I am reading Hartshorne's textbook on algebraic geometry.
I have just read the definition of the sheaf $\mathcal{O}(n)$ on $\text{Proj}(S)$ for a graded ring $S$. I want to first understand $\mathcal{O}(n)$ on $\Bbb P^1$, where $S=\Bbb{C}[x,y]$ with the standard grading.
Then by Proposition 5.11, we know that $\mathcal{O}(n)|_{D_+(x)}\cong \Bbb{C}[x,y](n)_{(x)}$ is the degree $0$ terms in the localised ring, after shifting the grading by $n$. I originally thought that we shift the grading by $n$, in which case $x^n,y^n$ have degree $0$ in $\Bbb{C}[x,y](n)$, and thus $x$ has degree $1-n$, and in the localisation, $1/x$ has degree $1-n-2$?
In the proof though it says that $S(n)_f$ has the degree $n$ elements of $S_f$, so it seems they localise with the old degree of $f$, back in $\Bbb{C}[x,y]$ rather than the degree in $\Bbb{C}[x,y](n)$.
Well what module is $\Bbb{C}[x,y](n)_{(x)}$ after all? I.e. what is $\mathcal{O}(n)|_{U_0}$ the $\widetilde{globalisation}$ of?
I guess it's the $\Bbb{C}[y/x]$-module spanned by $x^{-i}y^{n+i}$ for $i\geq -n$?