Let $R$ be a noetherian domain and let $M\to N$ be a map of finitely generated $R$-modules. Suppose I know that $M\otimes_R R_p/pR_p \to N\otimes_R R_p/pR_p$ is surjective for all primes $p\subset R$. Is it the case that $M\to N$ must be surjective? I know that this is true (with no assumptions) if I replace $R_p/pR_p$ with $R_p$, but the quotient has me thrown for a loop. I think I want to do something like the following: exactness with $p=(0)$ means that the kernel is supported on some proper closed subset of $\operatorname{Spec} R$, and then I can restrict to an irreducible component of this support and run the same proof again. But I'm not sure this works.

Followup: if this is true, can I weaken any of my assumptions?

Context: I'm attempting to prove that if I have a family $X\subset \Bbb P^n_R\to \operatorname{Spec} R$ where the fibers are projectively normal (as varieties over the residue field at each point), then $X$ is projectively normal as a subscheme of $\Bbb P^n_R$.


1 Answer 1


This is a direct application of Nakayama's lemma -- If $M \otimes \kappa(p) \to N \otimes \kappa(p)$ is surjective, then $M_p \to N_p$ is surjective. Since this holds for all primes $p$, $M \to N$ is surjective.

  • 1
    $\begingroup$ To be clear, the argument is that given a generating set for $N_p/pN_p$, we can take lifts to $M_p$, and the images of these elements in $N_p$ map down to the generating set for $N_p/pN_p$ and therefore generate $N_p$ (by the fourth version of Nakayama here), so $M_p\to N_p$ is surjective. So this means all I really needed was finiteness of $N$? $\endgroup$ Apr 22, 2021 at 21:06
  • $\begingroup$ This looks right to me! $\endgroup$ Apr 22, 2021 at 22:12
  • $\begingroup$ Great, thanks! Sorry for forgetting about Nakayama, lol... $\endgroup$ Apr 22, 2021 at 22:16
  • $\begingroup$ We all forget stuff sometimes :) Glad I could help $\endgroup$ Apr 23, 2021 at 1:24

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