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Surjective Homomorphisms of Coordinate Rings

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data
    I want to show that the homomorphism phi:A(X)->k+k given by taking f(x_1,....,x_n)-> (f(P_1),f(P_2)) is surjective. That is, given any (a,b) in k^2 (with addition and multiplication componentwise) I want to find a polynomial that has the property that f(P_1)=a and f(P_2)=b.

    The actual question is to show that if we take the coordinate ring of two points in k^n then the coordinate ring is isomorphic to k+k (direct product).

    2. Relevant equations



    3. The attempt at a solution
    I have everything but surjectivity. phi is obviously an injective homomorphism (ker phi = {0 polynomial}).
     
    Last edited: Dec 3, 2007
  2. jcsd
  3. Dec 3, 2007 #2
    Hmm, if the sum of the elements of P_1 and P_2 are distinct we can just define

    [tex]f(x_1,...,x_n)=a\left(\frac{x_1+...+x_n-P_2}{P_1-P_2}\right)+b\left(\frac{x_1+...+x_n-P_1}{P_2-P_1}\right)[/tex]

    where P_i means the sum of all the elements in P_i.

    I'm at a loss if the sum of the elements are the same.
     
  4. Dec 3, 2007 #3

    Hurkyl

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    I'm assuming:

    k -- a field
    X -- this is [itex]\mathbb{A}^n_k[/itex]; affine n-space over the field k.
    A(X) -- coordinate ring of X
    x_1, ..., x_n -- a set of generators for A(X)
    P_1, P_2 -- points in k^n (i.e. in [itex]\mathbb{A}^n(k)[/itex])

    And I'm assuming the codomain of phi is meant to be k^2.

    The basic idea is easy; you simply show that there exists two functions satisfying
    [tex]f(P_1) = 0[/tex]
    [tex]f(P_2) \neq 0[/tex]
    [tex]g(P_1) \neq 0[/tex]
    [tex]g(P_2) = 0[/tex]



    Now, I should point out that in the category of k-algebras, the direct sum is given by a tensor product: [itex]R \oplus S \cong R \otimes_k S[/itex]. In particular, [itex]k \oplus k \cong k[/itex]. And rings are dual to spaces; a direct sum of rings should correspond to a product of varieties. Conversely, the coordinate ring of a disjoint sum of varieties (e.g. a pair of points) should correspond to the direct product of the individual coordinate rings.

    So, if you really did mean to talk about k + k, then in what category are you talking about sums?


    That's cleraly wrong. I bet you can find a nonzero polynomial that has both P_1 and P_2 as a root! Try n=1 to make it easier.
     
    Last edited: Dec 3, 2007
  5. Dec 3, 2007 #4
    We're working in the coordinate ring, so all polynomials that have P_1 and P_2 as roots will be the zero polynomial (in A(X)).

    I did mean direct product, I'm sorry.
     
    Last edited: Dec 3, 2007
  6. Dec 3, 2007 #5

    Hurkyl

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    Er...

    You meant X to be the subvariety of [itex]\mathbb{A}^n_k[/itex] consisting only of the two points P_1 and P_2? You really should say these things. :tongue:
     
  7. Dec 3, 2007 #6
    I guess I didn't want to elaborate much because I just wanted an argument for surjectiveness.
     
  8. Dec 4, 2007 #7

    Hurkyl

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    But one can't make the argument if one doesn't know the pieces involved!


    Exercise: find a subvariety X of [itex]\mathbb{A}^n_k[/itex] (where k is algebraically closed) such that there does not exist a surjection [itex]\mathcal{O}(X) \to k^2[/itex].

    (I prefer [itex]\mathcal{O}(X)[/itex] to denote the ring of regular functions on X, rather than the notation A(X))


    (Hint: restate the problem either in terms of commutative algebra or in terms of geometry)
     
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