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Math Help - Spec(R), dense points

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by pascal4542 View Post
    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).
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