Singular points of a curve I want to solve the following exercise:

Let $F=Y^{2}-G(X)$, with $G\in\mathbb{C}[X]$ with degree $d\geq3$, and let $\mathcal{C}=\mathrm{V}(F)$.

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*Prove that the singular points of $\mathcal{C}$ are in the line $Y=0$.


*Prove that a point $(a,0)$ is singular for $\mathcal{C}$ if and only if $a$ is a multiple root of $G$.


*If $G=(5-X^{2})(4X^{4}-20X^{2}+25)$, compute the singular points of $\mathcal{C}$.


*Prove that $\mathcal{C}$ has a unique point at infinity and that this point is simple if and only if  $d=3$. Does the curve $\mathcal{C}$ have asymptotes?

And I thought this, but I need some help:

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*I know that the tangent lines are given by the form of lowest degree, which is $Y^2$. Then $Y=0$ is a double tangent line. Must all singular points be in the same tangent line?


*We have that $(a,0)$ is singular if and only if it is a zero of $F$, $\frac{\partial F}{\partial X}$ and $\frac{\partial F}{\partial Y}$. As $Y=0$, then $F(a,0)=G(a)$,  $\frac{\partial F}{\partial X}(a,0)=G'(a)$ and $\frac{\partial F}{\partial Y}(a,0)=0$ so it must be a multiple root of $G$. Is this correct?


*I haven't do this yet, but I know the algorithm to do it.


*Let $G(X)=a_0 X^d + a_1 X^{d-1}+\cdots +a_d$. We know that $P_\infty (\mathcal C)= \mathrm V (F^*, Z)$, so we compute $$F^* = Y^2 Z^{d-2}-(a_0 X^d + a_1 X^{d-1}Z+\cdots +a_d Z^d)$$ If $Z=0$, then $X=0$, so $P_\infty (\mathcal C)=\mathrm V(F^*, Z)=\{P=(0:1:0)\}$. Let's check the multiplicity:  $$\frac{\partial F^*}{\partial X}(P)=0,\quad \frac{\partial F}{\partial Y}(P)=0,$$ $$\frac{\partial F}{\partial Z}(P)=((d-2)Y^2Z^{d-3})(P)=(d-2)Z^{d-3}(P)\neq0 \Leftrightarrow d=3$$How can I know if it has aymptotes?
(Sorry for my English)
 A: *

*Your reasoning isn't quite right, because $G(X)$ might have terms of degree lower than $3$ as well, and the homogeneous component of $F(X)$ with the lowest degree only tells you whether $(0, 0)$ is a singular point and the equations of its tangent lines. Still, if $(X, Y)$ is a singular point, then $\frac{\partial F}{\partial Y} (X, Y) = 0$ implies that $Y = 0$. Therefore any singular point must lie on the line of equation $Y = 0$.

*Yes.

*Notice that it is enough to look for the points satisfying the condition in (2).

*The asymptotes of the curve are the tangent lines to its points at infinity which are not the line at infinity. Assuming $d = 3$, you have proved that there is one and only one point at infinity, namely $P = (0 : 1 : 0)$. From the dehomogenized polynomial: $$ F^*|_{Y = 1} = Z - (a_0 X^3 + a_1 X^2 Z + a_2 X Z^2 + a_3 Z^3) $$
we see that $P$ is indeed a simple point and its tangent line has equation $Z = 0$. Since this is precisely the line at infinity, it follows that $\mathcal C$ doesn't have any asymptote.
