Let \(X, P_0\) be an elliptic curve having an endomorphism \(f: X \to X\) of degree 2.
If we represent \(X\) as a 2-1 covering of \(\PP^1\) by a morphism \(\pi: X \to \PP^1\) ramified at \(P_0\), then as in (4.4), show that there is another morphism \(\pi': X \to \PP^1\) and a morphism \(g: \PP^1 \to \PP^1\), also of degree 2, such that \(\pi \circ f=g \circ \pi'\).
For suitable choices of coordinates in the two copies of \(\PP^1\), show that \(g\) can be taken to be the morphism \(x \mapsto x^2\).
Now show that \(g\) is branched over two of the branch points of \(\pi\), and that \(g^{-1}\) of the other two branch points of \(\pi\) consists of the four branch points of \(\pi'\). Deduce a relation involving the invariant \(\lambda\) of \(X\).
Solving the above, show that there are just three values of \(j\) corresponding to elliptic curves with an endomorphism of degree 2, and find the corresponding values of \(\lambda\) and \(j\). Answers: \(j=2^6 \cdot 3^3\); \(j=2^6 \cdot 5^3\); \(j=-3^3 \cdot 5^3\).
Solution
Hartshorne is assuming \(\characteristic k\ne2\) in this section.
1. The degree-\(2\) endomorphism \(f\) is unramified, and\[ \ker f=\{P_0,T\} \]for a nonzero point \(T\) of order \(2\).
Since \(\deg f=2\) and \(\characteristic k\ne2\), the morphism \(f\) is separable. Riemann–Hurwitz for \(f:X\to X\) gives \[ 0=2g(X)-2=2(2g(X)-2)+\deg R_f=\deg R_f, \] so \(f\) is unramified.
An endomorphism of an elliptic curve is a group homomorphism, so its kernel has two points. Write it as \(\{P_0,T\}\). Being a two-element subgroup, it has \(2T=P_0\).
2. There are degree-\(2\) morphisms\[ \pi':X\longrightarrow\PP^1, \qquad g:\PP^1\longrightarrow\PP^1 \]such that\[ \pi\circ f=g\circ\pi'. \]
Let \(\pi'\) be the morphism associated to \[ \abs{P_0+T}. \] Riemann–Roch gives \(h^0(P_0+T)=2\). It has no base point: if \(Q\) were a base point, then \[ h^0(P_0+T-Q)=h^0(P_0+T)=2, \] whereas a degree-\(1\) divisor on a genus-one curve has at most one independent section. Thus this degree-\(2\) divisor gives a base-point-free \(g^1_2\). Its deck involution is \[ \iota_T(R)=T-R, \] because \(R+(T-R)\sim P_0+T\).
The morphism \(\pi\), defined by \(\abs{2P_0}\), has deck involution \([-1](R)=-R\). Since \(f(T)=P_0\) and \(f\) is a homomorphism, \[ f(T-R)=-f(R). \] Hence \[ (\pi\circ f)(T-R)=\pi(-f(R))=\pi(f(R)). \] Thus \(\pi\circ f\) is constant on the fibres of \(\pi'\) and factors through a morphism \(g:\PP^1\to\PP^1\). Comparing degrees gives \[ 4=\deg(\pi\circ f)=\deg g\cdot2, \] so \(\deg g=2\).
3. After suitable choices of coordinates on the two copies of \(\PP^1\),\[ g(t)=t^2. \]
The degree-\(2\) map \(g\) is separable. Riemann–Hurwitz gives exactly two simple ramification points. Send them to \(0,\infty\) in the source and send their branch values to \(0,\infty\) in the target. Then \(g\) has a double zero at \(0\) and a double pole at \(\infty\), so \(g(t)=ct^2\). Rescaling the target coordinate makes \(c=1\).
4. The branch values of \(g\) are two of the four branch values of \(\pi\),and the branch values of \(\pi'\) are exactly the inverse images under \(g\) ofthe other two.
Because \(f\) is unramified, the branch values of \(\pi\circ f\) are exactly the branch values of \(\pi\), and every ramification index of the composite is \(2\).
Using \[ \pi\circ f=g\circ\pi', \] every branch value of \(g\) is therefore a branch value of \(\pi\). No branch value of \(\pi'\) can be a ramification point of \(g\): otherwise the corresponding point of \(X\) would have ramification index \(4\) for the composite. Thus the four branch values of \(\pi'\) lie over the two branch values of \(\pi\) that are not branch values of \(g\). Each of those two target points has two distinct inverse images under \(g\), giving exactly the four branch values of \(\pi'\).
5. Write the branch set of \(\pi\) as\[ \{0,\infty,1,\lambda\} \]so that the two branch values of \(g\) are \(0,\infty\), and choose\(\mu\in k\) with\[ \mu^2=\lambda. \]Then the Legendre parameter arising from \(\pi'\) can be taken to be\[ \lambda' = \left(\frac{\mu-1}{\mu+1}\right)^2, \]and \(\lambda'\) must belong to the \(\Sigma_3\)-orbit of \(\lambda\).
By step 3, \(g(t)=t^2\). Step 4 therefore says that the branch set of \(\pi'\) is \[ \{1,-1,\mu,-\mu\}. \] The fractional linear transformation \[ s = \frac{\mu-1}{\mu+1}\frac{t+1}{t-1} \] sends \[ -1\longmapsto0, \qquad 1\longmapsto\infty, \qquad \mu\longmapsto1, \] and sends \(-\mu\) to \[ \left(\frac{\mu-1}{\mu+1}\right)^2. \] Hence this last number is a Legendre parameter for the same elliptic curve \(X\). By Theorem IV.4.1 and Lemma IV.4.5, two Legendre parameters define isomorphic elliptic curves exactly when they differ by the \(\Sigma_3\) action. Thus \[ \lambda' \in \left\{ \lambda, 1-\lambda, \lambda^{-1}, 1-\lambda^{-1}, (1-\lambda)^{-1}, \lambda(\lambda-1)^{-1} \right\}. \] This is the required relation involving \(\lambda\).
6. Up to the \(\Sigma_3\)-action, the relation in step 5 has exactlythree classes of solutions:\[ \lambda=-1, \qquad \lambda^2-6\lambda+1=0, \qquad \lambda^2-\lambda+16=0. \]
5. Since\(\lambda\ne0,1\), we have \(\mu\ne0,\pm1\). Clearing denominators and removingthe excluded factors gives the equations\[ \begin{gathered} \mu^2+1=0,\\ \mu^2+2\mu-1=0, \qquad \mu^2-2\mu-1=0,\\ \mu^2+3\mu+4=0, \qquad \mu^2-3\mu+4=0,\\ 4\mu^2+3\mu+1=0, \qquad 4\mu^2-3\mu+1=0. \end{gathered} \]These are precisely the nonexcluded factors from the six equalities\[ \lambda'=\lambda, \ 1-\lambda, \ \lambda^{-1}, \ 1-\lambda^{-1}, \ (1-\lambda)^{-1}, \ \lambda(\lambda-1)^{-1}. \]
The first equation gives \(\lambda=-1\). Either of the next two gives \[ \lambda^2-6\lambda+1=0. \] The next two give \[ \lambda^2-\lambda+16=0. \] The last two give \[ 16\lambda^2-\lambda+1=0, \] whose reciprocal \(\lambda^{-1}\) satisfies \[ (\lambda^{-1})^2-\lambda^{-1}+16=0. \] Thus the last two lie in the same \(\Sigma_3\)-orbit as the preceding class. No other parameter classes occur.
7. The corresponding \(j\)-invariants are\[ \boxed{ 2^6\cdot3^3, \qquad 2^6\cdot5^3, \qquad -3^3\cdot5^3 }. \]
Use Hartshorne’s formula \[ j(\lambda) = 256\frac{(\lambda^2-\lambda+1)^3} {\lambda^2(\lambda-1)^2}. \]
For \(\lambda=-1\), \[ j=1728=2^6\cdot3^3. \]
If \(\lambda^2-6\lambda+1=0\), then \[ \lambda^2-\lambda+1=5\lambda, \qquad (\lambda-1)^2=4\lambda. \] Therefore \[ j = 256\frac{(5\lambda)^3}{4\lambda^3} = 8000 = 2^6\cdot5^3. \]
If \(\lambda^2-\lambda+16=0\), then \[ \lambda^2-\lambda+1=-15, \qquad \lambda(\lambda-1)=-16. \] Hence \[ j = 256\frac{(-15)^3}{(-16)^2} = -3375 = -3^3\cdot5^3. \]
Over the algebraically closed field one may choose representatives \[ \lambda=-1, \qquad \lambda=3\pm2\sqrt2, \qquad \lambda=\frac{1\pm3\sqrt{-7}}2, \] respectively. In small positive characteristic some of the three displayed integer \(j\)-values may coincide after reduction; the calculation above shows that no additional \(j\)-values occur.
8. Each of the three parameter classes is realized by a degree-\(2\)endomorphism.
Choose \(\mu\) satisfying one of the equations in step 6 and set \(\lambda=\mu^2\). Consider the smooth genus-one curves \[ X_\lambda: \qquad y^2=x(x-1)(x-\lambda) \] and \[ X_\mu': \qquad Y^2=(t^2-1)(t^2-\mu^2). \] The formulas \[ x=t^2, \qquad y=tY \] give a degree-\(2\) morphism \[ F:X_\mu'\longrightarrow X_\lambda, \] since \[ (tY)^2=t^2(t^2-1)(t^2-\mu^2). \]
By step 5, the Legendre parameter of \(X_\mu'\) is \[ \lambda' = \left(\frac{\mu-1}{\mu+1}\right)^2. \] For the solutions retained in step 6, \(\lambda'\) lies in the \(\Sigma_3\)-orbit of \(\lambda\). Hence Theorem IV.4.1 gives \[ X_\mu'\cong X_\lambda. \] Choose an isomorphism carrying the origin of \(X_\lambda\) to either point of \(X_\mu'\) lying above the origin of \(X_\lambda\) under \(F\); this is possible by composing any isomorphism with a translation on \(X_\mu'\). Its composition with \(F\) is then an endomorphism of \(X_\lambda\) of degree \(2\).
9. Q.E.D.
Steps <1>1–<1>3 prove parts (a)–(b), steps <1>4–<1>5 prove part (c), and steps <1>6–<1>8 give the three parameter classes and the three \(j\)-values of part (d), together with their realization.