Let \(X\) be an elliptic curve over \(\CC\), defined by the elliptic functions with periods \(1, \tau\). Let \(R\) be the ring of endomorphisms of \(X\).
If \(f \in R\) is a nonzero endomorphism corresponding to complex multiplication by \(\alpha\), as in (4.18), show that \(\deg f=\abs{\alpha}^2\).
If \(f \in R\) corresponds to \(\alpha \in \CC\) again, show that the dual \(\hat{f}\) of (Ex. 4.7) corresponds to the complex conjugate \(\bar{\alpha}\) of \(\alpha\).
If \(\tau \in \QQ(\sqrt{-d})\) happens to be integral over \(\ZZ\), show that \(R=\ZZ[\tau]\).
Solution
Put \[ \Lambda=\ZZ+\ZZ\tau, \qquad X(\CC)\cong\CC/\Lambda. \] Under this uniformization, \[ R = \{\alpha\in\CC:\alpha\Lambda\subseteq\Lambda\}. \]
1. If \(0\ne\alpha\in R\) and \(f=f_\alpha\), then\[ \boxed{\deg f=\abs{\alpha}^2.} \]
The kernel of \[ f_\alpha:\CC/\Lambda\longrightarrow\CC/\Lambda, \qquad z+\Lambda\longmapsto\alpha z+\Lambda \] is \[ \alpha^{-1}\Lambda/\Lambda. \] Multiplication by \(\alpha\) identifies this finite group with \[ \Lambda/\alpha\Lambda. \] Since the base field is \(\CC\), the nonzero isogeny \(f_\alpha\) is separable. Hence its degree is the number of points in its kernel, as in Exercise IV.4.7(c). Therefore \[ \deg f_\alpha = [\Lambda:\alpha\Lambda]. \]
Let \(A(\Lambda)\) denote the Euclidean area of a fundamental parallelogram of \(\Lambda\). Multiplication by \(\alpha\) scales area by \(\abs{\alpha}^2\), so \[ A(\alpha\Lambda)=\abs{\alpha}^2A(\Lambda). \] For a sublattice of finite index, \[ [\Lambda:\alpha\Lambda] = \frac{A(\alpha\Lambda)}{A(\Lambda)}. \] Combining the last three displays proves part (a).
2. If \(f_\alpha\) corresponds to \(\alpha\in R\), then its dual correspondsto \(\bar\alpha\).
If \(\alpha=0\), both the dual endomorphism and complex conjugate are zero. Assume \(\alpha\ne0\), and let \(\beta\in R\) correspond to \(\widehat f_\alpha\). By Exercise IV.4.7(c), \[ \widehat f_\alpha\circ f_\alpha=[\deg f_\alpha]_X. \] On the complex torus, composition multiplies the corresponding complex numbers. By step 1, \[ \beta\alpha = \deg f_\alpha = \abs{\alpha}^2 = \bar\alpha\alpha. \] Since \(\alpha\ne0\), cancellation gives \[ \beta=\bar\alpha. \] This proves part (b).
3. If \(\tau\) is integral over \(\ZZ\), then\[ \boxed{R=\ZZ[\tau].} \]
First let \(\alpha\in R\). Since \(1\in\Lambda\) and \(\alpha\Lambda\subseteq\Lambda\), one has \[ \alpha=\alpha\cdot1\in\Lambda. \] Thus \[ \alpha=a+b\tau \] for some \(a,b\in\ZZ\), so \[ R\subseteq\ZZ+\ZZ\tau. \]
Because \(\tau\in\QQ(\sqrt{-d})\setminus\QQ\) is integral, its monic minimal polynomial over \(\QQ\) has the form \[ T^2-uT+v \] with \(u,v\in\ZZ\). Hence \[ \tau^2=u\tau-v\in\ZZ+\ZZ\tau. \] It follows that \(\ZZ+\ZZ\tau\) is a subring of \(\CC\), namely \(\ZZ[\tau]\).
Conversely, if \(\alpha=a+b\tau\in\ZZ[\tau]\), then \[ \alpha\cdot1\in\Lambda, \qquad \alpha\tau=a\tau+b\tau^2\in\Lambda. \] Since \(1,\tau\) generate \(\Lambda\) over \(\ZZ\), this implies \[ \alpha\Lambda\subseteq\Lambda. \] Thus \(\alpha\in R\), proving the reverse inclusion and part (c).
4. Q.E.D.
Steps 1, 2, and 3 prove parts (a), (b), and (c), respectively.