Atiyah-MacDonald help with exercise 5.10

The solution is at the end of the exercise. You will need exercise 3.26 of the same book.

I copy-paste the solution (unless you have any question about it).

To prove $(a) \Rightarrow (c)$ [recall that : $(c)$ $\operatorname{Spec}(B_q) \rightarrow \operatorname{Spec}(A_p)$ is surjective for all $q$ and $p:=f^*(q)$], observe that $B_q$ is the direct limit of $B_t$ where $t \in B-q$; hence, by Chapter 3, Exercise 26, we have $f^*(\operatorname{Spec}(B_q)) = \bigcap_t f^*(\operatorname{Spec}(B_t))$. Since $\operatorname{Spec}(B_t)$ is an open neighborhood of $q$ in $\operatorname{Spec}(B)$, and since $f^*$ is open, it follows that $f^*(\operatorname{Spec}(B_t))$ is an open neighborhood of $p$ in $\operatorname{Spec}(A)$ and therefore contains $\operatorname{Spec}(A_{p})$.

EDIT: here is a direct self-contained proof (which is the above one but detailed and shortened). Let $q$ be an ideal of $B$ and $p' \subset A$ an ideal included in $p:= f^*(q)$. Recall that the set of primes of $B$ included in $q$ and sent to $p'$ is $\operatorname{Spec} B_q \otimes_A \operatorname{Frac}(A/p')$. So we have to show that the ring $$ B_q \otimes_A \operatorname{Frac}(A/p') = \varinjlim_{t \notin q} (B_t \otimes_A \operatorname{Frac}(A/p'))$$ is not zero, which is equivalent to prove that $B_t \otimes_A \operatorname{Frac}(A/p') \neq 0$ for all $t$ (cf. Exercises 2.20 and 2.21 of the book).

For $t \notin q$, $\operatorname{Spec} B_t$ is an open neighborhood of $q$, hence by $(a)$, $f^*(\operatorname{Spec} B_t)$ is an open neighborhood of $p$ so contains $p'$. Whence $\operatorname{Spec} B_t \otimes_A \operatorname{Frac}(A/p') \neq \emptyset$.