TITLE: Isomorphism of Proj schemes of graded rings, Hartshorne 2.14
QUESTION [9 upvotes]: This question is based on exercise $2.14$ of chapter $2$ of Hartshorne.

Suppose $\varphi:S\rightarrow T$ is a graded homomorphism of graded (commutative, unital) rings such that $\varphi_d := \varphi|_d$ is an isomorphism for all $d$ sufficiently large. Then I want to show that the natural morphism $ f: $Proj $T \rightarrow $Proj $S$  is an isomorphism by explicitly constructing an inverse.

The morphism is given on spaces by $P \mapsto \varphi^{-1}(P)$ and on sections by composing pointwise with $\varphi_P : S_{\varphi^{-1}(P)}\rightarrow T_P$.
I would like to show that this is an isomorphism by exhibiting an inverse homeomorphism to $f$ and showing the stalk maps are isomorphic. I appreciate that we can cover Proj $S$ with affine pieces and then show that the corresponding maps from the pullbacks to these pieces are isomorphisms, but I would like to know what $f^{-1}$ looks like explicitly. If there is a good way to see what $f^{-1}$ is by chasing through the local method of showing that $f$ is an isomorphism then I would also appreciate an explanation of that.
My candidate for $f^{-1}$ was $P \mapsto \sqrt{(\varphi(P))}=I$, the radical of the ideal generated by $\varphi(P)$. I think that I have shown that $I$ is homogeneous, doesn't contain $T_+$, $\varphi^{-1}(I) = P$ and is almost prime in the sense that if $a,b \in T$ are homogeneous and have degree at least $1$ then $ab \in I \implies a \in I$ or $b \in I$. But I think that in fact $I$ is not in general prime, since the degree $0$ component of $I$ is exactly $\sqrt{(\varphi(P_0))}$ in the ring $T_0$, where $P_0$ is the ideal of $S_0$ given by $S_0 \cap P$ and that for general rings $A$ and $B$, with $\rho:A\rightarrow B$, $P$ prime in $A$ doesn't imply $\sqrt{(\rho(P))}$ prime in $B$.

REPLY [3 votes]: In view of the complexity of the accepted answer, I want to sketch an alternate characterization of a point $\mathfrak q\in\operatorname{Proj} T$ lying over a point $\mathfrak p\in\operatorname{Proj} S$. First, as a matter of fact, $\ker\varphi\subset\mathfrak p$ for all $\mathfrak p\in\operatorname{Proj} S$, so (since $\varphi$ is a graded map) we may reduce to considering $S\subset T$ a subring with $S_d=T_d$ for $d\geq d_0$.
Let $\mathfrak p\in\operatorname{Proj} S$. Then $\mathfrak p\not\supset S_+$, and in fact $\exists w\in\bigoplus_{d\geq d_0}T_d$ homogeneous such that $w\notin\mathfrak p$. Put
$$\mathfrak q:=\{a\in T:aw\in\mathfrak p\}.$$
Then one can show that $\mathfrak q\in\operatorname{Proj} T$ and it is immediate that $\mathfrak q\cap S=\mathfrak p$.
So $f^{-1}(\mathfrak p)=\mathfrak q$ (showing injectivity of $f$ is not hard).