2. Preliminaries

The concept of Galois points for irreducible plane curve $C\subset \mathbb{P}^2$ was introduced by [Reference Miura and Yoshihara6], [Reference Yoshihara8], [Reference Fukasawa1]. In order to study the extension of an element in $G_P$ to $\mathrm{Bir}(\mathbb{P}^2)$ , we need the following lemma.

Proof. Let $F(X, Y, Z) = 0$ be the defining equation of $C$ of degree $d$ . We may fix $P=[1\;:\;0\;:\;0]\in \mathbb{P}^2$ and choose that $C$ is not the line $Z=0$ . Since $P$ has multiplicity $m_P$ , then the equation of $C$ is $F(X,Y,Z)=F_{m_p}(Y,Z)X^{d-m_p}+\ldots \ldots +F_{d-1}(Y,Z)X+F_{d}(Y,Z)$ , where $F_i(Y,Z)$ is a homogeneous polynomial of $Y$ and $Z$ of degree $i\,(m_P\leq i\leq d)$ and $F_{m_P}(Y,Z)\neq 0$ . Since $F(X,Y,Z)$ is irreducible in $k[X,Y,Z]$ and not a multiple of $Z$ , then $f=F(X,Y,1)\in k[X,Y]$ is also irreducible in $k[X,Y]$ . We can see $f$ as an irreducible polynomial in $\tilde{k}[X]$ with $deg_{X}(f)=d-m_P$ where $\tilde{k}=k(Y)$ . Hence, the extension $k(C)/K_{P}$ is isomorphic to $(k(Y)[X]/(f))/(k(Y))$ , and thus, it has the same degree equal to the degree of the irreducible polynomial $f\in \tilde{k}[X]$ , so $[k(C)\;:\;\tilde{k}]=deg_{X}(f)=d-m_P$ .

It is well known [Reference Hartshorne2, Ch. 1, Theorem 4.4] that for any two varieties $X$ and $Y$ , there is a bijection between the set of dominant rational maps $\varphi \;:\;X \dashrightarrow Y$ , and the set of field homomorphisms $\varphi ^{\star }\;:\;k(Y) \rightarrow k(X)$ . In particular, we obtain:

Lemma 2.5. Let $P=[1\;:\;0\;:\;0]$ , by taking an affine chart, a de Jonquières map with respect to $P$ is a special case of a Cremona transformation, of the form

\begin{equation*}\iota ^{-1}\circ \mathrm {Jonq}_P\circ \iota =\{(x,y)\mapsto (\frac {a x+b}{c x +d}, \frac {r_1(x) y+ r_2(x)}{r_3(x) y+ r_4(x)})\}\end{equation*}

and $ \iota \;:\;\mathbb{A}^2 \hookrightarrow \mathbb{P}^2,\, (x,y) \longmapsto [x\;:\;y\;:\;1]$ , where $a,b,c,d\in k$ with $a d-b c\neq 0$ and $r_1(x),r_4(x),r_2(x), r_3(x)\in k(x)$ with $r_1(x) r_4(x)-r_2(x) r_3(x)\neq 0$ .

Proof. We have the following commutative diagram

where $\iota \;:\;(x,y)\mapsto [x\;:\;y\;:\;1]$ and $\psi \;:\;x\mapsto [x\;:\;1]$ , which gives the equality

\begin{equation*}\iota ^{-1}\circ \mathrm {Jonq}_P\circ \iota =\{f\in \mathrm {Bir}(\mathbb {A}^2)|\exists \alpha \in \mathrm {Bir}(\mathbb {A}^1);\;\alpha \circ \pi _x=\pi _x\circ f\}.\end{equation*}

Let $f\in \iota ^{-1}\circ \mathrm{Jonq}_P\circ \iota$ given by $(x,y)\mapsto (f_1(x,y),f_2(x,y))$ , then $\pi _x\circ f\;:\;(x,y)\mapsto f_1(x,y)$ . Since $\alpha \circ \pi _x=\alpha (x)$ , it follows that $f_1(x,y)$ depends only on $x$ and is of the form $f_1(x,y)=(a x+b)/(c x +d)$ where $a,b,c,d\in k$ and $a d-b c\neq 0$ by Lemma 2.3. From Lemma 2.2, $f^*$ is subjective, so $k(x,y)= k((a x+b)/(c x +d),f_2(x,y))$ . To describe the second component $f_2(x,y)$ , let us define the birational map $\tau \;:\;(x,y)\mapsto ((d x-b)/(-c x +a),y)$ , hence $\tau \circ f\;:\;(x,y)\mapsto (x,f_2(x,y))$ is a birational map since both $f$ and $\tau$ are birationals. By Lemma 2.2, $k(x)(f_2(x,y))=k(x)(y)$ . Apply Lemma 2.3 over the field $k(x)$ , then $f_2(x,y)=(r_1(x) y+ r_2(x))/(r_3(x) y+ r_4(x))$ .

Lemma 2.6. Let $P,Q\in \mathbb{P}^2$ and $C,D\subset \mathbb{P}^2$ be two irreducible curves, if $\phi \in \mathrm{Bir}(\mathbb{P}^2)$ and $\phi |_{C}\;:\;C\dashrightarrow D$ is birational map, and there exists $\theta \in \mathrm{Aut}(\mathbb{P}^1)$ such that $\pi _{Q}\circ \phi =\theta \circ \pi _P$ , then $P$ is a Galois point of $C$ if and only if $Q$ is a Galois point of $D$ . Moreover, if $P$ is Galois, an element of $G_{P}$ extends an element of $\mathrm{Bir}(\mathbb{P}^2)$ (respectively $\mathrm{Jonq}_{P}$ ) if and only if its image in $G_{Q}$ extends an element of $\mathrm{Bir}(\mathbb{P}^2)$ (respectively $\mathrm{Jonq}_{Q}$ )

Proof. Since $\phi |_C$ is birational map from $C$ to $D$ , then $\phi ^{*}|_C \;:\;k(D)\rightarrow k(C)$ is an isomorphism. Moreover, as $\pi _Q\circ \phi =\theta \circ \pi _P$ , we have a commutative diagram

Therefore, $k(D)/\pi _{Q}^{*}(k(\mathbb{P}^1))$ is Galois if and only if $k(C)/\pi _{P}^{*}(k(\mathbb{P}^1))$ is Galois. In addition, $\phi$ conjugates $\mathrm{Jonq}_{P}$ to $\mathrm{Jonq}_{Q}$ and sends any element of $\mathrm{Bir}(\mathbb{P}^2)$ that preserves $C$ onto element of $\mathrm{Bir}(\mathbb{P}^2)$ that preserves $D$ .

Example 2.7. Let $P=Q\in \{[1\;:\;0\;:\;0],[0\;:\;1\;:\;0],[0\;:\;0\;:\;1]\},\,\phi \;:\;[X\;:\;Y\;:\;Z] \mapsto [Y Z\;:\;X Z\;:\;X Y]$ and $\theta \;:\;[Y\;:\;Z] \mapsto [ Z\;:\;Y]$ , let $C\subset \mathbb{P}^2$ be an irreducible curve not equal to $x=0,\,y=0$ or $z=0$ and let $D=\phi (C)$ , so we have the following diagram

Thus, if $P$ is a Galois point for $C$ , then $P$ becomes a Galois point for $D$ , this is a particular case of Lemma 2.6 corresponding to [Reference Miura3, Corollary 3].

3. Extensions of degree at most three

Lemma 3.1. Let $k$ be a field and let $L=k[x]/(x^3+a_2 x^2+a_1 x+a_0)$ where $f=x^3+a_2 x^2+a_1 x+a_0$ is a separable irreducible polynomial in $k[x]$ , then the field extension $L/K$ is Galois if and only if there exists an element $\sigma \in \mathrm{Gal}(L/K)$ of order $3$ such that,

\begin{equation*}\sigma \;:\;x \mapsto \frac {\alpha x+\beta }{\gamma x+\delta }\,\text {where}\, \alpha, \beta, \gamma,\delta \in k\,\text {with}\,\alpha \delta -\beta \gamma \neq 0.\end{equation*}

Proof. As $f$ is a separable irreducible polynomial of degree 3, the extension $L/K$ is separable of degree 3. It is then Galois if and only if there exists $\sigma \in \mathrm{Aut}(L/K)$ of order $3$ , so it remains to prove that we can choose $\sigma$ with the right form. If $\sigma \in \mathrm{Aut}(L/K)$ where, $\sigma \;:\;x\mapsto \nu _2 x^2+\nu _1 x+\nu _0$ and $\nu _i\in k$ for $i=0,1,2$ , so the question here is can we find $\{\alpha, \beta, \gamma,\delta \}\subset K$ with $\alpha \delta -\beta \gamma \neq 0$ such that the following equality holds?

(1) \begin{equation} \nu _2 x^2+\nu _1 x+\nu _0=\frac{\alpha x+\beta }{\gamma x+\delta }. \end{equation}

We can find a solution $\{\alpha = a_2 \nu _1 \nu _2 -a_1 \nu _2^2 + \nu _0 \nu _2 - \nu _1^2,\,\beta = a \nu _0 \nu _2 - a_0 \nu _2^2 - \nu _0\nu _2,\,\delta = a_2 \nu _2 - \nu _1,\,\gamma =\nu _2\}$ . We observe that $\alpha \delta -\beta \gamma \neq 0$ , otherwise we have $\sigma (x)\in K$ and this gives a contradiction as $x\notin K$ .

Proof. The equality $[k(C)\;:\;K_{P}]=d-m_{P}$ follows from Lemma 2.1. We may assume $P=[1\;:\;0\;:\;0]$ . Let $x = X/Z$ and $y = Y/Z$ be affine coordinates. Since the field extension $k(C)/\pi _{P}^{*}(k(\mathbb{P}^1))$ is of degree $d-m_{P}$ , then $k(C)=k(y)[x]/(f)$ , where $f\in k[x,y]$ is the equation of $C$ in these affine coordinates.

1. (1) If $d-m_p=1$ , then $k(C)=\pi _{P}^{*}(k(\mathbb{P}^1))$ and therefore $\pi _{P}^{*}\;:\;k(\mathbb{P}^1)\rightarrow k(C)$ is an isomorphism. Hence $\pi _{P}\;:\;C\dashrightarrow \mathbb{P}^1$ is birational.

2. (2) If $d-m_p=2$ , then the extension $k(C)/\pi _{P}^{*}(k(\mathbb{P}^1))$ is of degree $2$ and it is thus Galois if and only if it is separable. $k(C)/\pi _{P}^{*}(k(\mathbb{P}^1))$ is Galois $\Leftrightarrow$ there exists an element $\sigma \in G_{P}$ of order $2$ that permutes the roots of $f\,\Leftrightarrow \,f$ is separable $\Leftrightarrow$ the extension is separable. Furthermore, the element $\sigma \in G_P$ of order $2$ is given by $x\mapsto -x$ up to a suitable change of coordinates.

3. (3) If $d-m_{P}=3$ , the equation of the curve $C$ is given by $f=F_{d-3}(y,1)x^3+F_{d-2}(y,1)x^2+F_{d-1}(y,1)x+F_{d}(y,1)$ . We apply Lemma 3.1, replacing $k$ by $k(y)$ .

We now illustrate Theorem 3.2 in two examples.

Proof. The curve $C$ is birational to $\mathbb{P}^1$ via $\phi$ , with inverse $[X\;:\;Y\;:\;Z]\mapsto [X +Y\;:\; X +Z]$ . Define the projection by $\pi _P\;:\;[X\;:\;Y\;:\;Z]\mapsto [Y\;:\;Z]$ . Let $\psi =\pi _{P}\circ \phi$ , then $\psi \;:\;\mathbb{P}^1\rightarrow \mathbb{P}^1$ maps $[u\;:\;v]$ to $[u^3\;:\;v^3]$ , so the extension is Galois of degree $3$ with Galois group $G_{P}$ generated by $\sigma \;:\;x\mapsto \omega \cdot x$ , where $\omega$ is a primitive cubic root of unity. By Theorem 3.2, every element of order $3$ extends to an element of $\mathrm{Jonq}_{P}$ . Explicitly, $\sigma$ extends to the map that is given by

\begin{equation*}[X\;:\;Y\;:\;Z]\mapsto [\frac {\left (Y-\omega Z \right )X +Y Z(1- \omega )}{(\omega -1)X +Y\omega -Z}\;:\;Y\;:\;Z].\end{equation*}

Proof. Define the birational map $\phi \;:\;\mathbb{P}^1 \dashrightarrow C$ by $ \phi \;:\;[u\;:\;v] \mapsto [v^{3}\;:\; u^{3} \;:\; u^{2} v- v^3]$ with inverse $[X\;:\;Y\;:\;Z]\mapsto [Y\;:\;Z+X]$ . Define the projection by $\pi _P\;:\;[X\;:\;Y\;:\;Z]\mapsto [Y\;:\;Z]$ . Let $\psi =\pi _{P}\circ \phi$ , then $\psi \;:\;\mathbb{P}^1\rightarrow \mathbb{P}^1$ maps $[u\;:\;v]$ to $[u^{3} \;:\; u^{2} v- v^3]$ , so the extension is Galois of degree $3$ with Galois group $G_{P}$ generated by $\sigma \;:\;[u\;:\;v]\mapsto [u\;:\;u+v]$ . By Theorem 3.2, we know that every element of order $3$ extends to an element of $\mathrm{Jonq}_{P}$ . Explicitly $\sigma$ extends to the map that is given by $\sigma$ that is given by $[X\;:\;Y\;:\;Z]\mapsto [X +Y \;:\;Y\;:\;Z]$ .

4. Curves that are Cremona equivalent to a line

Proof. Let $\pi _1\;:\;X_1 \rightarrow \mathbb{P}^2$ be the blow-up of $\mathbb{P}^2$ at $P_1$ , and let $\pi _i\;:\;X_{i} \rightarrow X_{i-1}$ the blow-up of $X_{i-1}$ at $P_i\in X_{i-1}$ for $i\geq 2$ , $\mathcal{E}_{i}=\pi _i^{-1}(P_i)$ is a $(-1)$ -curve, where $\mathcal{E}_{i}^2=-1$ and $\mathcal{E}_{i}\cong \mathbb{P}^1$ . After blowing up $n$ points, let $\pi \;:\; Y\mapsto \mathbb{P}^2$ be the composition of the blow-ups $\pi _i$ , where $Y=X_n$ we choose enough points such that the strict transform of $C$ is smooth. By induction, we have $E_i=(\pi _{i+1}\circ \ldots ..\circ \pi _{n})^{*}(\mathcal{E}_i)$ , $Pic(Y)=\pi ^*(Pic(\mathbb{P}^2))\oplus \mathbb{Z}E_{1}\oplus \ldots \oplus \mathbb{Z}E_{n}$ , and $E_i^2=-1$ for every $i=1,\ldots,n,$ and $E_i\cdot E_j=0$ for every $i\neq j$ . Moreover,

\begin{equation*}K_{Y}=\pi _{n}^*\ldots .\; \pi _1^*(K_{\mathbb {P}^2})+\Sigma _{i=1}^{n}\pi _{n}^*\ldots .\; \pi _{i+1}^*(\epsilon _{i})=\pi ^*(K_{\mathbb {P}^2})+\Sigma _{i=1}^{n}E_{i}=-3\pi ^*(L)+\Sigma _{i=1}^{n}E_{i}.\end{equation*}

The strict transform $\tilde{C}\subset C$ is equivalent to $\tilde{C}=d \cdot \pi ^*(L)-\Sigma _{i=1}^{n}m_{P_i}(C)E_{i}$ . Hence we have $2 K_Y+\tilde{C}=(-6+d)\cdot \pi ^*(L)+\Sigma _{i=1}^{n}(2-m_{P_i}(C))E_{i}$ , so $\pi ^*(L)\cdot (2 K_{Y}+\tilde{C})=-6+d$ , thus $|2 K_{Y}+\tilde{C}|=\phi$ for every curve of degree $d\lt 6$ . [Reference Mohan Kumar and Murthy7, Corollary $2.4$ ] shows that $|2 K_{Y}+\tilde{C}|=\varnothing$ is equivalent to $\bar{\mathcal{K}}(C,\mathbb{P}^2)=-\infty$ , so $C$ is equivalent to a line by Theorem 4.4.

Proof. Let $\varphi \in \mathrm{Bir}(\mathbb{P}^2)$ that sends $C$ onto a line $L$ . For each $g\in G_{P},\,\varphi |_{C}\;:\;C\dashrightarrow L$ conjugates $g$ to an element of $\mathrm{Aut}(L)$ , that extends to $\hat{g}\in \mathrm{Aut}(\mathbb{P}^2)$ . Hence, $g$ extends to $\varphi ^{-1}\hat{g}\varphi \in \mathrm{Bir}(\mathbb{P}^2)$ .

Remark 4.7. Let $C$ be the smooth conic given by $C= \{[X \;:\; Y \;:\; Z] | Y^2=XZ\}\subset \mathbb{P}^2$ , then the natural embedding of $\mathrm{Aut}(\mathbb{P}^2, C)=\{g\in \mathrm{Aut}(\mathbb{P}^2)|g(C)=C\}= \mathrm{PGL}_2$ in $\mathrm{Aut}(\mathbb{P}^2) = \mathrm{PGL}_3$ is the one induced from the injective group homomorphism

\begin{equation*} \mathrm {GL}_2(k)\rightarrow \mathrm {GL}_3(k),\, \left [\begin {array}{c@{\quad}c} a & b \\[5pt] c & d \end {array}\right ]\mapsto \frac {1}{ad-bc}\left [\begin {array}{c@{\quad}c@{\quad}c} a^2 & ab&b^2 \\[5pt] 2ac & ad+bc&2bd\\[5pt] c^2&cd&d^2 \end {array}\right ] \end{equation*}

where $\rho \;:\;[u\;:\;v]\mapsto [u^2\;:\;uv\;:\;v^2]$ , and the following diagram commutes.

Lemma 4.8. Let $k$ be a field of characteristic $char(k)\neq 2$ containing a primitive fourth root of unity, and let $C$ be the irreducible curve defined by the equation

(2) \begin{equation} X^{4}-4 Z Y X^{2}-Z Y^{3}+2 Z^{2} Y^{2}-Y Z^{3} =0, \end{equation}

then the point $P=[1\;:\;0\;:\;0]$ is an outer Galois point of $C$ and the extension induced by the projection $\pi _P\;:\; C\dashrightarrow \mathbb{P}^1$ is Galois of degree $4$ . Furthermore, the group $G_{P}$ extends to $\mathrm{Bir}(\mathbb{P}^2)$ but not to $\mathrm{Jonq}_{P}$ .

Proof. Define the birational map $\phi \;:\;\mathbb{A}^1 \dashrightarrow C$ by $\phi \;:\;t\mapsto [t+t^3\;:\;t^4\;:\;1]$ with inverse $[X\;:\;Y\;:\;Z]\mapsto \left (X \left (Y +Z \right )\right )/(X^{2}-Y Z +Z^{2})$ . Hence $C$ is a rational irreducible curve of degree $4$ and therefore, $C$ is equivalent to a line by Lemma 4.5. Furthermore, every non-trivial element in $G_{P}$ extends to an element in $\mathrm{Bir}(\mathbb{P}^2)$ by Lemma 4.6. We will also prove it explicitly below. We have $K(C)=k(t)$ and define the projection by $\pi _P\;:\;[X\;:\;Y\;:\;Z]\mapsto [Y\;:\;Z]$ . Let $x = X/Z$ and $y = Y/Z$ be affine coordinates, so the affine equation $x^{4}-4 y x^{2}- y^{3}+2 y^{2}-y =0$ is defining the extension field $k(y)[x]/k(y)=k(t)/k(t^4)$ . Since $k$ contains the $4th$ root of unity, the extension is Galois of degree $4$ with basis $\{1,t, t^2,t^3\}$ , and we have the following diagram

where $\psi$ is given by $\psi \;:\;t\mapsto t^4$ . By contradiction, we prove that there is no de Jonquières map $f$ extending the action. Let us assume that there exists a de Jonquières map $g$ that extends the action, $i.e$ there exists $\tilde{\alpha },\tilde{\beta }, \tilde{\gamma },\tilde{\delta }\in k(y)$ with $\tilde{\alpha }\tilde{ \delta }-\tilde{\beta } \tilde{\gamma }\neq 0$ such that $g\;:\;(x,y)\mapsto (\frac{\tilde{\alpha } x+\tilde{\beta }}{\tilde{\gamma } x+\tilde{\delta }},y)$ . Since $g\circ \phi =\phi \circ \sigma$ , writing $\alpha =\tilde{\alpha }(t^4),\beta =\tilde{\beta }(t^4), \gamma =\tilde{\gamma }(t^4)$ and $\delta =\tilde{\delta }(t^4)$ where $\alpha,\beta,\gamma, \delta \in k(t^4)$ . We obtain the equation

\begin{equation*}\mathrm {i}t-\mathrm {i}t^3=\frac {\alpha (t+t^3)+\beta }{\gamma (t+t^3)+\delta }.\end{equation*}

This implies that $\beta = \beta (t)=-(\mathrm{i} t^{6}- \mathrm{i})\gamma t^{2}-(\mathrm{i} \delta +\alpha ) t^{3}+(\delta \mathrm{i}-\alpha )t\in k(t^4)$ and is then $(\mathrm{i} t^{4}- \mathrm{i})\gamma =0,$ $\mathrm{i} \delta +\alpha =0$ and $\delta \mathrm{i}-\alpha =0$ . This gives $\alpha =0,\gamma =0$ which is a contradiction. Viewing $C$ as an irreducible curve in $\mathbb{P}^2$ of degree $4$ , there are three singular points on the curve $ [0\;:\; 1\;:\; 1], \left [\mathrm{i} \sqrt{2}\;:\;-1\;:\; 1\right ], \left [\mathrm{i} \sqrt{2}\;:\; 1\;:\; -1\right ]$ . After suitable change of coordinates, $\sigma \;:\;\mathbb{P}^2\rightarrow \mathbb{P}^2$ is given by $[X\;:\;Y\;:\;Z]\mapsto \left [-\mathrm{i} \sqrt{2} X-\mathrm{i} \sqrt{2} Z\;:\; 2 X+Y-Z \;:\; 2X-Y+Z \right ]$ , this map sends the curve $C$ to $\tilde{C}$ , which is given by $\tilde{f}=X^{2} Y^{2}+6 X^{2} Y Z +X^{2} Z^{2}+4 Y^{2} Z^{2}=0$ and this new equation has $\{[1\;:\;0\;:\;0], [0\;:\;1\;:\;0],[0\;:\;0\;:\;1]\}$ as multiple points of order $2$ . After blowing up the three points $\{[1\;:\;0\;:\;0], [0\;:\;1\;:\;0],[0\;:\;0\;:\;1]\}$ in $\mathbb{P}^2$ and contract again, the strict transform curve is of degree $d'=2 \cdot d-m_1-m_2-m_3=2\cdot 4-2-2-2=2$ , so it is a conic given by the equation $F=4 X^{2}+Y^{2}+6 Y Z +Z^{2}=0$ . We change the coordinates using the following matrix

\begin{equation*}\left [\begin {array}{c@{\quad}c@{\quad}c} 4\mathrm {I} & 0 & -\mathrm {I} \\[5pt] 0 & 2\sqrt {2} & 0 \\[5pt] 8 & -6 \sqrt {2} & 2 \end {array}\right ]\end{equation*}

to send the conic to $Y^2-XZ=0$ and we extend $G_{P}$ explicitly using Remark 4.7.

5. Example where $\boldsymbol{G}_{\boldsymbol{P}}$ cannot be extended to $\mathrm{Bir}(\mathbb{P}^2)$

Proof. Let $deg(f)=d$ , and assume for contradiction that $d\gt 1$ . We take a commutative diagram where $\pi$ and $\eta$ are sequences of blow-ups

and we can assume that the strict transform of $C$ is smooth. As in Lemma 4.5 $\eta ^{*}(L)=d\cdot \pi ^{*}(L)-\Sigma m_iE_i,\,K_{X}=-3\pi ^{*}(L)+\Sigma E_i$ , and $\eta ^{*}(C)=\tilde{C}=deg(C)\cdot \pi ^{*}(L)-\Sigma m_{P_{i}}(C)E_i$ . Since $deg(C)=C\cdot L=\eta ^{*}(C)\cdot \eta ^{*}(L)=d\cdot deg(C)-\Sigma m_i\cdot m_{P_{i}}(C)$ , then $deg(C)(d-1)=\Sigma m_i\cdot m_{P_{i}}(C)\lt \Sigma m_i\cdot deg(C)/3$ and therefore $3(d-1)\lt \Sigma m_i$ Noether equality, which is a contradiction as $\Sigma m_i=3(d-1)$ : this equation follows from $\eta ^*(L)^2=L^2=1$ and from the adjunction formula $\eta ^*(L)\cdot (\eta ^*(L)+K_X)=-2$ , which gives $\eta ^*(L)\cdot K_X=-3$ .

Proof. The curve $C$ is birational to $\mathbb{P}^1$ via $\phi$ , with inverse $[X\;:\;Y\;:\;Z]\mapsto [X^{4} Y +4 X^{3} Y^{2}-2 X^{3} Z^{2}+6 X^{2} Y^{3}-2 X^{2} YZ^{2}+4 X Y^{4}+X Z^{4}+Y^{5}\;:\;Z (X^{4}+2 X^{3} Y+X^{2} Y^{2}-3 X^{2} Z^{2} -3 X Y Z^{2}+Z^{4})]$ . Define the projection by $\pi _P\;:\;[X\;:\;Y\;:\;Z]\mapsto [Y\;:\;Z]$ . Let $\psi =\pi _P\circ \phi$ then $\psi \;:\;\mathbb{P}^1\rightarrow \mathbb{P}^1$ maps $[u\;:\;v]$ to $[u^5:v^5]$ , hence the extension is Galois of degree $5$ with Galois group $G_{P}$ generated by $\sigma \;:\;x\mapsto \zeta \cdot x$ , where $\zeta$ is the $5$ th root of unity. We now prove that the curve $C$ does not have a point $P$ of multiplicity $m_P(C)\geq 3$ . By contradiction, we take a point $P=[P_0\;:\;P_1\;:\;P_2]$ of multiplicity $3$ , and then we take two distinct lines $a_{1}x+a_{2}y-a_{3}z=0$ and $b_{2}y-b_{3}z=0$ passing through the point $P$ . We take the preimage in $\mathbb{P}^1$ , so we get a common factor of degree at least $3$ . Let $f_{1}(u,v) = a_{1} \left (-u^{7}+u \,v^{6}\right )+a_{2} u^{5} \left (u^{2}+v^{2}\right )+a_{3} v^{5} \left (u^{2}+v^{2}\right )$ and $f_{2}(u,v) =\left (u^{2}+v^{2}\right ) \left (b_{2} u^{5}+b_{3} v^{5}\right )$ . We check now that it is not possible for the polynomials $f_1$ and $f_2$ to have a factor of degree $3$ in common. Assume first that $u +\mathrm{i} v$ divides both polynomials, so we should have $f_{1}(1,\mathrm{i})=f_{2}(1,\mathrm{i})=0$ implies to $a_{1}=0$ , hence $P=[1\;:\;0\;:\;0]$ is a smooth point and this gives a contradiction. If we assume that $u -\mathrm{i} v$ divides both polynomials, then we should have $f_{1}(1,-\mathrm{i})=f_{2}(1,-\mathrm{i})=0$ , again we have $a_{1}=0$ , so the factor of degree $3$ must divide $b_{2} u^{5}+b_{3} v^{5}$ . If we assume that $u$ divides the polynomial $f_{2}$ , then $b_{3}=0$ and $u^3$ should divide $f_{1}$ , but this is not true as $f_{1}=-u(u^6 - v^6)$ . If we assume that $v$ divides the polynomial $f_{2}$ , then $b_{2}=0$ and $v^3$ should divide $f_{1}$ , but this is not true as $f_{1}=uv^2(u^4 + v^4)$ . So the factor of degree $3$ must divide $b_{2} u^{5}+b_{3} v^{5}$ , $b_{2}\neq 0$ and $b_{3}\neq 0$ . Hence we can assume that $a_{1}=1$ and replace $f_{1}$ by $f_{1}-a_{3}f_{2}/b_{3}$ so we can put $a_{3}=0$ and up to multiple, we can assume that $b_{3}=1, a_{1}=1$ and $b_{2}=-\xi ^{5}$ . So $f_{2}=-u^{5} \xi ^{5}+v^{5}$ , $f_{1}=\left (-u^{7}+u \,v^{6}\right )+a_{2} u^{5} \left (u^{2}+v^{2}\right )$ this implies that the roots of $f_{2}$ are $(u, v)=(1,\xi \zeta ^i)$ , where $\zeta$ is a $5$ th root of unity. Since $f_{1}$ and $f_{2}$ should have three roots in common, therefore let $\{(1,\xi ),(1,\xi \zeta ),(1,\xi \rho )\}$ are the three roots in common where $\rho ^5=1$ and $\zeta ^5=1$ , and $\rho ^5\neq \zeta$ and they are not equal to $1$ , so $f_{1}$ should vanish on these three roots. This gives three equations $ q_{1}=\xi ^{6}-1+a_{2} \left (\xi ^{2}+1\right )=0,\,q_{2}=a_{2} \left (\xi ^{2} \zeta ^{2}+1\right )+\xi ^{6} \zeta -1=0,\,q_{3}=a_{2} \left (\rho ^{2} \xi ^{2}+1\right )+\rho \,\xi ^{6}-1=0$ , by solving this system in $a_{2}, \zeta$ and $\rho$ , we found that $\zeta =\rho =(\xi ^{4}+1)/(\xi ^{6}-1)$ , which is a contradiction as $\zeta \neq \rho$ . Finally, $f_{1}$ and $f_{2}$ cannot have a factor of degree $d\geq 3$ . Since $m_P(C)\lt 3$ for each $P\in \mathbb{P}^2$ , let us assume that there exists a birational map $g$ that extends the generator of the Galois group, then by Lemma 5.1, $g$ is a linear automorphism of $\mathbb{P}^2$ , so it is given by a matrix let us say $A\in \mathrm{PGL}_{3}$ , Since $g\circ \phi =\phi \circ \sigma$ , so we have

\begin{equation*}\left [\begin {array}{c}U V^6-U^7 \\[5pt] U^5(U^2+V^2)\\[5pt] V^5(U^2+V^2)\end {array}\right ]=A\cdot \left [\begin {array}{c}\zeta U V^6-\zeta ^2 U^7 \\[5pt] U^5(\zeta ^2 U^2+V^2)\\[5pt] V^5(\zeta ^2 U^2+V^2) \end {array}\right ]\end{equation*}

where $\zeta$ is the $5th$ root of unity. Since $ U V^6, U^7, U^5 V^2,$ and $V^7$ are linearly independent, after checking the calculation we found that $A$ should be diagonal, but $U^5(\zeta ^2 U^2+V^2)$ is not a multiple of $U^5(\zeta ^2 U^2+V^2)$ , then we have a contradiction.