For a Commutative Noetherian local ring $(R, \mathfrak m)$, let $K_0^{\mathfrak m}(R)$ denote the abelian Group generated by isomorphism classes of bounded chain complexes of finitely generated free modules with finite length homologies (i.e. the support of each homology is contained in $\{\mathfrak m\}$) subject to the following two relations: (1) $[P_{\bullet}]=0$ if $P_{\bullet}$ is exact. (2) $[P_{\bullet}]=[P'_{\bullet}]+[P''_{\bullet}]$ if there exists a short exact sequence of chain complexes (and chain maps) $0\to P'_{\bullet}\to P_{\bullet}\to P''_{\bullet}\to 0$ .
It can be easily shown that in $K_0^{\mathfrak m}(R)$, we have $[\sum P_{\bullet}]=-[P_{\bullet}]$ and $[P_{\bullet}]+[Q_{\bullet}]=[P_{\bullet}\oplus Q_{\bullet}]$ , so in particular, any element of $K_0^{\mathfrak m}(R)$ is of the form $[P_{\bullet}]$.
Now let $(R, \mathfrak m)$ be a Noetherian local domain of dimension $1$, with fraction field $F$.
So for any non-zero ideal $I$, the $R$-module $R/I$ has finite length, so in particular, for any $0\ne r\in R$, the bounded chain complex of f.g. free modules $...\to 0\to R \xrightarrow{.r} R\to 0\to ...$ has finite length homologies.
My claim is to show $K_0^{\mathfrak m}(R)\cong F^{\times}/R^{\times}$. So I define a map $F^{\times}\to K_0^{\mathfrak m}(R)$ by sending $a/b$ to
$[...\to 0\to R \xrightarrow{.a} R\to 0\to ...] - [...\to 0\to R \xrightarrow{.b} R\to 0\to ...]$ .
Now I can show that this map is well-defined , but I'm having trouble to show Surjectivity and that the kernel is $R^{\times}$ (the group of units of $R$).
Please help.
