If $V$ is a vector space over the field $K$ with basis ${v_1, v_2,…,v_n}$, then the symmetric algebra $S(V)= K[v_1,v_2,..,v_n]$. The question is: If $K$ is a commutative ring, then this equality is not true in general, why? What do the elements of $S(V)- K[v_1,v_2,..,v_n]$ look like? When can we get the equality?
Thank you for help
If $R$ is a commutative ring and $V$ is a free $R$-module then I believe the equality still holds. But if $R$ is not a field then there exist modules which are not free. While $S(V)$ can still be defined it will not be isomorphic to a polynomial ring $R[x_1, \ldots, x_n]$ for any $n$. The easiest example is $V = k$ as a module for $R = k[t]$ (where $t$ acts as $0$). The polynomial ring $R[x] \simeq k[x, t]$ has no torsion but $S(V)$ is annihilated by $t$.