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Suppose $\phi:S_\bullet \rightarrow R_\bullet$ is a morphism of graded rings that has degree $d$, i.e. $\phi$ maps $S_n$ to $R_{dn}$ for all $n$. Then $\phi$ induces a morphism \begin{equation} \Phi:\text{Proj}~R_\bullet \setminus V(\phi(S_+)) \rightarrow \text{Proj}~S_\bullet \end{equation}

If $V(\phi(S_+))$ is empty then $\Phi$ would be a morphism between two projective schemes. If both $R_\bullet$ and $S_\bullet$ are finitely generated in degree $1$, then does the line bundle $\mathcal{O}_{\text{Proj}S_\bullet}(1)$ pulls back to $\mathcal{O}_{\text{Proj}R_\bullet}(d)$?

If it is not true generally, is it true when $S_0=R_0=A$, i.e. they are both graded $A$-algebra? If it is still not true, what about we let $A=k$ is a field?

edit: Suppose with some new conditions, $\mathcal{O}_{\text{Proj}S_\bullet}(1)$ pulls back to $\mathcal{O}_{\text{Proj}R_\bullet}(d)$, then the global sections of $\mathcal{O}_{\text{Proj}S_\bullet}(1)$ pulls back to the global sections of $\mathcal{O}_{\text{Proj}R_\bullet}(d)$, is this the same as the map $S_1 \rightarrow R_d$?

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What is $O_{Proj S_*}(n) = O_S(n)$? Whatever it is, you can describe it and thus work with it by knowing the following facts:

  1. $O_S(n)$ is isomorphic to the structure sheaf on each open set of the cover $D(s)$, where $s \in S_n$, via some $\phi_s: O(n)|_{D(s)} \to O|_{D(s)}$. In fact the structure sheaf over $D(s)$ is just $S[s^{-1}]_0$, and $O_S(n)$ is just $S[s^{-1}]_n$. So $\phi_s$ can be taken to be multiplication by $s$, which is an isomorphism on this patch.

  2. The transition function $\phi_s \circ \phi_t^{-1}$ is precisely multiplication by $s/t$. (This follows immediately from 1.)

Main point of this: To understand a line bundle, we just need to know an open cover on which it trivializes, and the transition functions on the intersections.

We know how to pull back a line bundle, also.

Explicitly: If $O_V \to L$ is a trivialization over $V$, then the pullback by $\phi^*$ of this map gives a trivialization of $O_{\phi^{-1}(V)} \cong \phi^* O_V \to \phi^* L$.

The map $\psi: O_{\phi^{-1}(V)} \cong \phi^* O_V$ is the restriction of a canonical globally defined sheaf isomorphism between $\phi^* O_S$ and $O_T$, which is locally just "multiplication" $A \otimes_A B \to B$ - you can probably replace it with an equality without loss, but I personally am confused by that level of sloppiness - at least at this point in my mathematical life. Anyway it is not too difficult to keep track of it.

So let's just apply our understanding of the pullback to $O_S(n)$ and see what happens.

1) $\phi^{-1}(D(s)) = D(\phi(s))$.

So far so good - your condition that $V(\phi(S_+)) = \emptyset$ implies that this gives an open cover of $Proj T$ on which we know that $\phi^* O(n)$ trivializes.

(It may not be the case that $S_n$ surjects onto $S_{dn}$. That is okay, we still get a cover from the $D(\phi(s))$, as $s \in S_n$. Assuming your condition, which is that $V(\phi(S+)) = \emptyset$, for any homogeneous prime in $Proj(T)$ there is some $k$ and $s$ so that $\phi(s)^k \not \in P$. So this implies that $\phi(s) \not \in P$.)

2) Also, the trivialization $T_s: O_S|_{D(s)} \to O_S(n) |_{D(s)}$ becomes some $\phi^*(T_s) : \phi^* O_S \to \phi^*(O_S(n))$ over $D(\phi(s))$. Combing this with the isomorphism $\psi : O_T \to \phi^* O_S$, our trivialization over this chart is $\phi^*(T_s) \circ \psi|_{D(\phi(s)}$.

3) Claim: The transition functions become $\phi(s/t) = \phi(s) / \phi(t)$. Proof:

First we compute: $(\phi^*(T_t) \circ \psi))^{-1} \circ \phi^*(T_s) \circ \psi = \psi^{-1} \phi^* (T_s \circ T_t^{-1}) \psi = \psi^{-1} \_ \times \phi(s/t) \psi = \psi^{-1} (\_ \times \phi(s)/\phi(t)) \psi$. Then note: the isomorphism $\psi$ commutes with multiplication by elements of the structure sheaf (it is an isomorphism of $O_X$ algebras, in particular $O_X$ modules), so in the end we get that our transition function is just multiplication by $\phi(s) / \phi(t)$.

If you look back at the first paragraph, this is some cover and transition maps that describe $O_T(dn)$. So apparently the pullback is isomorphic as a line bundle to $O_T(dn)$.

I think your claim is correct - but maybe I am overlooking something subtle! This stuff is confusing!

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