Let the elliptic curve \(X\) be embedded in \(\PP^2\) so as to have the equation \(y^2=x(x-1)(x-\lambda)\). Show that any automorphism of \(X\) leaving \(P_0=(0,1,0)\) fixed is induced by an automorphism of \(\PP^2\) coming from the automorphism of the affine \((x, y)\)-plane given by \[ \begin{cases} x' = a x+b \\ y' = c y . \end{cases} \] In each of the four cases of (4.7), describe these automorphisms of \(\PP^2\) explicitly, and hence determine the structure of the group \(G=\Aut(X, P_0)\).
Solution
Put \[ f(x)=x(x-1)(x-\lambda), \] so on the affine chart \(z=1\) the curve is \[ y^2=f(x), \] and \[ P_0=[0:1:0] \] is the unique point at infinity.
1. At \(P_0\) the functions \(x\) and \(y\) have poles of orders \(2\) and \(3\),respectively, and\[ L(2P_0)=\langle1,x\rangle, \qquad L(3P_0)=\langle1,x,y\rangle. \]
The projective equation is \[ Y^2Z=X(X-Z)(X-\lambda Z). \] The line at infinity \(Z=0\) meets \(X\) only at \(P_0\), with intersection multiplicity \(3\). Hence the affine functions \(x=X/Z\) and \(y=Y/Z\) have no poles away from \(P_0\). The equation shows that their pole orders satisfy \[ 2\operatorname{ord}_{P_0}(y) = 3\operatorname{ord}_{P_0}(x), \] and the morphism \[ x:X\longrightarrow\PP^1 \] has degree \(2\), so \[ \operatorname{ord}_{P_0}(x)=-2, \qquad \operatorname{ord}_{P_0}(y)=-3. \]
Since \(g(X)=1\), Riemann–Roch gives \[ \ell(nP_0)=n \] for every \(n\ge1\). The displayed functions therefore give bases of the two spaces.
2. If\[ \sigma\in G=\Aut(X,P_0), \]then for some\[ a,c\in k^\times, \qquad b,d,e\in k, \]one has\[ \sigma^*x=ax+b, \qquad \sigma^*y=cy+dx+e. \]
Because \(\sigma\) fixes \(P_0\), pullback preserves the filtration by pole order at \(P_0\). Step 1 therefore gives \[ \sigma^*x\in L(2P_0)=\langle1,x\rangle. \] Its pole order is still exactly \(2\), so the coefficient of \(x\) is nonzero: \[ \sigma^*x=ax+b, \qquad a\ne0. \] Likewise \[ \sigma^*y\in L(3P_0)=\langle1,x,y\rangle, \] and exact pole order \(3\) forces the coefficient of \(y\) to be nonzero. This gives the asserted form.
3. In fact\[ \boxed{ \sigma^*x=ax+b, \qquad \sigma^*y=cy } \]with \(a,c\ne0\).
Apply \(\sigma^*\) to the equation \(y^2=f(x)\). Using step 2 gives \[ (cy+dx+e)^2=f(ax+b). \] The right-hand side belongs to \(k(x)\). Since \[ k(X)=k(x)\oplus k(x)y \] as a vector space over \(k(x)\), comparison of the coefficient of \(y\) gives \[ 2c(dx+e)=0. \] Here \(c\ne0\) and \(\operatorname{char}k\ne2\), hence \[ d=e=0. \]
4. Every \(\sigma\in G\) is induced by the projective linear automorphism\[ [X:Y:Z] \longmapsto [aX+bZ:cY:Z]. \]Conversely, this transformation preserves \(X\) precisely when\[ \tau(x)=ax+b \]permutes the set\[ \{0,1,\lambda\} \]and\[ c^2=a^3. \]
The displayed projective map is invertible because \(a,c\ne0\), and on the chart \(Z=1\) it is exactly \[ (x,y)\longmapsto(ax+b,cy). \] It also fixes \(P_0\).
It preserves the equation if and only if \[ c^2f(x)=f(ax+b). \] The roots of the left side are \(0,1,\lambda\), so this identity forces the affine map \(\tau(x)=ax+b\) to permute those three points. Conversely, if \(\tau\) permutes them, then both sides are cubic polynomials with the same three roots. Comparing leading coefficients gives \[ f(ax+b)=a^3f(x), \] so the equality is equivalent to \(c^2=a^3\).
Since \(k\) is algebraically closed and \(\operatorname{char}k\ne2\), there are exactly two choices of \(c\) for each admissible \(\tau\), differing by the involution \[ \iota:(x,y)\longmapsto(x,-y). \]
5. If\[ j(X)\ne0,1728, \]then\[ \boxed{G\cong\ZZ/2} \]and\[ G=\{1,\iota\}. \]
Corollary IV.4.7 says that in this case no nonidentity affine automorphism of \(\PP^1\) fixing \(\infty\) permutes the three finite branch points \(0,1,\lambda\). Thus step 4 leaves only the two lifts of the identity on the \(x\)-line, namely \[ (x,y)\longmapsto(x,y) \qquad\text{and}\qquad (x,y)\longmapsto(x,-y). \] The second has order \(2\).
6. Suppose\[ j(X)=1728, \qquad \operatorname{char}k\ne3. \]Then\[ \boxed{G\cong\ZZ/4}. \]
By Corollary IV.4.7 one has \[ \lambda\in\left\{-1,2,\frac12\right\}. \] In the three respective cases, the nontrivial affine symmetry of the finite branch set is \[ \tau(x)= \begin{cases} -x,&\lambda=-1,\\ 2-x,&\lambda=2,\\ 1-x,&\lambda=\frac12. \end{cases} \] In every case its linear coefficient is \(a=-1\). Choose \[ i\in k, \qquad i^2=-1. \] Then step 4 gives a lift \[ g:(x,y)\longmapsto(\tau(x),iy), \] induced on \(\PP^2\) by \[ [X:Y:Z]\longmapsto \begin{cases} [-X:iY:Z],&\lambda=-1,\\ [-X+2Z:iY:Z],&\lambda=2,\\ [-X+Z:iY:Z],&\lambda=\frac12. \end{cases} \] Now \[ g^2=\iota, \] so \(g\) has order \(4\). Corollary IV.4.7 gives \(|G|=4\), hence \[ G=\langle g\rangle\cong\ZZ/4. \]
7. Suppose\[ j(X)=0, \qquad \operatorname{char}k\ne3. \]Then\[ \boxed{G\cong\ZZ/6}. \]
Choose a primitive cube root of unity \[ \omega^3=1, \qquad \omega\ne1. \] Corollary IV.4.7 gives \[ \lambda=-\omega \qquad\text{or}\qquad \lambda=-\omega^2. \] The affine map cycling the three branch points is respectively \[ \tau(x)= \begin{cases} \omega^2x+1,&\lambda=-\omega,\\ \omega x+1,&\lambda=-\omega^2. \end{cases} \] Indeed, in either case \[ 0\longmapsto1\longmapsto\lambda\longmapsto0. \] Its coefficient \(a\) satisfies \(a^3=1\), so step 4 gives the order-\(3\) lift \[ r:(x,y)\longmapsto(\tau(x),y). \] On \(\PP^2\) this is \[ [X:Y:Z]\longmapsto \begin{cases} [\omega^2X+Z:Y:Z],&\lambda=-\omega,\\ [\omega X+Z:Y:Z],&\lambda=-\omega^2. \end{cases} \] The involution \(\iota\) commutes with \(r\), and \[ \langle r\rangle\cap\langle\iota\rangle=1. \] Thus \[ \langle r,\iota\rangle \cong \ZZ/3\times\ZZ/2 \cong \ZZ/6. \] Corollary IV.4.7 gives \(|G|=6\), so this subgroup is all of \(G\).
8. Finally suppose\[ \operatorname{char}k=3, \qquad j(X)=0=1728. \]Then, after taking\[ \lambda=-1, \]one has\[ \boxed{ G\cong(\ZZ/3)\rtimes(\ZZ/4), } \]where a generator of \(\ZZ/4\) acts on \(\ZZ/3\) by inversion.
In characteristic \(3\) the branch set is \[ \{0,1,-1\}=\FF_3. \] Its full affine symmetry group consists of \[ x\longmapsto ax+b, \qquad a\in\{1,-1\}, \quad b\in\FF_3, \] and has order \(6\).
Let \[ r:(x,y)\longmapsto(x+1,y). \] Since \[ (x+1)^3-(x+1)=x^3-x \] in characteristic \(3\), this preserves \[ y^2=x^3-x \] and has order \(3\). On projective coordinates, \[ r:[X:Y:Z]\longmapsto[X+Z:Y:Z]. \]
Choose \(i\in k\) with \(i^2=-1\) and put \[ s:(x,y)\longmapsto(-x,iy). \] This is induced by \[ s:[X:Y:Z]\longmapsto[-X:iY:Z]. \] Step 4 shows that it preserves \(X\), and \[ s^2=\iota, \] so \(s\) has order \(4\). On the \(x\)-line, reflection conjugates translation by \(1\) to translation by \(-1\), hence \[ srs^{-1}=r^{-1}. \] Therefore \[ \langle r,s\rangle = \left\langle r,s\mid r^3=s^4=1,\ srs^{-1}=r^{-1} \right\rangle \cong (\ZZ/3)\rtimes(\ZZ/4). \] It has \(12\) elements, represented uniquely by \[ r^a s^b, \qquad 0\le a<3, \quad 0\le b<4. \] Corollary IV.4.7 gives \(|G|=12\), so this is the full automorphism group.
9. Q.E.D.
Steps <1>1–<1>4 prove that every automorphism fixing \(P_0\) is induced by a projective transformation of the required form. Steps <1>5–<1>8 give the explicit transformations and the group structure in each of the four cases of Corollary IV.4.7.