Exercise 5.6.4 (Condition for a circle to be included in an algebraic curve)

Question

Let $f(x,y)$ be a polynomial of degree $d$ with real coefficients, and set $p(t) = f(2t/(1+t^2),(1-t^2)/(1+t^2))(1+t^2)^d$. Show that $p(t)$ is a polynomial of degree at most $2d$. Deduce that if $\mathscr{C}_f(\mathbb{R})$ has more than $2d$ distinct points in common with the circle $x^2 + y^2 =1$, then this circle is a subset of $\mathscr{C}_f(\mathbb{R})$.

Richard Ganaye · Accepted Answer

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

ewcommand{\N}{\mathbb{N}}

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

\begin{proof} 
By definition
\begin{align}
p(t) = f \left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} ight) (1+t^2)^d.
\end{align}
If we write $f(x,y) = \sum_{(i,j) \in A} a_{i,j}x^iy^j$, where $A \subset \N^2$ is a finite set such that $ i+j \leq d$ for all $(i,j) \in A$, then
\begin{align*}
p(t) &= (1+t^2)^d \sum_{(i,j) \in A} a_{i,j} \left(\frac{2t}{1+t^2} ight)^i \left(\frac{1-t^2}{1+t^2} ight)^j\
&= \sum_{(i,j) \in A} a_{i,j} (2t)^i (1-t^2)^j (1+t^2)^{d - i - j},
\end{align*}
so $p(t)$ is a polynomial, because in every term $i+j \leq d$. Moreover
$$\deg [(2t)^i (1-t^2)^j (1+t^2)^{d - i - j}]=  i +2j + 2(d-i-j) = 2d - i \leq 2d.$$
Therefore $\deg p \leq 2d.$

Let $\mathscr{C}$ denote the circle $x^2 + y^2 = 1$. Suppose that $\mathscr{C}_f(\mathbb{R})$ has more than $2d$ distinct points in common with the circle $x^2 + y^2 =1$. Assume for the sake of contradiction that $\mathscr{C} 
ot \subseteq \mathscr{C}_f(\R)$, so that there is some point $A = (a,b) \in \mathscr{C}$ such that $A 
ot \in \mathscr{C}_f(\R)$. By a rotation of the axes, we can choose a system of coordinates such that the coordinates of $A$ are $(0, -1)$. The equation of $\mathscr{C}$ is the same, and the new equation of $\mathscr{C}_f$ is $g$, where $\deg(g) = \deg(f)$, and $g(0,-1) 
e 0$.

(Explicitely, if $(\Vec{e}_1, \Vec{e}_2)$ is the old basis, and  $(\Vec{e'}_1,\Vec{e'}_2)$ the new basis, where $\Vec{e'}_1 = -b \Vec{e}_1 + a \Vec{e}_2,\ \Vec{e'}_2 = -a \Vec{e}_1 - b \Vec{e}_2$, then $g(X,Y) = f(-bX - a Y, aX - b Y)$, so that $g(0,-1) = f(a,b) 
e 0$.)

Replacing $g$ by $f$ in the notations, this brings the problem back to the case where $f(-1,0) 
e 0$, which we will assume from now on.

Then the parametrization of $\mathscr{C}$ starting from the point $A = (0,-1)$ with $x = t(y+1)$ gives 
\begin{align}
(x,y) \in \mathscr{C} \iff (x,y) = (0,-1) 	ext{ or } \exists t \in \R,\ x = \frac{2t}{1+t^2},\ y =  \frac{1-t^2}{1+t^2} .
\end{align}

Let $M = (x,y) \in \mathscr{C}_j(\R) \cap \mathscr{C}$ . Then $M 
e A$ since $f(0,-1) 
e 0$, thus there exists some $t \in \R$ such that $(x,y)  = \left (\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} ight) $ satisfies $f(x,y) = 0$, so that
$$f \left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} ight) = 0.$$

By equation (1), $t$ is a root of $p$. By (2), two distinct points correspond to distinct values of $t$.

By hypothesis, $\mathscr{C}_f(\mathbb{R})$ has more than $2d$ distinct points in common with the circle $\mathscr{C}$, which are distinct from $A$ because $A 
ot \in \mathscr{C}_f(\mathbb{R})$, thus $p$ has more that $2d$ roots. Since $\deg p \leq 2d$, this implies $p = 0$.

Then, by equation (1), for all $t \in \R$,
$$f \left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} ight) = 0.$$
This shows that
$$\mathscr{C} \setminus\{A\} \subseteq \mathscr{C}_f(\R).$$
Since $f$ is a continuous function,
$$f(0,-1) = \lim_{t	o \infty} f \left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} ight) = 0,$$
so $A \in \mathscr{C}_f(\R)$. But this is a contradiction, since by hypothesis $A 
ot \in \mathscr{C}_f(\R)$. This contradiction shows that $\mathscr{C} \subseteq \mathscr{C}_f(\R)$.

\end{proof}

Exercise 5.6.4 (Condition for a circle to be included in an algebraic curve)

Answers

Comments

Exercise 5.6.4 (Condition for a circle to be included in an algebraic curve)

Answers

Comments

Add answer