So this question arose from my attempt to do exercises II.6.10 and III.5.4, both from Hartshorne's algebraic geometry. I have almost finished, but am not quite sure how to make the last implication. My situation is as follows: I have a short exact sequence $$ 0 \longrightarrow \mathbb{Z}^{r-1} \longrightarrow K \longrightarrow K' \longrightarrow 0. $$ where $K$ is the Grothendieck group of coherent sheaves on $\mathbb{P}_{k}^{r}$ and $K$ is the Grothendieck group of coherent sheaves on $\mathbb{A}^{r}$. I want to show that $K \simeq \mathbb{Z}^{r}$. By induction, I have that the Grothendieck group for $\mathbb{P}^{r-1}$ is $\mathbb{Z}^{r-1}$. So my task is reduced to showing that the Grothendieck group for $\mathbb{A}^{r}$ is just $\mathbb{Z}$. This is apparently a fairly standard problem. Stated another way, I was to find the Grothendieck group of finitely-generated modules on $R=k[x_{1}, \ldots , x_{r}]$.
The approach I had planned to take was the following: Let $M$ be a finitely-generated module over $R$. Then we have a finite free resolution, $$ 0 \longrightarrow F_{k} \longrightarrow F_{k-1} \longrightarrow \cdots \longrightarrow F_{1} \longrightarrow F_{0} = M \longrightarrow 0 $$ Let $[M]$ denote the equivalence class of $M$ in the Grothendieck group. By this equivalence relation, we have $$ [M] = \sum_{i=1}^{k}(-1)^{i+1}[F_{i}] $$ Since each $F_{i}$ is free, and the Grothendieck group respects direct sums, we have that $$ [M] = \sum_{j} n_{j}[R] $$ Also define $$ \epsilon([M]) = \sum_{j}n_{j}. $$ I will call this value the resolvent trace (for lack of a better term). Then if $K(R)$ is the Grothendieck group for $R$, we have a map, $$ K(R) \longrightarrow \mathbb{Z} $$ defined by $$ M \longmapsto \epsilon([M]). $$ My question is about well-definedness.
Why is this process well-defined? In particular, surely a different choice of finite free resolution of $M$ would yield a different value of $\epsilon([M])$?
Potential remedy to the problem: Since $R$ is regular, it has a well-defined projective dimension being the length of any minimal projective resolution. Moreover by Quillen-Suslin, any such resolution must be free. So I can always choose my free resolution to be minimal of the same length, in this case $r$. But this still doesn't allow me to conclude that what I've called the "resolvent trace" will be the same.
The well-definedness in terms of respecting the equivalence relation defining the Grothendieck group should come from the Horseshoe Lemma, if I can resolve the above problem. Does this sound correct?
Thanks