# Intersection of plane curves

• Sep 5th 2010, 10:35 PM
NonCommAlg
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!)
• Nov 8th 2010, 07:23 PM
Bruno J.
I hadn't seriously attempted this problem until just now (I saw it last month and got scared (Worried)). 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.
• Nov 11th 2010, 01:30 PM
Sampras
Quote:

Originally Posted by Bruno J.
I hadn't seriously attempted this problem until just now (I saw it last month and got scared (Worried)). 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 \$?
• Nov 11th 2010, 03:42 PM
Bruno J.
Quote:

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?... :confused:
• Nov 11th 2010, 04:02 PM
Sampras
Quote:

Originally Posted by Bruno J.
Isn't that precisely what I said?... :confused:

No you said \$\displaystyle u(x)f(x,y) + v(x)f(x,y) = 1 \$.
• Nov 11th 2010, 04:22 PM
Bruno J.
Quote:

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.