\subsection{Square roots in two variables} 
\label{sec:TwoVariables}

The criterion to decide whether the square root of a polynomial in one variable is rationalizable or not relies on the computation of the geometric genus of the associated curve.
For surfaces, the situation is analogous with the role of the genus being played by the {\em Kodaira dimension}.

\begin{remark}
Recall that the Kodaira dimension of a projective variety $Y$ is an integer
$\kappa=\kappa(Y)\in \{ -1,0,1,...,\dim Y\}$.
Please note that some sources use $-\infty$ instead of $-1$.
For the precise definition, we refer to~\cite[Chapter 6]{Har77}.
\end{remark}

The surfaces arising in our context are mostly not smooth.
The theory of singular surfaces is extremely rich, and a detailed discussion of the topic goes well beyond the scope of this paper; for an overview, see~\cite{AVGZ85}.
In this subsection, we will only deal with surfaces with mild isolated singularities,
that is, surfaces with at most {\em rational simple} singularities
(also called {\em DuVal} or {\em ADE} singularities).
For the definition and their properties, we refer the reader to~\cite[Chapter 15]{AVGZ85}.
(We want to stress out that a ``rational'' simple singularity does \textit{not} need to be defined over $\QQ$;  
``rational'' only means that by resolving it one gets an exceptional divisor birationally equivalent to $\PP^1$.)

Throughout this subsection, $f$ will always denote a non-constant squarefree polynomial in $\CC [X,Y]$ of degree $d$ (cf. Corollary~\ref{c:SqFreePol} and Remark~\ref{r:Squarefree}). 
Following Definition~\ref{def:AssHyp}, we denote by $\overline{V}$ the hypersurface associated to $\sqrt{f(X,Y)}$,
that is,
the projective surface in $\PP^3$ with coordinates $(z,x,y,w)$ defined by
$$
\overline{V}\colon z^{d-2}w^2-z^{d}f(x/z,y/z)=0.
$$
Finally, recall that the notions of being rational and unirational are equivalent for surfaces, cf. Remark~\ref{r:UniRatEqRat}.


\begin{lemma}\label{l:EqvUniKodDeg}
Let $f, d$, and $\overline{V}$ be defined as above and assume $\overline{V}$ is smooth or has at most rational simple singularities.
Then the following statements are equivalent.
\begin{enumerate}[label=(\roman*)]
    \item $\overline{V}$ has Kodaira dimension $-1$.
    \item the degree of $f$ is at most three, i.e., $d\leq 3$;
    \item $\overline{V}$ is unirational;
    \item $\sqrt{f}$ is rationalizable.
\end{enumerate}
\end{lemma}
\begin{proof}
From Proposition~\ref{p:criterion}, we know that $iii)\iff iv)$.
We are left to show that $i), ii),$ and $iii)$ are equivalent.
Therefore, we will show that $i)\implies ii) \implies iii) \implies i)$.

{\em i) $\implies$ ii)} 
Assume $\overline{V}$ has at most rational simple singularities and $\kappa (\overline{V})=-1$.
Also, recall that $\overline{V}$ is a hypersurface in $\PP^3$ of degree $d$.
Since $\overline{V}$ has at most rational simple singularities, the canonical class is left unchanged after passing to the smooth model, 
and hence the Kodaira dimension of $\overline{V}$ is determined by its degree.
Hypersurfaces in $\PP^3$ of Kodaira dimension $-1$ have degree $d=1,2,3$ (in fact $\kappa(\overline{V})=0$ for $d=4$ and $\kappa(\overline{V})=2$ for $d\geq 5$, cf.~\cite[Section VI.1]{BHPV04}).

{\em ii) $\implies$ iii)}
 If $d=1,2$, then the statement follows from Corollary~\ref{c:SingProj} and Proposition~\ref{p:criterion}.
Assume $d=3$, then $\overline{V}$ is a cubic. 
If it is smooth, then it is rational (classic result by Clebsch, see~\cite{Cle66} for the original paper, or~\cite[Corollay V.4.7]{Har77} for a more modern statement and proof).
Rationality implies unirationality.
If $\overline{V}$ is singular, by assumption the singularities must be rational simple and hence of multiplicity $2=d-1$. 
Then the unirationality of $\overline{V}$ follows again from Corollary~\ref{c:SingProj} and Proposition~\ref{p:criterion}.

{\em iii) $\implies$ i)}
Assume $\overline{V}$ is unirational. As already noted, this is equivalent to saying that $\overline{V}$ is rational.
Then by the Enriques--Kodaira classification of surfaces it follows that $\kappa (\overline{V})=-1$ (see also~\cite[Theorem V.6.1]{Har77}). 
\end{proof}
\begin{remark}\label{r:SingSurf}
Unfortunately, Lemma~\ref{l:EqvUniKodDeg} is not very useful in practice: when $d\geq 4$,
the surface $\overline{V}$ has a non-simple singular point at $(0:0:0:1)$, so the result does not apply.
If $d=1,2$ we already know (unconditionally) that $\sqrt{f}$ is rationalizable  (cf. Proposition~\ref{p:criterion}).
The only interesting case is when $d=3$, as shown by the following corollary.
\end{remark}
\begin{corollary}\label{c:CubicSurface}
Assume $d=3$.
Then $\sqrt{f}$ is rationalizable if and only if $\overline{V}$ has no singular points of multiplicity $3$.
\end{corollary}
\begin{proof}
If $\overline{V}$ has a singular point of multiplicity $3$,
then it is the projective cone over a projective plane cubic curve (with the singular point being the vertex of the cone).
Hence $\overline{V}$ is ruled and not (uni)rational and so, by Proposition~\ref{p:criterion}, it follows that $\sqrt{f}$ is not rationalizable.

Conversely, assume that $\overline{V}$ has no singular points of multiplicity $3$.
As $\overline{V}$ is a cubic, this means that it is either smooth or has singular points of multiplicity $2$.
In the smooth case, the statement follows from Lemma~\ref{l:EqvUniKodDeg};
in the singular case, from Corollary~\ref{c:SingProj}.
\end{proof}
\begin{remark}\label{r:NonSimpleSing}
The assumption in Lemma~\ref{l:EqvUniKodDeg} for $\overline{V}$ to have at most rational simple singularities is strictly necessary, as shown by Corollary~\ref{c:CubicSurface}.
We have seen that if $d=3$ and $\overline{V}$ has a point of multiplicity $3$,
then $\overline{V}$ is a cone over a cubic curve, which is ruled but not (uni)rational, providing a counterexample to $ii)\implies iv)$.
In~\cite{IN04}, one can find a ruled quartic surface having Kodaira dimension $-1$ while not being (uni)rational, hence a counterexample to $i)\implies ii)$.
The implications $iii)\iff iv)$, $iii)\implies i)$, and $ii)\implies i)$ hold unconditionally.
\end{remark}


The guaranteed existence of non-simple singular points on $\overline{V}$ (cf. Remark~\ref{r:NonSimpleSing}) prevents us from getting much information about the rationalizability of $\sqrt{f}$.
However, we can also use the approach of Subsection~\ref{sec:DoubleCover}, which turns out to be much more suitable for the case of square roots in two variables, as shown in the theorem below.
Let $\overline{S}$ denote the double cover associated to $\sqrt{f}$ (cf. Definition~\ref{def:AssDoubleCover}).

\begin{theorem}\label{thm:DegreeCriterion}
Assume that $\overline{S}$ has at most rational simple singularities. 
Then $\sqrt{f}$ is rationalizable if and only if $d\leq 4$.
\end{theorem}
\begin{proof}
From Remark~\ref{r:UniRatEqRat} and Proposition~\ref{p:criterion}, $\sqrt{f}$ is rationalizable if and only if $\overline{S}$ (or, equivalently, $\overline{V}$) is unirational.


So now assume $\overline{S}$ has at most rational simple singularities. 
Let $T$ be the desingularization of $\overline{S}$.
By a result in~\cite{Hir64}, such a $T$ exists and is birationally equivalent to $\overline{S}$.
This means that $\overline{S}$ is (uni)rational if and only $T$ is.
As $\overline{S}$ has only rational simple singularities, the canonical divisor of $T$ equals the canonical divisor of $\overline{S}$. 
Hence, we can use the degree of $f$ to compute the canonical divisor of $\overline{S}$ and hence the Kodaira dimension of $T$. 

If $d=1,2$ then $\sqrt{f}$ is rationalizable by Proposition~\ref{c:SingProj}.

If $d=3,4$, then $T$ (and hence $\overline{S}$)  is birationally equivalent to a del Pezzo surface of degree $2$ (cf.~\cite[Theorem III.3.5]{Kol96}, where the degree of a del Pezzo surface is defined to be the self-intersection of the canonical divisor of the surface; notice that it does not need to coincide with the degree of the defining polynomial).
Del Pezzo surfaces are rational (cf.~\cite[Theorem IV.24.4]{Man86}).

We are left to show that if $d>4$, then $\sqrt{f}$ is not rationalizable. 
In order to see this, we prove that $\overline{S}$ is not (uni)rational.
Since $d>4$, we have that $T$ and, hence, $\overline{S}$ have Kodaira dimension greater than or equal to $0$ 
(in fact, their Kodaira dimension is $0$ if $d=5,6$ and $2$ if $d\geq 7$, cf.~\cite[Section V.22]{BHPV04}).
As $\overline{S}$ and $\overline{V}$ are birationally equivalent, they have the same Kodaira dimension.
Then Lemma~\ref{l:EqvUniKodDeg} implies that $\sqrt{f}$ is not rationalizable, proving the statement.
\end{proof}

\begin{remark}
At a first glance, Theorem~\ref{thm:DegreeCriterion} and  Lemma~\ref{l:EqvUniKodDeg}  contradict each other, as the square root of a polynomial $f$ of degree $d=4$ should be non-rationalizable, according to Lemma~\ref{l:EqvUniKodDeg}, but also rationalizable, 
according to Theorem~\ref{thm:DegreeCriterion}.
This contradiction does, however, not really exist:
as already noted in Remark~\ref{r:NonSimpleSing}, the hypersurface $\overline{V}$ associated to $f$ always has a non-simple singular point and, therefore, one cannot apply the implication $i)\implies ii)$ in Lemma~\ref{l:EqvUniKodDeg} (cf. Remark~\ref{r:NonSimpleSing}) needed to conclude that $\sqrt{f}$ is not rationalizable.
\end{remark}

\begin{remark}
In order to apply Theorem~\ref{thm:DegreeCriterion},
one needs to study the singularities of~$\overline{S}$.
To simplify this task, we wrote a {\tt Magma}~(cf.~\cite{Magma}) function.
For the source code of the function and a detailed explanation of how to apply it, see \cite{Besier:GitHub}.
Alternatively, one can use the {\tt classify2.lib} library of {\tt Singular} (cf.~\cite{singular}).
Both {\tt Magma} and {\tt Singular}  come with a free online calculator that one can use to perform the singularity classification.
\end{remark}

\begin{example}
In~\cite{Festi:092018}, the rationalizability of the square root
\begin{equation}\label{eq:Bhabha}
\sqrt{\frac{(X+Y)(1+XY)}{X + Y - 4XY + X^2Y + XY^2}}
\end{equation}
coming from the Bhabha scattering (cf. \cite{Henn:2013woa}) is studied.
Using Corollary~\ref{c:SqFreePol}, one immediately sees that this is equivalent to study the unirationality of the double cover $\overline{S}$ associated to the square root
$$
\sqrt{(X+Y)(1+XY)(X + Y - 4XY + X^2Y + XY^2)}.
$$
The surface $\overline{S}$ has only simple singularities as one can check either by hand or using our code, cf.~\cite{Besier:GitHub}.
Therefore, from Theorem~\ref{thm:DegreeCriterion}, it follows that the square root~\eqref{eq:Bhabha} is not rationalizable. 
\end{example}
{"config": "arxiv", "file": "2006.07121/06TwoVariables.tex"}