This question is on "computing" the Grothendieck group of the projective $n$-space with $m$ origins ($m\geq 1$). For any (noetherian) scheme $X$, let $K_0(X)$ be the Grothendieck group of coherent sheaves on $X$.

Firstly, let me sketch that $K_0(\mathbf{P}^n) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^n)$.

Let $X=\mathbf{P}^n$ be the projective $n$-space.

(I omit writing the base scheme in the subscript. In fact, you can take any noetherian scheme as a base scheme in the following, I think.)

Let $H\cong \mathbf{P}^{n-1}$ be a hyperplane with complement $U\cong \mathbf{A}^n$. By a well-known theorem on Grothendieck groups, we have a short exact sequence of abelian groups $$K_0(H) \rightarrow K_0(X) \rightarrow K_0(U) \rightarrow 0.$$ Now, let $i:H\longrightarrow X$ be the closed immersion. Then the first map in the above sequence is given by the "extension by zero", which in this case is just the K-theoretic push-forward $i_!$, or even better, just the direct image functor $i_\ast$. Now, there is a projection map $\pi:X\longrightarrow H$ such that $\pi\circ i = \textrm{id}_{H}$.

By functoriality of the push-forward, we conclude that $\pi_! \circ i_\ast = \pi_! \circ i_! = \textrm{id}_{K_0(H)}$.

Therefore, we may conclude that $i_\ast$ is injective and that we have a split exact sequence $$0 \rightarrow K_0(H) \rightarrow K_0(X) \rightarrow K_0(U) \rightarrow 0.$$ Thus, we have that $K_0(\mathbf{P}^n) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^n)$.

**Q1**: Let $\mathbf{P}^{n,m}$ be the projective $n$-space with $m$ origins ($m\geq 1$). For example, $\mathbf{P}^{n,1} = \mathbf{P}^n$. (Again the base scheme can be anything, I think.) Now, is it true that $$K_0(\mathbf{P}^{n,m}) \cong K_0(\mathbf{P}^{n-1,m}) \oplus K_0(\mathbf{A}^n)?$$

**Idea1**: Take a hyperplane $H$ in $\mathbf{P}^{n,m}$. Is it true that $H\cong \mathbf{P}^{n-1,m}$ and that its complement is $\mathbf{A}^n$? Also, even though the schemes are not separated, the closed immersion $i:H\longrightarrow \mathbf{P}^{n,m}$ is proper, right? Also, is the projection $\pi:\mathbf{P}^{n,m}\rightarrow H$ proper? If yes, the above reasoning applies. If no, how can one "fix" the above reasoning? I think that in this case one could still make sense out of $i_!$ and $\pi_!$ (even if they are not proper maps.)

**Idea2**: Maybe it is easier to show that $K_0(\mathbf{P}^{n,m}) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^{n,m})$, where $\mathbf{A}^{n,m}$ is the affine $n$-space with $m$ origins. Then one reduces to computing $K_0(\mathbf{A}^{n,m})$...

**Idea3**: One could also take $m=2$ as a starting case and look at the complement of one of the origins. Then we get a similar exact sequence as above and one could reason from there.

Which of these ideas do not apply and which do?

**Note**: Suppose that the base scheme is a field. Since $K_0(\mathbf{A}^n) \cong \mathbf{Z}$ and $K_0(\mathbf{P}^n) \cong \mathbf{Z}^{n+1}$, this would show that $$K_0(\mathbf{P}^{n,m}) \cong \mathbf{Z}^{n+m}.$$ More generally, if $S$ is the base scheme, $K_0(\mathbf{P}^{n,m}) \cong K_0(S)^{n+m}$.