# Map of $R$-modules is surjective iff it's surjective after tensoring with $R_p/pR_p$ for all primes $p\subset R$

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 $$\Bbb P^n_R$$.

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.
• 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 $N$? Apr 22, 2021 at 21:06