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Math Help - Maximal ideals of C[x,y]

  1. #1
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    Maximal ideals of C[x,y]

    Why maximal ideals in a ring of two-variable polynomials with complex coefficients are those generated by x - c, and y - d, for some complex c and d?


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  2. #2
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    Quote Originally Posted by Different View Post

    Why maximal ideals in a ring of two-variable polynomials with complex coefficients are those generated by x - c, and y - d, for some complex c and d?
    Thanks
    this is a special case of weak Nullstellensatz*. one side is trivial because \frac{\mathbb{C}[x,y]}{<x-c,y-d>} \simeq \mathbb{C}. the other side is much deeper: let \mathfrak{m} be a maximal ideal of R=\mathbb{C}[x,y]. then R/\mathfrak{m} is clearly a

    finitely generated \mathbb{C} algebra, which is also a field. a well-known result in commutative algebra says that a finitely generated domain over a field F is a field iff it's algebraic over F**. so R/\mathfrak{m}

    must be an algebraic extension of \mathbb{C}, which is possible only if R/\mathfrak{m}=\mathbb{C} because \mathbb{C} is algebraically closed. now let c,d \in \mathbb{C} be the image of x,y under the natural projection R \longrightarrow R/\mathfrak{m}=\mathbb{C}.

    then clearly x-c \in \mathfrak{m} and y-d \in \mathfrak{m}. thus <x-c,y-d> \subseteq \mathfrak{m}. but we already proved that <x-c,y-d> is a maximal ideal of R. therefore \mathfrak{m}=<x-c,y-d>. \ \Box


    * if F is any algebraically closed field, then maximal ideals of polynomial ring F[x_1,x_2, \cdots, x_n] are exactly the ideals <x_1-a_1, x_2-a_2, \cdots , x_n - a_n>, \ \ a_j \in F.

    ** see for example page 162 of the book "Graduate Algebra: Commutative View". the author is Louis Halle Rowen.
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    Thanks a lot. Funny enough, I understand the other direction, but could you explain why C[x,y] / <x - c, y - d> is isomorphic to C?

    I assume the isomorphism sends x to c and y to d....so, why does a polynomial f(x,y) s.t. f(c,d)=0 belongs to the ideal <x - c, y - d> ?
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  4. #4
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    Quote Originally Posted by Different View Post
    Thanks a lot. Funny enough, I understand the other direction, but could you explain why C[x,y] / <x - c, y - d> is isomorphic to C?

    I assume the isomorphism sends x to c and y to d....so, why does a polynomial f(x,y) s.t. f(c,d)=0 belongs to the ideal <x - c, y - d> ?
    right! you define f:\mathbb{C}[x,y] \longrightarrow \mathbb{C} by f(g(x,y))=g(c,d), which is obviously a surjective homomorphism. let I=<x-c,y-d>. it's clear that I \subseteq \ker f. now let g \in \ker f. then:

    g(x,y)=\sum a_{ij}x^iy^j=\sum a_{ij}(x-c+c)^i(y-d+d)^j \equiv \sum a_{ij}c^id^j =g(c,d)=0 \mod I. thus g \in I, i.e. \ker f \subseteq I.
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  5. #5
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    All clear now. Thanks!
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