Localization, saturation and Nilradical (help needed and proof verification) This question is from my commutative algebra assignment and I am struck with it. So, I am posting here for guidance.
Question: Let A be a commutative ring.

(a) Let $S\subseteq A$ be a multicatively closed subset. Then prove that $S^{-1}$ commutes with the nilradical i.e. $nil (S^{-1} A) = S^{-1} (nilA)$.

Attempt: I think that both sets are equal because if $x\in nil (S^{-1} A) $  => $x^n =(a/s)^n=0=(a)^n / (s.s... n times)= a^n /s=0$ which implies that $a^n=0$ for some $n \in \mathbb{N}$. So, $x\in S^{-1} (nil A)$.Conversaly, If $x\in S^{-1} (nil A)$  then there exists $a\in A$ and $n \in \mathbb{N}$ such that $\frac{a^n} {s}=0$ . Now s must be non zero and hence I can write $\frac{a^n} {s^n} =0$ which means that $x\in nil (S^{-1} A)$.
Is the proof fine?

(b) A prime ideal $p\in (Spec A, \subseteq ) $  iff $Spec(A_p)$ is singleton.

I think p always lie in Spec A if p is prime so there is nothing to prove in the case when $SpecA_p$ is singleton is given. Am i right?
Converse, I am not able to prove. Can you please give some hints on how to prove that $SpecA_p$ must  be singleton if p is a prime ideal.
Kindly help!

(c) If A is reduced and $p\in (Spec A, \subseteq )$ is minimal, then $A_p$ is a field.

A is reduced means that A has no non-zero nilpotent elements and p is minimal means that p  is not a proper subset of another member set of Spec A is called a minimal set. Let $x\in A_p$, I have to prove that x is a unit. But I don't have any ideas and I am struck.
Unfortunately, I am not so good in solving problems in Localization of rings and would very much appreciate help.
Thanks!
 A: You have the right idea for (a).  In the fist direction, it is clearer if you write $0=x^n=(a/s)^n=a^n/s^n$, hence $ta^n=0$ for some $t\in S$.  Then $x=(ta)/(st)$, and $(ta)^n=0$.
For the other direction you have $x=a/s$ where $a^n=0$.  Thus $x^n=(a/s)^n=a^n/s^n=0/1$.
For (b): Given an ideal $J\subseteq A$, define an ideal of $A_p$ by: $$J_p=\{a/s|\,\, a\in J, s\notin P\}.$$
For any ideal $J\subseteq A_p$, the set of numerators of its elements is an ideal $\hat{J}\subseteq A$ (note that if $x/s\in J$, then so is $x/1$).  Further $$J=\left(\hat{J}\right)_p$$
If $J\subseteq A_p$ is prime, then $\hat{J}$ must also be prime.
Thus Spec$(A_p)$ consists only of ideals of the form $q_p$, for prime ideals $q\subseteq p\subset A$.
You should show that if $x/t$ is in $q_p$ (so $x/t=y/s$ with $y\in q$) then $x\in q$. (Hint if $s\notin p$ then $s\notin q$, so if $sx\in q$ then $x\in q$.)
Deduce that $q_p$ is a prime ideal of $A_p$ for every  $q\subseteq p\subset A$.  Further deduce that the primes $q_p$ are distinct for distinct $q$ (if $q\neq q'$, then $\hat{q_p}=q\neq q'=\hat{q'_p}$), and conclude that we have a $1-1$ correspondence between primes in $p$ and primes in $A_p$.
Note: I have not seen the notation before, but from context Spec$(A,\subseteq)$ must mean "set of minimal prime ideals of $A$".
We have Spec$(A_p)$ is a singleton if and only if the set of primes contained in $p$ is a singleton, which is if and only if $p$ is minimal.
(c) Use (a) to show that nil$A_p$ is $\{0\}$.  Use (b) to show that $A_p$ only has one prime ideal: $p_p$.  Thus $p_p$ is the intersection of all prime ideals in $A_p$, which is nil$A_p=\{0\}$ (for example see here).  As maximal ideals are prime, $p_p=\{0\}$ is the only possible maximal ideal of $A_p$.
Every ring has a maximal ideal, so $\{0\}$ is a maximal ideal of $A_p$, hence $A_p$ is a field.
