I'm trying to better understand the symmetric algebra and how it behaves on the dual space of the algebra. Wikipedia states the following:
In mathematics, the symmetric algebra $S(V)$... corresponds to polynomials with indeterminates in $V$, without choosing coordinates. The dual, $S(V^∗)$ corresponds to polynomials on $V$.
I'm trying to understand this distinction better. I think that what they're saying is that $S(V)$ is the set of formal polynomials taking indeterminates in $V$, whereas $S(V^*)$ can be identified with the set of polynomial functions on $V$. I would appreciate a sanity check that I am seeing this right.
Here's how I'm viewing it so far:
For example, suppose your vector space $V$ is over $\Bbb F_2$, the finite field of order 2. Then for any vector $v$, the polynomial $v^2 - v$ is well-defined as a formal expression, and hence is an element of the symmetric algebra $S(V)$.
On the other hand, let $f$ be a linear functional in the dual space $V*$. Then we can likewise construct a formal polynomial $f^2-f$. However, if we actually treat this as a polynomial function on $V$, then for any $v$, $f(v)^2-f(v) = 0$, and hence can be identified with the zero polynomial function.
In other words, in this interpretation, $v^2-v$ exists as a member of $S(V)$, as it is a formal polynomial. However, within $S(V*)$, the polynomial function $f^2-f=0$.
Is this the correct way to view the distinction? Or is it more subtle than this?
No, this is not correct; $S(V)$ and $S(V^{\ast})$ are both polynomial algebras, and each is exactly as formal as the other. In particular, $f^2 \neq f$.
Over a finite field "polynomials on $V$" needs to be interpreted more carefully: a (finite-dimensional) vector space over $k$ can naturally be upgraded into an affine (group) scheme over $k$, whose set of $k$-points is $V$, and the ring of functions on this affine scheme (not on $V$ itself) is $S(V^{\ast})$. The cleanest way to describe this scheme is to describe its functor of points: if $k$ is the base field, then the functor of points of $V$-as-an-affine-scheme sends a commutative $k$-algebra $A$ to
$$V \otimes_k A.$$
Then it's an exercise to show that this functor is represented by $S(V^{\ast})$.