# Ring of polynomial functions

(Redirected from Polynomials on vector spaces)

In mathematics, the ring of polynomial functions on a vector space V over an infinite field k gives a coordinate-free analog of a polynomial ring. It is denoted by k[V]. If V has finite dimension and is viewed as an algebraic variety, then k[V] is precisely the coordinate ring of V.

The explicit definition of the ring can be given as follows. If $k[t_1, \dots, t_n]$ is a polynomial ring, then we can view $t_i$ as coordinate functions on $k^n$; i.e., $t_i(x) = x_i$ when $x = (x_1, \dots, x_n).$ This suggests the following: given a vector space V, let k[V] be the subring generated by the dual space $V^*$ of the ring of all functions $V \to k$. If we fix a basis for V and write $t_i$ for its dual basis, then k[V] consists of polynomials in $t_i$; it is a polynomial ring.

In applications, one also defines k[V] when V is defined over some subfield of k (e.g., k is the complex field and V is a real vector space.) The same definition still applies.

## Symmetric multilinear maps

Let $S^q(V)$ denote the vector space of multilinear linear functionals $\textstyle \lambda: \prod_1^q V \to k$ that are symmetric; $\lambda(v_1, \dots, v_q)$ is the same for all permutations of $v_i$'s.

Any λ in $S^q(V)$ gives rise to a homogeneous polynomial function f of degree q: let $f(v) = \lambda(v, \dots, v).$ To see that f is a polynomial function, choose a basis $e_i, \, 1 \le i \le n$ of V and $t_i$ its dual. Then

$\lambda(v_1, \dots, v_q) = \sum_{i_1, \dots, i_q = 1}^n \lambda(e_{i_1}, \dots, e_{i_q}) t_{i_1}(v_1) \cdots t_{i_q}(v_q)$.

Thus, there is a well-defined linear map:

$\phi: S^q(V) \to k[V]_q, \, \phi(\lambda)(v) = \lambda(v, \cdots, v).$

It is an isomorphism:[1] choosing a basis as before, any homogeneous polynomial function f of degree q can be written as:

$f = \sum_{i_1, \dots, i_q = 1}^n a_{i_0 \cdots i_q} t_{i_1} \cdots t_{i_q}$

where $a_{i_0 \cdots i_q}$ are symmetric in $i_0, \dots, i_q$. Let

$\psi(f)(v_1, \dots, v_q) = \sum_{i_1, \cdots, i_q = 1}^n a_{i_0 \cdots i_q} t_{i_1}(v_1) \cdots t_{i_q}(v_q).$

Then ψ is the inverse of φ. (Note: φ is still independent of a choice of basis; so ψ is also independent of a basis.)

Example: A bilinear functional gives rise to a quadratic form in a unique way and any quadratic form arises in this way.