3
$\begingroup$

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.

Diego

$\endgroup$
  • $\begingroup$ 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
  • $\begingroup$ @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
1
$\begingroup$

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).

$\endgroup$
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.