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Say you have a finite field $F$ of order $p^k$. Suppose that $f,g\in F[X_1,\dots,X_m]$, such that the degree of each $X_i$ is strictly less than $p^k$ in both $f$ and $g$. I'm putting this condition to avoid things like $f=X_1X_2X_3^{p^k}$ and $g=X_1X_2X_3$ which technically define the same polynomial function over $F$ since $X_i^{p^k}-X_i$ is in the kernel of the evaluation homomorphism, but are not equal in the polynomial ring.

Under this condition, if $f$ and $g$ define the same polynomial function over $F$, are they equal as polynomials? By equality of polynomial functions, I mean they are equal as sets of ordered pairs. I feel like restricting the degree of each indeterminate should force this to be so, but how can it actually be proven?

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How many scalar numbers define a polynomial with your degree conditions? How many scalar numbers define a function, polynomial or not, on $F^m$? Ergo? (Is this (homework)?) – Did Jul 16 '12 at 6:47
@did Thanks, so writing $p^k=q$, both sets have cardinality $q^{(q^m)}$, so every function in $m$ variables over $F$ is a polynomial function. So am I correct in thinking there is a 1-1 correspondence between polynomials of degree$<q$, and thus if $f$ and $q$ define the same polynomial function, then they must in fact be the same polynomial? By the way, this is not homework. – Nastassja Jul 16 '12 at 7:13
up vote 4 down vote accepted

Formalizing Jacob's answer (+1) and generalizing it to $m$ variables as in did's comment.

Let $F=GF(q)$ be the field of $q$ elements, and let $P=(a_1,a_2,\ldots,a_m)\in F^m$ be an arbitrary point. Let us fix an index $j$, $1\le j\le m$. The polynomial $$ f_j(x)=\prod_{a\in F, a\neq a_j}(x-a)=\frac{x^q-x}{x-a_j} $$ has the properties that $f_j(a_j)\neq0$, and $f_j(a)=0$ for all $a\in F, a\neq a_j.$ Therefore the product $$ f_P(x_1,x_2,\ldots,x_m):=f_1(x_1)f_2(x_2)\cdots f_m(x_m)=\prod_{j=1}^m f_j(x_j) $$ vanishes at all other points of $F^m$ except at $P$. Furthermore, the polynomial $f_P$ is of degree $\le q-1$ with respect to all the variables $x_j,j=1,2,\ldots,m.$

Let $V_m$ be the space of polynomials in $m$ variables over $F$ such that no variable appears with degree $\ge q$. Let $V_m'$ be the space of $F$-valued functions on $F^m$. There is a natural evaluation mapping $ev:V_m\rightarrow V_m'$. Our earlier calculations show that the functions $ev(f_P)$, with $P$ ranging over the points of $F^m$, form a basis of $V_m'$. Therefore $ev$ is surjective. Because $V_m$ and $V_m'$ both have dimension $q^m$ as vector spaces over $F$, the mapping $ev$ must also be injective (rank-nullity).

This answers your question in the affirmative.

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+1. (Quibble: The degree of $f_P$ in each variable seems to be exactly $q-1$, or am I mistaken?) – Did Jul 16 '12 at 9:35
Correct, @did. Inequality would suffice, but it is an equality. – Jyrki Lahtonen Jul 16 '12 at 10:37

I think the result you want to prove is the following. Let $\mathbb F$ be a finite field of order $q$ then the ring of functions from $\mathbb F$ to $\mathbb F$ is isomorphic to

$$\frac{\mathbb F[x]}{(x^q-x)}.$$

This is usually proved by taking the map from $\mathbb F[x]$ to the ring of functions and demonstrating that it's surjective with kernel $(x^q-x)$. Which is fairly straightforward.

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