\documentclass{amsart}
\input ellc/emac
\title{Homework 1}
\author{Topics in AG: Elliptic Curves}
\begin{document}
\maketitle
Due Monday, Feb. 23, 2004
\bigskip
\begin{enumerate}
\item Show that the map from the quartic $\cC: z^2 = x^4 + 1$
to the Weierstrass cubic $E_1: u^2 = v^3 - 4v$ is defined
for all points of a field except finitely many. Show that
the same is true for the inverse map. Thus the map
$\cC \lra E$ is an example of a birational isomorphism
(a map defined by rational functions, giving a bijection
between the (complex) points on the curves except for
finitely many exceptions).
\bigskip
\item Transform the Fermat quartic $\cC_2: z^2 = x^4 - 4$
into Weierstrass form $E_2: u^2 = v^3 + v$, and show that
the map you find is a birational isomorphism.
\bigskip
\item Consider the two `curves' $C_1: X^4 + Y^4 = Z^2$ and
$C_2: X^4 - 4Y^4 = Z^2$ occurring in the proof of FLT for
exponent $4$. In our proof we started with a rational point
$(x,y,z)$ on $C_1$ and obtained a rational point
$(a,b,x)$ on $C_2$. Express $x, y, z$ as rational
functions of $r, s, a$; this defines a rational map
$\phi: C_2 \lra C_1$.
Is the rational map that you get birational
(i.e., can the converse map $C_1 \lra C_2$ be expressed
using rational functions)?
\bigskip
\item Consider the following diagram:
$$ \begin{CD}
C_2 @>>> E_2 \\
@V{\phi}VV @. \\
C_1 @>>> E_1 \end{CD} $$
Since the horizontal maps are (essentially) bijective,
you can define a map $\psi: E_2 \lra E_1$ (i.e. rewrite
$\phi$ in terms of the new coordinates $u, v$). Do this.
\bigskip
\item Show that $7 = a^3 + b^3$ for {\em positive} rational
numbers. Hint: starting with the point $P = (2,1)$ on the
cubic $C:x^3 - y^3 = 7$, construct the tangent to $C$
at $P$ and compute their point of intersection $Q \ne P$.
\end{enumerate}
\end{document}