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 $n = \deg F = 3$.


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.

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