1
$\begingroup$

I have given the following elliptic curve $E:F(x,y) = 0$:

elliptic curve E
(Where $F(X,Y) := Y^2 + a_1XY + a_3Y - X^3 - a_2X^2 - a_4X - a_6$ with $a_1 = -1.5, a_2 = 3, a_3 = 1, a_4 = 0.5, a_6 = -1.5$.)

The curve $F$ is the black graph, $F_X$ is the red one and $F_Y$ is the blue one. (by $F_X$ I denote the partial derivative of $F$ w.r.t. $X$)

Could you please tell me, if my reasoning below is correct?

I have learned that a point p on a curve (i.e. $F(p) = 0$) is a multiple point if it is also a zero of all the first partial derivatives $F_X, F_Y$ and $F_Z$ of the homogenized equation $F(X,Y,Z)$. Now by looking at the graph we see that there is no point on the curve $F$ which is a zero of both $F_X$ and $F_Y$, because there is no point at which the three graphs $F$, $F_X$ and $F_Y$ intersect. It remains to check the points at infinity.

The curve has one point at infinity, namely $(0,1,0) =: \infty$ (because $-a^3 = F(a,b,0) = 0 \iff a = 0$). We calculate $F_X(\infty) = 0, F_Y(\infty) = 0$ but $F_Z(\infty) = 1$, so $\infty$ is also just a simple point.

Hence the curve has only simple points and is thus non-singular (smooth). We may verify that its genus is $1$ (as for all elliptic curves) via the equation $g = \frac{(n-1)(n-2)}{2}$ where $n = \deg F = 3$.

$\endgroup$
1
$\begingroup$

Just to say that by referring to $F$ as an elliptic curve, you are implicitly assuming that it's smooth as this forms part of the definition.

An easy way to see when this works is for medium Weierstrass equations. For $y^2=f(x)=x^2+a_2x^2+a_4x+a_6$, a point $P=(x,y)$ is singular if and only if $\partial F/\partial x=-f'(x)=0$ and $\partial F/\partial y=2y=0$. This is if and only if $y=f'(x)$, so when $f$ has a double root.

Try it with some obvious examples like $y^2=x^3$ and $y^2=x^3-3x+2$. You'll find a point of intersection at $(0,0)$ (resp. $(1,0)$) which is the unique singular point lying on the curve.

But yes! This gives a necessary and sufficient criterion for $F$ to have an affine singular point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.