$M_p\cong N_p$ implies $M_q\cong N_q$ for $p\subset q$?

Question

$f : M\rightarrow N$ is an $R$-module homomorphism and $p\subset q$ are two primes of $R$. Then

$f_p : M_p\cong N_p$ implies $f_q : M_q\cong N_q$?

It seems true since $q^c\subset p^c$. I got this question from Proposition 3.9 in AM Commutative Algebra, part of which says

$f:M\cong N$ iff $f_p:M_p\cong N_p$ for all primes $p$.

So I guess we only have to check the statement for minimal primes.

Answer 1

Let $R=\Bbb Z$, $M=0$ and $N=\Bbb Z/2\Bbb Z$. Also let $p=\{0\}$ and $q=2\Bbb Z$. Then $f$ must be the zero map. As $N_p=0$ and $N_q\ne0$, then $f_p$ is an isomorphism but $f_q$ isn't.