For a finite field $F$ any function $f\colon F\to F$ is given by a polynomial. My question is what happens when we are given a function of two or more variables? Is this necessarily a polynomial function of two or more variables?


Looking at the Lagrange Interpolation formula in the univariate case I try to argue for the 2-variable case this way.


So given some data $(a\_i,b\_j)\mapsto c\_{ij}$ . I can try to fit a many polynomial functions $F\_i$'s with $F\_i(a\_i,b\_j)=c\_{ij}$ for all $j $. The coefficients of $F\_i$'s vary as rational functions in the given data $c\_{ij}$.
As inversion is also a polynomial function in one variable can I assume I can get a polynomial of two variable to represent all functions of two variables?



Functions of several variables over finite fields
For $(a\_0,b\_0) \in F^2$, the two-variable polynomial


$$P\_{a\_0,b\_0} = \left ( \prod\_{a \in F \setminus \{a\_0\}} \frac{X-a}{a\_0 -a} \right ) \left ( \prod\_{b \in F \setminus \{b\_0\}} \frac{Y-b}{b\_0 -b} \right ) $$
yields the characteristic function of $\{(a\_0,b\_0)\}$, that is $1$ for $(a\_0,b\_0)$ and $0$ otherwise. 


These characteristic functions span the space of all functions from $F^2 $ to $F$ and the claim follows. 


Or concretely, for a function $f$ take $P\_f = \sum\_{(x,y)\in F^2} f(x,y)P\_{x,y}$.


The argument works for $F^n$ for any $n$ just as well.


