Ravi Vakil in Exercise 6.3.H asks:

If $S$ is a finitely generated graded $A$-algebra, describe a natural structure morphism $\mathrm{Proj}\:S \to \mathrm{Spec}\: A$.

My issue is "finitely generated" part, why does it matter if $S$ is finitely generated over $A$ or not?

Let $X=\mathrm{Proj}\: S$. The construction of this canonical morphism is by noting that given any homogeneous $f\in S$, there is an isomorphism $X_f\simeq \mathrm{Spec}S_{(f)}$, where $S_{(f)}$ is the degree zero part of the localization at $f$ of the graded ring $S$ (i.e. $(S_f)_0$) and $X_f$ is the locus of points where $f$ doesn't vanish. Then each $S_{(f)}$ is equipped naturally with an $A$-algebra structure $A\to S_{(f)}$. By noting that $X_f$ is a base for the Zariski topology of $X$, we can extend the sheaf on base to the structure sheaf of $X$. This puts an $A$-algebra structure on all $\Gamma(U, \mathscr{O}_X)$. With some gluing arguments then we find our morphism.

Alternatively one can use the $A$-algebra structure on $\Gamma(X, \mathscr{O}_X)$ to define the morphism $X\to \mathrm{Spec}\: A$. (there is a bijection between morphisms $X\to \mathrm{Spec}\:A$ and homomorphisms $A\to \Gamma(X, \mathscr{O}_X)$).

Doesn't this determine a morphism $X\to \mathrm{Spec}\: A$? Where does finitely generated play any role?

  • $\begingroup$ Is that the whole exercise? $\endgroup$ – Mariano Suárez-Álvarez Aug 23 '16 at 0:49
  • $\begingroup$ Yes the highlighted part is the whole exercise. The rest is me talking. $\endgroup$ – Hamed Aug 23 '16 at 0:55
  • 1
    $\begingroup$ He likes assuming f.g. in degree $1$ even when it doesn't matter. I'm sure if you look in EGA II this restriction isn't there. $\endgroup$ – Hoot Aug 23 '16 at 0:59
  • $\begingroup$ @Hoot - if memory serves (~20 years) you are correct about EGA II. $\endgroup$ – mathguy Aug 23 '16 at 1:24

I don't believe that the assumption plays any role at all, its only purpose being to set the reader in a known situation. Most authors begin explaining $\mathrm{Proj}$ in the context of finitely generated $A$-algebras, so assuming it in the question makes it "easier" to answer (i.e. one does not diverge into philosophical questions to see how to relax this assumption and just answers straightforwardly). An example of this behavior is also given in Hartshorne, for example (the "Hartshorne-projective" schemes are closed subschemes of $\mathbb P^n_X$, making them correspond to a finitely generated graded $A$-algebra when $X = \mathrm{Spec}(A)$).

This behavior exists mostly because people want to introduce projective schemes without going all out in Grothendieck-style generality to minimize confusion/keep people down to earth (whatever that means at this point). The point of assuming finitely generated in degree $1$ is that it gives a natural embedding to $\mathbb P^n$, which might be desirable if you want to write down some equations but is not always necessary to work out the theory. Your argument doesn't require the finitely generated hypothesis, as you noticed yourself.

Hope that helps,


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.