Weirstrass Normal Form, Canonical form and j-invariant of a curve.

Question

Given the following elliptical curve:

$y^{2}=x^{3}-4x^{2}+4$

How may I write it in:

1. Weirstrass Normal Form

2. The Canonical Form

3. and calculate the j-invariant

There are many authors suggesting different methods:

1. If we write it in the form: $y^{2}=x^{3}-27c_{4}x-54c_{6}$ ...

Then, $c_{4}=\frac {4}{27}$ and $c_{6}=-\frac {2}{27}$

3. j-invariant = $\frac {1728c_{4}^{3}}{c_{4}^{3}-c_{6}^{2}}=-2513$

Is this correct? Finally, how do I go about writing the equation in the Canonical Form?

Thank you in advance!

Answer 1

In such cases, it is very useful to use computer algebra systems (CAS) like sage - the CAS of my choice below. Some computations will still be done with bare hands below.

To get rid of the term in $x^2$ in the given equation, and since $\displaystyle x^3-4x^2+\dots = \left(x-\frac 43\right)^3+\dots$ it is useful to substitute $\displaystyle X=x-\frac 43$. So $\displaystyle x=X+\frac 43$. Let us take also $Y=y$. Then the given equation becomes $$ \begin{aligned} Y^2 &=\left( X+\frac 43\right)^3 - 4\left(X+\frac 43\right)^2+4\ ,\text{ i.e.} \\ Y^2 &=X^3 -\frac {16}3X -\frac {20}{27}\ . \\[3mm] &\qquad \text{To avoid denominators, we may multiply with $3^6$:} \\[3mm] (3^3 Y)^2 &= (3^2X)^3 - 16\cdot 3^3\;(3^2X)- 20\cdot 3^3\ . \\[3mm] &\qquad\text{ This is} \\[3mm] \eta^2 &= \xi^3 -432\xi - 540\ , \end{aligned} $$ after using the new variables $\xi=3^2X$, and $\eta=3^3Y$.

The two (affine) equations in $(X,Y)$, respectively in $(\xi,\eta)$ are both in short Weierstraß form, determining isomorphic elliptic curves.

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To compute the $j$ invariant, we can use each of the above curves.

$(A)$ Using $y^2=x^3-4x^2+4$: $$ \begin{aligned} a_1&=0\ ,\ a_2=-4\ ,\ a_3=0\ ,\ a_4=0\ ,\ a_6=4\ ,\\[3mm] b_2 &=a_1^2 + 4a_2 &&=-16\ ,\\ b_4 &=a_1a_3 + 2a_4 &&=0\ ,\\ b_6 &=a_3^2 + 4a_6 &&=16\ ,\\ b_8 &=a_1^2a_6 + 4a_2a_6 -a_1a_3a_4 + a_2a_3^2 - a_4^2 &&=-64\ ,\\ c_4 &=b_2^2 - 24b_4 &&=16^2\ ,\\ c_6 &=-b_2^3 +36b_2b_4 -216b_6&&=16^3-216\cdot 16=40\cdot 16=640\ ,\\ \Delta &= -b_2^2b_8 -8b_4^3 -27b_6^2 + 9b_2b_4b_6 &&=16^2\cdot 64-27\cdot 16^2=16^2\cdot 37\ ,\\ j &=\frac{c_4^3}{\Delta} &&=\frac{16^6}{16^2\cdot 37}=\frac{16^4}{37}=\frac{65536}{37}\ . \end{aligned} $$ Note that the first $b$-values can be also extracted from the given form by comparing $y^2=x^3-4x^2+4$ with $\displaystyle $y^2=x^3+\frac{b_2}4x^2+\frac{b_4}x+\frac{b_6}4$.

$(B)$ Using $\displaystyle Y^2=X^3 - \frac {16}3X -\frac{20}{27}$.

The above short form is in fact $\displaystyle Y^2=X^3 - \frac {c_4}{48}X -\frac{c_6}{864}$.

So $\displaystyle \frac {c_4}{48}=\frac {16}3$, giving $c_4=16\cdot 48/3=16^2$.

And $\displaystyle \frac {c_6}{864}=\frac {20}{27}$, giving $c_6=20\cdot 864/27=20\cdot 32=640$.

The value for $\Delta$ is computed in the same manner as in (A).

$(C)$ Using $\displaystyle \eta^2=\xi^3 - 432\xi -540$ we obtain $c_4=-432\cdot 48=-20736-256\cdot 3^4$, and $c_6=-540\cdot 864=640\cdot 3^6$. (The change of variables from $(X,Y)$ to $(\xi,\eta)$ introduces the powers of $3$.)

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Let us check the above in a dialog with sage:

sage: E = EllipticCurve([0, -4, 0, 0, 4])
sage: E1 = EllipticCurve([-16/3, -20/27])
sage: E2 = EllipticCurve([-432, -540])

sage: E.is_isomorphic(E1)
True
sage: E.is_isomorphic(E2)
True
sage: E.a_invariants()
(0, -4, 0, 0, 4)
sage: E.b_invariants()
(-16, 0, 16, -64)
sage: E.c_invariants()
(256, 640)

sage: E.discriminant()
9472
sage: E.c4()^3 / E.discriminant()
65536/37
sage: E.j_invariant()
65536/37

But we can also ask in a row:

sage: E1.a_invariants(), E1.b_invariants(), E1.c_invariants(), E1.discriminant(), E1.j_invariant()
((0, 0, 0, -16/3, -20/27),
 (0, -32/3, -80/27, -256/9),
 (256, 640),
 9472,
 65536/37)

sage: E2.a_invariants(), E2.b_invariants(), E2.c_invariants(), E2.discriminant(), E2.j_invariant()
((0, 0, 0, -432, -540),
 (0, -864, -2160, -186624),
 (20736, 466560),
 5033809152,
 65536/37)