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

Question

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 $\Bbb P^n_R$.

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.