Footnote 1:

Let X be the set of all filters on B. Let f(a) be the subset of all filters that contain a. Let N be the ideal generated by subsets of X of the form (f(a))c∩(f(~a))c. One can check that f(a) is in N if and only if a=0 (this is pretty easy: if f(a) is a subset of an intersection of the sets of the form (f(ai))c∩(f(~ai))c, then a will have to be orthogonal to the meet of all the ai. But swapping some of the ai with their negations, we see that a is orthogonal to every boolean combination of the ai, and hence is zero. Moreover, f preserves meets, so f embeds B into 2X/N.

Press the back button on your browser to go back to the post