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!
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.