Lecture 5: Wednesday, August 28, 2013

In the previous lecture, we introduced the notion of a non-singular projective curve. Today we discuss examples in detail by focusing on Elliptic Curves, Cubic Curves, and Quartic Curves.

Elliptic Curves

As always, we will denote K as either \mathbb Q, \mathbb R, or \mathbb C. We begin with a standard definition.

Proposition.
Consider the proejctive curve E: y^2 + a_1 \, x \, y + a_3 \, y = x^3 + a_2 \, x^2 + a_4 \, x + a_6, where the a_i \in K.


  • Using the substitution
    \left. \begin{aligned} x & = X - \frac {a_1^2 + 4 \, a_2}{12} \\ y & = Y - \frac {a_1}2 \, X + \frac {a_1^3 + 4 \, a_1 \, a_2 - 12 \, a_3}{12} \end{aligned} \right \} \quad \iff \quad \left \{ \begin{aligned} X & = x + \frac {a_1^2 + 4 \, a_2}{12} \\ Y & = y + \frac {a_1}2 \, x + \frac {a_3}2. \end{aligned} \right.
    we find a cubic curve in the form Y^2 = X^3 + A \, X + B, where
    \begin{aligned} A & = \frac {24 \left( a_1 \, a_3 + 2 \, a_4 \right) - \left( a_1^2 + 4 \, a_2 \right)^2}{48}, \\[5pt] B & = \frac {216 \left( a_3^2 + 4 \, a_6 \right) - 36 \left( a_1^2 + 4 \, a_2 \right) \left( a_1\, a_3 + 2 \, a_4 \right) + \left( a_1^2 + 4 \, a_2 \right)^3}{864}. \end{aligned}

  • E is a nonsingular curve if and only if 4 \, A^3 + 27 \, B^2 \neq 0.

A nonsingular curve as in the proposition above is called an elliptic curve. Note that the discriminant of the cubic f(X) = X^3 + A \, X + B is 4 \, A^3 + 27 \, B^2, so the cubic curve Y^2 = X^3 + A \, X + B is a nonsingular curve if and only if the cubic f(X) = X^3 + A \, X + B has distinct roots.

Proof: Following the ideas from the previous lecture, consider the homogeneous cubic polynomial
F(\tau_1,\tau_2,\tau_0) = \tau_2^2 \, \tau_0 + a_1 \, \tau_1 \, \tau_2 \, \tau_0 + a_3 \, \tau_2 \, \tau_0^2 - \tau_1^3 - a_2 \, \tau_1^2 \, \tau_0 - a_4 \, \tau_1 \, \tau_0^2 - a_6 \, \tau_0^3.
Its gradient involves the partial derivatives
\begin{aligned} \dfrac {\partial F}{\partial \tau_1} & = a_1 \, \tau_2 \, \tau_0 - 3 \, \tau_1^2 - 2 \, a_2 \, \tau_1 \, \tau_0 - a_4 \, \tau_0^2 \\[5pt] \dfrac {\partial F}{\partial \tau_2} & = 2 \, \tau_2 \, \tau_0 + a_1 \, \tau_1 \, \tau_0 + a_3 \, \tau_0^2 \\[5pt] \dfrac {\partial F}{\partial \tau_0} & = \tau_2^2 + a_1 \, \tau_1 \, \tau_2 + 2 \, a_3 \, \tau_2 \, \tau_0 - a_2 \, \tau_1^2 - 2 \, a_4 \, \tau_1 \, \tau_0 - 3 \, a_6 \, \tau_0^2 \end{aligned}
It suffices to consider either the “affine” points (x:y:1) or the point at infinity \mathcal O = (0:1:0). For the latter, we have \nabla F(0,1,0) = (0,0,1) \neq (0,0,0), showing that \mathcal O is never a singular point. For the former, we set \tau_0 = 1 and then use the substitutions above:
\begin{aligned} \dfrac {\partial F}{\partial \tau_1} & = a_1 \, Y - 3 \, X^2 - A \\[5pt] \dfrac {\partial F}{\partial \tau_2} & = 2 \, Y \\[5pt] \dfrac {\partial F}{\partial \tau_0} & = Y^2 + a_3 \, Y - \frac {a_1^2 + 4 \, a_2}{12} \left( 3 \, X^2 + A \right) - 2 \, A \, X - 3 \, B \end{aligned}
We seek all X and Y such that these three partial derivatives vanish. Clearly we must have 3 \, X^2 + A = 0, Y = 0, and 2 \, A \, X + 3 \, B = 0. It is easy to see that these equations have a solution if and only if 4 \, A^3 + 27 \, B^2 = 0. \square.

Let me mention in passing that (X_0 : Y_0 : 1) = (-3 \, B \ : \ 0 \ : \ 2 \, A) is the unique singular point on Y^2 = X^3 + A \, X + B if and only if 4 \, A^3 + 27 \, B^2 = 0 — assuming that not both A and B are zero, of course.

Examples

We discuss a few examples of cubic equations and whether they are singular or nonsingular.


  • Consider the curve E_1 : y^2 = x^3 + 1. We have A = 0 and B = 1. Then 4 \, A^3 + 27 \, B^2 = 27 \neq 0, so E_1 is indeed an elliptic curve.

  • Now consider the curve E_2 : y^2 = x^3 - x. We have A = -1 and B = 0. Then 4 \, A^3 + 27 \, B^2 = -4 \neq 0, so E_2 is also an elliptic curve.

  • Finally, consider the curve E_3 : y^2 = x^3 + x^2. We must place it in the proper form, so substitute x = X - (1/3) and y = Y. We find that Y^2 = X^3 - (1/3) \, X + (2/27), so we set A = - 1/3 and B = 2/27. In this case, 4 \, A^3 + 27 \, B^2 = 0, so E_3 is not an elliptic curve.

    We present another say to see E_3 is not an elliptic curve by explicitly finding a singular point. The homogenization of the polynomial f(x,y) = y^2 - x^3 - x^2 is the cubic polynomial F(\tau_1, \tau_2, \tau_0) = \tau_2^2 \, \tau_0 - \tau_1^3 - \tau_1^2 \, \tau_0 which has the gradient \nabla F(\tau_1, \tau_2, \tau_0) = \left( -3 \, \tau_1^2 - 2 \, \tau_1 \, \tau_0, \ 2 \, \tau_2 \, \tau_0, \ \tau_2^2 - \tau_1^2 \right). If P = (x:y:1) is a singular point, we must have the four equations
    \begin{aligned} -3 \, x^2 - 2 \, x & = 0 \\ 2 \, y & = 0 \\ y^2 - x^2 & =0 \\ y^2 - x^3 - x^2 & = 0 \end{aligned}
    It is easy to check that the only solution is x = y = 0, so P = (0:0:1) is indeed a singular point. Since there is a singular point on E_3 is it not a nonsingular projective curve.



There are really only four types of graphs for curves in the form Y^2 = X^3 + A \, X + B.

  • If 4 \, A^3 + 27 \, B^2 > 0, then there is one “egg”. Consider the graph of y^2 = x^3 + 1.

  • If 4 \, A^3 + 27 \, B^2 < 0 there are two. Consider the graph of y^2 = x^3 - x.

  • If 4 \, A^3 + 27 \, B^2 = 0 yet A, \, B \neq 0, then we have a “node.” Consider the graph of y^2 = x^3 + x^2.

  • If 4 \, A^3 + 27 \, B^2 = 0 because A = B = 0 when we have a “cusp.” Consider the graph of y^2 = x^3.


Cubic Curves

Here are some general observations about cubic polynomials. The most general polynomial of degree d = 3 is
\begin{aligned} f(x,y) & = \left( a_{30} \, x^3 + a_{21} \, x^2 \, y + a_{12} \, x \, y^2 + a_{03} \, y^3 \right) \\[5pt] & + \left( a_{20} \, x^2 + a_{11} \, x \, y + a_{02} \, y^2 \right) + \left( a_{10} \, x + a_{01} \, y \right) + a_{00}. \end{aligned}
(Note that there are 10 monomials in this expression.) The homogenization is the polynomial
\begin{aligned} F(\tau_1, \tau_2, \tau_0) & = \left( a_{30} \, \tau_1^3 + a_{21} \, \tau_1^2 \, \tau_2 + a_{12} \, \tau_1 \, \tau_2^2 + a_{03} \, \tau_2^3 \right) \\[5pt] & + \left( a_{20} \, \tau_1^2 + a_{11} \, \tau_1 \, \tau_2 + a_{02} \, \tau_2^2 \right) \tau_0 + \left( a_{10} \, \tau_1 + a_{01} \, \tau_2 \right) \tau_0^2 + a_{00} \, \tau_0^3 \end{aligned}
The “points at infinity” on the curve E: \, f(x,y) = 0 correspond to the roots of the equation a_{30} \, \tau_1^3 + a_{21} \, \tau_1^2 \, \tau_2 + a_{12} \, \tau_1 \, \tau_2^2 + a_{03} \, \tau_2^3 = 0. If E is nonsingular, then it has genus g(E) = (d-1) \, (d-2) / 2 = 1.

Quartic Curves

Now that we’ve seen cubic curves, let’s focus on quartic curves. We’ll show in some cases that these quartic curves are actually elliptic curves.

Proposition.
Consider the curve E: y^2 = a_4 \, x^4 + a_3 \, x^3 + a_2 \, x^2 + a_1 \, x + a_0. If (1) there exists a K-rational point (x_0, y_0) on E and (2) the quartic f(x) = a_4 \, x^4 + a_3 \, x^3 + a_2 \, x^2 + a_1 \, x + a_0 has distinct roots in \mathbb C, then E may also be expressed in the form Y^2 = X^3 + A \, X + B, where
\begin{aligned} A & = \dfrac {-a_2^2 + 3 \, a_1 \, a_3 - 12 \, a_0 \, a_4}3 \\ B & = \dfrac {2 \, a_2^3 - 9 \, a_1 \, a_2 \, a_3 + 27 \, a_0 \, a_3^2 + 27 \, a_1^2 \, a_4 - 72 \, a_0 \, a_2 \, a_4}{27}; \end{aligned}
such that 4 \, A^3 + 27 \, B^2 \neq 0. Moreover, the substitution between the two curves can be chosen to have coefficients in K.

Proof: The discriminant of the quartic 4 \, A^3 + 27 \, B^2 = -\text{disc}(f) \neq 0 is nonzero because f(x) has distinct roots. We break the proof into two cases for the point (x_0, y_0).

Case #1: y_0 \neq 0. Denote the polynomials
\begin{aligned}   f(u, v) & = a_4 \, u^4 + a_3 \, u^3 \, v + a_2 \, u^2 \, v^2 + a_1 \, u \, v^3 + a_0 \, v^4 \\[5pt]  \text{Hess}(f)(u, v) & = \left| \begin{matrix} \dfrac {\partial^2 f}{\partial u^2}(u, v) & \dfrac {\partial^2 f}{\partial u \, \partial v}(u, v) \\[10pt] \dfrac {\partial^2 f}{\partial v \, \partial u}(u, v) & \dfrac {\partial^2 f}{\partial v^2}(u, v) \end{matrix} \right| \\[5pt]  \text{Cov}(f)(u, v) & = \left| \begin{matrix} \dfrac {\partial f}{\partial u}(u, v) & \dfrac {\partial f}{\partial v}(u, v) \\[10pt] \dfrac {\partial \text{Hess}(f)}{\partial u}(u, v) & \dfrac {\partial \text{Hess}(f)}{\partial v}(u, v) \end{matrix} \right|\end{aligned}
Make the substitutions
\begin{aligned} x & = x_0 + y_0 \, \dfrac{ 2 \, (X-X_0)}{(Y-Y_0) + \mu \, (X-X_0)} \\[5pt] y & = x_0 \, \dfrac{(X-X_0)^3 + 2 \, Y_0 \, \bigl( (Y-Y_0) + \mu \, (X-X_0) \bigr) + 2 \, \nu \, (X-X_0)}{\bigl( (Y-Y_0) + \mu \, (X-X_0) \bigr)^2} \end{aligned}
in terms of the constants
\begin{aligned}  X_0 & = - \dfrac{\text{Hess}(f)(x_0, 1)}{36 \, f(x_0, 1)} \\[5pt]  Y_0 & = - \dfrac{\text{Cov}(f)(x_0, 1)}{288 \, f(x_0, 1)^2} \, y_0 \\[5pt]  \mu & = - \dfrac{\dfrac {\partial f}{\partial u}(x_0, 1)}{2 \, f(x_0, 1)} \, y_0 \\[5pt]  \nu & = - \dfrac { \begin{aligned} & (8 \, a_3^3 \, a_4 - 32 \, a_2 \, a_3 \, a_4^2 + 64 \, a_1 \, a_4^3) \, x_0^9 + (9 \, a_3^4 - 24 \, a_2 \, a_3^2 \, a_4 - 48 \, a_2^2 \, a_4^2 + 96 \, a_1 \, a_3 \, a_4^2 + 192 \, a_0 \, a_4^3) \, x_0^8 \\ & + (24 \, a_2 \, a_3^3 - 96 \, a_2^2 \, a_3 \, a_4 + 48 \, a_1 \, a_3^2 \, a_4 + 384 \, a_0 \, a_3 \, a_4^2) \, x_0^7 \\ & + (16 \, a_2^2 \, a_3^2 + 36 \, a_1 \, a_3^3 - 64 \, a_2^3 \, a_4 - 112 \, a_1 \, a_2 \, a_3 \, a_4 + 240 \, a_0 \, a_3^2 \, a_4 + 48 \, a_1^2 \, a_4^2 + 256 \, a_0 \, a_2 \, a_4^2) \, x_0^6 \\ & + (48 \, a_1 \, a_2 \, a_3^2 + 72 \, a_0 \, a_3^3 - 192 \, a_1 \, a_2^2 \, a_4 - 24 \, a_1^2 \, a_3 \, a_4 + 192 \, a_0 \, a_2 \, a_3 \, a_4 + 384 \, a_0 \, a_1 \, a_4^2) \, x_0^5 \\ & + (30 \, a_1^2 \, a_3^2 + 120 \, a_0 \, a_2 \, a_3^2 - 216 \, a_1^2 \, a_2 \, a_4 - 96 \, a_0 \, a_2^2 \, a_4 + 288 \, a_0 \, a_1 \, a_3 \, a_4 + 384 \, a_0^2 \, a_4^2) \, x_0^4 \\ &  + (-8 \, a_1^2 \, a_2 \, a_3 + 32 \, a_0 \, a_2^2 \, a_3 + 144 \, a_0 \, a_1 \, a_3^2 - 96 \, a_1^3 \, a_4 - 256 \, a_0 \, a_1 \, a_2 \, a_4 + 384 \, a_0^2 \, a_3 \, a_4) \, x_0^3 \\ & + (-12 \, a_1^3 \, a_3 + 48 \, a_0 \, a_1 \, a_2 \, a_3 + 144 \, a_0^2 \, a_3^2 - 240 \, a_0 \, a_1^2 \, a_4)  \, x_0^2 \\ & + (-24 \, a_0 \, a_1^2 \, a_3 + 96 \, a_0^2 \, a_2 \, a_3 - 192 \, a_0^2 \, a_1 \, a_4) \, x_0 + (a_1^4 - 8 \, a_0 \, a_1^2 \, a_2 + 16 \, a_0^2 \, a_2^2 - 64 \, a_0^3 \, a_4) \end{aligned}}{32 \, f(x_0, 1)^2}  \end{aligned}
It is straight-forward to check that Y^2 = X^3 + A \, X + B with A and B as above. In fact, Y_0^2 = X_0^3 + A \, X_0 + B, so that (X_0 : Y_0 : 1) is a K-rational point on the cubic curve.

Case #2: y_0 = 0. In this case, the quartic f(x) = a_4 \, x^4 + a_3 \, x^3 + a_2 \, x^2 + a_1 \, x + a_0 has a distinct K-rational root $x_0$. Say that we may factor f(x) in the form f(x) = a_4 \left( x - e_1 \right) \left( x - e_2 \right) \left( x - e_3 \right) \left( x - e_4 \right), where the e_i are complex roots. Without loss of generality, assume that e_4 \in K is the K-rational root, and note the identities

\begin{aligned} f(e_4) & = 0 \\ f'(e_4) & = a_4 \left( e_4 - e_1 \right) \left( e_4 - e_2 \right) \left( e_4 - e_3 \right) \\[5pt] f''(e_4) & = 2 \, a_4 \left[ \left( e_4 - e_1 \right) \left( e_4 - e_2 \right) + \left( e_4 - e_1 \right) \left( e_4 - e_3 \right) + \left( e_4 - e_2 \right) \left( e_4 - e_3 \right) \right]. \end{aligned}
Since e_4 \in K each of these expressions is also in K. We make the substitution
\left. \begin{aligned} x & = f'(e_4) \, \frac {6}{6 \, X - f''(e_4)} + e_4 \\[5pt] y & = f'(e_4) \, \frac {36 \, Y}{\left( 6 \, X - f''(e_4) \right)^2} \end{aligned} \right \} \quad \iff \quad \left \{ \begin{aligned} X & = f'(e_4) \, \dfrac 1{x - e_4} + \frac {f''(e_4)}6 \\ Y & = f'(e_4) \, \dfrac y{\left( x - e_4 \right)^2} \end{aligned} \right.
Note that these transformations have coefficients in K. \square.

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About Edray Herber Goins, Ph.D.

Edray Herber Goins grew up in South Los Angeles, California. A product of the Los Angeles Unified (LAUSD) public school system, Dr. Goins attended the California Institute of Technology, where he majored in mathematics and physics, and earned his doctorate in mathematics from Stanford University. Dr. Goins is currently an Associate Professor of Mathematics at Purdue University in West Lafayette, Indiana. He works in the field of number theory, as it pertains to the intersection of representation theory and algebraic geometry.
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2 Responses to Lecture 5: Wednesday, August 28, 2013

  1. Pingback: Lecture 15: Monday, September 23, 2013 | Lectures on Dessins d'Enfants

  2. Pingback: MA 59800 Course Syllabus | Lectures on Dessins d'Enfants

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