# Grothendieck group of $\mathbb P^n$

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?

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.
• See e.g. p. 142, Theorem 2.1 in Quillen's Higher Algebraic $K$-theory. – user3267 Jun 10 '17 at 17:44
• @slinshady Take the projective bundle of the vector bundle $k^{n+1} \to \text{Spec}(k)$ and apply the formula – 54321user Jun 10 '17 at 18:22
• thank you. This gave me some insight into higher K theory. I just dealt with the problem by Serre's theorem about coherent sheaves (so the $\mathcal O(k)$, all $k$ have to be generators) and the Koszul complex of the affine plane together with twisting gives us, that the $\mathcal O(k)$ are generate by those with $-n\leq k \leq 0$. – slinshady Jun 10 '17 at 18:47