I am trying to prove that the Grothendieck group $K_0(\mathbb P^n)$ is generated by $\{\mathcal O, \mathcal O(1),...,\mathcal O(n)\}$.
I already showed that this set generates a full sublattice of $K_0(\mathbb P^n)$. For this I showed via induction that the rank is bounded above by $n+1$ and then I defined the function:
$\phi:G_0(\mathbb P^n)\times G_0(\mathbb P^n) \rightarrow \mathbb Z$
$(\mathcal F,\mathcal G) \mapsto \chi(\mathcal F \otimes \mathcal G^{\vee})$
where $\chi$ denotes the Euler characteristic.
Using this function I defined the matrix $\varPhi$ by $\varPhi_{i,j}:=\phi(\mathcal O(i),\mathcal O(j))$.
How can I proceed from here that the $\mathcal O(i)$ actually generate the Grothendieck group?
Sincerely, slinshady
Edit: Maybe I should say that I only consider locally free $\mathcal O_{\mathbb P^n}$ -modules, as this is sufficient in sufficiently nice settings.