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Math Help - Intersection of plane curves

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    Intersection of plane curves

    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!)
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    MHF Contributor Bruno J.'s Avatar
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    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.
    Last edited by Bruno J.; November 11th 2010 at 05:22 PM.
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by Bruno J. View Post
    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 ?
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Sampras View Post
    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?...
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Isn't that precisely what I said?...
    No you said  u(x)f(x,y) + v(x)f(x,y) = 1 .
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Sampras View Post
    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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