# Surjective Homomorphisms of Coordinate Rings

1. Dec 3, 2007

### ZioX

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. Dec 3, 2007

### ZioX

Hmm, if the sum of the elements of P_1 and P_2 are distinct we can just define

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

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.

3. Dec 3, 2007

### Hurkyl

Staff Emeritus
I'm assuming:

k -- a field
X -- this is $\mathbb{A}^n_k$; 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 $\mathbb{A}^n(k)$)

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
$$f(P_1) = 0$$
$$f(P_2) \neq 0$$
$$g(P_1) \neq 0$$
$$g(P_2) = 0$$

Now, I should point out that in the category of k-algebras, the direct sum is given by a tensor product: $R \oplus S \cong R \otimes_k S$. In particular, $k \oplus k \cong k$. 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
4. Dec 3, 2007

### 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.

Last edited: Dec 3, 2007
5. Dec 3, 2007

### Hurkyl

Staff Emeritus
Er...

You meant X to be the subvariety of $\mathbb{A}^n_k$ consisting only of the two points P_1 and P_2? You really should say these things. :tongue:

6. Dec 3, 2007

### ZioX

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

7. Dec 4, 2007

### Hurkyl

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

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

(I prefer $\mathcal{O}(X)$ 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)