I'm trying to solve Exercise 6.4.F from Vakil's FOAG:

6.4.F. Exᴇʀᴄɪsᴇ.$\quad$Show that if $R_\bullet$ and $S_\bullet$ are the same finitely generated graded rings except in a finite number of nonzero degrees (make this precise!), then $\operatorname{Proj} R_\bullet \cong \operatorname{Proj} S_\bullet$.

First, I think what Vakil means when he says same finitely generated rings except in a finite number of nonzero degrees is that $R_{\bullet}$ and $S_{\bullet}$ have all their homogeneous pieces ($R_n$ and $S_n$ respectively) the same, except for finitely many $n$. Could someone please verify if this is right?

Second, assuming the above, here's how I would solve this exercise: since $S_n$ and $R_n$ are identical for all large $n$, if we consider $m \gg 0$, then we have that $R_{m\bullet}$ and $S_{m\bullet}$ are identical. Thus $\operatorname{Proj} S_{m\bullet}\cong\operatorname{Proj} R_{m\bullet}$. But Exercise 6.4.D shows that $\operatorname{Proj} S_{m\bullet}\cong \operatorname{Proj} S_{\bullet}$, and we have a similar isomorphism for $R_{\bullet}$. Thus, $\operatorname{Proj} S_{\bullet}\cong \operatorname{Proj} R_{\bullet}$ and we are done.

Now I'm also not sure if this solutions works, because I don't seem to be using the "finitely generated" hypothesis anywhere in my above "proof". I would be glad if someone would point out what I am missing.

  • $\begingroup$ “Same” probably means isomorphic. $\endgroup$
    – user5826
    Commented Sep 19, 2021 at 17:37
  • $\begingroup$ I'm making this a comment instead of an answer because I don't know what I'm talking about. But it seems like the preceding definition of $S_{n \bullet}$ is only made for finitely generated $S_\bullet$ in the first place. Could it be that the definition does not work, or is ill-behaved somehow, if $S_\bullet$ is not finitely generated? (Either way, it's possible that Exercise 6.4.D assumes $S_\bullet$ is finitely generated, because otherwise we haven't defined what $S_{n \bullet}$ is.) $\endgroup$ Commented Sep 19, 2021 at 17:51
  • 1
    $\begingroup$ @MishaLavrov You're correct that the previous definition is only made for fg $S_\bullet$, but the proposition is true in general. $\endgroup$
    – KReiser
    Commented Sep 19, 2021 at 18:12

1 Answer 1


Just to get this off the unanswered list, your solution is totally correct and does not need the finite generation hypothesis (see for instance here). As Misha Lavrov points out in the comments, Vakil has only made his definition of $S_{n\bullet}$ for finitely generated $S_\bullet$, which is why the question is stated as it is.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .