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 $\mathcal{C}=\mathrm{V}(F)$.

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

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

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

  4. 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:

  1. 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?

  2. 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?

  3. I haven't do this yet, but I know the algorithm to do it.
  4. 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)

  1. 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 $Y = 0$.

  2. Yes.

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

  4. 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.


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