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    Boolean Ring Question

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    Let R be a Boolean ring and let X be the set all maximal ideals of R. Put a topology on X by taking sets of the form Dr = {M ∈ X | r ∈/ M}, r is in R, as a basis for the open topology. Since the ideals are prime, Dr ∩ Ds = Drs, making the collection closed under finite intersections. Show that X is a Boolean space. Hint on way of showing that space
    is compact is to embed it in the product space 2R as the characteristic functions of the ideals; show it is a closed subset, then use the fact that a closed subset of a compact space is compact.

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    Recall that a topological space is called Boolean, if it is compact, Hausdorff and has a basis of clopen subsets (i.e. subsets which are both open and closed).
    Let us show that X is compact. Consider X's open cover X=⋃_(i∈I)▒X_i . Since〖(D_r)〗_(r∈I) is a basis of the topology of X, without loss of generality we can assume that each X_i is equal to some D_r. Let J={r∈R|D_r=X_i, for some i∈I}. Consider the ideal H generated by J. If H≠R, then there exists a maximal ideal M containing it. We have M∈X. Therefore M∈X_i for some I, and hence M∈D_r for some r∈J. Then r∉M, by the definition of D_r. But this ...

    Solution Summary

    This solution helps with a problem involving a Boolean ring.