Twisted structure sheaves for $X=\operatorname{Spec} k$

Question

Consider the twisted structure sheaves $\mathcal O_X(n), n\in \Bbb Z$ for X a closed subscheme of $\mathbb P^0_k$ for $k=\overline{k}$ a field.

By definition, $\mathcal O_X(1):= i^*{\mathcal O_{\Bbb P^0_k}(1)}$, where the only possible closed immersion $i:X\hookrightarrow \Bbb P^0_k$ I can think of in this situation is the identity.

Then $\mathcal O_X(1)= \mathcal O_{\Bbb P^0_k}(1) = \{0\}$ since all polynomials on $\Bbb P^0_k=\operatorname{Spec} k$ have degree $0$ except the $0$-polynomial which have all degrees (in particular degree $1$).

Then for $n\ne 0$ we have $\mathcal O_X(n)=\mathcal O_X^{\otimes n} = \{0\}$ and $\mathcal O_X(0)=k$ since the graded ring $k$ has only degree $0$ (i.e. constant) polynomials.

Edit

What confuses me is that $\mathcal O_X(n) \cong \mathcal O_X$ on distinguished open sets as we can read in Gathmann notes Example $7.1.2$ page 122.

Thank you for your help.

Answer 1

Since $\mathbb{P}^0_k =\operatorname{Spec}(k)$ is affine, quasicoherent sheaves correspond to modules, and closed immersions correspond to ideals.

The ideals of $k$ are 0 or $k$, so the closed immersions to $\operatorname{Spec}(k)$ are the identity and empty scheme.

On those schemes, the only invertible sheaf is the structure sheaf. Therefore, $O_X(n)=O_X$. You wrote $O_X(n)=O_X^{\oplus n}$ but it should be $O_X(n)=O_X^{\otimes n}$.