# Prime Rational Functions over a Field

Eva Goedhart  Department of Mathematics, Franklin and Marshall College
Philadelphia, PA US
<EMAIL_ADDRESS>,  OMAR KIHEL  Department of Mathematics and
Statistics, Brock University
St. Catharines, L2S 3A1, Canada
<EMAIL_ADDRESS>and  JESSE LARONE  Department of Mathematics and Statistics,
Brock University
St. Catharines, L2S 3A1, Canada
<EMAIL_ADDRESS>
###### Abstract.

The aim of this paper is to provide sufficient conditions for when a
polynomial or rational function over a field $K$ is prime using its order of
vanishing at infinity and the resultant.

###### Key words and phrases:

Prime polynomial; critical value; resultant.

###### 2010 Mathematics Subject Classification:

11C08

##  1\. Introduction

In 1922, J.F. Ritt [6] defined composite polynomials $f(x)$ over $\mathbb{C}$
as those for which there exist polynomials $g(x)$ and $h(x)$ in
$\mathbb{C}[x]$ such that $f=g\circ h$ where the degrees of $g(x)$ and $h(x)$
are each at least 2. Here the prime polynomials are those for which this does
not hold. Engstrom [3] and Levi [5] extended Ritt’s main results to
polynomials over fields of characteristic zero. Though many variations of
Ritt’s second theorem were published, Zannier [7] finished the work of Ritt
and Levi by extending Ritt’s second theorem to a field of any characteristic.
In 2000, Beardon [2] returned to $\mathbb{C}$ to study the role of critical
points and critical values in the prime decomposition of polynomials. A few
years later, Ayad [1] used critical points and critical values of polynomials
in $\mathbb{C}[x]$ to determine sufficient conditions for when a polynomial is
prime among other results. In 2015, Kihel and Larone [4] extended the
definitions of prime and composite polynomials to rational functions in
$\mathbb{C}(x)$. In this paper, we continue this work and extend Ayad’s
results to find conditions under which polynomials and rational functions are
prime over a field $K$.

In Section 2, we briefly discuss the decomposition of polynomial functions
over a finite field. In Section 3, for a field $K$ we define and consider
prime and composite rational functions in $K(x)$. We then provide sufficient
conditions on the order of vanishing at infinity for $f\in K(x)$ to be prime.
In Section 4, using the resultant and the discriminant, we give sufficient
conditions on the critical values of $f\in K(x)$ for $f$ to be prime.

##  2\. Functions over $\mathbb{F}_{q}$

Let $q$ be a power of a prime number, and denote by $\mathbb{F}_{q}$ the
finite field of $q$ elements. Any function
$\varphi:\mathbb{F}_{q}\to\mathbb{F}_{q}$ is equivalent to a polynomial
function. Thus, the set of all functions over $\mathbb{F}_{q}$ can be
described exactly as the ring

$A=\mathbb{F}_{q}[x]/(x^{q}-x)$

under addition and composition of functions. The units of $A$ with respect to
function composition are called the permutation polynomials of
$\mathbb{F}_{q}$. These are the polynomials that are bijections of
$\mathbb{F}_{q}$ into itself. There are exactly $q!$ such polynomials and they
can be computed readily, for example, using Lagrange’s interpolation formula.
Explicit classes of permutation polynomials are quite rare, with certain
exceptions such as linear polynomials, and as such they form an active area of
research.

######  Theorem 1.

Let $\varphi=\alpha\circ\beta$, where $\varphi,\alpha,\beta\in A$. Then
$\varphi$ is a permutation polynomial if and only if both $\alpha$ and $\beta$
are permutation polynomials.

###### Proof.

First, note that if both $\alpha$ and $\beta$ are permutation polynomials,
then so is $\varphi$. Next, assume that $\varphi$ is a permutation polynomial.
Then

$|\mathbb{F}_{q}|=|\varphi(\mathbb{F}_{q})|=|\alpha\big{(}\beta(\mathbb{F}_{q})\big{)}|\leq|\beta(\mathbb{F}_{q})|\leq|\mathbb{F}_{q}|$

from which we obtain $|\beta(\mathbb{F}_{q})|=|\mathbb{F}_{q}|$. Since
$\beta(\mathbb{F}_{q})\subseteq\mathbb{F}_{q}$, it follows that
$\beta(\mathbb{F}_{q})=\mathbb{F}_{q}$, so $\beta$ is a permutation
polynomial. It immediately follows that $\alpha=\varphi\circ\beta^{-1}$ is
also a permutation polynomial by the first part. ∎

We adopt the standard nomenclature of rings to describe the functions that
compose to the zero function in $A$.

######  Definition 2.1.

Let $\varphi\in A$. We say that $\varphi$ is a zero divisor if there exists
$\psi\in A\setminus\\{0\\}$ such that $\psi\circ\varphi=0$.

######  Theorem 2.

Every $\varphi\in A\setminus\\{0\\}$ is either a permutation polynomial or a
zero divisor.

###### Proof.

For $\varphi\in A\setminus\\{0\\}$, define the map $C_{\varphi}:A\to A$ by
$\psi\mapsto\psi\circ\varphi$. If $C_{\varphi}$ is injective, then it is
surjective as well, since $A$ is finite. Thus, there exists $\psi\in A$ such
that $(\psi\circ\varphi)=\mathds{1}$ where $\mathds{1}$ denotes the identity
function of $A$. Therefore by Theorem 1, $\varphi$ is a permutation
polynomial. If $C_{\varphi}$ is not injective, there exist distinct
$\alpha,\beta\in A$ such that $\alpha\circ\varphi=\beta\circ\varphi$. Setting
$\psi=\alpha-\beta$ yields
$\psi\circ\varphi=(\alpha\circ\varphi)-(\beta\circ\varphi)=0$, showing that
$\varphi$ is a zero divisor. ∎

As a consequence of the preceding result, we claim that there is no
interesting notion of irreducibility for the elements of the ring $A$. The
permutation polynomials are units, so they are considered trivial cases which
leaves us with the zero divisors of $A$. Letting $\varphi\in
A\setminus\\{0\\}$ be a zero divisor, there exists $\psi\in A\setminus\\{0\\}$
such that $\psi\circ\varphi=0$. Then for $\mathds{1}\in A$, the identity
function, $(\psi+\mathds{1})\circ\varphi=\varphi$ is a non-trivial
decomposition of $\varphi$ since $\psi+\mathds{1}\neq\mathds{1}$. In this
sense, every function over $\mathbb{F}_{q}$ is either a permutation polynomial
or is a composition of polynomial functions, and so is composite.

##  3\. Decompositions and the order at infinity

As described in the introduction, there are results regarding the
decomposition of polynomials and rational functions into indecomposables,
which have been referred to as prime polynomials and prime rational functions.
Ayad [1] studied this problem using valencies for polynomials over
$\mathbb{C}$, and Kihel and Larone [4] later considered some rational
functions over $\mathbb{C}$. In this section, we further discuss the topic and
improve upon a result of Ayad.

Throughout, let $K$ be a field and $\overline{K}$ an algebraic closure of $K$.
For any rational function $f\in K(x)\setminus\\{0\\}$, write $f=f_{1}/f_{2}$
with $f_{1},f_{2}\in K[x]$ such that $\gcd(f_{1},f_{2})=1$, and define as is
usual $\deg f=\max\\{\deg f_{1},\deg f_{2}\\}$. For convenience, we include
the following result regarding the degree of a composition.

######  Theorem 3 ([4], Proposition 2.3).

Let $g,h\in K(x)$. Then $\deg(g\circ h)=(\deg g)\cdot(\deg h)$.

It is straightforward to show that the rational functions of degree 1 form the
group of units in $K(x)$ under the operation of composition. This immediately
justifies the following definitions of prime and composite rational functions,
as well as provide a simple class of prime rational functions.

######  Definition 3.1.

Let $L/K$ be a field extension, and let $f\in K(x)$. If $f=g\circ h$ for some
$g,h\in L(x)$ with degrees at least 2, then $f$ is called composite over $L$.
Otherwise, $f$ is said to be prime over $L$.

As a direct corollary to Theorem 3, we have the following. A similar result is
indicated in [4, Corollary 2.4] for complex rational functions.

######  Theorem 4.

Let $f\in K(x)$. If $\deg f$ is a prime number, then $f$ is prime over
$\overline{K}$.

We recall the definitions used by Ayad [1, Section 2, p.146] over $\mathbb{C}$
here and interpret them over $K$.

######  Definition 3.2.

Let $f\in K[x]$ and let $a\in\overline{K}$. The smallest integer $i\geq 1$
such that $i$th derivative at $a$, $f^{(i)}(a)\neq 0$ is called the valency of
$f$ at $a$ and is denoted by $v_{f}(a)$. If $v_{f}(a)\geq 2$, $a$ is called a
critical point of $f$. An element $b\in\overline{K}$ is called a critical
value of $f$ if there exists a critical point $a$ of $f$ such that $f(a)=b$.

The above definition translates easily to $K(x)$, so we use it without further
modifications. Under the definition of the valency over $\mathbb{C}$, Ayad
proved the following result.

######  Theorem 5 ([1], Theorem 3(1)).

Let $f\in\mathbb{C}[x]$, and let $d$ be the greatest proper divisor of $\deg
f$. If $v_{f}(a)=p$ is prime with $p>d$ from some $a\in\mathbb{C}$, then $f$
is prime.

This theorem was proved using a relationship between the valencies of $f$ and
the degree of its derivative. For completeness, we include a general form of
this same relationship over $K$. This will help to clarify some of statements
that follow it.

######  Lemma 1.

Let $f\in K[x]$. Then

$(\deg f)-1\geq\sum_{a\in\overline{K}}\big{(}v_{f}(a)-1\big{)}$

where equality holds when $\deg f\neq 0$.

###### Proof.

Note that the only values of $v_{f}(a)-1$ which contribute to the sum are
those for which $v_{f}(a)\geq 2$. In particular, $v_{f}(a)-1$ counts the
multiplicity of $a$ as a root of the first derivative $f^{\prime}$, so the sum
cannot exceed the degree of $f^{\prime}$ which is itself less than or equal to
$(\deg f)-1$.

Let $f\in K[x]$ such that $\deg f\neq 0$, then either $\text{char}(K)=0$ or
$\gcd(\deg f,\text{char}(K))=1$. In either case, $\deg f^{\prime}=(\deg f)-1$,
so the sum of the multiplicities of the roots of $f^{\prime}$ is exactly
$(\deg f)-1$, and the sum of $v_{f}(a)-1$ over all $a\in\overline{K}$ must
equal $\deg f^{\prime}=(\deg f)-1$. ∎

We can refine Theorem 5 by finding a replacement for the valencies. Since
Ayad’s definition is in fact related to the order of vanishing at a point, we
simply look to that concept instead. That $v_{f}(a)-1$ counts the multiplicity
of $a$ as a root of $f^{\prime}$ makes it apparent that Lemma 1 relates to the
sum of the orders of zeros and poles of a rational function over all points in
$\mathbb{P}^{1}$ being 0.

Recalling that for $f\in K(x)\setminus\\{0\\}$ written as a ratio of
polynomials $f=f_{1}/f_{2}$, we have
$\operatorname{ord_{\infty}}:K(x)\setminus\\{0\\}\to\mathbb{Z}$ with
$\operatorname{ord_{\infty}}f=\deg f_{2}-\deg f_{1}$. Note that
$|\operatorname{ord_{\infty}}f|\leq\deg f$ and that
$\operatorname{ord_{\infty}}f$ remains unchanged when simplifying the ratio
$f_{1}/f_{2}$. Additionally, for a composite function $f\in K(x)$, we can
compose with the identity to obtain a property for
$\operatorname{ord_{\infty}}f$ which is similar to that of Theorem 3, which
states that the map
$\deg:\big{(}K(x)\setminus\\{0\\},\circ\big{)}\to\big{(}\mathbb{Z},\cdot\big{)}$
preserves operations. We first illustrate this with an example.

######  Example 3.1.

Consider the elements $f,g,h\in\mathbb{C}(x)$ given as

$f(x)=\frac{(x^{4}+1)^{3}(x^{4}+x^{2}+2)}{(x^{2}+1)^{4}},\qquad
g(x)=\frac{x+1}{x^{4}},\qquad\text{and }h(x)=\frac{x^{2}+1}{x^{4}+1}.$

One can readily verify that $f=g\circ h$, yet
$-\operatorname{ord_{\infty}}f=8\neq 3\cdot
2=\operatorname{ord_{\infty}}g\cdot\operatorname{ord_{\infty}}h$. Now,
consider a unit function $\mu(x)=1/x$ in $\mathbb{C}(x)$ with $\mu^{-1}=\mu$
and define $G=g\circ\mu^{-1}$ and $H=\mu\circ h$ so that

$G(x)=x^{4}+x^{3}\qquad\text{and}\qquad H(x)=\frac{x^{4}+1}{x^{2}+1}$

yield a composition $f=G\circ H$. However, for this new pair of functions we
note that
$\operatorname{ord_{\infty}}G\cdot\operatorname{ord_{\infty}}H=(-4)(-2)=8=-\operatorname{ord_{\infty}}f$.

The following result details the behaviour observed for the composition
$f=G\circ H$ in the example above.

######  Lemma 2.

Let $f\in K(x)$ and $L/K$ be an extension field. If $f=g\circ h$ for some
$g,h\in L(x)$ with $\operatorname{ord_{\infty}}h<0$, then
$-\operatorname{ord_{\infty}}f=\operatorname{ord_{\infty}}g\cdot\operatorname{ord_{\infty}}h$.

###### Proof.

Writing $h=h_{1}/h_{2}$ and

$g(x)=b\frac{\prod_{i=1}^{m_{1}}(x-\alpha_{i})}{\prod_{j=1}^{m_{2}}(x-\beta_{j})}$

with all $\alpha_{i},\beta_{j}\in\overline{K}$ yields

$f(x)=g\big{(}h(x)\big{)}=\frac{bh_{2}(x)^{\deg
g-m_{1}}\prod_{i=1}^{m_{1}}\big{(}h_{1}(x)-\alpha_{i}h_{2}(x)\big{)}}{h_{2}(x)^{\deg
g-m_{2}}\prod_{j=1}^{m_{2}}\big{(}h_{1}(x)-\beta_{j}h_{2}(x)\big{)}}.$

Since $\operatorname{ord_{\infty}}h=\deg h_{2}-\deg h_{1}<0$ by assumption, we
have

$\displaystyle\operatorname{ord_{\infty}}f$ $\displaystyle=$
$\displaystyle\big{(}(\deg g-m_{2})\deg h_{2}+m_{2}\deg
h_{1}\big{)}-\big{(}(\deg g-m_{1})\deg h_{2}+m_{1}\deg h_{1}\big{)}$
$\displaystyle=$ $\displaystyle(m_{1}-m_{2})(\deg h_{2}-\deg h_{1})$
$\displaystyle=$
$\displaystyle-\operatorname{ord_{\infty}}g\cdot\operatorname{ord_{\infty}}h$

as claimed. ∎

It is beneficial then to have decompositions $f=g\circ h$ for which
$\operatorname{ord_{\infty}}h<0$. To construct one such decomposition when
$\operatorname{ord_{\infty}}h\geq 0$, write $h=h_{1}/h_{2}$ so that dividing
$h_{1}$ by $h_{2}$ in $K[x]$ yields $a\in K$ and $r\in K[x]$ with
$h_{1}=ah_{2}+r$, and either $r=0$ or $\deg r<\deg h_{2}$. Then
$h=h_{1}/h_{2}=a+r/h_{2}$, and we define $\mu(x)=1/(x-a)$. This yields
$f=(g\circ\mu^{-1})\circ(\mu\circ h)$ where $H=\mu\circ h$ satisfies
$\operatorname{ord_{\infty}}H=\operatorname{ord_{\infty}}(h_{2}/r)=\deg r-\deg
h_{2}<0$. In Example 3.1, notice that for $h_{1}(x)=x^{2}+1$ and
$h_{2}(x)=x^{4}+1$, we have the trivial equation of $h_{1}(x)=0\cdot
h_{2}(x)+h_{1}(x)$ so that $a=0$ and $r=h_{1}$.

For a polynomial $f\in K[x]$, $-\operatorname{ord_{\infty}}f$ and $\deg f$ are
equal. Since $\deg$ retains some of its properties over $K[x]$ when extended
to $K(x)$, such as the one given in Theorem 3, it is natural to expect
$\operatorname{ord_{\infty}}$ to retain some properties as well. One such
interesting property related to the derivative is noted below.

######  Theorem 6.

Let $f\in K(x)$. If $\operatorname{ord_{\infty}}f$ is non-zero, then
$-\operatorname{ord_{\infty}}f^{\prime}=(-\operatorname{ord_{\infty}}f)-1$.

###### Proof.

For $f\in K(x)$ with $\operatorname{ord_{\infty}}f\neq 0$, we may write
$-\operatorname{ord_{\infty}}f=n_{1}-n_{2}$ for some
$n_{1},n_{2}\in\mathbb{Z}$ and

$f(x)=\frac{ax^{n_{1}}+r_{1}(x)}{x^{n_{2}}+r_{2}(x)}$

where $a\neq 0$ and $r_{1},r_{2}\in K[x]$ with $\deg r_{1}<n_{1}$ and $\deg
r_{2}<n_{2}$. The derivative of $f$ can be obtained by simplifying the product
rule

$\frac{\big{(}an_{1}x^{n_{1}-1}+r_{1}^{\prime}(x)\big{)}\big{(}x^{n_{2}}+r_{2}(x)\big{)}-\big{(}ax^{n_{1}}+r_{1}(x)\big{)}\big{(}n_{2}x^{n_{2}-1}+r_{2}^{\prime}(x)\big{)}}{\big{(}x^{n_{2}}+r_{2}(x)\big{)}^{2}}$

which we may write concisely as

$\frac{a(-\operatorname{ord_{\infty}}f)x^{n_{1}+n_{2}-1}+R_{1}(x)}{x^{2n_{2}}+R_{2}(x)}$

for some $R_{1},R_{2}\in K[x]$ where $\deg R_{1}<n_{1}+n_{2}-1$ and $\deg
R_{2}<2n_{2}$. Then
$-\operatorname{ord_{\infty}}f^{\prime}=(n_{1}+n_{2}-1)-(2n_{2})=n_{1}-n_{2}-1=(-\operatorname{ord_{\infty}}f)-1$.
∎

Finally, the refinement of Ayad’s result [1] in Theorem 5 is as follows.

######  Theorem 7.

Let $f\in K(x)$, and let $d$ be the greatest proper divisor of $\deg f$. If
$\operatorname{ord_{\infty}}f\neq 0$ is divisible by a prime $p>d$, then $f$
is prime over $\overline{K}$.

###### Proof.

Suppose that $f$ is composite. Then there exist $g,h\in\overline{K}(x)$ with
$\operatorname{ord_{\infty}}h<0$ such that $f=g\circ h$. By Lemma 2, either
$\operatorname{ord_{\infty}}h$ or $\operatorname{ord_{\infty}}g$ is divisible
by $p>d$, so that $|\operatorname{ord_{\infty}}h|\geq p>d$ or
$|\operatorname{ord_{\infty}}g|\geq p>d$. However, each also satisfies
$|\operatorname{ord_{\infty}}h|\leq\deg h\leq d$ or
$|\operatorname{ord_{\infty}}g|\leq\deg g\leq d$, which is a contradiction in
either case. ∎

This result does indeed extend Theorem 5. To see this, note that it is assumed
that the polynomial $f$ admits a prime $p>d$ and some $a$ such that
$v_{f}(a)=p$. In such a case, $f-f(a)$ has a zero of order $v_{f}(a)$ at $a$,
and defining the unit $\mu(x)=(ax+1)/x$ gives a rational function
$\big{(}f-f(a)\big{)}\circ\mu$ with
$-\operatorname{ord_{\infty}}\big{(}\big{(}f-f(a)\big{)}\circ\mu\big{)}=v_{f}(a)=p$.
Such a rational function is prime by Theorem 7, and so $f$ is prime as well
since it can be obtained from the composition with a unit.

##  4\. Decompositions and the resultant

In tis section, we first recall the resultant of two polynomials and the
discriminant of a polynomial in $K[x]$. Then, after we present a theorem
relating the discriminant to the resultant of composite functions for the
polynomial case, we generalize the notion to two rational functions of $K(x)$
and prove a similar result.

The resultant of the two polynomials $f,g\in K[x]$ given by
$f(x)=a_{n}x^{n}+\cdots+a_{0}$ and $g(x)=b_{m}x^{m}+\cdots+b_{0}$ is defined
as

$\text{Res}_{x}(f,g)=a_{n}^{m}b_{m}^{n}\prod_{\alpha,\beta}(\alpha-\beta)$

where $\alpha,\beta\in\overline{K}$ run over all of the roots of $f$ and $g$,
respectively. Three well-known properties of the resultant are as follows.
Given $f,g,h\in K[x]$, we have

  1. (1)

$\text{Res}_{x}(f,g)=(-1)^{\deg f\deg g}\text{Res}_{x}(g,f)$,

  2. (2)

$\text{Res}_{x}(f,gh)=\text{Res}_{x}(f,g)\text{Res}_{x}(f,h)$, and

  3. (3)

$\text{Res}_{x}(f,g)=0$ if and only $f$ and $g$ have a common root in
$\overline{K}$.

These properties all follow directly from the definition. The factorization of
one of the polynomials as $g(x)=b_{m}\displaystyle\prod_{\beta}(x-\beta)$ also
shows that

  1. (4)

$\text{Res}_{x}(f,g)=a_{n}^{m}\displaystyle\prod_{\alpha}g(\alpha)$.

The above property in turn shows that for a constant $c\in\overline{K}$ we
have

  1. (5)

$\text{Res}_{x}(f,cg)=c^{n}\text{Res}_{x}(f,g)$.

The discriminant of $f\in K[x]$ of degree $n$ and leading coefficient $a_{n}$
is given by

$D[f]=\frac{(-1)^{n(n-1)/2}}{a_{n}}\text{Res}_{X}(f,f^{\prime}).$

Letting $t$ be a new variable, as in Ayad [1], $b\in\overline{K}$ is a
critical value of $f$ if and only if it is a root of $D[f-t]$. The
multiplicity of a critical value is defined as its multiplicity as a root of
$D[f-t]$, and we call a critical value with multiplicity 1 a simple critical
value.

The following theorem, analogous to Ayad [1, Lemma 2], demonstrates how
$D[f-t]$ can be factored.

######  Theorem 8.

Let $f=g\circ h$ for $f,g,h\in K[x]$ and let $D[f-t]$ be the discriminant of
$f-t$. There exists $a\in K$ such that

$D[f-t]=aD[g-t]^{\deg h}\text{Res}_{x}\big{(}f(x)-t,h^{\prime}(x)\big{)}.$

###### Proof.

Let $n$ and $a_{n}$ be the degree and leading coefficient of $f$ respectively.
We have

$\displaystyle D[f-t]$
$\displaystyle=\frac{(-1)^{n(n-1)/2}}{a_{n}}\text{Res}_{x}\big{(}f(x)-t,f^{\prime}(x)\big{)}$
$\displaystyle=\frac{(-1)^{n(n-1)/2}}{a_{n}}\text{Res}_{x}\Big{(}g\big{(}h(x)\big{)}-t,g^{\prime}\big{(}h(x)\big{)}\Big{)}\text{Res}_{x}\Big{(}g\big{(}h(x)\big{)}-t,h^{\prime}(x)\Big{)}$

Write $g^{\prime}(x)=\displaystyle(\deg g)b\prod_{\beta}(x-\beta)$ where
$\beta\in\overline{K}$ runs over all roots of $g^{\prime}$, and $b\in K$ is
the leading coefficient of $g$. Then the properties (5), (1), and (2) of the
resultant can be applied here to yield

$\displaystyle\text{Res}_{x}\Big{(}g\big{(}h(x)\big{)}-t,g^{\prime}\big{(}h(x)\big{)}\Big{)}$
$\displaystyle=\text{Res}_{x}\Big{(}g\big{(}h(x)\big{)}-t,b(\deg
g)\prod_{\beta}\big{(}h(x)-\beta\big{)}\Big{)}$ $\displaystyle=(b(-1)^{\deg
g^{\prime}\deg h}\deg g)^{\deg
f}\prod_{\beta}\text{Res}_{x}\Big{(}h(x)-\beta,g\big{(}h(x)\big{)}-t\Big{)}.$

Letting $c$ denote the leading coefficient of $h(x)$, we have

$\displaystyle\text{Res}_{x}\Big{(}g\big{(}h(x)\big{)}-t,g^{\prime}\big{(}h(x)\big{)}\Big{)}$
$\displaystyle=(b(-1)^{\deg g^{\prime}\deg h}\deg g)^{\deg
f}\prod_{\beta}c^{\deg f}\big{(}g(\beta)-t\big{)}^{\deg h}$
$\displaystyle=(b(-1)^{\deg g^{\prime}\deg h}c^{\deg g^{\prime}}\deg g)^{\deg
f}\prod_{\beta}\big{(}g(\beta)-t\big{)}^{\deg h}.$

From the definition of the discriminant of $g-t$ and the resultant properties
(1) and (4), we have

$\displaystyle D[g-t]^{\deg h}$ $\displaystyle=\left[\frac{(-1)^{\deg g((\deg
g)-1)/2}}{b}\right]^{\deg
h}\Big{(}\text{Res}_{x}\big{(}g(x)-t,g^{\prime}(x)\big{)}\Big{)}^{\deg h}$
$\displaystyle=\left[\frac{(-1)^{\deg g((\deg g)-1)/2}}{b}\right]^{\deg
h}(b(-1)^{\deg g^{\prime}}\deg g)^{\deg
f}\prod_{\beta}\big{(}g(\beta)-t\big{)}^{\deg h}$ $\displaystyle=(-1)^{\deg
f(\deg g-1)/2+\deg f\deg g^{\prime}}b^{\deg f-\deg h}(\deg g)^{\deg
f}\prod_{\beta}\big{(}g(\beta)-t\big{)}^{\deg h},$

so that

$\displaystyle\text{Res}_{x}\Big{(}g\big{(}h(x)\big{)}-t,g^{\prime}\big{(}h(x)\big{)}\Big{)}$
$\displaystyle=\frac{(b(-1)^{\deg g^{\prime}\deg h}c^{\deg g^{\prime}}\deg
g)^{\deg f}D[g-t]^{\deg h}}{(-1)^{\deg f(\deg g-1)/2+\deg f\deg
g^{\prime}}b^{\deg f-\deg h}(\deg g)^{\deg f}}$ $\displaystyle=(-1)^{n\deg
g^{\prime}(\deg h-1)-n(\deg g-1)/2}b^{\deg h}c^{n\deg g^{\prime}}D[g-t]^{\deg
h}.$

Now, setting $a=\displaystyle\frac{(-1)^{n(n-1)/2-n(\deg g-1)/2+n\deg
g^{\prime}(\deg h-1)}b^{\deg h}c^{n\deg g^{\prime}}}{a_{n}}$ yields the
desired expression

$D[f-t]=aD[g-t]^{\deg
h}\text{Res}_{x}\Big{(}g\big{(}h(x)\big{)}-t,h^{\prime}(x)\Big{)}.$

∎

The next corollary follows directly from Theorem 8. We include the proof since
Ayad [1, Corollary 2] does not.

######  Corollary 1.

Let $f\in K[x]$ and let $D(t)$ be the discriminant of $f-t$. If $f(x)$ is
composite with a right composition factor of degree $k$, then there exist
polynomials $A,B\in K[t]$ such that $D(t)=A(t)^{k}B(t)$ and $\deg B\leq k-1$.
Moreover, if $\deg f\neq 0$, then $\deg B=k-1$.

###### Proof.

Let $f,g,h\in K[x]$ such that $f=g\circ h$ with $\deg h=k$. By applying
Theorem 8 and setting
$B(t)=a\text{Res}_{x}\Big{(}g\big{(}h(x)\big{)}-t,h^{\prime}(x)\Big{)}$ and
$A(t)=D[g-t]$, we have that $D(t)=A(t)^{k}B(t)$ with $A,B\in K[t]$.

Further, assuming that $\deg f$ is non-zero, then its divisor $k=\deg
h\leq\deg f$ is non-zero as well. This implies that $\deg B=\deg
h^{\prime}=(\deg h)-1=k-1$. ∎

######  Theorem 9.

Let $f\in K[x]$, and let $d$ be the greatest proper divisor of $\deg f$. If
$f$ has at least $d$ simple critical values, then $f$ is prime over $K$.

###### Proof.

Suppose that $f$ is composite with a right composition factor of degree $k$.
Then $2\leq k\leq d$, and we write the discriminant of $f-t$ as
$D(t)=A(t)^{k}B(t)$ as in Corollary 1. Since $f$ has at least $d$ simple
critical values, and each simple critical value is necessarily a root of
$B(t)$, we have $k\leq d\leq\deg B\leq k-1$, which is a contradiction. Thus,
$f$ is prime. ∎

Kihel and Larone [4] studied the resultant of the two rational functions
$f,g\in\mathbb{C}(x)$, which they defined as the resultant of their
numerators. This idea translates well to elements of $K(x)$, so we take the
same definition in $K(x)$. The following example illustrates that the result
is the expected one that coincides with the polynomial case.

######  Example 4.1.

Let

$f(x)=\frac{(x+1)^{4}}{x^{3}}\qquad\text{so}\qquad
f^{\prime}(x)=\frac{(x+1)^{3}(x-3)}{x^{4}}.$

The list of all critical points of $f$ are $(-1,0),(-1,0),(-1,0),(3,256/27)$.
Note that

$\text{Res}_{x}\big{(}f(x)-t,f^{\prime}(x)\big{)}=t^{3}(256-27T)$

has $0$ and $256/27$ as its roots with multiplicities 3 and 1 respectively.

Kihel and Larone’s proofs of Lemma 3.2 and Corollary 3.3 in [4] deal with only
$\mathbb{C}$ but immediately translate to $\overline{K}$ under the assumptions
made here, yielding the following result analogous to Theorem 8.

######  Theorem 10.

Let $f\in K(x)$ be composite over $\overline{K}$ with $f=g\circ h$. There
exist an integer $\ell\geq 0$ and polynomials $A,B\in\overline{K}[t]$ such
that

$\text{Res}_{x}(f(x)-t,f^{\prime}(x))=t^{\ell}A(t)^{\deg h}B(t)$

where $\deg B\geq 2(\deg h)-1$. Moreover, if $\gcd(\deg
f,\operatorname{ord_{\infty}}f)=1$ and $\operatorname{ord_{\infty}}f>d$ where
$d$ is the greatest proper divisor of $\deg f$, then $\ell>0$.

The polynomial $\text{Res}_{x}(f(x)-t,f^{\prime}(x))$ acts as the analog of
the discriminant for rational functions $f\in K(x)$ in the sense that
$b\in\overline{K}$ is a critical value of $f$ if and only if $b$ is a root of
$\text{Res}_{x}(f(x)-t,f^{\prime}(x))$. Extending the nomenclature, we call a
critical value with multiplicity 1 as a root of
$\text{Res}_{x}(f(x)-t,f^{\prime}(x))$ a simple critical value of $f$.

######  Corollary 2.

Let $f\in K(x)$, and let $d$ be the greatest proper divisor of $\deg f$. If
$f$ has at least $2d$ non-zero simple critical values, then $f$ is prime over
$\overline{K}$.

The proof is similar to that of Theorem 9 with instead the inequality

$2\deg h\leq 2d\leq\deg B\leq 2\deg h-1$

yielding the contradiction.

## References

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  * [4] O. Kihel and J. Larone, “Prime rational functions,” Acta Arith., 169 (2015), no. 1, 29–46.
  * [5] H. Levi, “Composite polynomials with coefficients in an arbitrary field of characteristic zero.” Amer. J. Math. 64 (1942), 389–400.
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