1. ## Spec(R), dense points

A point P in a topology for a set $X$ is called dense if P is contained in every non-empty open set of the topology. Alternatively, the closure of $\{P\}$ equals $X$.

Find and prove a necessary and sufficient condition for $\text{Spec}(R)$ to have a dense point. The condition should related to the nilradical. How many dense points can $\text{Spec}(R)$ have?

I am not seeing how to approach this problem right now. Any helpful hints will be very greatly appreciated. Thank you.

2. Originally Posted by pascal4542
A point P in a topology for a set $X$ is called dense if P is contained in every non-empty open set of the topology. Alternatively, the closure of $\{P\}$ equals $X$.

Find and prove a necessary and sufficient condition for $\text{Spec}(R)$ to have a dense point. The condition should related to the nilradical. How many dense points can $\text{Spec}(R)$ have?

I am not seeing how to approach this problem right now. Any helpful hints will be very greatly appreciated. Thank you.
This is my attempt.

The basic open set in X = Spec(R), denoted as $X_f$ for $f \in R$, is the collection of prime ideals in X = Spec(R) that do not contain f. For example, if $f \in R$ is a unit, then $X_f = X$ (Similary, $X_f$ is an empty set if f is nilpotent).

Thus, $X_f = |Spec(R)| \setminus V(\{f\})$, where V({f}) denotes the variety of {f}.

Any open set in Spec(R) is the union of the above basic open sets. Any open set is a union of an empty set and itself. Since $X_f$ is an empty set if f is a nilpotent, a set of all nilpotent elements, which is a nilradical, is a dense point of Spec(R) (Every prime ideal in Spec(R) contains a nilradical).