\documentclass{amsart}
\input ellc/emac

\title{Homework 1}
\author{Topics in AG: Elliptic Curves}
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\maketitle

Due Monday, Feb. 23, 2004

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\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} 

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