Let us work over a fixed but arbitrary field $k$. So for example, all vectorspaces and algebras are implicitly over $k$.
Given a vectorspace $V$, we can build two (potentially) different commutative algebras equipped with a canonical inclusion of $V$ as a subspace. Namely:
Let $S(V^*)$ denote the symmetric algebra of the dual of $V$.
Let $J(V)$ be the least subalgebra $A$ (of the algebra of all functions $k \leftarrow V$) satisfying $A \supseteq V^*.$
By the universal property of the symmetric algebra, there's an algebra homomorphism $$J(V) \leftarrow S(V^*).$$ Its clearly surjective.
Question. Is the above morphism always an isomorphism? If not, what is the usual terminology/notation for what I'm denoting $J(V)$?
I'm interested in both the finite-dimensional case and the general case.
Let $k$ be the ground field. I claim that $S(V^*)\to J(V)$ is a bijection if and only if the ground field is infinite.
First note the map is always surjective.
Then by Pedro Tamaroff's comment, if the ground field is finite of order $q$ for any $x\in S(V^*)$, then image of $x^q-x$ is zero, and hence in this case the map is not injective.
We now assume that $k$ is infinite and we first assume that $V$ is finite dimensional. Pick a basis $v_1,\ldots,v_n$ for $V$ and let $x_1,\ldots,x_n$ be its dual basis (i.e. $x_i(v_j)=1$ if $i=j$ and zero otherwise). Any element onf $S(V^*)$ can be expressed as a polynomial in the $x_i$. Let us have $g(x)\in S(V^*)$ and consider its image in $J(V)$, and call this image $h$. If $v=\sum_i a_i v_i\in V$, then $h(v)$ is given by evualating the polynomial $g$ at $(a_1,\ldots,a_n)$. As $k$ is infinite, as polynomials over infinite fields vanish identically if and only if they are zero, $g(x)\neq 0$ implies there is some $(a_1,\ldots,a_n)$ such that $g(a_1,\ldots,a_n)\neq 0$, and thus $h\neq 0$, so we have injectivity.
For infinite dimensional $V$, for any $s\in S(V^*)$, $s$ involves only finitely many linearly independent elements of $V^*$, $t_1,\ldots,t_m$. Consider a finite dimensional subspace of $V$, $W$, where $t_1|_W,\ldots,t_n|_W$ are linearly independent. Then the image of $s$ in $J(W)$ (a quotient of $J(V)$) is nonzero by the finite-dimensional case, so the image in $J(V)$ is nonzero, so again $S(V^*)\to J(V)$ is injective.
At some point I use the fact that if we have a polynomial over an infinite field, $f(x)$, then $f\equiv 0$ if and only if $f(a_1,\ldots,a_n)=0$ for all possible $a_i$. You seemed to express in the comments that you did not know how to do this. Can you see how to do this if $f$ is a polynomial in one variable? Then fixing all but one variable and letting the last vary, do you see how to do the general case by induction?