## 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 $K$. $\square$.