# Polynomial evaluation functions

1. Nov 21, 2005

### Pietjuh

There's this problem i have to solve, and I don't know if my approach is correct.

Problem:
Let K be an infinite field. Consider the ring R = K[X1,.....,Xn] of polynomials in n variables over K. Show that every polynomial in R gives rise to an unique function K^n --> K.

Idea to solution:
I know that for n=1, the lagrange interpolation formula gives this result. Then I'll try to use a proof with induction. Suppose you use the evaluation map on a variable Xi. The result gives you the ring in n-1 variables. By the induction hypotheses this ring gives rise to unique functions K^n-1 ---> K for every polynomial in this ring.

Now consider a polynomial P in R. Let's write the evaluation maps as f_i, this meaning the evaluation map in the variable X_i. Consider now the composition of g = f_1 o .... o f_n-1. By the induction hypothesis this a unique map from K^n-1 ---> K. We must now show that the map h = f_n o g, is unique. Essentially g is a polynomial in K[Xn], so by the lagrange interpolation formula this means that the evaluation map f_n is unique. So because g was unique, h = f_n o g is unique.

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I hope this approach is valid! Please give your comments on this!

2. Nov 21, 2005

### AKG

Maybe I'm missing something, but the answer seems to be incredibly simple. If p(x1, ..., xn) is a polynomial in R, then the unique function that p(x1, ..., xn) gives rise to can be denoted p* and can be defined by:

p*(k1, ..., kn) = p(k1, ..., kn)

for every (k1, ..., kn) in Kn, and p(k1, ..., kn) means what you think it means, i.e. replace the formal variables x1, ..., xn with the field elements k1, ..., kn and treat the thing now not as a formal expression involving formal variables, but as an expression involving field elements, and then actually evaluate the expression.

This is just an extremely complicated way of saying that if p(x) = x², then p*(2) = 4. In fact, the function Kn -> K induced by a polynomial p is so obvious that we're perfectly comfortable with saying that if p(x) = x², then p(2) = 4. Did I miss something, because I don't know what you're doing with Lagrange interpolation, and induction, etc.

3. Nov 21, 2005

### Pietjuh

The point is that somewhere you have to use the fact that you need a infinite field. Because it can go wrong for finite fields. Take for example F_2[X] = Z/2Z[X]. Take the polynomials F = X^2 + 1 and G=X+1. Then F(0) = 1 = G(0) and F(1) = 0 = G(1), so two different polynomials give rise to the same evaluatian map.
The trick is that the lagrange interpolation formula in infinite fields really says that for every polynomial in one variable you've got an unique evaluation map.

4. Nov 21, 2005

### AKG

I see. I don't exactly follow your argument. If

g = f1 o .... o fn-1

is a (unique) map from Kn to K, when what exactly are the fi? I mean we would have to have that

fn-1 : Kn -> ?,
fi : ? -> ?, for 1 < i < n-1, and
f1 : ? -> K.

But each of these evaluation maps, if they are the ones you're getting from the LIF, are K to K maps. Something is wrong in this construction of g. What you do know is that if p(x1, ..., xn) is an element of R, then the induced maps

pk2, ..., kn : K -> K

defined by:

pk2, ..., kn(k1) = p(k1, k2, ..., kn)

are unique. Then perhaps you want to define the maps

pkm, ..., kn : Km-1 -> K

by:

pkm, ..., kn(k1, ..., km-1) = pkm-1, ..., kn(k1, ..., km-2)

and build your induction that way.

Last edited: Nov 21, 2005
5. Nov 21, 2005

### Hurkyl

Staff Emeritus
This might be an easier way of looking at it: how many polynomials yield the zero function? (Or equivalently, what is the kernel of the map from polynomials to functions?)