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Math Help - maximal ideal in R[x,y]

  1. #1
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    maximal ideal in R[x,y]

    Show that every maximal ideal m \subset \mathbb{R}[x,y] is one of the following:
    * m=<x-a,y-b> for some a,b \in \mathbb{R};
    * m=<y-rx-s,x^2+vx+w> for some r,s,v,w \in \mathbb{R} with v^2-4w<0;
    * m=<x-t,y^2+vy+w> for some t,v,w \in \mathbb{R} with v^2-4w<0.
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  2. #2
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    Quote Originally Posted by Stiger View Post

    Show that every maximal ideal m \subset \mathbb{R}[x,y] is one of the following:

    * m=<x-a,y-b> for some a,b \in \mathbb{R};

    * m=<y-rx-s,x^2+vx+w> for some r,s,v,w \in \mathbb{R} with v^2-4w<0;

    * m=<x-t,y^2+vy+w> for some t,v,w \in \mathbb{R} with v^2-4w<0.
    this is a nice problem! i don't know what you know and what you don't, so i'll assume that you know that if k is a field, then any maximal ideal of k[x_1, \cdots, x_n] can be generated by n polynomials

    f_1, \cdots , f_n such that f_i is a polynomial in k[x_1, \cdots , x_i]. so if m is a maximal ideal of \mathbb{R}[x,y], then m=<f(x), g(x,y)>. now we may assume that f(x) \in \mathbb{R}[x] is irreducible, because if

    f(x)=f_1(x)f_2(x), then since m is prime, either f_1(x) \in m or f_2(x) \in m. if f_1(x) \in m, then <f_1(x),g(x,y)>=<f(x),g(x,y)>=m. this proves our claim that we may assume that f(x) is

    irreducible. thus either \deg f = 1, or \deg f is an even number. [note that f cannot be a constant because then m wouldn't be maximal] now if \deg f is even, then f is a product of some

    irreducible quadratic polynomials in \mathbb{R}[x]. thus in this case, we may assume that f is an irreducible quadratic polynomial in \mathbb{R}[x]. so you have two cases:

    case 1: \deg f = 1. we may assume that f(x)=x-a. now write g(x,y)=\sum_{j=0}^n u_j(x)y^j, where u_j(x) \in \mathbb{R}[x]. then dividing each u_j(x) by x-a gives us g(x,y)=(x-a)h(x,y) + p(y), for some

    h(x,y) \in \mathbb{R}[x,y], \ p(y) \in \mathbb{R}[y]. hence m=<x-a,g(x,y)>=<x-a,p(y)>. again, as we already showed, we may assume that either \deg p(y)=1, or p(y) is an irreducible quadratic

    polynomial in \mathbb{R}[y]. if \deg p(y)=1, then we'll get m=<x-a,y-b> and if p(y) is an irreducible quadratic polynomial, we'll get m=<x-a, y^2 + cy + d>, where of course c^2 - 4d < 0,

    because otherwise p(y) would have a real root and hence it wouldn't be irreducible anymore.

    case 2: f(x) is an irreducible quadratic polynomial. this case is left for you to try!
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