# Polynomials and function space over fields

Hi,

Can someone explain why the following is true? It seems to be an "accepted fact" everywhere I search, and I can't tell why.

Let F be a field. Let E be the function from F[x] to F^F, where F[x] is the set of all polynomials over F, and F^F is the set of all functions from F to F.
Then E is 1-to-1 if and only if F is infinite, and E is onto if and only if F is finite.

Why is this true?

For the first part, I can see that 1-to-1 would mean that the kernel of E is 0. So if F is finite, there should be some non-zero polynomial in F[x] which maps to the zero function in F^F.

For the second part, I can see that if F is finite, then F^F is finite. F[x] is of course infinite, and so it makes sense that E is onto, but can someone give a more conceptual reason why? Or is that the essence of the reasoning?

Thank you!

Kriti

matt grime
Homework Helper
What do you mean by 'the' function? You simply mean the inclusion right? Well, for the first case, if I take a polynomial and consider it as a function, how can it be the zero function (i.e. send every element of F to zero)? Only if it was the zero polynomial.

The second is also easy. Forget fields. Suppose I tell you, over the reals, that f(0)=1, f(1)=10 and f(3)=4, write down a polynomial through the points (0,1), (1,10) amd (3,4), you'd be able to do that easily. Well, the finite field question you ask is no harder.

Yes, by "the" function, I meant the natural function mapping each polynomial in F[x] to it's associated function in F^F.

For the first case, what you mentioned is true. I had figured that much myself, but that proves that F is infinite => ker E = 0 <=> E is 1-to-1. How do we go in the opposite direction? I.e., why is it that ker E = 0 => F is infinite?

Similarly, for the second part, what you said was the basic concept I understood. But why does this require a finite field then? Given any function in F^F, couldn't I find some polynomial that could interpolate it?

matt grime
Homework Helper
If a field is finite it has a characteristic. It is easy to show that (x+1)^p=x^p+1 when p is the characteristic of a finite field. This is more than enough of a prod to help you with the problems you're having. If that isn't enough, then does the word Frobenious help?

Here is a little something to add to what matt-grime was lecturing about.

Theorem: Let $$|F|= \infty$$ and $$f(x)$$ be a polynomial if $$f(\alpha)=0$$ for every $$\alpha \in F$$ then $$f(x)=0$$.

Proof: Let $$\deg f(x) = n$$ (assuming $$f(x)\not = 0$$) then $$f(x)$$ can have at most $$n$$ zeros. But it clearly does not for any $$\alpha \in F$$ is a zero. So $$f(x)$$ must be the trivial polynomial.

Note, if $$F$$ is finite and $$\deg f(x) > |F|$$ then the same conclusion can be drawn. But it need not to be always true. Consider $$F$$ a field of order 3. And $$f(x)=x^3-x$$ it maps all elements into zero.

mathwonk