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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.

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up vote 4 down vote accepted

Since by definition $\mathfrak{p}$ does not contain all of $T_+ = \oplus_{d>0} T_d$, there is some $s \in T_d$ for some $d>0$, such that $s \notin \mathfrak{p}$. You proved that $\mathfrak{p} \cap T_d = \mathfrak{q} \cap T_d$, so $s \notin \mathfrak{q}$ either.

Let $t \in \mathfrak{p} \cap T_0$. Then $t s \in \mathfrak{p} \cap T_{d} = \mathfrak{q} \cap T_{d}$, hence $t \in \mathfrak{q}$ or $s \in \mathfrak{q}$, and the latter is impossible.

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Thanks for the simple answer to a simple question! I think one has to use a similar trick when proving the surjectivity. – Justin Campbell May 12 '11 at 18:52

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