Serre's twisting sheaf

Remarks

Let $S = k[x_0, \dots, x_r]$ over a field $k$, with $r \ge 1$, so that $X = \mathbb{P}^r_k$. Then $\mathcal{O}_X(n)$ is an invertible sheaf, and its global sections are the homogeneous polynomials of degree $n$ for $n \ge 0$, and $0$ for $n < 0$. On $D_+(x_i)$, the section $x_i^n$ generates $\mathcal{O}_X(n)$, which gives the local isomorphism with $\mathcal{O}_X$. If $S$ is generated by $S_1$ as an $S_0$-algebra, then each $\mathcal{O}_X(n)$ is invertible and $\mathcal{O}_X(n) \otimes \mathcal{O}_X(m) \cong \mathcal{O}_X(n + m)$. Without that hypothesis $\mathcal{O}_X(1)$ need not be invertible. Take $S = k[x_0, x_1, x_2]$ with $x_0, x_1$ of degree $1$ and $x_2$ of degree $2$, and the point $p$ where $x_0 = x_1 = 0$. On $D_+(x_2)$, the sections of $\mathcal{O}_X(1)$ need the two generators $x_0$ and $x_1$ near $p$, so $\mathcal{O}_X(1)$ is not invertible at $p$. For a quasi-coherent sheaf $\mathcal{F}$, one writes $\mathcal{F}(n) = \mathcal{F} \otimes \mathcal{O}_X(n)$, the twist of $\mathcal{F}$ by $n$.