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Re: Elliptic Curves and Mathematica

• To: mathgroup at smc.vnet.net
• Subject: [mg40836] Re: Elliptic Curves and Mathematica
• From: "Richard I. Pelletier" <bitbucket at attbi.com>
• Date: Mon, 21 Apr 2003 06:58:36 -0400 (EDT)
• References: <b7qolq$3ij$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

[[ This message was both posted and mailed: see
   the "To," "Cc," and "Newsgroups" headers for details. ]]

In article <b7qolq$3ij$1 at smc.vnet.net>, "flip"
<flip_alpha at safebunch.com> wrote:

> Hello,
> 
> I am working on Elliptic Curves and was wondering if these are doable in
> Mathematica?  I can do them manually, but was wondering if it is possible to
> show in
> Mathematica).

[posted and emailed]

hi flip,

the nb pasted at the end shows 1 problem i worked when i took a class
in elliptic curves.  

the problem was (in case i mistype and this does not match the nb, use
the nb): given the points A = (0,2) and B=(-2,0) on the elliptic curve
y^2 - (x-1)(x^2-4) = 0. compute 2A, 3A, A+B, 2A+B, and 3A+B.

the nb uses david park's drawgraphics package at his site 

http://home.earthlink.net/~djmp/

which makes it very easy to add lines, points, and text to a graph of
an elliptic curve. there is a drawing at the beginning showing the
curve and points A,B; and a drawing at the end showing the curve, all
the points i computed, and the lines that determined them.

hope this helps.

rip

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A. Given the points A,B on the elliptic curve C, compute: 2A, 3A, \
A+B, 2A+B, 3A+B.\
\>", "Subsection"],

Cell[BoxData[{
    \(\(A = {0, 2};\)\), "\[IndentingNewLine]", 
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- 
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let's get 2A. we get the tangent line at A, and we get the \
intersections:\
\>", "Text"],

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we see the double intersection at A, and the new one is the 1st. \
write it to change the sign of the y-component, though in this case
it's 0:\
\
\>", "Text"],

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    \(twoA = \({x, \(-y\)} /. y -> \(-y\)\) /. soln[\([1]\)]\)],
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but that's interesting. because 2A is the intersection, we don't \
get a third point: the line thru 2A and A is the tangent line at A.\
\>", \
"Text"],

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the first point is 2A, the other two are A. i really don't like 3A \
= 2A, so i guess that A is the intersection, so i change the sign:\
\>", \
"Text"],

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    \(threeA = {x, \(-y\)} /. soln[\([2]\)]\)], "Input"],

Cell["\<\
let's go for A+B. here's the line thru the two points, and the \
three intersections:\
\>", "Text"],

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    \(twoAandB == {x, \(-y\)} /. soln[\([2]\)]\)], "Input"],

Cell["finally, 3A+B = 2A+(A+B):", "Text"],

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    \(line3AB = twoPtLine[twoA, AandB]\), "\[IndentingNewLine]", 
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Cell[BoxData[
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i.e. it looks like we have a tangent at A+B. i have to remind \
myself to flip the sign:\
\>", "Text"],

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    \(threeAandB = {x, \(-y\)} /. soln[\([2]\)]\)], "Input"],

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Cell[TextData[{
  "we see 7 distinct points, and add the point at \[Infinity], 0. can i
build \
the group table?\ni think 2B = 0 (vertical tangent at B), 4A=0
(vertical \
tangent at 2A). then A+3A=0 (follows, but also they lie on vertical
line), \
and (2A+B)+(2A+B)=0 (follows, but also vertical tangent at 2A+B), and
finally \
(3A+B)+(A+B) = 0 (follows, but also they lie on vertical line.\nso, we
have \
an element of order 4, and an element of order 2. this is an abelian
group, \
so it's ",
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-- 
NOTE: email address is rip1 AT attbi DOT com

• Follow-Ups:
  • Re: Re: Elliptic Curves and Mathematica
    • From: Murray Eisenberg <murraye@attbi.com>