Definition of algebra of polynomial functions on a vector space. In his lectures nots on Topology, Prof. Nacinovich gives as an example of topological space the Zariski Topology on a vector space. To this end, he says

Let $V$ a finite dimensional vector space over a field $\Bbb K$. Let $\mathscr{A}(V)$ be the $\Bbb K$-algebra of the polynomial functions on $V$ with values in $\Bbb K$. Let us denote by $\mathfrak{C}$ the class of sets of the form
$$ C(F) = \{ v \in V : f(v) = 0 \quad \forall f \in F\} $$
when $F$ ranges among the subsets of $\mathscr{A}(V)$.

Does anyone know the definition of polynomial functions on a vector space? I have tried to read it on Wikipedia but it is not that clear.
 A: Formally, the polynomial functions on a finite-dimensional vector space $V$ are given by the symmetric algebra $S(V^{\ast})$ on the dual vector space $V^{\ast}$. The idea is to think of the elements of $V^{\ast}$ as the homogeneous polynomial functions on $V$ of degree $1$ and then take sums and products of these. You can check that if you pick a basis $e_1, \dots e_n$ of $V$ then $S(V^{\ast})$ can be identified with the algebra of polynomials $k[f_1, \dots f_n]$ on the dual basis of $e_i$.
Note that when the underlying field $k = \mathbb{F}_q$ is finite a polynomial function on $V$ is not faithfully determined by its action on the underlying set of $V$ (for any polynomial $f(x)$, the polynomial $f(x^q)$ determines the same function); this commonly trips up beginners. However, it is faithfully determined by its action on the tensor products $V \otimes_k L$ as $L$ ranges over all extensions (even all finite extensions suffices) of $k$. This is a small hint of the functor of points approach to algebraic geometry.
