# Thread: Intersection of plane curves

1. ## Intersection of plane curves

Let $\displaystyle f(x,y)$ and $\displaystyle g(x,y)$ be polynomials with real coefficients. Show that if $\displaystyle f,g$ have no common factors, then the curves $\displaystyle f(x,y)=0$ and $\displaystyle g(x,y)=0$ will intersect in at most finitely

many points. ("at most finitely many" means either none or finitely many!)

2. I hadn't seriously attempted this problem until just now (I saw it last month and got scared ). But I just solved it without too much trouble!

Let $\displaystyle K=\mathbb{R}(x)$, the field of rational functions in $\displaystyle x$. Consider $\displaystyle f,g$ as elements of $\displaystyle K[y]$ (which is a P.I.D.). Then it is easy to see that $\displaystyle f,g$ are relatively prime in $\displaystyle K[y]$. Hence we can find rational functions $\displaystyle u(x), v(x)$ such that $\displaystyle u(x)f(x,y)+v(x)g(x,y)=1$. Multiplying through by a common denominator, we find that there exist polynomials $\displaystyle p(x), q(x), d(x)$ such that $\displaystyle p(x)f(x,y)+q(x)g(x,y)=d(x)$. The right side of this equation can vanish for only a finite number of possible values of $\displaystyle x$, hence there are only a finite number of values of $\displaystyle x$ which are candidates for an intersection point of the curves $\displaystyle f(x,y)=0$ and $\displaystyle g(x,y)=0$. Reversing the roles of $\displaystyle x,y$, we arrive at the same conclusion for $\displaystyle y$; hence there are only a finite number of possible intersection points of the two curves.

3. Originally Posted by Bruno J.
I hadn't seriously attempted this problem until just now (I saw it last month and got scared ). But I just solved it without too much trouble!

Let $\displaystyle K=\mathbb{R}(x)$, the field of rational functions in $\displaystyle x$. Consider $\displaystyle f,g$ as elements of $\displaystyle K[y]$ (which is a P.I.D.). Then it is easy to see that $\displaystyle f,g$ are relatively prime in $\displaystyle K[y]$. Hence we can find rational functions $\displaystyle u(x), v(x)$ such that $\displaystyle u(x)f(x,y)+v(x)f(x,y)=1$. Multiplying through by a common denominator, we find that there exist polynomials $\displaystyle p(x), q(x), d(x)$ such that $\displaystyle p(x)f(x,y)+q(x)f(x,y)=d(x)$. The right side of this equation can vanish for only a finite number of possible values of $\displaystyle x$, hence there are only a finite number of values of $\displaystyle x$ which are candidates for an intersection point of the curves $\displaystyle f(x,y)=0$ and $\displaystyle g(x,y)=0$. Reversing the roles of $\displaystyle x,y$, we arrive at the same conclusion for $\displaystyle y$; hence there are only a finite number of possible intersection points of the two curves.

Do you mean to say that we can find rational functions $\displaystyle u(x), v(x)$ such that $\displaystyle u(x)f(x,y) + v(x)g(x,y) = 1$ since $\displaystyle \gcd(f(x,y), g(x,y)) = 1$?

4. Originally Posted by Sampras
Do you mean to say that we can find rational functions $\displaystyle u(x), v(x)$ such that $\displaystyle u(x)f(x,y) + v(x)g(x,y) = 1$ since $\displaystyle \gcd(f(x,y), g(x,y)) = 1$?
Isn't that precisely what I said?...

5. Originally Posted by Bruno J.
Isn't that precisely what I said?...
No you said $\displaystyle u(x)f(x,y) + v(x)f(x,y) = 1$.

6. Originally Posted by Sampras
No you said $\displaystyle u(x)f(x,y) + v(x)f(x,y) = 1$.
Haha! Oops. I read it twice and I didn't see it! Obviously, that's a typo.