If E is non-singular then it has non-singular weierstrass equation

Question

Let $E$ be a non-singular projective curve of genus one. There exist regular functions $x,y$ on $E$ satisfying a Weiestrass equation

$$ y^2 = x^3 + ax + b$$

Is this equation necessarily non-singular?

According to Milne (Modular Functions and Modular Forms, p. 47) "The fact that E is nonsingular implies that $\Delta = 4a^3 - 27b^2 \neq 0$" Could you explain why is this the case?

Answer 1

I was given an answer today:

My question was, "is a planar model of the form y^2 = x^3 + ax + b of E non-singular?"

The answer is yes.

Sketch of proof:

Suppose it wasn't. The planar curve self-intersects at some point. Given such point we can construct a morphism from the complex projective line $\mathbb{P}^1(\mathbb{C})$ to $E$, which turns out to be an isomorphism. But $\mathbb{P}^1(\mathbb{C})$ has genus $0$, which is a contradiction.

The map constructed gives the third point of intersection of $E$ with the line through the nonsingular point and with slope $z\in \mathbb{P}^1(\mathbb{C})$.

Answer 2

The equation $y^2=x^3+ax+b$ is not necessarily non-singular - take $a=0,b=0$ to get $y^2=x^3$ which has a singular point at the origin. The equation $y^2=x^3+ax+b$ is non-singular if and only if $\Delta \neq 0$ and here is the proof:

Let $F(x,y)=y^2-(x^3+ax+b)$. Then

• $E$ is singular $\iff$
• there is a (geometric) point $(x_0, y_0)$ of $E$ such that $\partial F/\partial x (x_0, y_0)=0$ and $ \partial F/\partial y (x_0, y_0)=0$ $\iff$
• $4a^2-27b^3=0$

Since $\partial F/\partial x = -(3x^2+ax)$ and $\partial F/\partial y = 2y$, (and assuming we are not in characteristic 2) we get from the latter that if a singular point $(x_0,y_0)$ exists if and only if $2y_0=0$ so $y_0$=0 and $3x_0^2+a = 0$ and so $a=-3x_0^2$. Since $y_0^2=x_0^3+ax_0+b$ and $y_0=0$, so $x_0^3 + ax_0+b=0$. Substituting $\bf a=-3x_0^2$ we have $x_0^3-3x_0^3+b=0$. So $\bf b=2x_0^3$. Then $4a^3=4(-3x_0^2)^3=-108x_0^6$ and $27b^2=27(2x_0^3)^2=108x_0^6$. So the presence of a singular point $(x_0, y_0)$ gives $4a^3-27b^2=0$. Conversely, if $4a^3-27b^2=0$, then either $a=b=0$ (giving the cuspidal cubic mentioned at the beginning), otherwise set $x_0= \frac{b/2}{a/-3}$. Then arguing backwards shows that $(x_0, 0)$ is a singular point.

A shortcut is to say that $\Delta$ is the discriminant of $x^3+ax+b$ and so $\Delta$ is non-zero if an only if $x^3+ax+b$ has no double roots, which amounts to $\frac{\partial}{\partial x}(x^3+ax+b) =0 $ and $F=0$ at $(x_0, 0)$, and this is equivalent to $\partial F/\partial x (x_0, y_0)=0$ and $ \partial F/\partial y (x_0, y_0)=0$ (at least in char $\neq 2$)