# Roots of polynomials over division rings

Alina G. Goutor, Sergey V. Tikhonov  Goutor: Belarusian State University,
Nezavisimosti Ave., 4, 220030, Minsk, Belarus<EMAIL_ADDRESS>Tikhonov:
Belarusian State University, Nezavisimosti Ave., 4, 220030, Minsk, Belarus
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study properties of polynomials over division rings.
Moreover, we present formulas for finding roots of some polynomials.

##  1\. Introduction and preliminary results

In this paper, we study polynomials over division rings. Let ${\mathcal{D}}$
be an associative division ring. Let also ${\mathcal{D}}[x]$ denote the
polynomial ring in one variable $x$ over ${\mathcal{D}}$, where $x$ commutes
elementwise with ${\mathcal{D}}$. The coefficients of such polynomials may not
commute with elements of the ring. Polynomials in ${\mathcal{D}}[x]$ are added
in the obvious way, and multiplied according to the rule

$(a_{n}x^{n}+\dots+a_{0})(b_{m}x^{m}+\dots+b_{0})=(c_{m+n}x^{m+n}+\dots+c_{0}),$

where $c_{k}=\sum_{i+j=k}a_{i}b_{j}$. For references on polynomial rings over
division rings, see [10, Ch. 5, §16] and [7]. The degree of
$P(x)\in{\mathcal{D}}[x]$ is defined in the usual way. For a polynomial

$P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}\in{\mathcal{D}}[x]$

and an element $a\in{\mathcal{D}}$, we define $P(a)$ to be the element

$a_{n}a^{n}+a_{n-1}a^{n-1}+\dots+a_{1}a+a_{0}.$

An element $a\in{\mathcal{D}}$ is said to be a (right) root of $P(x)$ if
$P(a)=0$. The noncommutative form of the Remainder Theorem says that an
element $a\in{\mathcal{D}}$ is a root of a nonzero polynomial $P(x)$ iff $x-a$
is a right divisor of $P(x)$ in ${\mathcal{D}}[x]$ (see, e.g., [10, Prop.
16.2]).

For $a\in{\mathcal{D}}$, the set

$[a]:=\\{dad^{-1}|d\in{\mathcal{D}}\setminus\\{0\\}\\}$

will be called the conjugacy class of $a$. The centralizer of $a$ is defined
as

$Z(a):=\\{b\in{\mathcal{D}}|ab=ba\\}.$

Over a field, a polynomial of degree $n$ has at most $n$ distinct roots. Over
a division ring this is no longer true, but by Gordon-Motzkin theorem ([10,
Th. 16.4]), a polynomial of degree $n$ in ${\mathcal{D}}[x]$ has roots in at
most $n$ conjugacy classes of ${\mathcal{D}}$, moreover, if
$P(x)=(x-a_{1})\dots(x-a_{n})$, where $a_{1},\dots,a_{n}\in{\mathcal{D}}$,
then any root of $P(x)$ is conjugate to some $a_{i}$. Note that from
$P(x)=L(x)R(x)\in{\mathcal{D}}[x]$ it does not follow that $P(a)=L(a)R(a)$
(see Proposition 6 bellow). In particular, if $a$ is a root of $L(x)$, then
$a$ is not necessarily a root of $P(x)$.

The problem of finding the roots of a polynomial over a division ring has been
investigated in ring theory and applied mathematics. The most studied is the
case of polynomials over Hamilton’s quaternion algebra $\mathbb{H}$. In
analogy to field theory, the notion of a (right) algebraically closed division
ring ${\mathcal{R}}$ is defined. This is equivalent to saying that every
polynomial in ${\mathcal{R}}[x]$ splits completely into a product of linear
factors in ${\mathcal{R}}[x]$. By Niven-Jacobson theorem ([10, Th. 16.14]),
the quaternion division algebra over a real-closed field is algebraically
closed. Baer’s theorem ([10, Th. 16.15]) says that noncommutative centrally
finite (right) algebraically closed division rings are precisely the division
rings of quaternions over real-closed fields. In [8], a formula was found for
roots of any quadratic polynomial in $\mathbb{H}[x]$. This formula was
generalized to any quaternion algebra in [2] and [3]. In [11], it was shown
that the roots of any polynomial in $\mathbb{H}[x]$ are roots of the real
companion polynomial. In [9], it was presented an algorithm for finding all
roots of a polynomial in $\mathbb{H}[x]$ using the real companion polynomial.
In [5], a few of these results were generalized to the case of any central
division algebra. Recall that a central division algebra is a division algebra
which is finite dimensional over its center. A complete method for finding the
roots of all polynomials over an octonion division algebra was described in
[4].

In [6], it was presented the following explicit formula describing roots of a
product of linear factors in $\mathbb{H}[x]$.

######  Theorem 1.

([6, Th. 4]). Let $P(x)=(x-q_{n})\dots(x-q_{1})$, where
$q_{1},\dots,q_{n}\in\mathbb{H}$. If the conjugacy classes $[q_{k}]$ are
distinct, then the polynomial $P(x)$ has exactly $n$ roots $\zeta_{k}$ which
are related to the elements $q_{k}$ as follows:

$\zeta_{k}=\overline{P}_{k}(q_{k})q_{k}(\overline{P}_{k}(q_{k}))^{-1};k=1,\dots,n,$
$P_{k}(x):=\left\\{\begin{array}[]{l}1,\mbox{ if }k=1,\\\
(x-q_{k-1})\dots(x-q_{1}),\mbox{ otherwise }\\\ \end{array}\right.$

and $\overline{P}_{k}(x)$ is the conjugate polynomial of $P_{k}(x)$.

In theorem 9 below, we generalize this formula for the case of any division
ring. This is the first aim of this paper.

Let $F$ be the center of a division ring ${\mathcal{D}}$. If $a$ is a root of
a polynomial $f(x)\in F[x]$, then any element from the conjugacy class $[a]$
is a root of $f(x)$. The conjugacy class $A$ is called algebraic over $F$ if
one (an hence all) of its elements is algebraic over $F$. If $A$ is algebraic
over $F$, then the minimal polynomial of $A$ is, by definition, the minimal
polynomial of any element from $A$.

In the case of a quadratic minimal polynomial, there is the following

######  Theorem 2.

[10, Lm. 16.17]. Let ${\mathcal{D}}$ be a division ring with center $F$, and
let $A$ be a conjugacy class of ${\mathcal{D}}$ which has a quadratic minimal
polynomial $\lambda(x)$ over $F$. If $P(x)\in{\mathcal{D}}[x]$ has two roots
in $A$, then $P(x)\in{\mathcal{D}}[x]\lambda(x)$ and $P(x)$ vanishes
identically on $A$.

This means that a polynomial over a quaternion division algebra may have two
different types of roots: isolated and spherical roots. A root $q$ of $P(x)$
is called spherical if $q$ is not central and for every $d\in[q]$ we have
$P(d)=0$. A root $q$ is called isolated if the conjugacy class $[q]$ contains
no other root of $P(x)$.

The second aim of this paper is to show that in the case of a conjugacy class
with minimal polynomial of bigger degree the situation is completely
different. More precisely, we proved the following

######  Theorem 3.

Let ${\mathcal{D}}$ be a noncommutative division ring with the center $F$,
$a\in{\mathcal{D}}$ an algebraic over $F$ element with minimal polynomial
$\lambda(x)$ of degree $n>2$. Then there exists a quadratic polynomial
$P(x)\in{\mathcal{D}}[x]$ such that

1\. $P(x)$ has infinitely many roots in the conjugacy class $[a]$,

2\. there are infinitely many elements in $[a]$ which are not roots of $P(x)$,

3\. $\lambda(x)$ does not divide $P(x)$.

##  2\. Proof of Theorem 3

In the proof of Theorem 3 we will use the following statements.

######  Theorem 4.

([10, Th. 16.6]). Let ${\mathcal{D}}$ be a division ring with center $F$ and
$A$ a conjugacy class of ${\mathcal{D}}$ which is algebraic over $F$ with
minimal polynomial $f(x)\in F[x]$. A polynomial $P(x)\in{\mathcal{D}}[x]$
vanishes identically on $A$ iff $P(x)\in{\mathcal{D}}[x]f(x)$.

######  Theorem 5.

([10, Th. 16.11] and [7, Th. 4]). If a polynomial $P(x)\in{\mathcal{D}}[x]$
has two distinct roots in a conjugacy class of ${\mathcal{D}}$, then it has
infinitely many roots in that class.

######  Proposition 6.

([10, Pr. 16.3]). Let ${\mathcal{D}}$ be a division ring and let
$P(x)=L(x)R(x)\in{\mathcal{D}}[x]$. Let $d\in{\mathcal{D}}$ be such that
$h:=R(d)\neq 0$. Then

$P(d)=L(hdh^{-1})R(d).$

In particular, if $d$ is a root of $P(x)$ but not of $R(x)$, then $hdh^{-1}$
is a root of $L(x)$.

Proof of Theorem 3. Let $a\in{\mathcal{D}}$ be an element with minimal
polynomial $\lambda(x)$ of degree $n>2$. Let also $d\in{\mathcal{D}}$ be an
element such that $d$ does not commute with $a$. Let $q=dad^{-1}$ and
$b=(q-a)q(q-a)^{-1}$. Then $q\neq a$ and $q\in[a]$. By Proposition 6, $q$ is a
root of the polynomial $P(x):=(x-b)(x-a)$.

Since $a$ is also a root of $P(x)$, then by Theorem 5, $P(x)$ has infinitely
many roots in $[a]$. Moreover, since the degree of $\lambda(x)$ is bigger than
2, then $\lambda(x)$ does not divide $P(x)$. Hence by Theorem 4, $P(x)$ does
not vanish identically on $[a]$.

Suppose that $tat^{-1}\in[a]$ is not a root of $P(x)$. This means that
$tat^{-1}\neq a$ and

$(tat^{-1}-a)tat^{-1}(tat^{-1}-a)^{-1}\neq b$

by Proposition 6.

Note that

$(tat^{-1}-a)tat^{-1}(tat^{-1}-a)^{-1}=(ta-at)a(ta-at)^{-1}.$

Let $z\in Z(a)$, $t_{1}=t+z$ and $q_{1}=t_{1}at_{1}^{-1}$. Then

$(q_{1}-a)q_{1}(q_{1}-a)^{-1}=(t_{1}at_{1}^{-1}-a)t_{1}at_{1}^{-1}(t_{1}at_{1}^{-1}-a)^{-1}=(t_{1}a-at_{1})a(t_{1}a-at_{1})^{-1}=$
$=((t+z)a-a(t+z))a((t+z)a-a(t+z))^{-1}=(ta-at)a(ta-at)^{-1}\neq b.$

Thus $q_{1}$ is not a root of $P(x)$ by Proposition 6.

Now let $z_{1}\in Z(a)$, $z_{1}\neq z$. Note that the centralizer $Z(a)$ is
infinite by [7, Th. 3]. Let also $t_{2}=t+z_{1}$ and $q_{2}=t_{2}at_{2}^{-1}$.
Then $q_{2}$ is also not a root of $P(x)$.

Assume that $q_{1}=q_{2}$. Then

$(t+z)a(t+z)^{-1}=(t+z_{1})a(t+z_{1})^{-1}\Leftrightarrow$
$(t+z_{1})^{-1}(t+z)a=a(t+z_{1})^{-1}(t+z)\Leftrightarrow$
$(t+z_{1})^{-1}(t+z_{1}+(z-z_{1}))a=a(t+z_{1})^{-1}(t+z_{1}+(z-z_{1}))\Leftrightarrow$
$a+(t+z_{1})^{-1}(z-z_{1})a=a+a(t+z_{1})^{-1}(z-z_{1})\Leftrightarrow$
$(t+z_{1})^{-1}(z-z_{1})a=a(t+z_{1})^{-1}(z-z_{1})\Leftrightarrow(t+z_{1})^{-1}a(z-z_{1})=a(t+z_{1})^{-1}(z-z_{1})\Leftrightarrow$
$(t+z_{1})^{-1}a=a(t+z_{1})^{-1}\Leftrightarrow
a(t+z_{1})=(t+z_{1})a\Leftrightarrow at=ta.$

This gives a contradiction since $t$ does not commute with $a$. Then
$q_{1}\neq q_{2}$. Hence any $z\in Z(a)$ defines the element
$(t+z)a(t+z)^{-1}\in[a]$ which is not a root of $P(x)$ and all such elements
are distinct. Since the centralizer $Z(a)$ is infinite, then there are
infinitely many elements in $[a]$ which are not roots of $P(x)$. ∎

######  Remark 7.

In the notation of the proof of Theorem 3, let

$b_{1}=(tat^{-1}-a)tat^{-1}(tat^{-1}-a)^{-1}.$

The polynomials $(x-b)(x-a)$ and $(x-b_{1})(x-a)$ have infinitely many roots
in the conjugacy class $[a]$, but $a$ is the unique common root of these
polynomials.

##  3\. Roots of polynomials

It seems to us that the following lemma may be a known result, but we have not
found an exact reference. For the reader’s convenience, we provide a proof
here.

######  Lemma 8.

Let ${\mathcal{D}}$ be a division ring. Let also

$P(x)=(x-d_{n})\dots(x-d_{1}),$

where $d_{1},\dots,d_{n}\in{\mathcal{D}}$. If the conjugacy classes $[d_{k}]$
are distinct, then the polynomial $P(x)$ has exactly $n$ roots and any root of
$P(x)$ is conjugate to some $d_{i}$.

Proof. By Gordon-Motzkin theorem ([10, Th. 16.4]), the roots of $P(x)$ lie in
$n$ conjugacy classes of ${\mathcal{D}}$ and any root of $P(x)$ is conjugate
to some $d_{i}$. Let $d\in{\mathcal{D}}$ be a root of $P(x)$, then
$P(x)=L(x)(x-d)$ for some $L(x)\in{\mathcal{D}}[x]$. By Proposition 6, all
roots of $P(x)$ different from $d$ are conjugate to roots of $L(x)$. Since the
conjugacy classes $[d_{k}]$ are distinct and $\text{\rm deg}(L(x))=n-1$, then
by Gordon-Motzkin theorem, $L(x)$ has no roots in $[d]$. Thus $P(x)$ has only
one root in each conjugacy class. ∎

######  Theorem 9.

Let ${\mathcal{D}}$ be a division ring with center $F$. Let also

$P(x)=(x-d_{n})\dots(x-d_{1}),$

where $d_{1},\dots,d_{n}\in{\mathcal{D}}$. Assume that $d_{1},\dots,d_{n-1}$
are algebraic over $F$. Let also $f_{i}(x)$ be the minimal polynomial of
$d_{i}$, $i=1,\dots,n-1$. If the conjugacy classes $[d_{k}]$ are distinct,
then the polynomial $P(x)$ has exactly $n$ zeros $\zeta_{k}$ which are related
to the elements $d_{k}$ as follows:

$\zeta_{k}=P_{k}(d_{k})d_{k}(P_{k}(d_{k}))^{-1};k=1,\dots,n,$
$P_{k}(x):=\left\\{\begin{array}[]{l}1,\mbox{ if }k=1,\\\ S_{1}(x)\dots
S_{k-1}(x),\mbox{ otherwise, }\\\ \end{array}\right.$

where $S_{i}(x)\in{\mathcal{D}}[x]$ is such that $f_{i}(x)=S_{i}(x)(x-d_{i})$,
$i=1,\dots,n-1$.

Proof. Since $S_{i}(x)$ has coefficients in the field $F(d_{i})$, then

$f_{i}(x)=S_{i}(x)(x-d_{i})=(x-d_{i})S_{i}(x)$

for $i=1,\dots,n-1$. Note that

$P(x)P_{k}(x)=(x-d_{n})\dots(x-d_{1})S_{1}(x)\dots S_{k-1}(x)=$
$=(x-d_{n})\dots(x-d_{k})f_{k-1}(x)\dots f_{1}(x).$

Since $f_{k-1}(x)\dots f_{1}(x)\in F[x]$, then $d_{k}$ is a root of the
polynomial $P(x)P_{k}(x)$. Note that for $i=1,\dots,k-1$,
$d_{k}\not\in[d_{i}]$, so $d_{k}$ is not a root of $f_{i}(x)$ by Dickson’s
Theorem ([10, Th. 16.8]). Hence $d_{k}$ is not a root of $f_{k-1}(x)\dots
f_{1}(x)$. Then $d_{k}$ is not a root of $P_{k}(x)$. Indeed,

$(x-d_{k-1})\dots(x-d_{1})P_{k}(x)=(x-d_{k-1})\dots(x-d_{1})S_{1}(x)\dots
S_{k-1}(x)=f_{k-1}(x)\dots f_{1}(x)$

and if $d_{k}$ is a root of $P_{k}(x)$, then $d_{k}$ is a root of
$f_{k-1}(x)\dots f_{1}(x)$.

Hence by Proposition 6, $P_{k}(d_{k})d_{k}(P_{k}(d_{k}))^{-1}$ is a root of
$P(x)$ for $k=1,\dots,n$.

By Lemma 8, $P(x)$ has no other roots.

∎

In the notation of Theorem 9, we have the following

######  Corollary 10.

Let ${\mathcal{D}}$ be a division ring with center $F$,
$d_{1},d_{2}\in{\mathcal{D}}$ such that the conjugacy classes $[d_{1}]$ and
$[d_{2}]$ are distinct. Assume that $d_{1}$ is algebraic over $F$. Let also
$f(x)$ be the minimal polynomial of $d_{1}$ and $S(x)\in{\mathcal{D}}[x]$ such
that $f(x)=S(x)(x-d_{1})$. Then

$(x-d_{2})(x-d_{1})=(x-d)(x-S(d_{2})d_{2}(S(d_{2}))^{-1}),$

where
$d=(d_{1}-S(d_{2})d_{2}(S(d_{2}))^{-1})d_{1}(d_{1}-S(d_{2})d_{2}(S(d_{2}))^{-1})^{-1}$.

Proof. Let $P(x):=(x-d_{2})(x-d_{1})$. By Theorem 9,
$d_{3}:=S(d_{2})d_{2}(S(d_{2}))^{-1}$ is a root of $P(x)$. Then $x-d_{3}$ is a
right divisor of $P(x)$ and $P(x)=(x-d)(x-d_{3})$ for some
$d\in{\mathcal{D}}$. Since $d_{1}$ is a root of $P(x)$ and $d_{1}\neq d_{3}$,
then by Proposition 6,

$d=(d_{1}-d_{3})d_{1}(d_{1}-d_{3})^{-1}=(d_{1}-S(d_{2})d_{2}(S(d_{2}))^{-1})d_{1}(d_{1}-S(d_{2})d_{2}(S(d_{2}))^{-1})^{-1}.$

∎

######  Remark 11.

The formula from the previous corollary allows to change the order of factors
in products of monic linear polynomials. This formula generalizes formulas for
Hamilton’s quaternion algebra from [12, Lm. 1] (see also [6, Th.7]).

######  Example 12.

Let $F$ be a field, $\text{\rm char}(F)\neq 2$. Let also ${\mathcal{Q}}$ be a
quaternion division algebra over $F$. Assume that
$d_{1},d_{2}\in{\mathcal{Q}}$, $[d_{1}]\neq[d_{2}]$. If $d_{1}\not\in F$, then
the minimal polynomial of $d_{1}$ is $(x-\overline{d}_{1})(x-d_{1})$, where
$\overline{d}_{1}$ is the conjugate of $d_{1}$. Hence in the notation of
Corollary 10, $S(x)=x-\overline{d}_{1}$. Then

$S(d_{2})d_{2}(S(d_{2}))^{-1}=(d_{2}-\overline{d}_{1})d_{2}(d_{2}-\overline{d}_{1})^{-1}.$

Simple computations show that

$(d_{1}-S(d_{2})d_{2}(S(d_{2}))^{-1})d_{1}(d_{1}-S(d_{2})d_{2}(S(d_{2}))^{-1})^{-1}=(d_{2}-\overline{d}_{1})d_{1}(d_{2}-\overline{d}_{1})^{-1}.$

Thus

$(x-d_{2})(x-d_{1})=(x-hd_{1}h^{-1})(x-hd_{2}h^{-1}),$

where $h=d_{2}-\overline{d}_{1}$ (compare with the formula from [12, Lm. 1]).

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