Again let \(X\) be an elliptic curve over \(\CC\) determined by the elliptic functions with periods \(1, \tau\), and assume that \(\tau\) lies in the region \(G\) of (4.15B).
If \(X\) has any automorphisms leaving \(P_0\) fixed other than \(\pm 1\), show that either \(\tau=i\) or \(\tau=\omega\), as in (4.20.1) and (4.20.2). This gives another proof of the fact (4.7) that there are only two curves, up to isomorphism, having automorphisms other than \(\pm 1\).
Now show that there are exactly three values of \(\tau\) for which \(X\) admits an endomorphism of degree 2. Can you match these with the three values of \(j\) determined in (Ex. 4.5)? Answers: \(\tau=i\); \(\tau=\sqrt{-2}\); \(\tau=\frac{1}{2}(-1+\sqrt{-7})\).
Solution
Put \[ \Lambda=\ZZ+\ZZ\tau, \qquad \tau=x+iy, \qquad y>0. \] Hartshorne’s region \(G\) satisfies \[ -\frac12\le x<\frac12, \qquad \abs{\tau}\ge1, \] with the boundary convention in Theorem IV.4.15B. In particular, \[ y^2 = \abs{\tau}^2-x^2 \ge 1-\frac14 = \frac34. \]
1. Let \(\alpha\in R=\Endo(X,P_0)\) correspond to multiplication by \(\alpha\) on \(\CC/\Lambda\).Write\[ \alpha=a+b\tau, \qquad \alpha\tau=c+d\tau, \]with \(a,b,c,d\in\ZZ\).If \(\alpha\notin\ZZ\), then \(b\ne0\) and\[ \boxed{ b\tau^2+(a-d)\tau-c=0. } \]
Proposition IV.4.18 says precisely that \[ R=\{\alpha\in\CC:\alpha\Lambda\subseteq\Lambda\}. \] Since \(1,\tau\) is a \(\ZZ\)-basis of \(\Lambda\), the two displayed integral expressions for \(\alpha\) and \(\alpha\tau\) are necessary and sufficient. Eliminating \(\alpha\) gives the boxed quadratic equation. If \(b=0\), then \(\alpha=a\in\ZZ\).
2. If \(\deg f_\alpha\le2\) and \(\alpha\notin\ZZ\), then\[ \boxed{b=\pm1.} \]
By Exercise IV.4.11(a), \[ \deg f_\alpha=\abs{\alpha}^2. \] Since \[ \Im(\alpha)=by, \] we have \[ b^2y^2\le\abs{\alpha}^2\le2. \] The bound \(y^2\ge3/4\) therefore gives \[ b^2\le\frac83<4. \] Thus \(0<\abs b<2\), so \(b=\pm1\).
3. After replacing \(\alpha\) by \(-\alpha\) if necessary, assume \(b=1\).Then there is an integer\[ t=d-a \]such that\[ \tau^2-t\tau-c=0, \]and\[ \boxed{ t=2x, \qquad -c=\abs{\tau}^2, \qquad \abs{\alpha}^2=a^2+at-c. } \]Moreover \(t\in\{-1,0\}\) for the points of \(G\) relevant below.
With \(b=1\), step 1 gives \[ \tau^2+(a-d)\tau-c=0, \] which is the first assertion. Since this polynomial has real integral coefficients and \(\tau\notin\RR\), its other root is \(\bar\tau\). Vieta’s formulas give \[ \tau+\bar\tau=t, \qquad \tau\bar\tau=-c. \] Thus \(t=2x\) and \(-c=\abs\tau^2\). Also \[ \abs{a+\tau}^2 = a^2+a(\tau+\bar\tau)+\tau\bar\tau = a^2+at-c. \] Because \(-1/2\le x<1/2\) and \(t=2x\in\ZZ\), one has \(t\in\{-1,0\}\). The excluded right boundary \(x=1/2\) is precisely the boundary convention making the representative in \(G\) unique.
4. If \(X\) has an automorphism fixing \(P_0\) other than \(\pm1\), then\[ \boxed{ \tau=i \quad\text{or}\quad \tau=\omega=\frac{-1+\sqrt{-3}}2. } \]
An automorphism has degree \(1\). If its multiplier \(\alpha\) were an integer, then \[ 1=\abs\alpha^2=\alpha^2, \] so \(\alpha=\pm1\). Thus a new automorphism has \(\alpha\notin\ZZ\), and steps <1>2–<1>3 apply. Since \[ 1=\abs\alpha^2=a^2+at-c, \] we have \[ c=a^2+at-1. \]
If \(t=0\), then \(c=a^2-1\). The quadratic \[ \tau^2-c=0 \] must have nonreal roots, so its discriminant \(4c\) is negative. Hence \(a=0\), \(c=-1\), and \[ \tau^2+1=0. \] The root in the upper half-plane is \(\tau=i\).
If \(t=-1\), then \[ c=a^2-a-1, \] and nonreality gives \[ 1+4c=4a^2-4a-3<0. \] Thus \(a=0\) or \(a=1\); in either case \(c=-1\), so \[ \tau^2+\tau+1=0. \] The root in the upper half-plane is \[ \tau=\frac{-1+\sqrt{-3}}2=\omega. \]
Conversely, multiplication by \(i\) preserves \(\ZZ+\ZZ i\), and multiplication by \(\omega\) preserves \(\ZZ+\ZZ\omega\). Both have norm \(1\) and are different from \(\pm1\). This proves part (a).
5. If \(X\) has an endomorphism of degree \(2\), then\[ \boxed{ \tau=i, \qquad \tau=\sqrt{-2}, \qquad \tau=\frac{-1+\sqrt{-7}}2. } \]
No integer multiplier has degree \(2\), since an integer \(n\) has degree \(n^2\). Hence \(\alpha\notin\ZZ\), and steps <1>2–<1>3 apply. Now \[ 2=\abs\alpha^2=a^2+at-c, \] so \[ c=a^2+at-2. \]
If \(t=0\), then \(c=a^2-2\). Nonreality of the roots gives \[ 4c=4a^2-8<0, \] so \(a\in\{-1,0,1\}\). For \(a=\pm1\) one has \(c=-1\) and therefore \[ \tau=i. \] For \(a=0\) one has \(c=-2\) and therefore \[ \tau=\sqrt{-2}. \]
If \(t=-1\), then \(c=a^2-a-2\). The discriminant condition is \[ 1+4c = 4a^2-4a-7 <0, \] so \(a=0\) or \(a=1\). In either case \(c=-2\), and hence \[ \tau^2+\tau+2=0. \] The root in the upper half-plane is \[ \tau=\frac{-1+\sqrt{-7}}2. \] There are no other possibilities.
6. Each of the three values in step 5 actually has a degree-\(2\) endomorphism.
For \(\tau=i\), multiplication by \[ 1+i \] preserves \(\ZZ+\ZZ i\) and has squared absolute value \(2\).
For \(\tau=\sqrt{-2}\), multiplication by \(\tau\) preserves \(\ZZ+\ZZ\tau\) because \(\tau^2=-2\), and \[ \abs{\tau}^2=2. \]
For \[ \tau=\frac{-1+\sqrt{-7}}2, \] one has \(\tau^2+\tau+2=0\). Hence multiplication by \(\tau\) preserves \(\ZZ+\ZZ\tau\), and again \[ \abs{\tau}^2=\tau\bar\tau=2. \] Exercise IV.4.11(a) therefore gives degree \(2\) in all three cases.
7. The three period values match the three \(j\)-values of Exercise IV.4.5 as follows:\[ \boxed{ \begin{array}{c|c|c} \tau & \text{CM discriminant} & j\\ \hline i & -4 & 1728=2^6\cdot3^3\\ \sqrt{-2} & -8 & 8000=2^6\cdot5^3\\ \dfrac{-1+\sqrt{-7}}2 & -7 & -3375=-3^3\cdot5^3. \end{array} } \]
For \(\tau=i\), Example IV.4.20.1 already gives \[ j(i)=1728. \] The other two lattices have endomorphism rings containing the quadratic orders of discriminants \(-8\) and \(-7\), respectively: their defining equations are \[ \tau^2+2=0 \] and \[ \tau^2+\tau+2=0. \] The corresponding class-number-one Hilbert class polynomials are \[ H_{-8}(T)=T-8000, \qquad H_{-7}(T)=T+3375. \] Thus the associated singular moduli are \(8000\) and \(-3375\). These are exactly the two remaining \(j\)-values found in Exercise IV.4.5.
8. Q.E.D.
Step 4 proves part (a), steps <1>5–<1>6 prove the complete list in part (b), and step 7 matches the three period representatives with the three \(j\)-values from Exercise IV.4.5.