Surjective Homomorphisms of Coordinate Rings

[ZioX]

Homework Statement

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).

Homework Equations

The Attempt at a Solution

I have everything but surjectivity. phi is obviously an injective homomorphism (ker phi = {0 polynomial}).

 

[ZioX]

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.

 

[Hurkyl]

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?

I have everything but surjectivity. phi is obviously an injective homomorphism (ker phi = {0 polynomial}).

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.

 

[ZioX]

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.

 

[Hurkyl]

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.

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.

 

[ZioX]

I guess I didn't want to elaborate much because I just wanted an argument for surjectiveness.

 

[Hurkyl]

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)