# Noether’s normalization in skew polynomial rings

Elad Paran<EMAIL_ADDRESS>Thieu N. Vo<EMAIL_ADDRESS>The Open
University of Israel  Fractional Calculus, Optimization and Algebra Research
Group, Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi
Minh City, Vietnam

###### Abstract

We study Noether’s normalization lemma for finitely generated algebras over a
division algebra. In its classical form, the lemma states that if $I$ is a
proper ideal of the ring $R=F[t_{1},\ldots,t_{n}]$ of polynomials over a field
$F$, then the quotient ring $R/I$ is a finite extension of a polynomial ring
over $F$. We prove that the lemma holds when $R=D[t_{1},\ldots,t_{n}]$ is the
ring of polynomials in $n$ central variables over a division algebra $D$. We
provide examples demonstrating that Noether’s normalization may fail for the
skew polynomial ring $D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ with
respect to commuting automorphisms $\sigma_{1},\ldots,\sigma_{n}$ of $D$. We
give a sufficient condition for $\sigma_{1},\ldots,\sigma_{n}$ under which the
normalization lemma holds for such ring. In the case where $D=F$ is a field,
this sufficient condition is proved to be necessary.

##  1 Introduction

Let $D$ be a division algebra, and suppose that $R$ is a quotient ring of the
polynomial ring $D[x_{1},\ldots,x_{n}]$ in $n$ central variables over $D$ by a
proper two-sided ideal $I$. We shall call such a ring $R$ a centrally finitely
generated extension of $D$. The motivating question for this paper is the
following: What can be said of the structure of such rings? In the case where
$D$ is a field, this is a classical question, and the answer is given by
Noether’s Normalization Lemma [6, Theorem 13.3], which states that $R$ itself
must be a finite extension of a polynomial ring over $D$.

If $D$ is an arbitrary division algebra and $R$ is a simple (left)
$D[x_{1},\ldots,x_{n}]$-module (equivalently, $R$ is a quotient module of
$D[x_{1},\ldots,x_{n}]$ by a maximal left ideal $I$), a theorem of Amitsur and
Small [4, Theorem 1] states that $R$ is finite over a polynomial ring in
central variables over $D$. Our first observation is closely related, stating
the following generalization of Noether’s lemma:

######  Theorem 1.1.

Let $R=D[x_{1},\ldots,x_{n}]/I$ be a centrally finitely generated extension of
$D$. Then $R$ is finite as a left module over a subring which is isomorphic to
a polynomial ring in central variables over $D$.

Let us emphasize that in the mentioned theorem of Amitsur-Small, the ideal $I$
is one-sided, but maximal, while in Theorem 1.1 the ideal $I$ is an arbitrary
two-sided ideal.

Next, we consider a more general situation: Given commuting111Throughout this
work, whenever we refer to a sequence of objects as “commuting”, we mean
“pairwise commuting”. automorphisms $\sigma_{1},\ldots,\sigma_{n}$ of a
division algebra $D$, we denote by
$D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ the corresponding
multivariate skew polynomial ring (see §2 below for a brief survey of basic
facts concerning such rings). We say that a ring extension $S$ of $D$ is
automorphically finitely generated, if $S$ is a quotient ring of
$D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ by a proper two-sided
ideal, for some commuting automorphisms $\sigma_{1},\ldots,\sigma_{n}$ of $D$.
The main goal of this paper is to study the structure of such rings. To that
end, we introduce the following terminology:

######  Definition 1.2.

Let $D$ be a division algebra.

  1. 1.

We say that a ring extension $S$ of $D$ is automorphically normalizable over
$D$, if $S$ is finite as a left module over a subring which is isomorphic to
$D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$, for some commuting
automorphisms $\sigma_{1},\ldots,\sigma_{n}$ of $D$. If these automorphisms
can all be chosen as the identity, we say that $S$ is centrally normalizable
over $D$.

  2. 2.

A tuple $(\sigma_{1},\ldots,\sigma_{n})$ of commuting automorphisms in
$\mathbf{Aut}(D)$ is called automorphically (resp. centrally) normalizable
over $D$ if every quotient ring of
$D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ is automorphically (resp.
centrally) normalizable over $D$.

Using this terminology, Theorem 1.1 above states that every centrally finitely
generated extension over a division algebra $D$ is centrally normalizable over
$D$. In the case where $D$ is centrally finite (that is, of finite dimension
over its center), we deduce from Theorem 4.6 that Noether’s normalization
holds for arbitrary subrings of polynomial rings that are finitely generated
over $D$:

######  Theorem 1.3.

Let $D$ be a centrally finite division algebra and let
$R=D[t_{1},\ldots,t_{n}]$ be the ring of polynomials in $n$ central variables
over $D$. Then for every finite subset $A$ of $R$, the ring $S=D[A]$ generated
by $A$ over $D$ is centrally normalizable over $D$.

The key point in the proof of this result is that such a finitely generated
subring $S$ must be centrally finitely generated, which a priory is not
obvious – we obtain this a consequence of a theorem of Wilczyński.

Noether’s normalization lemma, in its classical form, can be stated as
follows: The tuple $(\mathbf{id}_{F},\ldots,\mathbf{id}_{F})$ of finitely many
identity maps of a field $F$ is centrally normalizable over $F$. We aim to
identify those tuples of automorphisms of a division algebra $D$ that are
automorphically normalizable over $D$.

Given a division algebra $D$, we denote its center by $\mathcal{Z}(D)$, and
given an automorphism $\sigma$ of $D$, we denote by $D_{\sigma}$ the fixed
ring of $D$. We prove:

######  Theorem 1.4.

Let $D$ be a division algebra and let $\sigma$ be an automorphism of $D$, such
that the field $F=\mathcal{Z}(D)\cap D_{\sigma}$ is infinite. Then the
constant tuple $(\sigma,\ldots,\sigma)$, of any length, is automorphically
normalizable over $D$.

As consequence of Theorem 1.4, we have the following corollary:

######  Corollary 1.5.

Let $D$ be a division algebra and let $\sigma_{1},\ldots,\sigma_{n}$ be
commuting automorphisms of $D$ such that
$F=\mathcal{Z}(D)\cap\big{(}\cap_{i=1}^{n}D_{\sigma_{i}}\big{)}$ is infinite.
If $\sigma_{1}^{d_{1}}=\ldots=\sigma_{n}^{d_{n}}$ for some positive integers
$d_{1},\ldots,d_{n}$, then the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is
automorphically normalizable over $D$. In particular, if
$\sigma_{1},\ldots,\sigma_{n}$ are of finite orders, then the tuple
$(\sigma_{1},\ldots,\sigma_{n})$ is centrally normalizable over $D$.

This result gives a sufficient condition for a tuple of automorphisms of a
division algebra $D$ to be automorphically normalizable over $D$. The authors
do not know whether this condition is generally also a necessary one, see
Question 5.14 below. However, in the case where $D$ is a field, we show that
it is, provided that the condition that
$\mathcal{Z}(D)\cap\big{(}\cap_{i=1}^{n}D_{\sigma_{i}}\big{)}$ is infinite
holds. That is, we prove:

######  Theorem 1.6.

Let $F$ be a field and let $\sigma_{1},\ldots,\sigma_{n}$ be commuting
automorphisms of $F$ such that $\cap_{i=1}^{n}F_{\sigma_{i}}$ is infinite.
Then the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is automorphically
normalizable over $F$ if and only if
$\sigma_{1}^{d_{1}}=\ldots=\sigma_{n}^{d_{n}}$ for some positive integers
$d_{1},\ldots,d_{n}$.

For general division algebras, we give the following example where automorphic
normalizability fails: If $\sigma\in\mathbf{Aut}(D)$ is of infinite inner
order, then the tuple $(\sigma,\sigma^{-1})$ is not automorphically
normalizable over $D$ (Proposition 5.7 below). Another negative example is
given by Proposition 5.11 below. These examples lead to some natural open
questions, see Question 5.8 and Question 5.12.

Finally, we note that the notion of automorphic normalizability can be used to
formulate the following generalization of Zariski’s lemma:

######  Proposition 1.7.

Let $S$ be a ring extension of a division algebra $D$ such that $S$ is
automorphically normalizable over $D$. If $S$ is a division algebra, then $S$
is finite as a left module over $D$.

The results discussed above concerning automorphic normalizability yield
various situations to which this version of Zariski’s lemma applies.

The paper is organized as follows: In §2 we provide basic facts concerning
skew polynomials rings, automorphically finitely generated extensions,
normalizability, and the proof of Proposition 1.7. In §3 we study
normalizability of constant tuples of automorphisms, and prove Theorem 1.4 and
Corollary 1.5. In §4 we study central normalizability and prove Theorem 1.1
and Theorem 1.3. Finally, in §5 we study general tuples of automorphisms,
introduce the above mentioned negative examples where normalizability does not
hold, and prove Theorem 1.6.

##  2 Preliminaries

Throughout this paper all rings are assumed to be associative with unity. By a
division algebra or a division ring, we mean a ring in which every nonzero
element has a multiplicative inverse. If $D$ is a division algebra of finite
dimension over its center, one says that $D$ is a centrally finite division
algebra [10, Definition 14.1].

###  2.1 Automorphically finitely generated ring extensions

Let $R$ be a (possibly non-commutative) ring and let $\sigma$ be an
automorphism of $R$. The skew polynomial ring $R[t,\sigma]$ consists of all
polynomials in the variable $t$, with the usual addition, and with
multiplication determined by the rule $ta=a^{\sigma}t$ for all $a\in R$ [9,
Chapter 2], [19, Chapter 1]. The study of skew polynomial rings is prolific,
both from a theoretical point of view, and for their various applications, see
for example [3, 5, 7, 8, 12, 13, 14, 15, 22].

More generally, if $\sigma_{1},\ldots,\sigma_{n}$ are automorphisms of $R$,
one forms the multivariate skew polynomial ring
$R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$, in which the variables
$t_{1},\ldots,t_{n}$ commute pairwise, and multiplication is determined by the
rule $x_{i}a=a^{\sigma_{i}}x_{i}$ for all $1\leq i\leq n$ and $a\in R$. If one
further assumes that the $\sigma_{i}$ commute pairwise, then one may view
$R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ as an iterated skew
polynomial ring:

$R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]=R[t_{1},\ldots,t_{n-1};\sigma_{1},\ldots,\sigma_{n-1}][t_{n},\sigma_{n}],$

by extending $\sigma_{n}$ trivially to an automorphism of
$R[t_{1},\ldots,t_{n-1};\sigma_{1},\ldots,\sigma_{n-1}]$, given by
$\sigma_{n}(t_{i})=t_{i}$ for all $1\leq i\leq n-1$ (see [23, Theorem 2.4]).
If the ring $R$ is (left) Noetherian then so is
$S=R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$. Indeed, one can view
$S$ as
$R[t_{1},\ldots,t_{n-1};\sigma_{1},\ldots,\sigma_{n-1}][t_{n};\sigma_{n}]$, by
extending $\sigma_{n}$ to an automorphism of
$R[t_{1},\ldots,t_{n-1};\sigma_{1},\ldots,\sigma_{n-1}]$ given by
$\sigma_{n}(t_{i})=t_{i}$ for each $1\leq i<n$. Then, by applying inductively
the skew Hilbert basis theorem given in [9, page 17], we get that $S$ is also
(left) Noetherian. As a consequence, if a ring extension of $S$ is finite over
$S$ (i.e. finitely generated as a left $S$-module) then it is left integral
over $S$ (see [21, Lemma 2.2]).

In the case where $\sigma_{1},\ldots,\sigma_{n}$ are the identity
automorphism, the ring $R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ is
the usual polynomial ring $R[t_{1},\ldots,t_{n}]$ in $n$ central variables
over $R$.

######  Definition 2.1.

Let $S/R$ be a ring extension. We say that $S$ is automorphically finitely
generated over $R$, if $S$ is isomorphic to a quotient ring
$R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]/I$, where:

  1. 1.

$\sigma_{1},\ldots,\sigma_{n}$ are commuting automorphisms of $R$.

  2. 2.

$I$ is a two-sided ideal of
$R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$, satisfying $R\cap
I=\\{0\\}$.

We say that $S$ is centrally finitely generated over $R$, if $S$ is
automorphically finitely generated, and the automorphisms
$\sigma_{1},\ldots,\sigma_{n}$ can be chosen to be the identity automorphism.

Note that in the case where $R$ is a division algebra, the condition that
$R\cap I=\\{0\\}$ is equivalent to $I$ being a proper two-sided ideal. Thus in
the case where $R$ is a field, a centrally finitely generated extension of $R$
is simply a finitely generated extension of $R$, in the classical sense.

######  Definition 2.2 (see [16, 10.1.3, p. 366]).

Let $S/R$ be a ring extension. An element $a\in S$ is automorphic over $R$ if
there exists an automorphism $\sigma\in\mathbf{Aut}(R)$ such that
$ab=b^{\sigma}a$ for all $b\in R$. In this case, we say that $a$ is
automorphic over $R$ with respect to $\sigma$. If, furthermore,
$\sigma=\mathbf{id}_{R}$, we say that $a$ centralizes $R$.

######  Remark 2.3.

In general, if $S/R$ is a ring extension, an element $a\in S$ can be
automorphic over $R$ with respect to several different automorphisms in
$\mathbf{Aut}(R)$. However, if $\mathbf{Ann}_{R}(a):=\\{r\in R\,|\,ra=0\\}$ is
equal to $\\{0\\}$, then the automorphism $\sigma$ is uniquely determined by
$a$. Indeed, if $a$ is automorphic over $R$ with respect to two automorphisms
$\sigma,\tau$, then for all $b\in R$ we have $b^{\sigma}a=ab=b^{\tau}a$, hence
$(b^{\sigma}-b^{\tau})a=0$, hence $b^{\sigma}-b^{\tau}\in\mathbf{Ann}_{R}(a)$,
hence $b^{\sigma}=b^{\tau}$.

In the case where $S=R=D$ is a division algebra, every element of $D$ is
automorphic over $D$:

######  Example 2.4.

Let $D$ be a division algebra and $a$ an arbitrary nonzero element of $D$. We
denote by $\mathbf{in}_{a}$ the inner automorphism of $D$ with respect to $a$
which is defined as $\mathbf{in}_{a}(r)=ara^{-1}$ for all $r\in D$. Then $a$
is automorphic over $D$ with respect to $\mathbf{in}_{a}$, since
$ar=ara^{-1}\cdot a=\mathbf{in}_{a}(r)a$.

######  Lemma 2.5.

Let $S/R$ be a ring extension. Let $a_{1},\ldots,a_{n}$ be $n$ commuting
elements of $S$, automorphic over $R$ with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}\in\mathbf{Aut}(R)$. Then the natural embedding
$R\xhookrightarrow{}S$ can be uniquely extended to a ring homomorphism
$\phi\colon R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]\to S$ such that
$\phi(t_{i})=a_{i}$ for all $1\leq i\leq n$.

###### Proof.

We construct $\phi$ by repeatedly using [9, Proposition 2.4, p. 37] as
follows. First, let $\phi$ be the natural embedding of $R$ to $S$. Since
$a_{1}r=\sigma_{1}(r)a_{1}$ for all $r\in R$, it follows from [9, Proposition
2.4, p. 37] that $\phi$ can be extended to a ring homomorphism from
$R[t_{1};\sigma_{1}]$ to $S$ with $\phi(t_{1})=a_{1}$. Next, we write
$R[t_{1},t_{2};\sigma_{1},\sigma_{2}]=(R[t_{1};\sigma_{1}])[t_{2};\sigma_{2}]$
(using [23, Theorem 2.4], as mentioned before) by extending $\sigma_{2}$ to an
automorphism of $R[t_{1};\sigma_{1}]$ given by $\sigma_{2}(a_{1})=a_{1}$. It
follows that $a_{2}r=\sigma_{2}(r)a_{2}$ for all $r\in R[a_{1};\sigma_{1}]$.
Applying [9, Proposition 2.4, p. 37] again, we extend $\phi$ to a homomorphism
from $R[t_{1},t_{2};\sigma_{1},\sigma_{2}]$ to $S$ with $\phi(t_{2})=a_{2}$.
We continue this process until we reach a ring homomorphism $\phi$ from the
skew polynomial ring $R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ to
$S$ such that $\phi(t_{i})=a_{i}$ for all $i$. Clearly, $\phi$ is uniquely
determined by the images of the variables $t_{i}$. ∎

Let $S/R$ be a ring extension and $A$ a subset of $S$. The smallest subring of
$S$ containing $R\cup A$, which will be denoted by $R[A]$, is the ring
generated by $R\cup A$ inside $S$. In case $A=\\{a_{1},\ldots,a_{n}\\}$ is a
finite set, we shall usually write $R[a_{1},\ldots,a_{n}]$ instead of
$R[\\{a_{1},\ldots,a_{n}\\}]$.

######  Proposition 2.6.

A ring extension $S/R$ is automorphically (centrally) finitely generated, if
and only if there exist commuting elements $a_{1},\ldots,a_{n}$ in $S$,
automorphic over $R$ (which centralize $R$) with respect to commuting
automorphisms $\sigma_{1},\ldots,\sigma_{n}$, such that
$S=R[a_{1},\ldots,a_{n}]$.

###### Proof.

Suppose first that $a_{1},\ldots,a_{n}\in S$ are automorphic commuting
elements, with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}$, such that $S=R[a_{1},\ldots,a_{n}]$. Apply
Lemma 2.5 to construct a homomorphism $\phi\colon
R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]\to S$ such that
$\phi(t_{i})=a_{i}$ for all $i$ and $\phi(r)=r$ for all $r\in R$. Consider the
set of all elements of $S$ of the form:

$\sum_{i_{1},\ldots,i_{n}\geq
0}r_{i_{1},\ldots,i_{n}}a_{1}^{i_{1}}\cdot\ldots\cdot a_{n}^{i_{n}},$

with finitely many non-zero $r_{i_{1},\ldots,i_{n}}\in R$. Since the $a_{i}$
are commuting, this set is clearly a subring of $S$ containing $R$ and
$a_{1},\ldots,a_{n}$, hence it equals $S=R[a_{1},\ldots,a_{n}]$. Thus every
element of $S$ is of the presented form, hence it is the image via $\phi$ of a
corresponding element

$\sum_{i_{1},\ldots,i_{n}\geq
0}r_{i_{1},\ldots,i_{n}}t_{1}^{i_{1}}\cdot\ldots\cdot t_{n}^{i_{n}}$

of $R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$. Thus $\phi$ is an
epimorphism onto $S$, hence

$S=R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]/\ker(\phi).$

Since $\phi$ fixes $R$ point-wise, we have $R\cap\ker(\phi)=\\{0\\}$. Thus $S$
is an automorphically finitely generated extension of $R$. Moreover, if
$a_{1},\ldots,a_{n}$ centralize $R$, then we may assume that
$\sigma_{1},\ldots,\sigma_{n}$ are the identity automorphism, hence
$S=R[t_{1},\ldots,t_{n}]/\ker(\phi)$, which means that $S$ is a centrally
finitely generated extension of $R$.

Conversely, if $S$ is an automorphically finitely generated extension of $R$,
then by definition there exist commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}$ of $R$, such that
$S=R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]/I$, for some ideal $I$
satisfying $R\cap I=\\{0\\}$ . Then we may view $R$ as a subring of $S$, via
the canonical map $r\mapsto r+I$. For each $1\leq i\leq n$, let
$a_{i}=t_{i}+I$. Then $a_{1},\ldots,a_{n}$ generates $S$ over $R$ and are
commuting, since $t_{1},\ldots,t_{n}$ are commuting. We have

$a_{i}r=(t_{i}+I)(r+I)=t_{i}r+I=r^{\sigma_{i}}t_{i}+I=r^{\sigma_{i}}a_{i}$

for all $1\leq i\leq n$ and $r\in R$. Thus $a_{1},\ldots,a_{n}$ are
automorphic elements over $R$, with respect to automorphisms
$\sigma_{1},\ldots,\sigma_{n}$. Moreover, if $\sigma_{1},\ldots,\sigma_{n}$
are the identity automorphism, then $a_{1},\ldots,a_{n}$ centralize $R$, by
the presented equality. ∎

######  Notation 2.7.

If $S/R$ is an automorphically finitely generated ring extension, let
$a_{1},\ldots,a_{n}\in S$ be commuting elements, automorphic over $R$, such
that $S=R[a_{1},\ldots,a_{n}]$, as in Proposition 2.6, with
$\sigma_{1},\ldots,\sigma_{n}$ being corresponding commuting automorphisms of
$R$. Then we write $S=R[a_{1},\ldots,a_{n};\sigma_{1},\ldots,\sigma_{n}]$. If
in addition $\sigma_{1}=\ldots=\sigma_{n}$, then we write
$S=R[a_{1},\ldots,a_{n};\sigma_{1}]$. If furthermore,
$\sigma_{1}=\ldots=\sigma_{n}=\mathbf{id}_{R}$ (the case where $S$ is
centrally finitely generated over $R$), then we write
$S=R[a_{1},\ldots,a_{n}]$.

Note that, with the above notation, we have
$S=R[a_{1},\ldots,a_{n};\sigma_{1},\ldots,\sigma_{n}]=R[a_{1},\ldots,a_{n}]$,
that is, the ring $S$ is generated over $R$ by the set
$\\{a_{1},\ldots,a_{n}\\}$. In particular, if $S$ is the skew polynomial ring
$R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$, then
$S=R[t_{1},\ldots,t_{n}]$ (indeed, as a ring, the skew polynomial ring $S$ is
generated by the variables $t_{1},\ldots,t_{n}$). The notation
$S=R[a_{1},\ldots,a_{n};\sigma_{1},\ldots,\sigma_{n}]$ includes the extra
information concerning the automorphisms that correspond to the automorphic
elements $a_{1},\ldots,a_{n}$.

######  Remark 2.8.

Let $S/R$ be a ring extension. If two non zero-divisors $a_{1},\,a_{2}\in S$
are automorphic over $R$ with respect to automorphisms
$\sigma_{1},\,\sigma_{2}$, respectively, and if $a_{1}$ commutes with $a_{2}$,
then $\sigma_{1}$ also commutes with $\sigma_{2}$. Indeed, for every $b\in R$,
we have

$b^{\sigma_{1}\circ\sigma_{2}}a_{1}a_{2}=(b^{\sigma_{2}})^{\sigma_{1}}a_{1}a_{2}=a_{1}a_{2}b=a_{2}a_{1}b=(b^{\sigma_{1}})^{\sigma_{2}}a_{2}a_{1}=b^{\sigma_{2}\circ\sigma_{1}}a_{2}a_{1}=b^{\sigma_{2}\circ\sigma_{1}}a_{1}a_{2},$

which implies that
$b^{\sigma_{1}\circ\sigma_{2}}=b^{\sigma_{2}\circ\sigma_{1}}$, since
$a_{1}a_{2}$ is also a non-zero divisor.

###  2.2 Substitution maps and algebraic independence

######  Definition 2.9.

Let $S/R$ be a ring extension. Let $a_{1},\ldots,a_{n}$ be $n$ commuting
elements of $S$, automorphic over $R$ with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}\in\mathbf{Aut}(R)$. Let

$\phi\colon R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]\to S$

be the homomorphism given by Lemma 2.5. We call $\phi$ a substitution map at
the point $a=(a_{1},\ldots,a_{n})$. For a polynomial $f\in
R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$, we define the
substitution of $f$ at $a=(a_{1},\ldots,a_{n})$, denoted by $f(a)$ or
$f(a_{1},\ldots,a_{n})$, to be the image $\phi(f)$ in $S$.

######  Example 2.10.

Let $S=R=D$ be a division algebra and let $a_{1},\ldots,a_{n}$ be commuting
elements of $D^{\times}$. Then each $a_{i}$ is automorphic over $D$ with
respect to the inner automorphism $\sigma_{i}=\mathbf{in}_{a_{i}}$ (see
Example 2.4), and these inner automorphisms are commuting. Thus we have a
substitution homomorphism $\phi\colon
D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]\to D$, given by
$\phi(f)=f(a_{1},\ldots,a_{n})=\sum
r_{\mathbf{i}}a_{1}^{i_{1}}\cdot\ldots\cdot a_{n}^{i_{n}}$ for all $f=\sum
r_{\mathbf{i}}t_{1}^{i_{1}}\cdot\ldots\cdot t_{n}^{i_{n}}\in
D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$. We note that in the case
of one variable, i.e. $n=1$, the substitution here coincides with the
$\sigma$-substitution $D[t_{1},\sigma_{1}]\to D$ studied in [12, §2]. Indeed,
this follows immediately from the fact that $\sigma_{1}(a_{1})=a_{1}$. We
note, however, that the $\sigma$-substitution considered in [12] is generally
not a homomorphism, while in our situation it is.

######  Example 2.11.

Let $\sigma_{1},\ldots,\sigma_{n}$ be commuting automorphisms of a division
algebra $D$, let $R=S=D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$. We
extend each $\sigma_{i}$ to an automorphism of $R$, determined by
$\sigma_{i}(t_{j})=t_{j}$ for all $i,j$. If $f_{1},\ldots,f_{n}\in R$ are
commuting and automorphic over $D$ with respect to automorphisms
$\sigma_{1},\ldots,\sigma_{n}$, then they induce the substitution map
$f\mapsto f(f_{1},\ldots,f_{n})$ from $R$ to $R$.

######  Remark 2.12.

In Definition 2.9, the requirement that $a_{1},\ldots,a_{n}\in S$ are
commuting and automorphic over $R$ with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}$, is essential. Indeed, suppose that
$a_{1},\ldots,a_{n}$ are some elements of $S$. Further Assume that there is a
ring homomorphism $\phi$ from
$R[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ to $S$ such that
$\phi(r)=r$ for all $r\in R$ and $\phi(t_{i})=a_{i}$ for all $i$. Then one
verifies directly that the elements $a_{1},\ldots,a_{n}\in S$ are commuting
and automorphic over $R$ with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}$.

######  Definition 2.13.

Let $S/R$ be a ring extension and let $a_{1},\ldots,a_{n}\in S$ be commuting
elements. We say that $a_{1},\ldots,a_{n}$ are (left) algebraically
independent over $R$ if the set

$\left\\{a_{1}^{i_{1}}\cdot\ldots\cdot
a_{n}^{i_{n}}\right\\}_{\mathbf{i}=(i_{1},\ldots,i_{n})\in(\mathbb{N}\cup\\{0\\})^{n}}$

of monomials in $a_{1},\ldots,a_{n}$ is (left) linearly independent over $R$.
Otherwise, we say that $a_{1},\ldots,a_{n}$ are (left) algebraically dependent
over $R$.

If $S$ is a commutative ring, then the above definition coincides with the
usual definition of algebraic independence.

######  Lemma 2.14.

Let $S$ be a ring extension of a division algebra $D$. Let
$a_{1},\ldots,a_{n}$ be $n$ commuting elements of $S$, automorphic over $D$,
with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}\in\mathbf{Aut}(D)$. Then $a_{1},\ldots,a_{n}$
are algebraically dependent over $D$ if and only if there exists a nonzero
polynomial $f\in D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ such that
$f(a_{1},\ldots,a_{n})=0$.

###### Proof.

The claim follows from the definitions: If $a_{1},\ldots,a_{n}$ are
algebraically dependent, then there are monomials $M_{1},\ldots,M_{k}$ in
$a_{1},\ldots,a_{n}$ and elements $b_{1},\ldots,b_{k}\in D$, not all zero,
such that $b_{1}M_{1}+\ldots+b_{k}M_{k}=0$. Consider now the homomorphism
$\phi\colon D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]\to S$ given by
Lemma 2.5. Since the $a_{i}$ commute, the monomials $M_{1},\ldots,M_{k}$ are
images via $\phi$ of respective monomials $N_{1},\ldots,N_{k}$ in
$t_{1},\ldots,t_{n}$, and the $N_{i}$ are distinct since the $M_{i}$ are
distinct. Thus $f=b_{1}N_{1}+\ldots+b_{k}N_{k}$ is a non-zero polynomial in
$D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$, and we have
$\phi(f)=\sum b_{i}M_{i}=0$.

Conversely, if $f\in D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ is a
non-zero polynomial then we may write it as $b_{1}N_{1}+\ldots+b_{k}N_{k}$,
where the $N_{i}$ are monomials in $t_{1},\ldots,t_{k}$ and the $b_{i}$ are
not all zero. If $\phi(f)=0$ then $\sum b_{i}\phi(N_{i})=0$, and so the
$\phi(N_{i})$ are left linearly dependent over $D$. But each $\phi(N_{i})$ is
a monomial expression in $a_{1},\ldots,a_{n}$, by the definition of $\phi$,
hence the $a_{i}$ are algebraically dependent. ∎

######  Lemma 2.15.

Let $S$ be a ring extension of a division algebra $D$. Let
$a_{1},\ldots,a_{n}$ be $n$ commuting elements of $S$, automorphic over $D$,
with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}\in\mathbf{Aut}(D)$. If $a_{1},\ldots,a_{n}$ are
algebraically independent over $D$, then the subring $D[a_{1},\ldots,a_{n}]$
of $S$ is isomorphic to the skew polynomial ring
$D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$.

###### Proof.

It follows from Lemma 2.5 and Definition 2.9 that the substitution map
$\phi\colon f\mapsto f(a_{1},\ldots,a_{n})$ from the skew polynomial ring
$R=D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ to $S$ is a ring
homomorphism. Since $a_{1},\ldots,a_{n}$ are algebraically independent over
$D$, $\mathbf{ker}(f)=0$. Thus, $\phi$ is injective. Hence, $\phi$ induces an
isomorphism from $R$ to its image $\mathbf{Im}(\phi)=D[a_{1},\ldots,a_{n}]$,
which is a subring of $S$. ∎

###  2.3 Normalizability

We now begin our study of the “normalizability” property for automorphically
finitely generated extensions of division algebras.

######  Definition 2.16.

Let $S$ be a ring extension of a division algebra $D$. We say that $S$ is
automorphically normalizable over $D$ if there exists a non-negative integer
$n$ and commuting elements $a_{1},\ldots,a_{n}\in S$, automorphic over $D$
with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}\in\mathbf{Aut}(D)$, such that:

  1. 1.

$a_{1},\ldots,a_{n}$ are left algebraically independent over $D$, and

  2. 2.

$S$ is finite as a left module over the subring $D[a_{1},\ldots,a_{n}]$ of
$S$.

If in addition $\sigma_{1}=\ldots=\sigma_{n}=\mathbf{id}_{D}$, that is if
$a_{1},\ldots,a_{n}$ centralize $D$, then we say that $S$ is centrally
normalizable over $D$.

Note that, in the above definition, the subring $D[a_{1},\ldots,a_{n}]$ of $S$
is isomorphic to the skew polynomial ring
$D[a_{1},\ldots,a_{n};\sigma_{1},\ldots,\sigma_{n}]$ (see Lemma 2.15). Thus,
the automorphically normalizable extensions of $D$ are precisely those which
are finite extensions of skew polynomial rings of the form
$D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$.

######  Definition 2.17.

Let $D$ be a division algebra. A tuple $(\sigma_{1},\ldots,\sigma_{n})$ of
commuting automorphisms of $D$ is called automorphically (resp. centrally)
normalizable over $D$ if every automorphically finitely generated extension of
the form $S=D[a_{1},\ldots,a_{n};\sigma_{1},\ldots,\sigma_{n}]$ of $D$ is
automorphically (resp. centrally) normalizable over $D$. Equivalently,
$(\sigma_{1},\ldots,\sigma_{n})$ is automorphically normalizable over $D$ if
every quotient ring of $D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$ by
a proper two-sided ideal is automorphically (resp. centrally) normalizable
over $D$.

######  Remark 2.18.

In the above definition, the normalizability of a tuple is clearly independent
of the order of the automorphisms. That is, if $\pi$ is a permutation in
$S_{n}$, then the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is automorphically
(centrally) normalizable over $D$ if and only if the tuple
$(\sigma_{\pi(1)},\ldots,\sigma_{\pi(n)})$ is automorphically (centrally)
normalizable over $D$.

We concludes this section with the following generalization of Zariski’s lemma
(Proposition 1.7 of the introduction):

######  Proposition 2.19.

Let $S$ be a ring extension of a division algebra $D$ such that $S$ is
automorphically normalizable over $D$. If $S$ is a division algebra, then $S$
is finite over $D$.

###### Proof.

Since $S$ is automorphically normalizable over $D$, there exist a non-negative
integer $n$ and commuting elements $a_{1},\ldots,a_{n}$ of $S$, automorphic
over $D$ with respect to commuting automorphisms
$\sigma_{1},\ldots,\sigma_{n}\in\mathbf{Aut}(D)$, such that: the subring
$R=D[a_{1},\ldots,a_{n}]$ of $S$ is isomorphic to the skew polynomial ring
$D[t_{1},\ldots,t_{n};\sigma_{1},\ldots,\sigma_{n}]$, and $S$ is finite over
$R$.

We claim that $n=0$. Assume by the contrary, that $n>0$, then $a_{1}$ is not
invertible in $R$ (but $a_{1}$ is invertible in $S$). By a noncommutative form
of Hilbert’s Basis Theorem [9, Theorem 1.17, page 17], it follows that $R$ is
left Noetherian. Therefore, by [21, Lemma 2.3], $S$ is integral over $R$. In
particular, the element $a_{1}^{-1}\in S$ is integral over $R$. Thus, there
exists a positive integer $k$ and elements $r_{0},\ldots,r_{k-1}\in R$ such
that

$a_{1}^{-k}+r_{k-1}a_{1}^{-(k-1)}+\ldots+r_{1}a_{1}^{-1}+r_{0}=0.$

By multiplying both sides by $a_{1}^{k-1}$, we obtain

$a_{1}^{-1}=-\left(r_{k-1}+\ldots+r_{1}a_{1}^{k-2}+r_{0}a^{k-1}\right),$

which is an element in $R$. This contradicts to the fact that $a_{1}$ is not
invertible in $R$. Thus $n=0$ and $S$ is finite over $R=D$. ∎

##  3 Normalizability of constant tuples

In the following, given a division algebra $D$, we denote by
$D[t_{1},\ldots,t_{d};\sigma]$ we mean the ring of skew polynomials,
$D[t_{1},\ldots,t_{d};\sigma,\ldots,\sigma]$, for any
$\sigma\in\mathbf{Aut}(D)$. In the case where $\sigma=\mathbf{id}_{D}$, we
will drop $\sigma$ and write $D[t_{1},\ldots,t_{d}]$ to denote the ring of
polynomials in $d$ central variables over $D$ (see Notation 2.7). Let
$D_{\sigma}$ be the fixed subring of $D$ via $\sigma$, i.e.
$D_{\sigma}=\\{r\in D\,|\,\sigma(r)=r\\}$. We also denote
$F=\mathcal{Z}(D)\cap D_{\sigma}$, where $\mathcal{Z}(D)$ the center of $D$.
The next result is a special case of [20, Theorem 1.2]. The proof here is a
simplified version of the proof of that theorem, which is simpler for the case
considered here. That proof, in turn, is an adaptation of Michalek’s proof of
the combinatorial Nullstellensatz, given in [17].

######  Lemma 3.1.

Let $D$ be a division algebra. Let $f\in D[t_{1},\ldots,t_{d}]$ be a nonzero
polynomial of total degree $m=\sum_{i=1}^{d}{k_{i}}$, where each $k_{i}$ is a
non-negative integer, such that the coefficient of
$t_{1}^{k_{1}}\cdot\ldots\cdot t_{d}^{k_{d}}$ in $p$ is non-zero. Let
$A_{1},\ldots,A_{d}$ be finite subsets of $F=\mathcal{Z}(D)$ such that
$|A_{i}|>k_{i}$ for $i=1,2,\ldots,d$. Then there is a point in
$A_{1}\times\dots\times A_{d}$ at which $f$ does not vanish.

###### Proof.

Note that elements in $F$ centralize $D$, therefore, the substitution of
points in $A_{1}\times\dots\times A_{d}$ in polynomials in
$D[t_{1},\ldots,t_{d}]$ is well-defined, according to Example 2.11. We will
prove the lemma by induction on the degree $m$ of $f$. In case $m=0$, $f$ is a
non-zero constant in $D$ and the assertion holds trivially.

Let us consider the case when $m>0$ and suppose that we have proven the claim
for all polynomials of degree smaller than $m$. Suppose that that $f$ vanishes
at all tuples in $A_{1}\times\dots\times A_{d}$. Without loss of generality,
we can assume that $k_{1}>0$. Choose $a_{1}\in A_{1}$ and apply right-hand
division with remainder to write

$f=g\cdot(t_{1}-a_{1})+h,\quad\text{with }h\in D[t_{2},\ldots,t_{n}].$ (1)

Since in $f$ there appears a monomial of the form $\lambda
x_{1}^{k_{1}}\cdot\ldots\cdot t_{d}^{k_{d}}$ with $\lambda\neq 0$, it follows
that in $g$ there appears a monomial of the form $\lambda
t_{1}^{k_{1}-1}\cdot\dots\cdot t_{n}^{k_{n}}$, and clearly
$\deg(g)=\deg(f)-1$.

Given an arbitrary point $a\in\\{a_{1}\\}\times A_{2}\times\dots\times
A_{d}\subseteq F^{d}$, since each coordinate of $a$ is automorphic over $D$
with respect to $\sigma$, we may substitute $a$ into Eq. (1) and get that
$f(a)=h(a)=0$. Therefore, $h$ vanishes on $A_{2}\times\dots\times A_{d}$.

Again, given any point
$b=(b_{1},a)=(b_{1},a_{2},\ldots,a_{d})\in\left(A_{1}\setminus\\{a_{1}\\}\right)\times
A_{2}\times\ldots\times A_{d}$, we may substitute it into Eq. (1) one more
time and get that $g(b)(b_{1}-a_{1})=f(b)-h(a)=0$, thus $g(b)=0$. Therefore,
$g$ vanishes on $\left(A_{1}\setminus\\{a_{1}\\}\right)\times
A_{2}\times\ldots\times A_{d}$. By the induction hypothesis, $g$ must be equal
to zero. But then $f=h\in D[t_{2},\ldots,t_{d}]$, which contradicts to the
assumption that $f$ contains a monomial of the form $\lambda
x_{1}^{k_{1}}\cdot\ldots\cdot t_{d}^{k_{d}}$ with $\lambda\neq 0$ and
$k_{1}>0$. Thus, $f$ cannot vanish at all tuples in $A_{1}\times\ldots\times
A_{d}$. ∎

######  Lemma 3.2.

Let $D$ be a division algebra. Let $f\in D[t_{1},\ldots,t_{d}]$ be a non-zero
polynomial. If $F=\mathcal{Z}(D)\cap D_{\sigma}$ is infinite, then there exist
$a_{1},\ldots,a_{d}\in F$ such that $f(a_{1},\ldots,a_{d})\neq 0$.

###### Proof.

Let $\lambda t_{1}^{k_{1}}\cdot\ldots\cdot t_{d}^{k_{d}}$ with be a monomial
appearing in $f$ with $\lambda\neq 0$ and $\deg(f)=k_{1}+\ldots+k_{d}$. Choose
any subsets $A_{1},\ldots,A_{d}$ of $F$ such that $|A_{i}|>k_{i}$ for each
$i$, and apply Lemma 3.1. ∎

######  Lemma 3.3.

Let $D$ be a division algebra. Let $f\in D[t_{1},\ldots,t_{d}]$ be a non-zero
homogeneous polynomial. If $F=\mathcal{Z}(D)\cap D_{\sigma}$ is infinite, then
there exist $a_{1},\ldots,a_{d-1}\in F$ such that
$f(a_{1},\ldots,a_{d-1},1)\neq 0$.

###### Proof.

Apply Lemma 3.2 for $f(t_{1},\ldots,t_{d-1},1)$. ∎

######  Lemma 3.4.

Let $D$ be a division algebra. Let $R=D[t_{1},\ldots,t_{d};\sigma]$ be a skew
polynomial ring, and let $a_{1},\ldots,a_{d-1}\in F=\mathcal{Z}(D)\cap
D_{\sigma}$. Then the map $\phi:R\to R$ defined by
$\phi(f(t_{1},\ldots,t_{d}))=f(t_{1}+a_{1}t_{d},\ldots,t_{d-1}+a_{d-1}t_{d},t_{d})$
is a ring homomorphism.

###### Proof.

For all $i=1,\ldots,n-1$ and $r\in D$, we have

$(t_{i}+a_{i}t_{d})r=r^{\sigma}t_{i}+a_{i}r^{\sigma}t_{d}=r^{\sigma}t_{i}+r^{\sigma}a_{i}t_{d}=r^{\sigma}(t_{i}+a_{i}t_{d}).$

Therefore, the elements $t_{1}+a_{1}t_{d},\ldots,t_{d-1}+a_{d-1}t_{d},t_{d}$
are all automorphic over $D$ with respect to $\sigma$. The claim then follows
by Example 2.11. ∎

######  Lemma 3.5.

Let $D$ be a division algebra. Let $f\in D[t_{1},\ldots,t_{d};\sigma]$ be a
nonzero skew polynomial. If $F=\mathcal{Z}(D)\cap D_{\sigma}$ is infinite,
then there exist $a\in D$ and $a_{1},\ldots,a_{d-1}$ in $F$ such that the
polynomial

$g=a\cdot f(t_{1}+a_{1}t_{d},\ldots,t_{d-1}+a_{d-1}t_{d},t_{d})$

is a monic polynomial in $t_{d}$ with coefficients in
$D[t_{1},\ldots,t_{d-1};\sigma]$. (Note that we may view
$D[t_{1},\ldots,t_{d};\sigma]$ as
$D[t_{1},\ldots,t_{d-1};\sigma][t_{d};\sigma]$, as discussed in the beginning
of §2.1).

###### Proof.

Assume that
$f=\sum\limits_{\mathbf{i}\in\mathbb{N}}b_{\mathbf{i}}t_{1}^{i_{1}}\cdot\ldots\cdot
t_{n}^{i_{d}}$ is of degree $m$. Let $a_{1},\ldots,a_{d-1}$ be elements in $F$
that we will determine later. According to Lemma 3.4, the substitution
$f(t_{1}+a_{1}t_{d},\ldots,t_{d-1}+a_{d-1}t_{d},t_{d})$ is well-defined. For
each $k=1,\ldots,d-1$, since $a_{k}$ is in $\mathcal{Z}(D)\cap D_{\sigma}$, we
have

$\displaystyle(t_{k}+a_{k}t_{d})^{i_{k}}$
$\displaystyle=(a_{k}t_{d})^{i_{k}}+{\text{ terms of lower degree in }i_{k}}$
$\displaystyle=a_{k}\cdot\sigma(a_{k})\cdot\ldots\cdot\sigma^{i_{k}-1}(a_{k})t_{n}^{i_{k}}+{\text{
terms of lower degree in }i_{k}}$
$\displaystyle=a_{k}^{i_{k}}t_{n}^{i_{k}}+{\text{ terms of lower degree in
}i_{k}}.$

Therefore,

$\displaystyle f(t_{1}+a_{1}t_{d},$
$\displaystyle\ldots,t_{d-1}+a_{d-1}t_{d},t_{d})=\sum\limits_{\mathbf{i}}b_{\mathbf{i}}\left(\prod\limits_{k=1}^{d-1}(t_{k}+a_{k}t_{d})^{i_{k}}\right)\cdot
t_{d}^{i_{d}}$
$\displaystyle=\sum\limits_{\mathbf{i}}b_{\mathbf{i}}\left(\prod\limits_{k=1}^{d-1}\left(a_{k}^{i_{k}}t_{d}^{i_{k}}+{\text{
terms of lower degree in }t_{d}}\right)\right)\cdot t_{d}^{i_{d}}$
$\displaystyle=\sum\limits_{\mathbf{i}}b_{\mathbf{i}}\left(\prod\limits_{k=1}^{d-1}a_{k}^{i_{k}}\right)\cdot
t_{d}^{i_{1}+\ldots+i_{d}}+{\text{ terms of lower degree in }t_{d}}$
$\displaystyle=f_{m}(a_{1},\ldots,a_{d-1},1)t_{d}^{m}+\text{ terms of lower
degrees in }t_{d}.$

Here,
$f_{m}=\sum\limits_{\mathbf{i}}b_{\mathbf{i}}\left(\prod\limits_{k=1}^{d-1}u_{k}^{i_{k}}\right)$
is a homogeneous polynomial in the central polynomial ring
$D[u_{1},\ldots,u_{d}]$ (thus the substitution $f_{m}(a_{1},\ldots,a_{d-1},1)$
is well-defined). Note that $f_{m}$ is identical to the homogeneous part of
degree $m$ of $f$, except that the variables in $f_{m}$ centralize $D$. By
Lemma 3.3, there exist nonzero elements $a_{1},\ldots,a_{d-1}$ in $F$ such
that $f_{m}(a_{1},\ldots,a_{d-1},1)\neq 0$. Set
$a=f_{m}(a_{1},\ldots,a_{d-1},1)^{-1}$. Then the polynomial $g=a\cdot
f(t_{1}+a_{1}t_{d},\ldots,t_{d-1}+a_{d-1}t_{d},t_{d})$ is monic in $t_{d}$. ∎

######  Lemma 3.6.

Let $D$ be a division algebra and let $D\subseteq S\subseteq T$ be ring
extensions. If $S$ is automorphically (centrally) normalizable over $D$ and if
$T$ is finite over $S$, then $T$ is also automorphically (centrally)
normalizable over $D$.

###### Proof.

Since $S$ is automorphically (centrally) normalizable over $D$, there exist
commuting elements $a_{1},\ldots,a_{n}$ in $S$ that are automorphic over $D$
(which centralize $D$), algebraically independent over $D$, and such that $S$
is finite over $R=D[a_{1},\ldots,a_{n}]$. Since $T$ is finite over $S$ and $S$
is finite over $R$, $T$ is also finite over $R$. Thus $T$ is automorphically
(centrally) normalizable over $D$. ∎

We can now prove Theorem 1.4 of the introduction:

######  Theorem 3.7.

Let $D$ be a division algebra and let $\sigma$ be an automorphism of $D$ such
that the field $F=\mathcal{Z}(D)\cap D_{\sigma}$ is infinite. Then any
constant tuple $(\sigma,\ldots,\sigma)$ is automorphically normalizable over
$D$.

###### Proof.

Let $S=D[z_{1},\ldots,z_{n};\sigma]$ be an arbitrary automorphic extension of
$D$. We must prove that $S$ is automorphically normalizable over $D$. We will
do so by induction on $n$. For $n=0$ we have $S=D$ and there is nothing to
prove.

Assume we have proven the claim up to $n-1$, and let us prove it for $n$.
Since $z_{1},\ldots,z_{n}$ are automorphic over $D$ with respect to $\sigma$
over $D$, the substitution map $f\mapsto f(z_{1},\ldots,z_{n})$ for skew
polynomials in $D[t_{1},\ldots,t_{n};\sigma]$ is well-defined and is a ring
homomorphism, by Lemma 2.5. If the elements $z_{1},\ldots,z_{n}$ are
algebraically independent over $D$, then it follows from Lemma 2.15 that $S$
is isomorphic to the skew polynomial ring $D[t_{1},\ldots,t_{n};\sigma]$. Thus
$S$ is automorphically normalizable over $D$ (by choosing $d=n$ and
$a_{i}=z_{i}$ in Definition 2.16).

Let us consider the case when the elements $z_{1},\ldots,z_{n}$ are
algebraically dependent over $D$. Then there is a nonzero skew polynomial
$f\in D[t_{1},\ldots,t_{n};\sigma]$ such that $f(z_{1},\ldots,z_{n})=0$. By
Lemma 3.5, there exist an element $a\in D$, a tuple $(b_{1},\ldots,b_{n-1})\in
F^{n-1}$, and a non-negative integer $m$ such that
$g=af(t_{1}+b_{1}t_{n},\ldots,t_{n-1}+b_{n-1}t_{n},t_{n})$ is a monic
polynomial in $t_{n}$, of the form

$g=t_{n}^{m}+\text{ terms in }(t_{1},\ldots,t_{n})\text{ of lower degree in
}t_{n}.$ (2)

Note that, since the elements
$t_{1}+b_{1}t_{n},\ldots,t_{n-1}+b_{n-1}t_{n},t_{n}$ are commuting and
automorphic over $D$ with respect to $\sigma$, the substitution of $f$ at
$(t_{1}+b_{1}t_{n},\ldots,t_{n-1}+b_{n-1}t_{n},t_{n})$ is well-defined (see
Example 2.11).

Set
$\tilde{z}_{1}=z_{1}-b_{1}z_{n},\ldots,\tilde{z}_{n-1}=z_{n-1}-b_{n-1}z_{n}$
and $\tilde{z}_{n}=z_{n}$. Then the elements $\tilde{z}_{i}$ are commuting and
automorphic over $D$ with respect to $\sigma$. We can evaluate the skew
polynomial $g$ at these elements and obtain

$g(\tilde{z}_{1},\ldots,\tilde{z}_{n})=af(\tilde{z}_{1}+b_{1}\tilde{z}_{n},\ldots,\tilde{z}_{n-1}+b_{n-1}\tilde{z}_{n},\tilde{z}_{n})=af(z_{1},\ldots,z_{n})=0.$

Therefore, by substituting $(t_{1},\ldots,t_{n})$ at
$(\tilde{z}_{1},\ldots,\tilde{z}_{n})$ in Eq. (2), we get that

$0=\tilde{z}_{n}^{m}+\text{ terms in
}(\tilde{z}_{1},\ldots,\tilde{z}_{n})\text{ of lower degree in
}\tilde{z}_{n}.$

Thus $\tilde{z}_{n}$ is left integral over the sub-algebra
$R=D[\tilde{z}_{1},\ldots,\tilde{z}_{n-1}]$ of $S$. As a consequence,
$S=D[z_{1},\ldots,z_{n}]=D[\tilde{z}_{1},\ldots,\tilde{z}_{n}]$ is finite over
$R$. By the induction hypothesis, $R$ is automorphically normalizable over
$D$. It follows from Lemma 3.6 that $S$ is also automorphically normalizable
over $D$. ∎

We can now prove Corollary 1.5 of the introduction:

######  Corollary 3.8.

Let $D$ be a division algebra and $\sigma_{1},\ldots,\sigma_{n}$ be commuting
elements in $\mathbf{Aut}(D)$ such that
$F=\mathcal{Z}(D)\cap\big{(}\cap_{i=1}^{n}D_{\sigma_{i}}\big{)}$ is infinite.
If $\sigma_{1}^{d_{1}}=\ldots=\sigma_{n}^{d_{n}}$ for some positive integers
$d_{1},\ldots,d_{n}$, then the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is
automorphically normalizable over $D$.

In particular, if $\sigma_{1},\ldots,\sigma_{n}$ are of finite orders, then
the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is centrally normalizable over $D$.

###### Proof.

Let $S=D[z_{1},\ldots,z_{n};\sigma_{1},\ldots,\sigma_{n}]$ be an arbitrary
automorphic extension of $D$. Set $\sigma=\sigma_{1}^{d_{1}}$, and for each
$i=1,\ldots,n$, we set $w_{i}=z_{i}^{d_{i}}$. Then for every $b\in D$, we have

$\displaystyle
w_{i}b=z_{i}^{d_{i}}b=\sigma_{i}^{d_{i}}(b)z_{i}^{d_{i}}=\sigma(b)w_{i}.$

Therefore, $w_{i}$ is automorphic over $D$ with respect to $\sigma$. Since,
the elements $z_{i}$ are commuting, the elements $w_{i}$ are also commuting.
Thus, the ring $R=D[w_{1},\ldots,w_{n};\sigma]$ is automorphically finitely
generated over $D$. Note that $F\subseteq\mathcal{Z}(D)\cap D_{\sigma}$.
Therefore, $\mathcal{Z}(D)\cap D_{\sigma}$ is an infinite set. It follows from
Theorem 3.7 that $R$ is automorphic normalizable over $D$. Observe that $S$ is
finite over $R$, therefore, by Lemma 3.6, $S$ is automorphically normalizable
over $D$. Hence, the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is automorphically
normalizable over $D$. ∎

The above corollary gives a sufficient condition for a tuple of automorphisms
in $\mathbf{Aut}(D)$ to be automorphically normalizable over $D$. The question
whether this sufficient condition is “necessary” is still open. In Theorem
5.13 below, we will prove that this condition is indeed “necessary” in the
case where $D=F$ is a field.

##  4 Tuples of identity maps and central extensions

In Theorem 3.7, if the tuple of automorphisms is the identity tuple
$(\mathbf{id}_{D},\ldots,\mathbf{id}_{D})$, (i.e. $\sigma=\mathbf{id}_{D}$),
we will prove that the “infinite” condition can be dropped

######  Lemma 4.1.

Let $D$ be a division algebra, let $R=D[t_{1},\ldots,t_{n}]$, and let
$d_{1},\ldots,d_{n-1}$ be non-negative integers. Then the map $\phi:R\to R$
defined by
$\phi(f(t_{1},\ldots,t_{n}))=f(t_{1}+t_{n}^{d_{1}},\ldots,t_{d-1}+t_{n}^{d_{n-1}},t_{n})$
is a ring homomorphism.

###### Proof.

This follows directly from Example 2.11. ∎

We will also need the following lemma, which is similar to [18, (41.1), page
44] and [6, Lemma 13.2].

######  Lemma 4.2.

Let $D$ be a division algebra and let $f\in D[t_{1},\ldots,t_{n}]$ be a
nonzero polynomial. Set $d=1+\deg f$ where $\deg f$ is the total degree of
$f$. Then there exists $a\in D$ such that the polynomial

$g=a\cdot
f\left(t_{1}+t_{n}^{d^{n-1}},t_{2}+t_{n}^{d^{n-2}},\ldots,t_{n-1}+t_{n}^{d},t_{n}\right)$

is a monic polynomial in $t_{n}$ with coefficients in
$D[t_{1},\ldots,t_{n-1}]$.

###### Proof.

We write $f=\sum_{I\in\mathcal{S}}a_{I}t_{1}^{i_{1}}t_{2}^{i_{2}}\ldots
t_{n}^{i_{n}}$, where $\mathcal{S}\subset\mathbb{N}^{n}$ is the support of $f$
and each $I=(i_{1},\ldots,i_{n})\in\mathcal{S}$ is a multi-index. Since
$t_{1}+t_{n}^{d^{n-1}},t_{2}+t_{n}^{d^{n-2}},\ldots,t_{n-1}+t_{n}^{d}$ are
elements in $F[t_{1},\ldots,t_{n}]$, we can apply the substitution

$\displaystyle
f\left(t_{1}+t_{n}^{d^{n-1}},t_{2}+t_{n}^{d^{n-2}},\ldots,t_{n-1}+t_{n}^{d},t_{n}\right)$
$\displaystyle\hskip
56.9055pt=\sum\limits_{I\in\mathcal{S}}\left(a_{I}t_{n}^{i_{1}d^{n-1}+i_{2}d^{n-2}+\ldots+i_{n-1}d+i_{n}}+\text{
terms of lower degrees in }t_{n}\right).$

We see that the coordinates of $I$ are non-negative integers between $0$ and
$d-1$, and the exponent $i_{1}d^{n-1}+i_{2}d^{n-2}+\ldots+i_{n-1}d+i_{n}$ in
the right hand side of the above expression is exactly the $d$-adic
representation. Therefore, among $I$ in $\mathcal{S}$, there is only one
multi-index, say $I_{0}$, that reaches the maximal exponent. Thus
$a_{I_{0}}^{-1}f\left(t_{1}+t_{n}^{d^{n-1}},t_{2}+t_{n}^{d^{n-2}},\ldots,t_{n-1}+t_{n}^{d},t_{n}\right)$
is a monic polynomial in $t_{n}$. ∎

We can now prove Theorem 1.1 of the introduction:

######  Theorem 4.3.

Let $D$ be a division algebra. The tuple
$(\mathbf{id}_{D},\ldots,\mathbf{id}_{D})$ of identity maps is centrally
normalizable over $D$. Equivalently, every centrally finitely generated
extension over $D$ is centrally normalizable over $D$.

###### Proof.

The proof is the same as the proof of Theorem 3.7, except using the
transformation $t_{i}\mapsto t_{i}+t_{n}^{r^{i}}$ and Lemma 4.2 instead of the
transformation $t_{i}\mapsto t_{i}+b_{i}t_{n}$ for $i=1,\ldots,n-1$ and Lemma
3.5 (here $r=1+\deg f$). ∎

In the case where $D$ is a centrally finite division algebra (i.e. $D$ is
finite over its center $F=\mathcal{Z}(D)$), and where $S$ is a subring of a
ring of central polynomials over $D$, one can further drop the “central”
condition in the hypothesis of Theorem 4.3 (see Proposition 4.6 below).

######  Lemma 4.4.

Let $D$ be a centrally finite division algebra with center $F$, and let
$R=D[t_{1},\ldots,t_{n}]$. For any polynomial $p\in R$, there exist
$p_{1},\ldots,p_{m}\in F[t_{1},\ldots,t_{n}]$ such that
$D[p]=D[p_{1},\ldots,p_{m}]$ inside $R$.

###### Proof.

Write $p=\sum_{I}p_{I}t^{I}$, where $t=(t_{1},\ldots,t_{n})$ and $I$ is a
multi-index. Let $v_{1},\ldots,v_{m}$ be a basis for $D$ over $F$ and write
$p_{I}=\sum_{i=1}^{m}p_{I,i}v_{i}$ with $\\{p_{I,i}\\}\subseteq F$. Let
$p_{i}=\sum p_{I,i}x^{I}\in F[t_{1},\ldots,t_{n}]$, $1\leq i\leq m$. Since the
variables $t_{1},\ldots,t_{n}$ are central in $R$, we have

$p=\sum_{I}\big{(}\sum_{i}p_{I,i}v_{i}\big{)}t^{I}=\sum_{i}\big{(}\sum_{I}p_{I,i}t^{I}\big{)}v_{i}=\sum_{i}p_{i}v_{i}.$

Hence $p\in D[p_{1},\ldots,p_{m}]$.

By [2, Corollary 4] (a result previously discovered by Wilczyński [24, Theorem
4.1]), there exist elements $b_{st}^{i}\in F$, $1\leq i,s,t,\leq m$, such that

$p_{I,i}=\sum_{s,t=1}^{m}b_{st}^{i}v_{s}p_{I}v_{t}.$

We thus have

$p_{i}=\sum_{I}p_{I,i}t^{I}=\sum_{I}\sum_{s,t}b_{st}^{i}v_{s}p_{I}t^{I}v_{t}=\sum_{s,t}b_{s,t}^{i}v_{s}(\sum_{I}p_{I}t^{I})v_{t}=\sum_{s,t}b_{s,t}^{i}v_{s}pv_{t}.$

Hence $p_{i}\in D[p]$ for $1\leq i\leq m$. ∎

######  Corollary 4.5.

Let $D$ be a centrally finite division algebra, let $R=D[t_{1},\ldots,t_{n}]$
and let $A$ be a finite set of polynomials in $R$, and let $S=D[A]$. Then $S$
is centrally finitely generated over $D$.

###### Proof.

Write $A=\\{p_{1},\ldots,p_{k}\\}$. By Proposition 4.4, there exist
polynomials $p_{ij}\in F[t_{1},\ldots,t_{n}]$, such that
$D[p_{i}]=D[p_{i1},\ldots,p_{im}]$. It follows that
$S=D[p_{i1},\ldots,p_{im},\ldots,p_{k1},\ldots,p_{km}]$. ∎

From Corollary 4.5 we can deduce Noether’s normalization for finitely
generated subrings of rings of polynomials in central variables over a
centrally finite division algebra $D$ (Theorem 1.3 of the introduction). The
key point in the proof is that such rings must be centrally finitely generated
over $D$ (which follows from Lemma 4.4).

######  Proposition 4.6.

Let $D$ be a centrally finite division algebra and let
$R=D[t_{1},\ldots,t_{n}]$. Then for every finite subset $A$ of $R$, the ring
$S=D[A]$ is centrally normalizable over $D$.

###### Proof.

By Corollary 4.5, $S$ is centrally finitely generated over $D$, hence the
claim follows from Theorem 4.3. ∎

A simple application of Theorem 4.3 is the following Nullstellensatz for two-
sided ideals in $\mathbb{H}[t_{1},\ldots,t_{n}]$, where $\mathbb{H}$ is the
real quaternion algebra. The proof is similar to the proof via Zariski’s lemma
of the classical Nullstellensatz for algebraically closed fields.

######  Proposition 4.7.

Let $R=\mathbb{H}[t_{1},\ldots,t_{n}]$ the ring of polynomials in $n$ central
variables over $\mathbb{H}$. Let $M$ be a two-sided ideal in $R$, which is
maximal as a left ideal. Then there exists $a_{1},\ldots,a_{n}\in\mathbb{R}$
such that $t_{1}-a_{1},\ldots,t_{n}-a_{n}$ generate $M$.

###### Proof.

The assumption that $M$ is maximal as a left ideal means that $R/M$ is a
division algebra, which contains $\mathbb{H}$ as a subring. Moreover, $R/M$ is
centrally finitely generated over $\mathbb{H}$ by $t_{1}+M,\ldots,t_{n}+M$. It
follows from Proposition 4.3 that $R/M$ is centrally normalizable over
$\mathbb{H}$. Thus, by Proposition 2.19, $R/M$ is a finite extension of
$\mathbb{H}$. This implies that $R/M=\mathbb{H}$, by the theorem of Frobenius,
and hence there exist $a_{1},\ldots,a_{n}\in\mathbb{H}$ such that
$t_{1}-a_{1},\ldots,t_{n}-a_{n}\in M$. By [1, Lemma 2.1] and [1, Proposition
2.2] (the proof of both of these claims is short and elementary), this means
that $a_{i}a_{j}=a_{j}a_{i}$ for all $1\leq i,j\leq n$ and that $M$ is
generated by $t_{1}-a_{1},\ldots,t_{n}-a_{n}$. Finally, since $M$ is two-sided
ideal, we must have $a_{1},\ldots,a_{n}\in\mathbb{R}$. Indeed, if say
$a_{1}\notin\mathbb{R}$, then there exist $0\neq b\in\mathbb{H}$ such that
$a_{1}b\neq ba_{1}$. Then
$(t_{1}-a_{1})b=bt_{1}-a_{1}b=b(t_{1}-b^{-1}a_{1}b)\in M$, hence
$(t_{1}-b^{-1}a_{1}b)\in M$. But this means that $t_{1}-b^{-1}a_{1}b$ must
vanish at $a_{1}$, that is, $a_{1}=b^{-1}a_{1}b$, hence $ba_{1}=a_{1}b$. ∎

We note that [1, Proposition 2.6] goes further and characterizes all maximal
left ideals in $\mathbb{H}[t_{1},\ldots,t_{n}]$, not only the two-sided ones.

##  5 Normalizability of general tuples

In the previous sections, we have shown that constant tuples of automorphisms
that are automorphically normalizable over a division algebra. In this
section, we consider arbitrary tuples (of finite length) and give some
negative examples. In the case where our base ring $D$ is a field, we give a
necessary and sufficient condition for such a tuple to be automorphically
normalizable.

###  5.1 Basic properties

######  Proposition 5.1.

If a tuple $(\sigma_{1},\ldots,\sigma_{n})$ of commuting automorphisms in
$\mathbf{Aut}(D)$ is not automorphically normalizable over $D$, then the tuple
$(\sigma_{1}^{d_{1}},\ldots,\sigma_{n}^{d_{n}})$ is also not automorphically
normalizable over $D$, for any sequence of positive integers
$d_{1},\ldots,d_{n}$.

###### Proof.

Assume that the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is not automorphically
normalizable over $D$. Then there exists an automorphic extension
$S=D[z_{1},\ldots,z_{n};\sigma_{1},\ldots,\sigma_{n}]$ of $D$ that is not
automorphically normalizable over $D$. Observe that
$R=D[z_{1}^{d_{1}},\ldots,z_{n}^{d_{n}};\sigma_{1}^{d_{1}},\ldots,\sigma_{n}^{d_{n}}]$
is an automorphic extension of $D$ and $S$ is finite over $R$ (with the left
linear basis being the set of monomials in $z_{1},\ldots,z_{n}$ of degrees at
most $d_{1},\ldots,d_{n}$, respectively). Therefore, according to Lemma 3.6,
$R$ is not automorphically normalizable over $D$. As a consequence, the tuple
$(\sigma_{1}^{d_{1}},\ldots,\sigma_{n}^{d_{n}})$ is not automorphically
normalizable over $D$. ∎

######  Proposition 5.2.

Let $\sigma_{1},\ldots,\sigma_{n}$ be commuting automorphisms in
$\mathbf{Aut}(D)$. If the tuple $(\sigma_{1},\ldots,\sigma_{r})$ is not
automorphically normalizable over $D$ for some a integer $r\leq n$, then so is
the tuple $\Sigma=(\sigma_{1},\ldots,\sigma_{n})$.

###### Proof.

Let $S=D[z_{1},\ldots,z_{r};\sigma_{1},\ldots,\sigma_{r}]$ be an automorphic
extension of $D$ which is not automorphically normalizable over $D$. Set
$z_{r+1}=\ldots=z_{n}=0$. Then the ring
$D[z_{1},\ldots,z_{n};\sigma_{1},\ldots,\sigma_{n}]=S$ is of course an
automorphic extension of $D$ and it is not automorphically normalizable over
$D$. Therefore, $\Sigma$ is not automorphically normalizable over $D$. ∎

By Propositions 5.1 and 5.2, if we have identified a tuple of commuting
automorphisms of $D$ that is not automorphically normalizable over $D$, we can
generate infinitely many additional tuples that are not automorphically
normalizable over $D$ by prolonging the tuple, or by introducing positive
powers to each automorphism in the tuple. However, we have not seen any
negative examples to normalizability thus far. In the following, we provide
two such examples, in Propositions 5.7 and 5.11.

###  5.2 First negative example - the tuple $(\sigma,\sigma^{-1})$

Let $D$ be a division algebra. Recall that for a nonzero element $c\in D$, the
notation $\mathbf{in}_{c}$ stands for the inner automorphism of $D$ with
respect to $c$ (Example 2.4).

######  Definition 5.3 (see [11, p. 43]).

An automorphism $\sigma\in\mathbf{Aut}(D)$ is called of finite inner order if
$\sigma^{k}$ is an inner automorphism for some positive integer $k$ (i.e.
$\sigma^{k}=\mathbf{in}_{c}$ for some nonzero element $c\in D$). Otherwise, we
say that $\sigma$ is of infinite inner order.

A finite-order automorphism of $\mathbf{Aut}(D)$ is always of finite inner
order (with the inner automorphism to be the identity map), but the converse
does not always hold:

######  Example 5.4.

Let $D=\mathbb{H}$ be the real quaternion algebra. Let $\alpha$ be an
irrational number and set $u=e^{\mathbf{i}\alpha}$, where $\mathbf{i}$ is the
imaginary unit of $\mathbb{C}$. Then for any integer $k$, the number
$u^{k}=e^{\mathbf{i}k\alpha}$ is not a real number. Consider the inner
automorphism $\sigma=\mathbf{in}_{u}$ of $\mathbb{H}$, i.e.
$\sigma(a)=u^{-1}au$ for any $a\in\mathbb{H}$. Of course, $\sigma$ is of
finite inner order (taking $n=1$ in the above definition). However, $\sigma$
is not of finite order. Indeed, if there was an integer $n$ such that
$\sigma^{n}=\mathbf{in}_{u^{n}}$ is the identity, then $u^{n}$ would belong to
the center $\mathcal{Z}(\mathbb{H})=\mathbb{R}$, a contradiction.

######  Lemma 5.5.

Let $S/D$ be a ring extension. Assume that elements $f_{1},\ldots,f_{d}\in S$
are automorphic and left linearly independent over $D$. Then the sum
$f=f_{1}+\ldots+f_{d}$ is automorphic over $D$ if and only if
$f,f_{1},\ldots,f_{d}$ are automorphic over $D$ with respect to the same
automorphism.

###### Proof.

Assume that $f_{i}$ is automorphic over $D$ with respect to
$\pi_{i}\in\mathbf{Aut}(D)$ for each $i=1,\ldots,d$. The “if” part is trivial.

For the “only if” part, let us suppose that $f$ is automorphic over $D$ with
respect to $\pi\in\mathbf{Aut}(D)$. Then, for every $b\in D$, we have

$\displaystyle 0=fb-\pi(b)f$
$\displaystyle=(f_{1}+\ldots+f_{d})b-\pi(b)(f_{1}+\ldots+f_{d})$
$\displaystyle=(\pi_{1}(b)-\pi(b))f_{1}+\ldots+(\pi_{n}(b)-\pi(b))f_{n}.$

Since $f_{i}$s are left linearly independent over $D$, we must have
$\pi_{1}(b)=\pi(b)$ for all $i$. Thus, $\pi_{i}=\pi$ for all $i$. In other
words, $f,f_{1},\ldots,f_{d}$ are automorphic over $D$ with respect to the
same automorphism $\pi$. ∎

We will now give the first negative example for the normalizability of an
automorphic extension of a division algebra. This example is chosen to be a
subring of a division algebra $D(t;\sigma)$, which is the (left) fraction
division algebra of the skew polynomial ring $D[t;\sigma]$. We note that that
this (left) fraction division algebra exists since skew polynomial rings are
Ore domains (see [9, Corollary 6.7]).

######  Lemma 5.6.

Let $\sigma\in\mathbf{Aut}(D)$ be an automorphism. Let
$S=D[t,t^{-1};\sigma,\sigma^{-1}]$ be the subring generated by $t$ and
$t^{-1}$ of the division algebra $D(t;\sigma)$. Then $S$ is automorphically
normalizable over $D$ if and only if $\sigma$ is of finite inner order.

###### Proof.

Let us assume that $\sigma$ is of finite inner order, i.e. there exist a
positive integer $k$ such that $\sigma^{k}=\mathbf{in}_{c}$, where
$\mathbf{in}_{c}$ is the inner automorphism of $D$ with respect to a nonzero
element $c\in D$ (see Example 2.4). We will prove that $S$ is automorphically
normalizable over $D$. Indeed, let $u=t^{-k}+c^{-2}t^{k}$ be an element in
$S$. It is clear that $u$ is algebraically independent over $D$. In addition,
for every $b\in D$, we have

$\displaystyle ub$
$\displaystyle=t^{-k}b+c^{-2}t^{k}b=\sigma^{-k}(b)t^{-k}+c^{-2}\sigma^{k}(b)t^{k}$
$\displaystyle=c^{-1}bct^{-k}+c^{-1}bc^{-1}t^{k}=c^{-1}bc(t^{-k}+c^{-2}t^{k})=\mathbf{in}_{c^{-1}}(b)u.$

This means that $u$ is automorphic over $D$ with respect to the inner
automorphism $\mathbf{in}_{c^{-1}}$. Furthermore, we see that
$c^{-2}t^{2k}-ut^{k}+1=0$. Thus, $t$ is left integral over $D[u]$, and
$D[t,t^{-k}]=D[t,u]$ is finite over $D[u]$. Since $S=D[t,t^{-1}]$ is finite
over $D[t,t^{-k}]$, $S$ is also finite over $D[u]$. Hence, $S$ is
automorphically normalizable over $D$.

Next, let us assume that $\sigma$ is of infinite inner order. We will prove
that $S$ cannot be automorphically normalizable over $D$. Assume that there
exist a non-negative integer $d$ and commuting elements $a_{1},\ldots,a_{d}$
in $S$ that are algebraically independent over $D$, automorphic over $D$ with
respect to a commuting automorphisms
$\sigma_{1},\ldots,\sigma_{d}\in\mathbf{Aut}(D)$, and such that $S$ is finite
over the subring $R=D[a_{1},\ldots,a_{d};\sigma_{1},\ldots,\sigma_{d}]$, and
$R$ is a skew polynomial ring over $D$.

First, we claim that an element $a$ in $S$ is automorphic over $D$ only if it
is of the form $a=bt^{k}$ for some $b\in D$ and an integer $k$. Indeed, as an
element of $S$, $a$ must be of the form $a=\sum_{k\in\mathbb{Z}}b_{k}t^{k}$
for some $b_{k}\in D$, where the sum is taken with finitely many non-zero
coefficients. Each nonzero summand $b_{k}t^{k}$ in this presentation of $a$ is
automorphic over $D$ with respect to the automorphism
$\mathbf{in}_{b_{k}}\circ\sigma^{k}$. Now let us assume that $a$ is
automorphic over $D$. According to Lemma 5.5, all nonzero summands appearing
in $a$ must be automorphic over $D$ with respect to the same automorphism.
However, since $\sigma$ is of infinite inner order, we have
$\mathbf{in}_{b_{k}}\circ\sigma^{i}\neq\mathbf{in}_{b_{j}}\circ\sigma^{j}$ for
all integers $i\neq j$ such that $b_{i}$ and $b_{j}$ are both nonzero. Thus
there can be only one non-zero summand in the given presentation of $a$, that
is $a=b_{k}t^{k}$ for some integer $k$, as claimed.

Thus, since $a_{1},\ldots,a_{d}$ are automorphic over $D$, each $a_{i}$ is of
the form $b_{i}t^{k_{i}}$ with $b_{i}\in D$. Then, on the one hand, $d$ must
be nonzero, since $S$ is infinite over $D$. On the other hand, $d$ cannot be
greater than $1$, since any two such nonzero terms
$b_{i}t^{k_{i}},b_{j}t^{k_{j}}$ are algebraically dependent over $D$ (recall
that $R=D[a_{1},\ldots,a_{d};\sigma_{1},\ldots,\sigma_{d}]$ is a skew
polynomial ring over $D$) . Thus $d=1$.

Suppose that $S$ is finite over $R=D[a_{1}]$. Write $a_{1}=bt^{k}\in S$ for
some nonzero element $b\in D$ and a nonzero integer $k$. Without loss of
generality, we assume further that $b=1$ and $k>0$. Thus $R=D[t^{k}]$. Observe
that $R$ is left Noetherian and $S$ is finite over $R$. It follows from [21,
Lemma 2.2] that $S$ is left integral over $R$. In particular, $t^{-1}$ is left
integral over $R$, i.e. there exist a positive integer $m>0$ and elements
$f_{0},\ldots,f_{m-1}\in R$ such that

$\displaystyle t^{-m}+f_{m-1}t^{-m+1}+\ldots+f_{1}t^{-1}+f_{0}=0.$

By multiplying both side by $t^{m-1}$, we obtain

$\displaystyle t^{-1}=-\left(f_{m-1}+\ldots+f_{1}t^{m-2}+f_{0}t^{m-1}\right)$

which is an element of $D[t]$, a contradiction. Hence, $S$ cannot be finite
over $R$, and so $S$ is not automorphically normalizable over $D$. ∎

######  Proposition 5.7.

If $\sigma\in\mathbf{Aut}(D)$ is of infinite inner order, then the tuple
$(\sigma,\sigma^{-1})$ is not automorphically normalizable over $D$.

###### Proof.

Straightforward from Lemma 5.6. ∎

######  Question 5.8.

What can be said concerning normalizability of the tuple
$(\sigma,\sigma^{-1})$ when $\sigma$ is of finite inner order? Corollary 3.8
gives a partial answer when $\sigma$ is of finite-order. In the case where
$\sigma$ is of finite inner order but not of finite order, the authors do not
know.

###  5.3 Second negative example

To give our second negative example, we first need the following lemma.

######  Lemma 5.9.

Let $\sigma_{1}$ and $\sigma_{2}$ be two commuting automorphisms in
$\mathbf{Aut}(D)$. In the skew polynomial ring
$T=D[t_{1},t_{2};\sigma_{1},\sigma_{2}]$, let $I=(t_{1}t_{2})$ be the two-
sided ideal generated by $t_{1}t_{2}$. Let $S=T/I$ be the quotient ring. Then
every two commuting elements of $S$ are algebraically dependent over $D$.

###### Proof.

Let $z_{1}$ and $z_{2}$ be the images of $t_{1}$ and $t_{2}$ in $S$,
respectively. Then $z_{1}z_{2}=0$ and
$S=D[z_{1},z_{2};\sigma_{1},\sigma_{2}]$. Assume that $x_{1}$ and $x_{2}$ are
two arbitrary elements of $S$ which are the images of two skew polynomials
$f_{1}$ and $f_{2}$ in $T$, respectively (i.e. $x_{i}=f_{i}+I$, $i=1,2$). Let
$d$ be the maximal total degrees of $f_{1}$ and $f_{2}$, and let $N$ be a
positive integer that we will select later. We denote by $T_{dN}$ the set of
polynomials in $T$ of total degree at most $dN$, and let $S_{dN}$ be the image
of $T_{dN}$ in $S$. We consider the following two sets:

$\mathcal{A}=\\{x_{1}^{k_{1}}x_{2}^{k_{2}}\,|\,0\leq k_{1}+k_{2}\leq N\\},$

which is the set of all monomials in $x_{1}$ and $x_{2}$ of total degree at
most $N$, thus is a subset of $S_{dN}$, and

$\mathcal{B}=\\{1,z_{1},z_{2},z_{1}^{2},z_{2}^{2},\ldots,z_{1}^{dN},z_{2}^{dN}\\},$

which is a left linear basis of $S_{dN}$ as a vector space over $D$. If there
are two pairs of non-negative integers $(i_{1},i_{2})$ and $(j_{1},j_{2})$
with $i_{1}+i_{2},\,j_{1}+j_{2}\leq dN$ such that
$x_{1}^{i_{1}}x_{2}^{i_{2}}=x_{1}^{j_{1}}x_{2}^{j_{2}}$, then the two
monomials $x_{1}^{i_{1}}x_{2}^{i_{2}}$ and $x_{1}^{j_{1}}x_{2}^{j_{2}}$ are
linearly dependent over $D$. Thus $x_{1}$ and $x_{2}$ is algebraically
dependent over $D$.

Let us consider the case when $x_{1}^{i_{1}}x_{2}^{i_{2}}\neq
x_{1}^{j_{1}}x_{2}^{j_{2}}$ for all pairs of nonnegative integers
$(i_{1},i_{2})\neq(j_{1},j_{2})$ with $i_{1}+i_{2},\,j_{1}+j_{2}\leq dN$. In
this case, we have

$\displaystyle\frac{|\mathcal{A}|}{|\mathcal{B}|}$
$\displaystyle=\frac{\binom{N+2}{2}}{2dN+1}=\frac{(N+2)(N+1)}{2(2dN+1)}\geq\frac{(N+2)(N+1)}{2(2d+1)N}\geq\frac{N+2}{2(2d+1)}.$

Therefore, by selecting $N>4d$, we will have $|\mathcal{A}|>|\mathcal{B}|$,
thus $\mathcal{A}$ is left linearly dependent over $D$. Hence, $x_{1}$ and
$x_{2}$ are algebraically dependent over $D$. ∎

We now have:

######  Lemma 5.10.

Let $\sigma_{1}$ and $\sigma_{2}$ be two commuting automorphisms in
$\mathbf{Aut}(D)$. In the skew polynomial ring
$T=D[t_{1},t_{2};\sigma_{1},\sigma_{2}]$, let $I=(t_{1}t_{2})$ be the two-
sided ideal generated by $t_{1}t_{2}$. Then the quotient ring $S=T/I$ is
automorphically normalizable over $D$ if and only if
$\sigma_{1}^{k_{1}}\circ\sigma_{2}^{-k_{2}}$ is an inner automorphism of $D$
for some positive integers $k_{1},k_{2}$.

###### Proof.

Let $z_{1}$ and $z_{2}$ be the images of $t_{1}$ and $t_{2}$ in $S$,
respectively. Then $z_{1}z_{2}=0$ and
$S=D[z_{1},z_{2};\sigma_{1},\sigma_{2}]$. We consider the following two cases.

  * Case 1: $\sigma_{1}^{k_{1}}\circ\sigma_{2}^{-k_{2}}$ is an inner automorphism of $D$ for some positive integers $k_{1},k_{2}$. This means that $\sigma_{1}^{k_{1}}\circ\sigma_{2}^{-k_{2}}=\mathbf{in}_{c}$ for some nonzero element $c\in D$. We need to prove that $S$ is automorphically normalizable over $D$. Indeed, set $u=z_{1}^{k_{1}}+cz_{2}^{k_{2}}$. Then, it is clear that $u$ is algebraically indepedent over $D$. In addition, for every $r\in D$, we have

$\displaystyle ur=z_{1}^{k_{1}}r+cz_{2}^{k_{2}}r$
$\displaystyle=\sigma_{1}^{k_{1}}(r)z_{1}^{k_{1}}+c\sigma_{2}^{k_{2}}(r)z_{2}^{k_{2}}=\sigma_{1}^{k_{1}}(r)z_{1}^{k_{1}}+\mathbf{in}_{c}\circ\sigma_{2}^{k_{2}}(r)\cdot
cz_{2}^{k_{2}}$
$\displaystyle=\sigma_{1}^{k_{1}}(r)(z_{1}^{k_{1}}+cz_{2}^{k_{2}})=\sigma_{1}^{k_{1}}(r)u.$

Therefore, $u$ is automorphic over $D$ with respect to $\sigma_{1}^{k_{1}}$.
Furthermore, since $z_{1}^{k_{1}+1}=uz_{1}$, $z_{1}$ is left integral over
$D[u]$. As a consequence,
$D[z_{1},z_{2}^{k_{2}}]=D[z_{1},cz_{2}^{k_{2}}]=D[z_{1},u]$ is finite over
$D[u]$. Since $S=D[z_{1},z_{2}]$ is finite over $D[z_{1},z_{2}^{k_{2}}]$, $S$
is also finite over $D[u]$. Hence, $S$ is automorphically normalizable over
$D$.

  * Case 2: $\sigma_{1}^{k_{1}}\circ\sigma_{2}^{-k_{2}}$ is not an inner automorphism of $D$ for arbitrary positive integers $k_{1},k_{2}$. We will prove that $S$ is not automorphically normalizable over $D$. Assume not, then according to Lemma 5.9, there exist an automorphic element $x\in S$ such that $x$ is algebraically independent over $D$ and that $S$ is finite over $D[x]$.

As an element of $S$, $x$ can be written in the form

$\displaystyle x=a_{0}+\sum\limits_{i\geq 1}a_{i}z_{1}^{i}+\sum\limits_{j\geq
1}b_{j}z_{2}^{j},$ (3)

where the sums are finite and the nonzero terms are left linearly independent
over $D$. The terms $a_{i}z_{1}^{i}$ and $b_{j}z_{2}^{j}$ are automorphic over
$D$ with respect to the automorphisms $\mathbf{in}_{a_{i}}\circ\sigma_{1}^{i}$
and $\mathbf{in}_{b_{j}}\circ\sigma_{2}^{j}$, respectively. Observe that
$\mathbf{in}_{a_{i}}\circ\sigma_{1}^{i}\neq\mathbf{in}_{b_{j}}\circ\sigma_{2}^{j}$
for all positive integers $i,j$ and nonzero elements $a_{i},b_{j}$ of $D$.
Otherwise, $\sigma_{1}^{i}\circ\sigma_{2}^{-j}=\mathbf{in}_{a_{i}^{-1}b_{j}}$
is an inner automorphism of $D$, which is a contradiction. This means that, in
Eq. (3), no nonzero term in the first sum is automorphic over $D$ with respect
to the same automorphism with any nonzero term of the second sum. Therefore,
as a consequence of Lemma 5.5, in order $x$ to be automorphic over $D$, one of
the two sums in Eq. (3) must be cancelled. Thus, either $x\in D[z_{1}]$ or
$x\in D[z_{2}]$.

Without loss of generality, let us assume that $x\in D[z_{1}]$. Then
$D[x]\subseteq D[z_{1}]$. Since $S$ is finite over $D[x]$, $S$ is also finite
over $D[z_{1}]$. According to [21, Lemma 2.2], $S$ is left integral over
$D[z_{1}]$. In particular, $z_{2}$ is left integral over $D[z_{1}]$, i.e.
there exist a positive integer $d$ and elements $f_{0},\ldots,f_{d-1}\in
D[z_{1}]$ such that

$\displaystyle z_{2}^{d}+f_{d-1}z_{2}^{d-1}+\ldots+f_{1}z_{2}+f_{0}=0.$

Since $z_{1}z_{2}=0$, by replacing $f_{1},\ldots,f_{d-1}$ by their free
coefficients, we can assume that $f_{1},\ldots,f_{d-1}\in D$. Then

$f_{0}=-(z_{2}^{d}+f_{d-1}z_{2}^{d-1}+\ldots+f_{1}z_{2})$

is an element in $D[z_{1}]\cap D[z_{2}]=D$. Thus, $z_{2}$ is left integral
over $D$, which is impossible. Hence, $S$ is not automorphically normalizable
over $D$.

∎

######  Proposition 5.11.

Let $\sigma_{1}$ and $\sigma_{2}$ be two commuting automorphisms in
$\mathbf{Aut}(D)$. If $\sigma_{1}^{k_{1}}\circ\sigma_{2}^{-k_{2}}$ is not an
inner automorphism for every positive integers $k_{1},k_{2}$, then the tuple
$(\sigma_{1},\sigma_{2})$ is not automorphically normalizable over $D$.

###### Proof.

Straightforward from Lemma 5.10. ∎

######  Question 5.12.

What can be said concerning the normalizability of the tuple
$(\sigma_{1},\sigma_{2})$ if $\sigma_{1}^{k_{1}}\circ\sigma_{2}^{-k_{2}}$ is
an inner automorphism for some positive integers $k_{1},k_{2}$? Corollary 3.8
gives a partial answer for the case where
$\sigma_{1}^{k_{1}}\circ\sigma_{2}^{-k_{2}}=\mathbf{id}_{D}$.

In the case where $D=F$ is a field, we give a necessary and sufficient
condition for a tuple of commuting automorphisms in $\mathbf{Aut}(F)$ to be
automorphically normalizable in the following theorem (Theorem 1.6 of the
introduction).

######  Theorem 5.13.

Let $F$ be a field and $\sigma_{1},\ldots,\sigma_{n}$ be commuting
automorphisms in $\mathbf{Aut}(F)$ such that $\cap_{i=1}^{n}F_{\sigma_{i}}$ is
infinite. Then the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is automorphically
normalizable over $F$ if and only if
$\sigma_{1}^{d_{1}}=\ldots=\sigma_{n}^{d_{n}}$ for some positive integers
$d_{1},\ldots,d_{n}$.

###### Proof.

The “if” part is a direct consequence of Corollary 3.8. For the “only if”
part, let us assume that the tuple $(\sigma_{1},\ldots,\sigma_{n})$ is
automorphically normalizable over $F$. According to Proposition 5.2 and Remark
2.18, the shorter tuple $(\sigma_{1},\sigma_{i})$ is also automorphically
normalizable over $F$ for every $i=2,\ldots,n$. Then it follows from
Proposition 5.11 that there exist positive integers $k_{i,1}$ and $k_{i,2}$
such that $\sigma_{1}^{k_{i,1}}\circ\sigma_{i}^{-k_{i,2}}$ is an inner
automorphism. However, every inner automorphism on $F$ is the identity map.
Therefore, $\sigma_{1}^{k_{i,1}}\circ\sigma_{i}^{-k_{i,2}}=\mathbf{id}_{F}$,
or equivalently, $\sigma_{1}^{k_{i,1}}=\sigma_{i}^{k_{i,2}}$, for all
$i=2,\ldots,n$. Set $d_{1}=\prod_{i=1}^{n}k_{i,1}$, then we have
$\sigma_{1}^{d_{1}}=\sigma_{i}^{\frac{d_{1}k_{i,2}}{k_{i,1}}}$. Thus, for each
$i=2,\ldots,n$, if we set $d_{i}=\frac{d_{1}k_{i,2}}{k_{i,1}}$, then
$d_{1},\ldots,d_{n}$ are positive integers and
$\sigma_{1}^{d_{1}}=\ldots=\sigma_{n}^{d_{n}}$ as expected. ∎

######  Question 5.14.

Can one find a necessary and sufficient condition for the above theorem to
hold, when the field $F$ is replaced by a division algebra ?

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