I'm reading the 1980 paper "Graph Algebras" by KH Kim, L Makar-Limanov, J Neggers, and FW Roush. It is a prerequisite to the solution to the isomorphism problem for graph groups aka right-angled Artin groups, or RAAGs, aka (free) partially commutative groups.
Let $K$ be any field, and $G$ any simple graph. The graph algebra $K(G)$ is the algebra over field $K$ with generators $V(G)$, the vertices of $G$, subject to exactly the relations $v_iv_j - v_jv_i = [v_i,v_j] = 0$ for $v_iv_j \in E(G)$ (the paper equivalently defines it by $v_iv_j \not\in E(G)$ but it seems that convention fell out of favor). Indeed this is the algebra over a field analogue of graph groups. Elements of $K(G)$ are polynomials in $V(G)$ under the usual addition and multiplication.
Which elements other than of course the copy of $K$ have inverses? I believe no others have them, but can't prove it.
In the usual polynomial ring over a field like $\mathbb{Q}[x]$, the proof is that for any product $(c_0 + c_1x + \ldots + c_nx^n)(d_0 + d_1x + \ldots + d_mx^m)$, $c_n, d_m \not= 0$, the highest degree term $c_n d_m x^{n+m}$ must have coefficient 0, leading to $c_n = d_m = 0$ which isn't true. However, I don't know if this reasoning extends to the above algebra.
This is relevant for the beginning of the proof of the paper's Lemma 5. If my claim is true, then any automorphism of $K(G)$ must map $K$ to $K$, so indeed we may always compose with an isomorphism from Lemma 4 (there is a typo that says Lemma 1, but of course they mean 4). I came to this question wondering why $f^{-1}(k_i)$ must be in $K$?
Thank you as always for any and all help!
