p}}^{X}:K(X)\to\mathbb{Z}\cup\\{\infty\\}$ and
$v_{\mathfrak{p}}^{Y}:K(Y)\to\mathbb{Z}\cup\\{\infty\\}$ and the ramification
index $e_{\mathfrak{p}}$. Lemma 9.3 gives

(15) $e_{\mathfrak{p}}d=\ell,\quad\text{where we define
$\ell=\operatorname{\text{\rm lcm}}(d,e(B/xB))$.}$

Consider $C_{\mathfrak{p}}^{Y}\subset Y$ and $C_{\mathfrak{p}}^{X}\subset X$
as in 9.2; note that $C_{\mathfrak{p}}^{Y}=\mathbb{V}_{+}(x)\subset Y$ and let
$U=\mathbb{D}_{+}(x)=Y\setminus C_{\mathfrak{p}}^{Y}$. Let $R=B_{(x)}$ and
note that $R$ is a subring of $K(Y)$. Note that if $f\in R\setminus\\{0\\}$
then $\operatorname{{\rm div}}_{U}(f)\geq 0$; it follows that
$\operatorname{{\rm div}}_{\mathfrak{p}}^{Y}(f)\leq 0$, otherwise we would
have $\operatorname{{\rm div}}_{Y}(f)\geq 0$ and $\operatorname{{\rm
div}}_{Y}(f)\neq 0$, which is absurd. Consequently, the map

$\deg_{R}:R\to\mathbb{N}\cup\\{-\infty\\},\quad\text{$\deg_{R}(f)=-v_{\mathfrak{p}}^{Y}(f)$
for all $f\in R$}$

is a well-defined degree function. Consider the associated graded ring
$\operatorname{\text{\rm Gr}}(R)$ determined by $(R,\deg_{R})$; recall that
$\operatorname{\text{\rm Gr}}(R)=\bigoplus_{n\in\mathbb{N}}R_{\leq n}/R_{<n}$
where for each $n\in\mathbb{N}$ we set

$R_{\leq n}=\big{\\{}\,r\in R\,\mid\,\deg_{R}(r)\leq
n\,\big{\\}}\quad\text{and}\quad R_{<n}=\big{\\{}\,r\in
R\,\mid\,\deg_{R}(r)<n\,\big{\\}}\quad\text{(so $R_{<0}=0$).}$

Since $v_{\mathfrak{p}}^{Y}:K(Y)\to\mathbb{Z}\cup\\{\infty\\}$ is surjective
and $\operatorname{\text{\rm Frac}}R=K(Y)$, it follows that the gcd of the
subset $\deg_{R}(R\setminus\\{0\\})$ of $\mathbb{N}$ is equal to $1$; so
$e(\operatorname{\text{\rm Gr}}R)=1$. We claim (see Notation 9.5):

(16) $\operatorname{\text{\rm
Gr}}(R)\cong\big{(}B/xB\big{)}^{\langle\ell\rangle}.$

Let us prove (16). Given $f\in R\setminus\\{0\\}$, let
$S(f)=\big{\\{}\,n\in\mathbb{N}\,\mid\,fx^{n}\in B\,\big{\\}}$ and note that
$S(f)$ is a nonempty subset of $\mathbb{N}$. Clearly, if $n\in S(f)$ then
$n+1\in S(f)$. We claim that

(17) $\text{for each $f\in R\setminus\\{0\\}$,}\quad
S(f)=\big{\\{}\,n\in\mathbb{N}\,\mid\,n\geq
e_{\mathfrak{p}}\deg_{R}(f)\,\big{\\}}.$

Indeed, let $f\in R\setminus\\{0\\}$. Given any $n\in\mathbb{N}$, we have
$n\in S(f)\iff fx^{n}\in B\iff\operatorname{{\rm div}}_{X}(fx^{n})\geq 0$.
Consider the $\mathbb{Q}$-linear map $D\mapsto D^{*}$ from
$\operatorname{\text{\rm Div}}_{\mathbb{Q}}(Y)$ to $\operatorname{\text{\rm
Div}}_{\mathbb{Q}}(X)$ defined in 9.2; then
$(C_{\mathfrak{p}}^{Y})^{*}={e_{\mathfrak{p}}}C_{\mathfrak{p}}^{X}$, and (14)
gives $(\operatorname{{\rm div}}_{Y}(f))^{*}=\operatorname{{\rm div}}_{X}(f)$.
Since

$\textstyle\operatorname{{\rm div}}_{X}(fx^{n})=\operatorname{{\rm
div}}_{X}(f)+n\operatorname{{\rm div}}_{X}(x)=\operatorname{{\rm
div}}_{X}(f)+nC_{\mathfrak{p}}^{X}=(\operatorname{{\rm
div}}_{Y}(f)+\frac{n}{e_{\mathfrak{p}}}C_{\mathfrak{p}}^{Y})^{*},$

and since $D^{*}\geq 0\Leftrightarrow D\geq 0$ for any
$D\in\operatorname{\text{\rm Div}}_{\mathbb{Q}}(Y)$, we have

(18) $\textstyle n\in S(f)\iff\operatorname{{\rm
div}}_{Y}(f)+\frac{n}{e_{\mathfrak{p}}}C_{\mathfrak{p}}^{Y}\geq 0\iff
v_{\mathfrak{p}}^{Y}(f)\geq-\frac{n}{e_{\mathfrak{p}}}\iff\deg_{R}(f)\leq\frac{n}{e_{\mathfrak{p}}},$

which proves (17) (the middle equivalence in (18) follows from
$\operatorname{{\rm div}}_{U}(f)\geq 0$; see the paragraph after (15)). Now
(17) implies that if $n\in\mathbb{N}$ and $f\in R\setminus\\{0\\}$ then:

$\displaystyle f\in R_{\leq n}$ $\displaystyle\Leftrightarrow\deg_{R}(f)\leq
n\Leftrightarrow e_{\mathfrak{p}}n\in S(f)\Leftrightarrow
fx^{e_{\mathfrak{p}}n}\in B;$ $\displaystyle f\in R_{<n}$
$\displaystyle\Leftrightarrow\deg_{R}(f)<n\Leftrightarrow
e_{\mathfrak{p}}n>\min S(f)\Leftrightarrow e_{\mathfrak{p}}n-1\in
S(f)\Leftrightarrow fx^{e_{\mathfrak{p}}n-1}\in B$
$\displaystyle\Leftrightarrow fx^{e_{\mathfrak{p}}n}\in xB.$

So the following holds for each $n\in\mathbb{N}$:

(19) each $f\in R_{\leq n}$ satisfies $\big{(}fx^{e_{\mathfrak{p}}n}\in B$ and
$fx^{e_{\mathfrak{p}}n}\in xB\Leftrightarrow f\in R_{<n}\big{)}$.

Statement (19) implies that, for each $n\in\mathbb{N}$, the ${\rm\bf
k}$-linear map $\theta_{n}:R_{\leq n}\to B_{de_{\mathfrak{p}}n}$, $f\mapsto
fx^{e_{\mathfrak{p}}n}$, is well-defined and induces an injective ${\rm\bf
k}$-linear map $\bar{\theta}_{n}:R_{\leq
n}/R_{<n}\to(B/xB)_{de_{\mathfrak{p}}n}$. Note that $\bar{\theta}_{n}$ is also
surjective. Indeed, consider $\xi\in(B/xB)_{de_{\mathfrak{p}}n}$. Then
$\xi=h+xB$ for some $h\in B_{de_{\mathfrak{p}}n}$. Obviously, if $h=0$ then
$\xi\in\operatorname{\text{\rm im}}(\bar{\theta}_{n})$. Assume that $h\neq 0$
and let $f=h/x^{e_{\mathfrak{p}}n}\in R\setminus\\{0\\}$; then
$e_{\mathfrak{p}}n\in S(f)$ by definition of $S(f)$, so $e_{\mathfrak{p}}n\geq
e_{\mathfrak{p}}\deg_{R}(f)$ by (17), so $\deg_{R}(f)\leq n$, i.e., $f\in
R_{\leq n}$; we have $\theta_{n}(f)=h$, so $\bar{\theta}_{n}(f+R_{<n})=\xi$.
This shows that, for every $n\in\mathbb{N}$, the ${\rm\bf k}$-linear map
$\bar{\theta}_{n}:R_{\leq n}/R_{<n}\to(B/xB)_{\ell n}$ is bijective (note that
$\ell n=de_{\mathfrak{p}}n$ by (15)). The family
$\big{(}\bar{\theta}_{n}\big{)}_{n\in\mathbb{N}}$ defines an injective
${\rm\bf k}$-linear map $\bar{\theta}:\operatorname{\text{\rm Gr}}(R)\to B/xB$
which is easily seen to preserve multiplication. Moreover, the image of
$\bar{\theta}$ is the subring $\big{(}B/xB)^{(\ell)}$ of $B/xB$. If we define
$S_{n}=(B/xB)_{\ell n}$ for all $n\in\mathbb{N}$, then
$\big{(}B/xB)^{\langle\ell\rangle}=\bigoplus_{n\in\mathbb{N}}S_{n}$, so
$\bar{\theta}:\operatorname{\text{\rm
Gr}}(R)\to\big{(}B/xB)^{\langle\ell\rangle}$ is a degree-preserving
isomorphism of graded ${\rm\bf k}$-algebras. This proves (16).

It follows from (16) that $\operatorname{\text{\rm Gr}}(R)$ is ${\rm\bf
k}$-affine (because $(B/xB\big{)}^{\langle\ell\rangle}$ is ${\rm\bf k}$-affine
by Lemma 2.6), so part (b) of Theorem 1.7 of [Dai12] implies that $\deg_{R}$
is tame over ${\rm\bf k}$ (in the terminology of [Dai12], this means that for
every ${\rm\bf k}$-derivation $D:R\to R$, the set
$\big{\\{}\,\deg_{R}(Dr)-\deg_{R}(r)\,\mid\,r\in R\setminus\\{0\\}\,\big{\\}}$
has a greatest element). As is well known, this implies that each element $D$
of $\operatorname{\text{\rm LND}}(R)\setminus\\{0\\}$ gives rise to a well-
defined $\operatorname{\text{\rm Gr}}(D):\operatorname{\text{\rm
Gr}}(R)\to\operatorname{\text{\rm Gr}}(R)$ which belongs to
$\operatorname{\text{\rm HLND}}(\operatorname{\text{\rm
Gr}}R)\setminus\\{0\\}$. So if $\operatorname{\text{\rm LND}}(R)\neq\\{0\\}$
then $\operatorname{\text{\rm HLND}}(\operatorname{\text{\rm
Gr}}R)\neq\\{0\\}$, which (by (16)) implies $\operatorname{\text{\rm
HLND}}\big{(}\big{(}B/xB\big{)}^{\langle\ell\rangle}\big{)}\neq\\{0\\}$, which
implies that $\big{(}B/xB\big{)}^{\langle\ell\rangle}$ is non-rigid. Since the
graded rings $\big{(}B/xB\big{)}^{\langle\ell\rangle}$ and