Question about some details of a proof 
i) Why it's a unit can prove this proposition 
ii)see picture
 A: Some hints: 
1) First note that in a Noetherian ring every ideal contains a power of its radical. Hence there exists some $n>0$ such that $p^n \subset q$. Now if $x \in S \cap p$ then $x^n \in S \cap q$ because $S$ is multiplicatively closed. This implies that $S^{-1}q = S^{-1} A$.
2) If $A$ is an integral domain, i.e. it has no zero divisors, then the canonical map $A \rightarrow S^{-1}A$ is injective and so we may say that $q \subset S^{-1} A$. But the correct relation in general is simply $S^{-1} q \subset S^{-1} A$.
3) In general, $S^{-1}q$ is not maximal, it is just primary. To prove this claim, we just check that the definition of primary is true. So let $(x_1/s_1)(x_2/s_2) \in S^{-1} q$ and suppose that $\frac{x_1}{s_1} \not\in S^{-1}q$. This latter condition implies that $s x_1 \not\in q$ for any $s \in S$. Then $x_1 x_2 / s_1 s_2 \in S^{-1} q$ and so $\frac{x_1 x_2}{s_1 s_2} = \frac{y}{t}$ where $y \in q, t \in S$. Then there exists some $s_3 \in S$ such that $s_3 t x_1 x_2 = s_1 s_2 s_3 y$. The right hand side is inside $q$ and so $s_3 t x_1 x_2 \in q$. Now $(s_3 t) x_1$ can not be inside $q$, because this would contradict the implication of our hypothesis that $s x_1 \not\in q, \forall s \in S$. Thus $x_2^n \in q$ for some $n$ and so $(x_2/s_2)^n \in S^{-1}q$.  
