Exercise 10.12

Question

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\def\dcro{\mbox{]\hspace{-.15em}]}}

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Show that the affine curve defined by $x^2 + y^2 + x^2 y^2 = 0$ has two points at infinity and that both are singular.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

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ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

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ewcommand{e}{\,\mathrm{Re}\,}

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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
The homogeneous equation of this curve is
$$\overline{f} (t,x, y) =  x^2 t^2 + y^2 t^2 + x^2 y^2, $$
where $t=0$ is the equation of the line at infinity.

The point $\overline{a} = [u_0, u_1,u_2]$ is a point at infinity if $u_0 = 0$. This gives the equation $$\overline{f} (0,u_1,u_2) = u_1^2u_2^2 = 0,$$
 where $u_1 
e 0$ or $u_2
e 0$ (otherwise $u_0 = u_1 = u_2 = 0$, and $[u_0, u_1,u_2]$ is not a projective point).
 
 If $u_1 
e 0$, then $u_2 = 0$, and if $u_2 
e 0$, then $u_1 = 0$.
 
 Therefore $\overline{a} = [0, u_1,0] = [0,1,0]$, or $\overline{a} = [0,0,u_2] = [0,0,1]$.
 
 $p = [0,1,0]$ and $q = [0,0,1]$ are the two points at infinity of the curve.
 
 $$\frac{\partial \overline{f}}{\partial t} =  2 t (x^2 + y^2), \qquad \frac{\partial \overline{f}}{\partial x} = 2x (t^2 +y^2),\qquad  \frac{\partial \overline{f}}{\partial y} = 2y (t^2 + x^2).$$
Therefore
$$\frac{\partial \overline{f}}{\partial t}(0,1,0) = \frac{\partial \overline{f}}{\partial x}(0,1,0)=\frac{\partial \overline{f}}{\partial y}(0,1,0) = 0,$$
and
$$\frac{\partial \overline{f}}{\partial t}(0,0,1) = \frac{\partial \overline{f}}{\partial x}(0,0,1)=\frac{\partial \overline{f}}{\partial y}(0,0,1) = 0.$$
This proves that the two points at infinity $p,q$ are singular.
\end{proof}

Exercise 10.12

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Exercise 10.12

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