Intersection of plane curves

**NonCommAlg** · September 5th 2010, 10:35 PM

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

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

**Bruno J.** · November 8th 2010, 07:23 PM

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

**Sampras** · November 11th 2010, 01:30 PM

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

**Bruno J.** · November 11th 2010, 03:42 PM

Originally Posted by Sampras

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

Isn't that precisely what I said?...

**Sampras** · November 11th 2010, 04:02 PM

Originally Posted by Bruno J.

Isn't that precisely what I said?...

No you said $u(x)f(x,y) + v(x)f(x,y) = 1$ .

**Bruno J.** · November 11th 2010, 04:22 PM

Originally Posted by Sampras

No you said $u(x)f(x,y) + v(x)f(x,y) = 1$ .

Haha! Oops. I read it twice and I didn't see it! Obviously, that's a typo.

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