This question is from my assignment in algebraic geometry and I am struck on it.
Question: Let X =Spec A be the prime spectrum ( with Zariski Topology) of a ring A.
(a) Prove that for every $x\in X$ , the closure of x = {$y\in X| $ $p_x\subseteq p_y$}
Attempt: A point $x\in X$ is a prime ideal in A and is denoted by $p_x$. Closure of x = x $\cup $ all limit points of x. I think all the limit points of X are x only as x is a singleton set. So, I don't have any idea on how I can prove the assertion given in the statement. Can you please give a hint as I think there is something wrong with the way I am trying to attempt the question?
(b) A point $x\in X$ is closed iff the prime ideal $p_x$ corresponding to x is a maximal ideal in A.
Let x is closed and $p_x$ is not maximal ideal but a prime ideal. I don't know which result to use to derive a contradiction?
Let on the converse , the prime ideal $p_x$ corresponding to x is a maximal ideal in A, then x not be a closed set, again I am not sure what result to use to prove what is asked.
I would like to say that I haven't used any property of Zariski topology till now. I understand properties of Zariski topology but I am not sure what exact property I can use it here.
(c) Spm A is dense in Spec A iff $N_A = M_A$, where $N_A$ and $M_A$ denote the nil -radical and Jacobson radical ideal of A.
Again, In this part I am not sure how can I proceed in any of the direction.
Kindly just give some hints only. I would like to complete the solution by myself.
Thanks!