Questions about Grothendieck groups. I have a question of the exercise 26 on page 88 of the book introduction to commutative algebra by Atiyah and Macdonald. In 26(iii), let $A$ be a field. Then finitely generated $A$-modules are finite dimensional $A$-vector spaces. Two finitely dimensional vector spaces are isomorphic if and only if they have the same dimension. Therefore $F(A)$ is isomorphic to $\mathbb{Z}_{\geq 0}$. But $\mathbb{Z}_{\geq 0}$ is not a group. It is said that $F(A)$ is isomorphic to $\mathbb{Z}$. How to prove this? How to prove that $F(A)$ is isomorphic to $\mathbb{Z}$ for the case that $A$ is a principal ideal domain. Thank you very much.

 A: It is not $F(A)$ that is isomorphic to $\mathbb{Z}$, but rather $K(A)$.  In general, $F(A)$ does not have a group structure, and the Grothendieck group $K(A)$ is the most free way of imposing a group structure on $F(A)$ such that exact sequences behave as specified.
An element of $C$ is a finite formal sum $\sum_{k \geq 0} n_k [V_k]$ where $V_k$ is a $k$-dimensional vector space over the field $A$ and $n_k \in \mathbb{Z}$.  If you consider that element in $C/D = K(A)$, then since you can always find an exact sequence
$$
0 \rightarrow V_{k-1} \rightarrow V_k \rightarrow V_1 \rightarrow 0
$$
(where the subscript denotes the dimension of the vector space), it is not hard to show by induction that $[V_k] = k [V_1]$ in $K(A)$.  Thus every element in $K(A)$ has the form $\sum_{k \geq 0} n_k k [V_1]$ and the map sending this element to $\sum_{k \geq 0} n_k k \in \mathbb{Z}$ is easily seen to be an isomorphism (exercise).
Hint for the PID case:  Use the exact sequence
$$
0 \rightarrow A \xrightarrow{p} A \rightarrow A / (p) \rightarrow 0
$$
to show that $[A / (p)] = 0$ in $K(A)$.  What does every finitely generated module over a PID look like?  Use this, the above observation and the argument in the case when $A$ is a field to prove that $K(A) \cong \mathbb{Z}$.
