Let $A$ be a ring, $B = A[T_0,\dots, T_d]$, and $X = \textrm{Proj } B$. Then at every point $x \in X$,
$$\mathcal{O}_X (n)_x \cong \mathcal{O}_{X,x}$$
Let $x$ correspond to a homogeneous prime ideal $p$.
So by definition, $\mathcal{O}_X(n) = B(n)^\tilde{}$ thus taking stalks, we get $(\tilde{B(n)})_p = B(n)_{(p)}$. The latter is represented by $a/s$ where $a$ is a homogeneous polynomial of degree $n + \deg (s)$.
On the other hand, $\mathcal{O}_{X,x} = B_{(p)}$ which is represented by quotients of homogeneous polynomials that are of the same degree.
- How does one see that these are isomorphic? Should we take a degree $0$ quotient and multiply by $x_0 \dots x_d$ until it is of degree $n$?
- How general is this result? Does it hold for all such $X$? Does it hold for general twisted sheaf of $\mathcal{O}_X$-modules?
In general, let $R$ be a graded ring and let $\operatorname{Proj}R=X$, we can define \begin{equation} \forall n\in\mathbb{Z},\,\mathcal{O}_X(n)=\widetilde{R(n)} \end{equation} where for any homogeneous element $f\in R$ of degree $m$ \begin{equation} \mathcal{O}_X(n)_{|D_+(f)}=R_{(f)}(n)=R_{f,0}(n)=\left\{\frac{x}{f^k}\in R_f\mid x\in R_{km}\right\}(n)=\left\{\frac{x}{f^k}\in R_f\mid x\in R_{km+n}\right\}; \end{equation} then: \begin{equation} \forall\mathfrak{p}\in X,\,\left(\mathcal{O}_X(n)\right)_{\mathfrak{p}}=\lim_{\overrightarrow{f\notin\mathfrak{p}}}\mathcal{O}_X(n)_{|D_+(f)}=\dots=R_{\mathfrak{p},0}(n)\simeq R_{\mathfrak{p},0}=\mathcal{O}_{X,\mathfrak{p}}, \end{equation} where an isomorphism of rings is: \begin{equation} \varphi_a:\frac{x}{f^k}\in R_{\mathfrak{p},0}(n)\to\frac{x}{f^ka^n}\in R_{\mathfrak{p},0} \end{equation} where $a\in R_{\mathfrak{p},1}$.
Very important remark: $\mathcal{O}_X$ and $\mathcal{O}_X(n)$ are isomorphic as quasi-coherent $\mathcal{O}_X$-modules on $X$ associated to the ring $R$, but they are not isomorphic as quasi-coherent $\mathcal{O}_X$-modules on $X$ associated to the graded ring $R$!
Are you agree? Is it all clear?