The short version, I am looking for a proof that, as vector spaces, a filtered algebra $A$ is isomorphic to its associated graded algebra $G(A)$. I.e. for filtration $\{0\}=F_0\subset F_1\subset \dots \subset A$, and $G_n=F_n/F_{n-1}$ then $G(A):=\bigoplus_n G_n \cong A$ as vector spaces.
The motivation is the following, I am trying to understand the connection between the following two statements: for lie algebra $\mathfrak{g}$ and universal enveloping algebra $U(\mathfrak{g})$,
- Poincare-Birkhoff-Witt theorem: the associated graded algebra $G(U(\mathfrak{g}))$ is isomorphic to $S(\mathfrak{g})$.
- for basis $x_1,\dots,x_n$ of $\mathfrak{g}$, then $\{x_1^{k_1}\dots x_n^{k_n}| k_i\in \mathbb{N}\}$ ($\otimes$'s omitted) is a basis set for $U(\mathfrak{g})$
I believe that with a choice of basis $x_i$ for $\mathfrak{g}$, there is an isomorphism of algebras $S(\mathfrak{g})\cong k[x_1,\dots,x_n]$. Then assuming the statement at the top is true we have as vector spaces: $U(\mathfrak{g})\cong G(U(\mathfrak{g}))$. So taking everything as vector spaces: $$U(\mathfrak{g})\cong G(U(\mathfrak{g}))\cong S(\mathfrak{g})\cong k[x_1,\dots,x_n]$$ The notation $x_1^{k_1}\dots x_n^{k_n}$ of the basis for $U(\mathfrak{g})$ coincides with that of the usual basis for $k[x_1,\dots,x_n]$, hinting/implying one basis maps to the other through these isomorphisms? I am very unfamiliar with this area, so I'd appreciate it if someone could confirm or deny if these intuitions are correct?