I have a simple question about Exercise II.2.14(c) in Hartshorne's book. The claim is that if $\varphi : S \to T$ is a graded homomorphism which induces isomorphisms on all homogeneous pieces of sufficiently high degree, then the induced morphism $f : \text{Proj } T \to \text{Proj } S$ is an isomorphism (in particular, defined on all of $\text{Proj } T$).
When one is showing that $f$ is injective, I think the proof should go something like this: if $\mathfrak{p},\mathfrak{q} \in \text{Proj } T$ satisfy $\varphi^{-1}(\mathfrak{p}) = \varphi^{-1}(\mathfrak{q})$, then $\mathfrak{p} \cap T_d = \mathfrak{q} \cap T_d$ for sufficiently large $d$. So if $t \in \mathfrak{p}$ is homogeneous of positive degree, then $t^n \in \mathfrak{p} \cap T_d = \mathfrak{q} \cap T_d$ for large $n$, which implies $t \in \mathfrak{q}$.
But what happens with elements of degree zero? A couple of solutions to this exercise exist online, and they also seem to gloss over this point.