# Twisting relative proj (exercise from Vakil)

I'm stuck on another problem (17.2.G) from Vakil's notes, and I'm wondering if somebody could get me started.

Specifically, we are given a scheme $X$, an invertible sheaf $\mathscr{L}$ on $X$ and a quasicoherent sheaf $\mathscr{S}$ of graded algebras on $X$. We define $\mathscr{S}' = \oplus_{n=0}^{\infty}{(\mathscr{S}_n\otimes \mathscr{L}^{\otimes n})}$. How do we get a natural map \begin{align*} \mathcal{Proj}\mathscr{S}'\leftarrow \mathcal{Proj}\mathscr{S}? \end{align*} The problem asks for more (in particular the map should be an isomorphism, where we might have to use that $\mathscr{S}$ is generated in degree 1), but I'm mainly stuck on just defining the map.

I've have tried writing down the transition maps explicitly for the case where $X=\mathbb{P}^1$, $\mathscr{L}=\mathcal{O}(1)$, and $\mathscr{S}=\pi_{*}\mathcal{O}_{\mathbb{P}^1\times\mathbb{A}^2}$, where $\pi$ is the projection $\mathbb{P}^{1}\times \mathbb{A}^2\rightarrow \mathbb{P}^1$, but I'm still missing something silly. (not homework)

• Dear Dtseng, I don't think your example really fits the set-up, at least not in the way you were imagining. In fact $\pi_*\mathcal O_{\mathbb P^1 \times \mathbb P^1}$ is just $\mathcal O_{\mathbb P^1}$; in particular, it is not graded. Jun 13, 2014 at 21:34
• Oops, you're right. I tried to fix the example.
– DCT
Jun 13, 2014 at 23:36

Remember that relative Proj is constructed by choosing an affine open cover $\{U_i\}$ of $X$, say $U_i =$ Spec $A_i$, considering the graded rings $B_i = \oplus_n B_{i,n} = \oplus_n \Gamma(U_i,\mathcal S_n)$, forming the absolute Proj of each $B_i$, and then gluing these Proj's on the overlaps of the $U_i$.
So why don't you try choosing a cover $U_i$ which trivializes $\mathcal L$, and seeing what happens?