# Rationalizability of field extensions

Dino Festi, Andreas Hochenegger

###### Abstract.

In this paper we generalize the definition of rationalizability for square
roots of polynomials introduced by M. Besier and the first author to field
extensions. We then show that the rationalizability of a set of field
extensions is equivalent to the rationalizability of the compositum of the
field extensions, providing a new strategy to prove rationalizability of sets
of square roots of polynomials.

##  1\. Introduction

The argument of Feynman integrals in high-energy physics often contains a
square root. This square root makes the numerical evaluation of the integral
computationally more expensive, preventing the achievement of higher precision
that could lead to more incisive experiments. This issue can often be overcome
with a change of variables in the integrals making the square root disappear
and leaving only a rational expression. Unfortunately, such change of
variables does not always exist and stating its (non-)existence can be a
cumbersome problem for physicists with no strong background in arithmetic
geometry as well as for experienced geometers. This challenge led Marco Besier
and the first author to introduce the concept of _rationalizability_ of square
roots of polynomials, in [1]. Roughly speaking, a square root of a polynomial
is rationalizable if there is a rational change of variables making the square
root disappear, i.e., turning the polynomial into a square. In the same paper,
some practical criteria for the rationalizability of square roots of
polynomials in one and two variables are given. Finally, the concept of
rationalizability for _sets_ (also called _alphabets_ in that paper) of square
roots of polynomials is introduced. On the one hand, there is a practical
strategy to disprove rationalizability of a given set (cf. [1, Proposition 47
and Remark 48]), but on the other hand the original paper does not offer a
solid strategy to _prove_ the rationalizability (cf. [1, Remarks 48 and 52]).

In this paper, we try to fill in this gap. To do so, we introduce the notion
of rationalizability for field extensions, generalizing the concept of
rationalizability for square roots.

######  Definition 1.1.

Let $K$ be any field. We say that a field extension $K\subset L$ is
_rationalizable_ if there exists a non-zero homomorphism $\psi\colon L\to K$.

For $i\in\\{1,...,m\\}$ let $L_{i}$ be field extensions of $K$, with
$\iota_{i}\colon K\hookrightarrow L$ the natural inclusion. We say that the
set $\\{L_{1},...,L_{m}\\}$ is _rationalizable_ if for each $i$ there is a
non-zero homomorphism $\psi_{i}\colon L_{i}\to K$ such that

$\psi_{1}\circ\iota_{1}=...=\psi_{m}\circ\iota_{m}\;.$

######  Example 1.2.

Consider $K={\mathbb{C}}(x)$, $p=z^{2}-(x-1)\in K[z]$ and

$L:=K(\sqrt{x-1})=\frac{K[z]}{(p)}.$

Then $K\subset L$ is rationalizable. Indeed consider the map $\psi\colon L\to
K$ defined as the identity on the elements of ${\mathbb{C}}$ and by $x\mapsto
x^{2}+1,z\mapsto x$. With this definition, the map $\phi\colon K\to K$ sends
$x-1$ to $x^{2}$, (cf. [1, Remark 52]).

######  Example 1.3.

Consider $K={\mathbb{C}}(x)$ and set
$L_{1}:=K(\sqrt{x}),L_{2}:=K(\sqrt{1-x})$. Then the set of field extensions
$\\{L_{1},L_{2}\\}$ is rationalizable. Indeed, for $i=1,2$, consider the
homomorphisms $\psi_{i}\colon L_{i}\to K$ defined as follows.

$\displaystyle\psi_{1}\colon$ $\displaystyle\hskip 10.0pt\sqrt{x}\hskip
11.0pt\mapsto\hskip 3.0pt\frac{2x^{2}}{1+x^{2}}$ $\displaystyle\psi_{2}\colon$
$\displaystyle\sqrt{1-x}\hskip 3.0pt\mapsto\hskip
3.0pt\frac{1-x^{2}}{1+x^{2}}$

One can easily check that

$\psi_{1}\circ\iota_{1}=\psi_{2}\circ\iota_{2}=:\phi\;,$

where $\phi\colon K\to K$ is the homomorphism defined by

$\phi\colon x\mapsto\left(\frac{2x^{2}}{1+x^{2}}\right)^{2}\;.$

Later we will see that the rationalizability of the set considered in this
example does not depend on the special form of the polynomials but only on
their degree, cf. Section 4.

Our main result is the following theorem, which shows that the
rationalizability of a set of field extensions is equivalent to the
rationalizability of the compositum of the field extensions.

######  Theorem 1.4.

Let $K$ be a field and $p_{1},\ldots,p_{m}\in K[z]$ irreducible; let $L_{i}$
be the extension of $K$ obtained by adjoining a root $\alpha_{i}$ of $p_{i}$
to $K$, that is, $L_{i}=K(\alpha_{i})\cong K[z]/(p_{i})$; let $L$ be the
compositum of the $L_{i}$’s, i.e., $L=K(\alpha_{1},...,\alpha_{m})$. Then the
extension $K\subset L$ is rationalizable if and only if the family of
extensions $\\{L_{1},...,L_{m}\\}$ is.

This allows us to develop a new strategy to prove rationalizability of sets of
square roots of polynomials (cf. Section 3.2).

We proceed as follows. In Section 2 we recall the notion of rationalizability
for square roots of polynomials and some well-known facts about field
extensions. In Section 3 we prove Theorem 1.4 and we outline a strategy to
prove rationalizability of sets of square roots. Finally, in Section 4 we
apply Theorem 1.4 to some sets of square roots of polynomials. The results
thus obtained, joint with [1, Proposition 47], led us to formulate the
following conjecture.

######  Conjecture 1.5.

The set of square roots $\left\\{\sqrt{f_{1}},...,\sqrt{f_{m}}\right\\}$ is
rationalizable if and only if for every non-empty subset
$J\subseteq\\{1,...,m\\}$ the square root

$\sqrt{\prod_{j\in J}f_{j}}$

is rationalizable.

### Acknowledgments

We thank Marco Besier for stimulating conversations and useful remarks.

##  2\. Preliminaries

In this section we recall the notion of rationalizability of square roots
introduced in [1] and we extend it to higher order roots; we also recall some
basic results in field theory. Proofs are omitted or referred to the original
sources.

###  2.1. Rationalizability of square roots

Let $\mathbf{k}$ be any field and consider the $\mathbf{k}$-algebra
$Q:=\mathbf{k}(x_{1},\ldots,x_{n})$.

######  Definition 2.1.

Consider $q\in Q$. We say that the square root $\sqrt{q}$ is _rationalizable_
if there exists $\phi\colon Q\to Q$ such that $\phi(q)$ is a square in $Q$.

We say that a family of square roots $\\{\sqrt{q_{1}},\ldots,\sqrt{q_{m}}\\}$
is _rationalizable_ if there exists $\phi\colon Q\to Q$ such that
$\phi(q_{i})$ is a square in $Q$ for $i\in\\{1,\ldots,m\\}$.

We recall the main theorems of [1] as they will serve as a starting point for
the theory developed here. The notation and statements are slightly changed to
better fit the content of this article.

######  Theorem 2.2 ([1, Theorem 1]).

The following statements hold:

  1. (1)

There exists a squarefree polynomial $f\in R=\mathbf{k}[x_{1},...,x_{n}]$ such
that $\sqrt{q}$ is rationalizable if and only if $\sqrt{f}$ is.

  2. (2)

The square root $\sqrt{f}$ is rationalizable if and only if the variety
$V(z^{2}-f)\subset\operatorname{Spec}R[z]={\mathbb{A}}^{n+1}_{\mathbf{k}}$ is
unirational.

  3. (3)

Assume $\mathbf{k}$ is algebraically closed and let $d$ denote the degree of
$f$. If $d=1,2$, or if $\overline{V}$ has a singular point of multiplicity
$d-1$, then $\sqrt{f}$ is rationalizable.

######  Corollary 2.3.

Consider $f\in\mathbf{k}[x]$. Then the square root of $f$ is rationalizable if
and only if $V(z^{2}-f)\subset{\mathbb{A}}^{2}$ has geometric genus zero.

###  2.2. Rationalizability of roots of higher order

One can naturally generalize the concept of rationalizability of square roots
to roots of higher order. Although it is a very natural step, this subsection
will not be needed in the rest of the paper and non-interested reader can
safely skip it. Also in this subsection, let $\mathbf{k}$ be any field and $Q$
the $\mathbf{k}$-algebra $\mathbf{k}(x_{1},\ldots,x_{n})$. Consider $q\in Q$
of degree $d$ and an integer $e\geq 2$.

######  Definition 2.4.

We say that the root $\sqrt[e]{q}$ is _rationalizable_ if there exists a
homomorphism of $\mathbf{k}$-algebras $\phi\colon Q\to Q$ such that $\phi(q)$
is an $e$-power in $Q$, that is, $\phi(q)=h^{e}$, for some element $h\in Q$.
We say that a family of roots
$\\{\sqrt[e_{1}]{q_{1}},...,\sqrt[e_{m}]{q_{m}}\\}$ is rationalizable if there
exists a homomorphism of $\mathbf{k}$-algebras $\phi\colon Q\to Q$ such that
for every $i=1,...,m$ we have $\phi(q_{i})=h_{i}^{e_{i}}$ for some $h_{i}\in
Q$

Working exactly as in [1, §2], one can prove the following result.

######  Proposition 2.5.

The following statements hold:

  1. (1)

There exists a polynomial $f\in R=\mathbf{k}[x_{1},...,x_{n}]$ such that its
factorization does not contain any $e$-th power and $\sqrt[e]{q}$ is
rationalizable if and only if $\sqrt[e]{f}$ is.

  2. (2)

The root $\sqrt[e]{f}$ is rationalizable if and only if the variety
$V(z^{e}-f)\subset\operatorname{Spec}R[z]={\mathbb{A}}^{n+1}_{\mathbf{k}}$ is
unirational.

######  Remark 2.6.

Assume $f\in{\mathbb{C}}[x]$ and consider the root $\sqrt[e]{f}$. Then, up to
birational transformations of ${\mathbb{P}}^{1}$, we can always assume that
the curve $C=V(z^{e}-f)$ is given by the equation

$C\colon z^{e}=x^{\ell_{0}}\cdot\prod_{i=1}^{m}(x-a_{i})^{\ell_{i}}$

where $\sum_{i=0}^{m}\ell_{i}\equiv 0\bmod e$. Then, following [5, §2], we
have that its geometric genus is

$g(C)=\frac{(e-1)(s-2)}{2},$

where

$s=\begin{cases}m&\textrm{if }m\equiv 0\bmod e\;,\\\
m+1&\textrm{otherwise}\;.\end{cases}$

In view of Section 2.2, this yields that the root $\sqrt[e]{f}$ is
rationalizable if and only if $s=2$, i.e., $m=e=2$ or $m=1$.

###  2.3. Field theory

Before a further generalization of the notion of rationalizability, we recall
a few well-known facts about algebraic field extensions. For more information
and details see, for example, [2, Chapter 3].

######  Definition 2.7.

The field extension $K\subset L$ is called _separable_ if for every element
$\alpha\in L$, the minimal polynomial of $\alpha$ over $K$ is separable, i.e.,
it has no repeated roots in any field extension of $K$.

######  Proposition 2.8 ([2, Proposition 3.6.2]).

If $K$ is a field of characteristic $0$, then every algebraic extension is
separable.

######  Proposition 2.9 (Primitive element theorem).

Let $K\subset L$ be a separable field extension. Then there is an irreducible
element $p\in K[z]$ such that $L\cong K[z]/(p)$.

For a proof of this result we refer to [2, Proposition 3.6.12]. In the
examples in this article we will only consider fields of characteristic $0$
and hence separable. Therefore, using the above proposition, all the
considered field extensions $K\subset L$ will always be of the form
$L=K[z]/(p)$ with an irreducible polynomial $p$.

######  Remark 2.10.

Notice that it is also possible to write $L=K(\alpha)$ where $\alpha\in L$ is
a root of $p$ (cf. [2, Proposition 3.2.6]). The element $\alpha$ (or $p$) is
called a _primitive element_ of the extension $K\subset L$.

######  Example 2.11.

Let $f\in K$ be a squarefree element. Then the polynomial $z^{2}-f\in K[z]$ is
irreducible and $L=K[z]/(z^{2}-f)$ is a field extension of $K$. We can also
write $L=K(\sqrt{f})$, where $\sqrt{f}$ denotes a square root of $f$.

##  3\. Rationalizability of field extensions

In this section we prove the main theorem of the paper, Theorem 1.4, and we
outline how this theorem can be used to prove rationalizability of sets of
square roots.

For the whole section, let $\mathbf{k}$ be an algebraically closed field of
characteristic $0$ and set $R:=\mathbf{k}[x_{1},\ldots,x_{n}]$ and
$Q:=\operatorname{Frac}(R)=\mathbf{k}(x_{1},\ldots,x_{n})$, with $n\geq 1$; as
before, $K$ will denote any field and $K\subset L$ a field extension; the the
embedding $K\hookrightarrow L$ will be denoted by $\iota$.

Let us recall the definition of rationalizability for field extensions given
in the introduction.

######  Definition 3.1.

We say that a field extension $K\subset L$ is _rationalizable_ if and only if
there is a non-zero homomorphism $\psi\colon L\to K$.

######  Remark 3.2.

By taking the composition $\phi=\psi\circ\iota$, we get a commutative diagram

${K}$${K}$${L}$$\scriptstyle{\phi}$$\scriptstyle{\iota}$$\scriptstyle{\psi}$

Note that both $\phi$ and $\psi$ are automatically injective.

######  Definition 3.3.

Let $K$ be a field and let $\iota_{i}\colon K\hookrightarrow L_{i}$ be field
extensions for $i\in\\{1,\ldots,m\\}$. We say that the set of field extensions
$\\{L_{1},\ldots,L_{m}\\}$ is rationalizable if for each $i=1,...,m$ the field
extension $K\subset L_{i}$ is rationalizable and there exists a rationalizing
homomorphism $\psi_{i}\colon L_{i}\to K$ such that

$\psi_{1}\circ\iota_{1}=...=\psi_{m}\circ\iota_{m}=:\phi\;.$

In other words, for each $i\in\\{1,\ldots,m\\}$ we have the following
commutative diagram:

${K}$${K}$${L_{i}}$$\scriptstyle{\phi}$$\scriptstyle{\iota_{i}}$$\scriptstyle{\psi_{i}}$

######  Remark 3.4.

The central point of the definition is, that we can choose the homomorphisms
$\psi_{i}\colon L_{i}\to K$ in such a way, that we get the same homomorphism
$\phi=\psi_{i}\circ\iota_{i}$, independently of $i\in\\{1,\ldots,m\\}$.

###  3.1. First properties

In this subsection we show that the rationalizability of a square root in the
sense of [1] is indeed a special case of the rationalizability of a field
extension.

######  Proposition 3.5.

Consider $f\in R=\mathbf{k}[x_{1},\ldots,x_{n}]$ squarefree. Then $\sqrt{f}$
is rationalizable if and only if the field extension $Q\subset Q[z]/(z^{2}-f)$
is rationalizable.

Note that $f$ being squarefree implies the irreducibility of $z^{2}-f$,
assuring that $Q[z]/(z^{2}-f)$ is indeed a field.

###### Proof.

Assume $\sqrt{f}$ to be rationalizable: there is a $\phi\colon Q\to Q$ with
$\phi(f)=h^{2}$, $h\in Q$. Then we define $\psi\colon Q[z]/(z^{2}-f)\to Q$ by
mapping $q\in Q$ to $\psi(q)\coloneqq\phi(q)$ and by setting $\psi(z)\coloneqq
h$. This is indeed a well-defined homomorphism, as
$\psi(z^{2}-f)=(\psi(z))^{2}-\psi(f)=0$. Since $\psi$ is non-zero, we conclude
that $Q\subset Q[z]/(z^{2}-f)$ is rationalizable.

Conversely, assume $Q\subset Q[z]/(z^{2}-f)$ to be rationalizable, so there is
a non-zero homomorphism $\psi\colon Q[z]/(z^{2}-f)$. Using the composition

$\phi\coloneqq\psi\circ\iota\colon Q\hookrightarrow Q[z]/(z^{2}-f)\to Q\;,$

we see that $\phi(f)=\psi(z)^{2}$ is a square. Hence the square root of $f$ is
rationalizable. ∎

The following lemma is a generalization of [1, Lem. 2.7 & Cor. 2.8], but will
not be used later on.

######  Lemma 3.6.

If $Q^{\prime}=Q[z]/(q)$ for some irreducible $q\in Q[z]$, then there is an
irreducible $p\in R[z]$ such that $Q^{\prime}\cong Q[z]/(p)$.

If $q$ is monic, than $p$ can also be chosen to be monic.

###### Proof.

As $R$ is a unique factorization domain, there is a minimal $u\in R$ such that
$p\coloneqq u\cdot q\in R[z]$, by clearing denominators. Moreover, $p$ is
irreducible in $R[z]$, as any non-trivial factorization of $p$ would give rise
to a factorization of $q$. As $u$ is a unit in $Q$, we get that
$Q^{\prime}\cong Q[z]/(p)$.

In the case that $q$ is monic, that is, $q=z^{n}+c_{n-1}z^{n-1}+\cdots+c_{0}$,
we get that $p=uq=uz^{n}+uc_{n-1}z^{n-1}+\cdots+uc_{0}$ is not monic anymore.
By defining the monic
$\tilde{p}=z^{n}+\tilde{c}_{n-1}z^{n-1}+\cdots+\tilde{c}_{0}$ with
$\tilde{c}_{i}=c_{i}\cdot u^{n-i}$, we get that $\tilde{p}(uz)=u^{n}p(z)$. As
$Q[z]/(\tilde{p})\to Q[z]/(p),z\mapsto uz$ is an isomorphism, we find that
$Q^{\prime}\cong Q[z]/(\tilde{p})$ with $\tilde{p}\in R[z]$ monic. ∎

Note that the above proof uses only that $R$ is a unique factorization domain.

We give a geometric interpretation of rationalizability of a field extension
in the spirit of [1, §2.2].

######  Proposition 3.7.

Let $p$ be a polynomial in $R[z]$. Then $Q\subset Q[z]/(p)$ is rationalizable
if and only if $V(p)\subset\operatorname{Spec}R[z]$ is unirational.

###### Proof.

Recall that $\operatorname{Spec}R={\mathbb{A}}^{n}$ is rational, as
$Q=\mathbf{k}(x_{1},\ldots,x_{m})$.

If $Q\subset Q[z]/(p)$ is rationalizable, we have an (injective) homomorphism
$\psi\colon Q[z]/(p)\to Q$. Note that
$Q[z]/(p)=\operatorname{Frac}(R[z]/(p))=k(V(p))$, the function field of
$V(p)$, so

$\iota\circ\psi\colon\operatorname{Frac}(R[z]/(p))\hookrightarrow\mathbf{k}(x_{1},\ldots,x_{m})$

implies that $V(p)$ is unirational.

Conversely, suppose that $V(p)$ is unirational, that is, there is an injective
homomorphism

$k(V(p))=Q[z]/(p)=\operatorname{Frac}(R[z]/(p))\hookrightarrow\mathbf{k}(x_{1},\ldots,x_{m})\cong
Q.$

This shows that $Q\subset Q[z]/(p)$ is rationalizable. ∎

######  Remark 3.8.

The two implications can be generalized relaxing some hypotheses on $R$.

The first implication, i.e., that if $Q\subset Q[z]/(p)$ is rationalizable
then $V(p)$ is unirational, holds also for any $R$ such that
$\operatorname{Spec}R$ is unirational.

The converse implication holds also for an $R$ such that
$\mathbf{k}(x_{1},\ldots,x_{m})\hookrightarrow Q=\operatorname{Frac}(R)$.

In particular, the statement of Section 3.1 is true for any
$Q=\operatorname{Frac}(R)$ where $R$ is such that $\operatorname{Spec}R$ is
rational.

Finally, also the rationalizability of a family
$\\{\sqrt{f_{1}},\ldots,\sqrt{f_{m}}\\}$ can be understood in the context of
field extensions. Indeed, by the same arguments as in Section 3.1, one can
check that this definition generalizes the rationalizability of a set of
square roots.

######  Proposition 3.9.

Consider $f_{1},...,f_{m}\in R$ squarefree polynomials and define the field
extensions $Q\subset\frac{Q[z]}{(z^{2}-f_{i})}=:L_{i}$. Then the set
$\\{\sqrt{f_{1}},...,\sqrt{f_{m}}\\}$ is rationalizable if and only if the set
$\\{L_{1},...,L_{m}\\}$ is.

###### Proof.

As in Section 3.1. ∎

###  3.2. The main theorem

In this subsection we prove the main result of the paper, Theorem 1.4, and we
highlight its impact on the study of rationalizability of families, cf.
Section 3.2. For the convenience of the reader we re-state the theorem here.

######  Theorem 3.10.

Let $K$ be a field and $p_{1},\ldots,p_{m}\in K[z]$ irreducible; let $L_{i}$
be the extension of $K$ obtained by adjoining a root $\alpha_{i}$ of $p_{i}$
to $K$, that is, $L_{i}=K(\alpha_{i})\cong K[z]/(p_{i})$; let $L$ be the
compositum of the $L_{i}$’s, i.e., $L=K(\alpha_{1},...,\alpha_{m})$. Then the
extension $K\subset L$ is rationalizable if and only if the family of
extensions $\\{L_{1},...,L_{m}\\}$ is.

###### Proof.

Suppose the field $L$ is rationalizable, so we have a commutative diagram:

${K}$${K}$${L}$$\scriptstyle{\phi}$$\scriptstyle{\iota}$$\scriptstyle{\psi}$

Note that for each $i\in\\{1,\ldots,m\\}$ we find that $K\subset L_{i}\subset
L$ (by construction of $L$). For every $i=1,...,m$, let $\jmath_{i}$ denote
the inclusion $L_{i}\hookrightarrow L$. Then, defining $\psi_{i}\colon
L_{i}\to K$ as the composition $\psi\circ\jmath_{i}$, we see that the family
$\\{L_{1},\ldots,L_{m}\\}$ is rationalizable:

${K}$${K}$${L_{i}}$${L}$$\scriptstyle{\phi}$$\scriptstyle{\imath_{i}}$$\scriptstyle{\imath}$$\scriptstyle{\jmath_{i}}$$\scriptstyle{\psi_{i}}$$\scriptstyle{\psi}$

Conversely, suppose the family of field extensions
$\\{L_{i}=K(\alpha_{i})\\}_{i\in\\{1,...,m\\}}$ is rationalizable with
homomorphisms $\phi\colon K\to K$ and $\psi_{i}\colon L_{i}\to K$. Note that
the homomorphisms $\psi_{i}$ are completely determined by $\psi_{i}|_{K}=\phi$
and the image of $\psi_{i}(\alpha_{i})$. We inductively construct a sequence
of field extensions

$K\subset M_{1}\subset M_{2}\subset\cdots\subset M_{m}$

that fit into the following commutative diagram:

${K}$ ${K}$ ${M_{1}}$ $\ddots$ ${M_{m}}$
$\scriptstyle{\phi}$$\scriptstyle{\pi_{1}}$$\scriptstyle{\pi_{m}}$

As the starting point, we set $M_{1}\coloneqq L_{1}=K(\alpha_{1})$ and
$\pi_{1}\coloneqq\psi_{1}$. Suppose now that $K\subset
M_{1}\subset\cdots\subset M_{i}$ are constructed. Consider now $p_{i+1}\in
K[z]$ as a polynomial in $M_{i}[z]$. We can distinguish two cases:

  * •

if $\alpha_{i+1}$ is in $M_{i}$: set $M_{i+1}\coloneqq M_{i}$ and
$\pi_{i+1}\coloneqq\pi_{i}$.

  * •

if $\alpha_{i+1}$ is not in $M_{i}$: Set $M_{i+1}\coloneqq
M_{i}(\alpha_{i+1})$ and define $\pi_{i+1}$ by setting
$\pi_{i+1}|_{M_{i}}\coloneqq\pi_{i}$ and
$\pi_{i+1}(\alpha_{i+1})\coloneqq\psi_{i+1}(\alpha_{i+1})$.

By construction, $M_{m}=L$. Hence we get a homomorphism
$\psi\coloneqq\pi_{m}\colon L\to K$ and the commutative diagram,

${K}$${K}$${L=M_{m}}$$\scriptstyle{\phi}$$\scriptstyle{\psi=\pi_{m}}$

i.e., the extension $K\subset L$ is rationalizable. ∎

######  Remark 3.11.

Note that $L\cong K[z]/(p)$ for some irreducible $p\in K[z]$, a so-called
primitive element of the extension. So the above proposition reduces the
question about the rationalizability of a family of field extensions to the
rationalizability of a single field extension which, in turn, is equivalent to
the unirationality of the variety $V(p)$. This will prove very useful in the
study of rationalizability of families of square (or higher order) roots and
represents a more solid approach to the problem than the one given in [1,
Remark 52].

On the other hand, this result is not very handy when it comes to _disprove_
rationalizability of sets of square roots; for this goal [1, Proposition 47]
stays a more practical tool.

##  4\. Applications and further discussions

In this section we use Theorem 1.4 to deduce some results about
rationalizability of families of square roots; we then apply these new results
to some examples taken from [1], recovering the same conclusions with shorter
arguments.

The computation of a primitive element of a field extension can be done by
hand (if the degree is small enough) or with the aid of any computer algebra
system. For some of the computations we used the algebra software Magma [3]
and Singular [4], as they allow the computation of the geometric genus of a
curve.

As before $\mathbf{k}$ denotes an algebraically closed field of characteristic
$0$ and we define $R:=\mathbf{k}[x_{1},\ldots,x_{n}]$ and
$Q:=\operatorname{Frac}(R)=\mathbf{k}(x_{1},\ldots,x_{n})$, with $n\geq 1$.

######  Corollary 4.1.

For any two linear polynomials $f,g\in Q$ the field extension $Q\subset
Q(\sqrt{f},\sqrt{g})$ is rationalizable.

###### Proof.

By Theorem 1.4 $Q\subset Q(\sqrt{f},\sqrt{g})$ is rationalizable if and only
if the family of extensions
$\\{\frac{Q[z]}{(z^{2}-f)},\frac{Q[z]}{(z^{2}-g)}\\}$ is, which in turn, by
Section 3.1, is equivalent to saying that the set of square roots
$\\{\sqrt{f},\sqrt{g}\\}$ is rationalizable. So in order to prove the
statement it is enough to prove that $\\{\sqrt{f},\sqrt{g}\\}$ is
rationalizable. In order to do so we will use the strategy already explained
in [1, Remark 52].

First note that without loss of generality, we can always assume $f=x_{1}$.
This implies that $\sqrt{f}$ is rationalized, for instance, by the map
$\phi_{1}\colon Q\to Q$, sending $x_{1}\mapsto x_{1}^{2}$ and acting as the
identity on all the other variables. In this way we obtain $\phi_{1}(f)=x^{2}$
and $\phi_{1}(g)=:g^{\prime}$ is a polynomial of degree $2$. Consider the
square root $\sqrt{g^{\prime}}$ and notice that it is rationalizable (cf. [1,
Theorem 1]). Let $\phi_{2}\colon Q\to Q$ be the rationalizing homomorphism
then we have $\phi_{2}(g^{\prime})=h^{2}$ for some $h\in Q$. Consider then the
composition $\phi:=\phi_{2}\circ\phi_{1}\colon Q\to Q$ and note that
$\phi(f)=(\phi_{2}(x))^{2}$ and $\phi(g)=h^{2}$, proving that
$\\{\sqrt{f},\sqrt{g}\\}$ is rationalizable. ∎

######  Corollary 4.2.

Let $a_{1},...,a_{m}\in{\mathbb{C}}$ be pairwise distinct complex numbers. The
set ${\mathcal{A}}:=\\{\sqrt{x-a_{1}},...,\sqrt{x-a_{m}}\\}$ is rationalizable
if and only if $m\leq 2$.

###### Proof.

If $m=1$, then ${\mathcal{A}}$ is trivially rationalizable (cf. [1, Theorem
1]).

If $m=2$, then Section 4 implies that ${\mathcal{A}}$ is rationalizable.

If $m\geq 3$, then the polynomial $p:=\prod_{i=1}^{m}(x-a_{i})$ is of degree
$m\geq 3$ with $m$ distinct roots. It follows that the square root $\sqrt{p}$
is not rationalizable (cf. [1, Theorem 2]) and hence, in turn, ${\mathcal{A}}$
is not rationalizable by [1, Proposition 47]. ∎

######  Remark 4.3.

We observe that Section 4 can also be proved using Section 2.2.

######  Example 4.4 (c.f. [1, Example 52]).

The set $\\{\sqrt{x-1},\sqrt{x-2}\\}$ is rationalizable by Section 4.

######  Corollary 4.5.

Let $a,b,c\in{\mathbb{C}}$ three pairwise distinct complex numbers. Then the
set

${\mathcal{A}}:=\\{\sqrt{(x-a)(x-b)},\sqrt{(x-a)(x-c)},\sqrt{(x-b)(x-c)}\\}$

is rationalizable.

###### Proof.

By applying an affine transformation of
${\mathbb{A}}^{1}\subset{\mathbb{P}}^{1}$ we can assume without loss of
generality that $a=0$ and $b=1$ and write $c=\lambda$. Set

$\displaystyle f_{1}$ $\displaystyle:=(x-a)(x-b)=x(x-1),$ $\displaystyle
f_{2}$ $\displaystyle:=(x-a)(x-c)=x(x-\lambda),$ $\displaystyle f_{3}$
$\displaystyle:=(x-b)(x-c)=(x-1)(x-\lambda).$

and, for $i=1,2,3$, consider the field extension

$L_{i}:=\frac{{\mathbb{C}}(x)[z]}{(z^{2}-f_{i})}\cong{\mathbb{C}}(x)(\sqrt{f_{i}}).$

Then the rationalizability of ${\mathcal{A}}$ is equivalent to the
rationalizability of the set $\\{L_{1},L_{2},L_{3}\\}$. From Theorem 1.4 this
is equivalent to the rationalizability of the compositum field
$L={\mathbb{C}}(x)(\sqrt{f_{1}},\sqrt{f_{2}},\sqrt{f_{3}})$.

First notice that, as

$\sqrt{f_{3}}=\sqrt{(x-1)(x-\lambda)}=\frac{1}{x}\sqrt{x(x-1)}\sqrt{x(x-\lambda)}=\frac{1}{x}\sqrt{f_{1}}\sqrt{f_{2}},$

the compositum field $L$ is equal to just
${\mathbb{C}}(x)(\sqrt{f_{1}},\sqrt{f_{2}})$. It is then immediate to see that
$[L:{\mathbb{C}}(x)]=4$. Let $p\in{\mathbb{C}}(x)[z]$ be the irreducible
polynomial of the primitive element of $L$ over ${\mathbb{C}}(x)$, that is,
$p$ is an irreducible polynomial of degree $4$ such that

$L\cong\frac{{\mathbb{C}}(x)[z]}{(p)}.$

Notice that in general, by possibly multiplying by an element of
${\mathbb{C}}(x)$, we can always assume that $p\in{\mathbb{C}}[x,z]$. In
particular, in our case

$p=z^{4}+(-4x+2\lambda+2)xz^{2}+(\lambda^{2}-2\lambda+1)x^{2}.$

From Section 3.1 we know that $L$ is rationalizable if and only if $V(p)$ is
unirational. The variety $V(p)\subset{\mathbb{A}}^{2}$ is a curve and its
genus is $0$, proving the statement. ∎

With the same argument used to prove Section 4 one can easily prove also the
following result.

######  Corollary 4.6.

Let $f$ be a linear polynomial in ${\mathbb{C}}(x,y)\setminus\\{x,y\\}$. Then
the following set is rationalizable.

${\mathcal{A}}:=\\{\sqrt{xy},\sqrt{xf},\sqrt{yf}\\}$

###### Proof.

Notice that without loss of generality we may (and do) assume that $f=x+ay+b$
with $a,b\in{\mathbb{C}}$. Using Theorem 1.4 we know that ${\mathcal{A}}$ is
rationalizable if and only if the extension
$L:={\mathbb{C}}(x,y)(\sqrt{xy},\sqrt{xf},\sqrt{yf})\supset{\mathbb{C}}(x,y)$
is. Proceeding as in the proof of Section 4, one can show that

$L\cong\frac{{\mathbb{C}}(x,y)[z]}{(p)},$

with

$\displaystyle p\,=\,$ $\displaystyle
x^{4}+2(a-1)x^{3}y+2bx^{3}+(a-1)^{2}x^{2}y^{2}+2b(a-1)x^{2}y+$
$\displaystyle-2x^{2}z^{2}+b^{2}x^{2}-2(a+1)xyz^{2}-2bxz^{2}+z^{4}\;.$

From Section 3.1 we know that $L$ is rationalizable if and only if $S=V(p)$ is
rational. Let $\overline{S}\subset{\mathbb{P}}^{3}$ be the projective closure
of $S$. One can easily verify that $\overline{S}$ is a quartic surface with a
singular locus given by the union of two lines $L_{1}$ and $L_{2}$ taken with
multiplicity $2$; the two lines intersect in one point. Using Urabe’s
classification of non-normal quartic surfaces (cf. [6, §1]) we then deduce
that $\overline{S}$ is rational, concluding the proof. ∎

The following example shows that the (rationalizable) situation of Corollaries
4 and 4 involving polynomials of degree two is very special.

######  Example 4.7 (c.f. [1, Example 49]).

In order to study the set

${\mathcal{A}}:=\\{\sqrt{X},\sqrt{1+4X},\sqrt{X(X-4)}\\}$

consider the following family of field extensions of $Q=\mathbf{k}(x)$:

$\\{Q[z]/(z^{2}-x),\ Q[z]/(z^{2}-(4x+1)),\ Q[z]/(z^{2}-x(x-4))\\}.$

The compositum of these field extensions is $L=Q[z]/(p)$ with

$\begin{split}p=&z^{8}+(-24x+12)z^{6}+(144x^{2}-72x+86)z^{4}+\\\
&+(-256x^{3}+96x^{2}+88x+300)z^{2}+1296x^{2}+1800x+625\end{split}$

where the roots of $p$ are $\pm\sqrt{x}\pm\sqrt{4x+1}\pm\sqrt{x-4}$.

The geometric genus of the curve $\operatorname{Spec}\mathbf{k}[x,z]/(p)$ is
one, hence $Q\subset Q^{\prime}$ and therefore ${\mathcal{A}}$ is not
rationalizable.

Notice that Corollaries 4, 4, 4 and 4 are all positive examples of the reverse
of [1, Proposition 47], leading us to formulate Section 1. The next natural
step would be to generalize the conjecture to sets of field extensions. We
refrain from doing this as at the moment we do not have strong evidences in
this direction.

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