1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook