Let $\displaystyle\varphi:\bigoplus_{d=0}^\infty S_d\rightarrow \bigoplus_{d=0}^\infty T_d$ be a morphism of commutative graded rings (with identity) and suppose that there exists $d_0$ such that $\varphi_d:S_d\rightarrow T_d$ is a bijection for all $d\geq d_0$. Can we say something about the ring homomorphism $\varphi_0:S_0\rightarrow T_0$?, for example, does it take prime ideals into prime ideals?.

Actually, my aim is to prove that there exists a bijection between the set $\mbox{Proj}\ S$ of homogeneous prime ideals of $S$ which do not contain $S_+$ and the set $\mbox{Proj}\ T$ homogeneous prime ideals of $T$ which do not contain $T_+$. One side is easy: take $f:\mbox{Proj}\ T\rightarrow \mbox{Proj}\ S$ the map $f(\mathfrak{q})=\varphi^{-1}(\mathfrak{q})$. I have problem to define an inverse map of this map.

Any help will be strongly appreciated.


  • $\begingroup$ Of course nothing can be said about $\phi_0$. $\endgroup$ – Martin Brandenburg Feb 20 '13 at 20:14
  • $\begingroup$ For all you know, we have $S_d=T_d=0$ for all $d>0$! $\endgroup$ – Mariano Suárez-Álvarez Feb 20 '13 at 20:15
  • $\begingroup$ Why? I can't see this. $\endgroup$ – Martin Brandenburg Feb 20 '13 at 21:01
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    $\begingroup$ @MartinBrandenburg, For all we know, we might have that, and in that case the fact that $\phi$ induces an isomorphism in large degree is useless to say anything about what it does in degree zero. $\endgroup$ – Mariano Suárez-Álvarez Feb 21 '13 at 5:31

Your map $f$ is not well-defined (when $\phi$ is not surjective), since it may happen that $\phi^{-1}(\mathfrak{q})$ contains the irrelevant ideal. Instead of defining $f$ globally, you can use the affine covering of the Proj scheme. It suffices to prove that a) the ring homomorphism $S_{(f)} \to T_{(\phi(f))}$ induced by $\phi$ is an isomorphism for every homogeneous $f \in S$ of positive degree, b) the basic open subsets $D(\phi(f))$ cover $\mathrm{Proj}(T)$.

Hint for b): If $g \in T$ is homogeneous of positive degree, choose some $n \geq 1$ such that $g^n$ has degree $\geq d_0$.

Hint for a): This is easy when you have already digested the idea of b).

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    $\begingroup$ Actually $f$ is well defined. Define $U=\{\mathfrak{q}\in \mbox{Proj}\ T:\mathfrak{q}\nsupseteq \varphi(S_+)\}$. Then $f$ is clearly defined in $U$. The condition that $\varphi_d$ is a bijection for all $d\geq d_0$ ensures that $U=\mbox{Proj}\ T$. This is because if some homogeneous prime ideal contains $\varphi(S_+)$ then it contains $\bigoplus_{d\geq d_0} T_d$ and therefore it contains $T_+$. $\endgroup$ – Diego Feb 20 '13 at 23:18
  • $\begingroup$ Thanks. When one wants to construct the morphism directly, it doesn't suffice to give the set-map, one also has to construct the sheaf portion, and then show that it is an isomorphism. I think that my argument is simpler. $\endgroup$ – Martin Brandenburg Feb 21 '13 at 10:41
  • $\begingroup$ Just a comment on Martin's answer related to $S_0$ and $T_0$. To show the isomorphism between $Proj T \rightarrow Proj S$, we show locally by taking an affine cover $D_+(T_i^d)$ for $d$ large. To show that $S_{(f)} \rightarrow T_{(\phi(f))}$ is an isomorphism: for $\frac{t}{f^n}\in T_{(\phi(f))}$ if $n\geq 1$ then $t$ is in high degree; for $\frac{t_0}{1}$, it is equal to $\frac{t_0\cdot f }{f}$ thus in the image of $S_{(f)}$, although it might not be in the image of $S_0.$ This confusion led me to this post : ) $\endgroup$ – Ivan Jan 8 at 21:45

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