I'd just like some verification of the following proof/computation, thanks in advance.
Let $k$ be a field. Claim: $K_0(k[x])$ the Grothendieck group is isomorphic to $\mathbb{Z}$. By the Quillen-Suslin theorem every finitely generated projective module over a polynomial ring is free. Therefore, the isomorphism type of projective modules over $k[x]$ is determined by its rank. Since the monoid structure we endow these isomorphism classes with is direct summation, this means that the monoid on the isomorphism classes of projective modules is completely determined by addition of ranks. This monoid is therefore clearly isomorphic to $(\mathbb{N}, +)$ so the Grothendieck group $K_0(k[x]) \cong \mathbb{Z}$ the group completion of the natural numbers.
Just so that this question is marked answered.
Claim: $K_0(k[x])$ for $k$ a field is $\mathbb{Z}$.
Proof: By the Quillen-Suslin theorem every finitely generated projective module over a polynomial ring is free. Therefore, the isomorphism type of projective modules over $k[x]$ are determined by their rank. The direct sum of two such modules is the monoid operation. This makes this monoid obviously isomorphic to $(\mathbb{N}, +)$. So $K_0(k[x])$ is isomorphic to the group completion of $\mathbb{N}$ which is $\mathbb{Z}$.